[
  {
    "id": 40001,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ be positive real numbers. Prove that\n$$\n\\begin{aligned}\n\\frac{1}{a+b+\\frac{1}{a b c}+1}+ & \\frac{1}{b+c+\\frac{1}{a b c}+1}+\\frac{1}{c+a+\\frac{1}{a b c}+1} \\\\\n& \\leq \\frac{a+b+c}{a+b+c+1}\n\\end{aligned}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "With the notation $d=\\frac{1}{a b c}$ the inequality becomes\n$$\n\\frac{1}{a+b+d+1}+\\frac{1}{b+c+d+1}+\\frac{1}{c+a+d+1}+\\frac{1}{a+b+c+1} \\leq 1 .\n$$\nLet $x=\\sqrt[4]{a}$, $y=\\sqrt[4]{b}$, $z=\\sqrt[4]{c}$, $w=\\sqrt[4]{d}$. Then\n$$\nx y z w=\\sqrt[4]{a b c d}=1\n$$\nand it suffices to prove that\n$$\n\\begin{equation*}\n\\sum \\frac{1}{x^{4}+y^{4}+z^{4}+z y x w} \\leq 1 . \\tag{1}\n\\end{equation*}\n$$\nBut $x^{4}+y^{4}+z^{4} \\geq x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}$ so\n$$\n\\begin{aligned}\nx^{4}+y^{4}+z^{4} & =\\frac{x^{4}+y^{4}+z^{4}+x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}{2} \\\\\n& =\\frac{x^{4}+y^{2} z^{2}}{2}+\\frac{y^{4}+z^{2} x^{2}}{2}+\\frac{z^{4}+x^{2} y^{2}}{2} \\\\\n& \\geq x^{2} y z+y^{2} z x+z^{2} x y=x y z(x+y+z)\n\\end{aligned}\n$$\nwhere we used the AM-GM inequality three times. The same result follows from the rearrangement inequality. Hence\n$$\n\\begin{gathered}\n\\frac{1}{x^{4}+y^{4}+z^{4}+x y z w} \\leq \\frac{1}{x y z(x+y+z)+x y z w} \\\\\n=\\frac{1}{x y z(x+y+z+w)}=\\frac{w}{x+y+z+w}\n\\end{gathered}\n$$\nand so\n$$\n\\sum \\frac{1}{x^{4}+y^{4}+z^{4}+x y z w} \\leq \\sum \\frac{w}{x+y+z+w}=1\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40002,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geq 2$ be a positive integer and let $x_{n}$ be a positive real root to the equation $x(x+1) \\ldots(x+n)=1$. Prove that\n$$\nx_{n}<\\frac{1}{\\sqrt{n!H_{n}}}\n$$\nwhere $H_{n}=1+\\frac{1}{2}+\\ldots+\\frac{1}{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because $x_{n}$ satisfies the equation we have $x_{n}^{2} \\cdot S_{n-1}< 1$, where $S_{n-1}$ is the $n-1$ symmetric sum of $1,2, \\ldots, n$. We can write\n$$\nS_{n-1}=n!\\left(1+\\frac{1}{2}+\\ldots+\\frac{1}{n}\\right)=n!\\cdot H_{n}\n$$\nand the above inequality is equivalent to $x_{n}^{2}<\\frac{1}{n!H_{n}}$, hence $x_{n}<\\frac{1}{\\sqrt{n!H_{n}}}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40003,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n \\geq 2$ for which $\\sqrt[n]{3^{n}+4^{n}+5^{n}+8^{n}+10^{n}}$ is an integer.",
    "options": [],
    "answer": "3",
    "solution": "We have $3^{3}+4^{3}+5^{3}+8^{3}+10^{3}=12^{3}$, that is $n=3$ satisfies the property. We shall prove that this is the unique solution. Consider the function $h:[2, \\infty) \\rightarrow \\mathbb{R}$,\n$$\nh(x)=\\left(\\frac{3}{12}\\right)^{x}+\\left(\\frac{4}{12}\\right)^{x}+\\left(\\frac{5}{12}\\right)^{x}+\\left(\\frac{8}{12}\\right)^{x}+\\left(\\frac{10}{12}\\right)^{x}\n$$\nwhich is a strictly decreasing function with $h(3)=1$. It follows that for $n>3$, we have $h(n)<h(3)=1$. That is\n$$\n\\sqrt[n]{3^{n}+4^{n}+5^{n}+8^{n}+10^{n}}<12\n$$\nOn the other hand, it is clear that\n$$\n10<\\sqrt[n]{3^{n}+4^{n}+5^{n}+8^{n}+10^{n}}\n$$\nhence if $\\sqrt[n]{3^{n}+4^{n}+5^{n}+8^{n}+10^{n}}$ is an integer, then it must be 11. We get $3^{n}+4^{n}+5^{n}+8^{n}+10^{n}=11^{n}$, relation which is not possible since the left hand side is an even number.\nAssume that\n$$\n\\sqrt[n]{3^{n}+4^{n}+5^{n}+8^{n}+10^{n}}=k\n$$\nfor some positive integer $k$, that is $3^{n}+4^{n}+5^{n}+8^{n}+10^{n}=k^{n}$. Since the left hand side is an even number, it follows that $2 \\mid k^{n}$, hence $2^{n} \\mid k^{n}$. We obtain $2^{n} \\mid 3^{n}+5^{n}$.\nOn the other hand, we have\n$$\n3^{n}+5^{n} \\equiv (-1)^{n}+1 \\quad (\\bmod 4) \\equiv 2 \\quad (\\bmod 4)\n$$\nif $n$ is even, hence there are no even integers satisfying the property.\nIf $n$ is odd, then we have\n$$\n3^{n}+5^{n}=(3+5)\\left(3^{n-1}-3^{n-2} \\cdot 5+\\ldots+5^{n-1}\\right)=2^{3} \\cdot a\n$$\nwhere $a$ is an odd number. It follows $n \\geq 3$, and we get the unique solution $n=3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40004,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that\n$$\n2 f(x) = f(x + y) + f(x + 2y)\n$$\nfor all $x \\in \\mathbb{R}$ and for all $y \\geq 0$.",
    "options": [],
    "answer": "All constant functions",
    "solution": "Without loss of generality, we may assume that $f(0) = 0$. Let $y > 0$ and $n$ a positive integer. For $x = n y$ we get\n$$\n2 f(n y) = f((n + 1) y) + f((n + 2) y).\n$$\nThe sequence $a_n = f(n y), n = 0, 1, 2, \\ldots$, satisfies the second order linear recursive relation\n$$\na_{n+2} = -a_{n+1} + 2 a_n\n$$\nwith $a_0 = f(0) = 0, a_1 = f(y)$. The characteristic equation is $t^2 + t - 2 = 0$, having the roots $t_1 = 1, t_2 = -2$. It follows that $a_n = \\alpha + \\beta (-2)^n$, where $\\alpha + \\beta = 0$ and $\\alpha - 2 \\beta = f(y)$. We get\n$$\nf(n y) = a_n = \\frac{1 - (-2)^n}{3} \\cdot f(y)\n$$\nIn particular, we have $f(4 y) = -5 f(y)$. On the other hand, from (3), we obtain $f(2 y) = -f(y)$, hence $f(4 y) = -f(2 y) = f(y)$, for any $y \\geq 0$. That is $f(y) = 0$ for any $y \\geq 0$.\nLet $x \\in \\mathbb{R}$. Then\n$$\n2 f(x) = f(x + |x|) + f(x + 2|x|) = 0\n$$\nhence $f(x) = 0$. The desired functions are all constant functions.\n\nA simple checking shows that any constant function is solution to the functional equation.\nIf $x_1, x_2 \\in \\mathbb{R}, x_1 > x_2$, we can find $x \\in \\mathbb{R}$ and $y \\geq 0$ such that $x_1 = x + 4y$ and $x_2 = x + y$. Indeed, solving the system in $x$ and $y$, we get\n$$\nx = \\frac{1}{3}(4 x_2 - x_1), \\quad y = \\frac{1}{3}(x_1 - x_2)\n$$\nReplacing $y$ by $2y$ in the functional equation we obtain\n$$\n2 f(x) = f(x + 2y) + f(x + 4y),\n$$\nhence we get\n$$\nf(x + y) + f(x + 2y) = f(x + 2y) + f(x + 4y)\n$$\nthat is\n$$\nf(x + y) = f(x + 4y), \\quad x \\in \\mathbb{R}, y \\geq 0\n$$\nFrom (1) it follows $f(x_1) = f(x_2)$, for any $x_1, x_2 \\in \\mathbb{R}$, hence $f$ is constant function.\n\nA simple checking shows that any constant function is solution to the functional equation.\nAs in the previous solution, we have\n$$\n2 f(x) = f(x + y) + f(x + 2y) = f(x + 2y) + f(x + 4y),\n$$\nhence, we get relation (1). Taking $x = -y, y \\geq 0$, from (1) we obtain $f(0) = f(3y), y \\geq 0$, that is $f$ is constant on $[0, \\infty)$. Taking $x = -3y$, it follows\n$$\nf(-3y) = f(y) = 0, \\quad y \\geq 0\n$$\nhence $f$ is constant on $(-\\infty, 0]$. Therefore, all solutions are given by the constant functions.\n\nA simple checking shows that any constant function is solution to the functional equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40005,
    "subject": "Mathematics (Multi-modal)",
    "question": "The quadrilateral $ABCD$ has $AD = DC = CB < AB$ and $AB \\parallel CD$. Points $E$ and $F$ lie on the sides $CD$ and $BC$ such that $\\widehat{ADE} = \\widehat{AEF}$. Prove that:\n\na. $4CF \\leq CB$.\n\nb. If $4CF = CB$, then $AE$ is the angle bisector of $\\widehat{DAF}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\n$\\widehat{FEC} = 180^\\circ - \\widehat{AEF} - \\widehat{DEA}$\n$$\n= 180^\\circ - \\widehat{ADE} - \\widehat{DEA} = \\widehat{DAE}\n$$\nFrom $AD = DC = CB < AB$ and $AB \\parallel CD$ it follows that $\\widehat{ADC} = \\widehat{DCB}$, hence triangles $ADE$ and $ECF$ are similar. This yields\n$$\n\\frac{AD}{EC} = \\frac{AE}{EF} = \\frac{DE}{CF} . \\tag{1}\n$$\nThis leads to\n$$\nAD \\cdot CF = EC \\cdot DE \\leq \\frac{1}{4}(EC + DE)^2 = \\frac{1}{4} CD^2,\n$$\nhence $4CF \\leq CB$, because $AD = DC = CB$.\n\n![](attached_image_1.png)\n\nb.\nIf $4CF = CB$, then the inequality\n$$\nEC \\cdot DE \\leq \\frac{1}{4}(EC + DE)^2\n$$\nbecomes an equality, that is $CE = ED$.\n\nRelation (1) becomes $\\frac{AD}{DE} = \\frac{AE}{EF}$. Since $\\widehat{ADE} = \\widehat{AEF}$, triangles $ADE$ and $AEF$ are similar.\n\nThen $\\widehat{DAE} = \\widehat{EAF}$, so $AE$ bisects the angle $\\widehat{DAF}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40006,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a point in the interior of triangle $A B C$. Lines $A P, B P, C P$ intersect sides $B C, C A, A B$ at $L, M, N$, respectively. Prove that\n$$\nA P \\cdot B P \\cdot C P \\geq 8 P L \\cdot P M \\cdot P N .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K_{A}, K_{B}, K_{C}$ be the areas of triangle $B P C$, $C P A$, $A P B$, respectively. Let $A'$ and $P'$ be the projections of $A$ and $P$ on side $B C$. Triangles $A L A'$ and $P L P'$ are similar, hence we have\n$$\n\\frac{A L}{P L}=\\frac{A A'}{P P'}=\\frac{A A' \\cdot B C}{P P' \\cdot B C}=\\frac{K}{K_{A}}, \\tag{1}\n$$\nwhere $K$ is area of triangle $A B C$. From (1) we get\n$$\n\\frac{A P}{P L}=\\frac{K_{B}+K_{C}}{K_{A}}\n$$\n![](attached_image_1.png)\nIn analogous way we obtain\n$$\n\\frac{B P}{P M}=\\frac{K_{C}+K_{A}}{K_{B}}, \\quad \\frac{C P}{P N}=\\frac{K_{A}+K_{B}}{K_{C}} . \\tag{2}\n$$\nThe inequality is equivalent to\n$$\n\\frac{A P}{P L} \\cdot \\frac{B P}{P M} \\cdot \\frac{C P}{P N} \\geq 8\n$$\nthat is\n$$\n\\left(K_{A}+K_{B}\\right)\\left(K_{B}+K_{C}\\right)\\left(K_{C}+K_{A}\\right) \\geq 8 K_{A} K_{B} K_{C} . \\tag{3}\n$$\nInequality (3) follows by applying three times the inequality $x+y \\geq 2 \\sqrt{x y}$, where $x, y>0$.\n\n![](attached_image_1.png)\n\nWe have equality if and only if $K_{A}=K_{B}=K_{C}=\\frac{1}{3} K$, hence if and only if $P=G$, the centroid of triangle $A B C$.\nWe will use so-called Van Aubel relation, that is\n$$\n\\frac{A P}{P L}=\\frac{A N}{N B}+\\frac{A M}{M C} . \\tag{1}\n$$\n![](attached_image_2.png)\nIn order to prove (1), we use Menelaos Theorem for triangle $A B L$ and collinear points $C, P, N$, and for triangle $A C L$ and collinear points $B, P, M$. We get\n$$\n\\frac{C L}{C B} \\cdot \\frac{N B}{N A} \\cdot \\frac{P A}{P L}=1 \\text{ and } \\frac{B L}{B C} \\cdot \\frac{M C}{M A} \\cdot \\frac{P A}{P L}=1 . \\tag{2}\n$$\nFrom (2) we obtain\n$$\n\\frac{C L}{C B} \\cdot \\frac{A P}{P L}+\\frac{B L}{B C} \\cdot \\frac{A P}{P L}=\\frac{A N}{N B}+\\frac{A M}{M C},\n$$\nhence relation (1) since $C L+B L=B C$.\nWriting the similar relations to (1) for Cevians $B M$ and $C N$, we have\n$$\n\\frac{B P}{P M}=\\frac{B L}{L C}+\\frac{B N}{N A} \\text{ and } \\frac{C P}{P N}=\\frac{C L}{L B}+\\frac{C M}{M A} . \\tag{3}\n$$\nIt follows\n$$\n\\frac{A P}{P L} \\cdot \\frac{B P}{P M} \\cdot \\frac{C P}{P N}=\\left(\\frac{A N}{N B}+\\frac{A M}{M C}\\right)\\left(\\frac{B L}{L C}+\\frac{B N}{N A}\\right)\\left(\\frac{C L}{L B}+\\frac{C M}{M A}\\right)\n$$\nand the inequality follows from AM-GM inequality for two numbers.\n\n",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40007,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that $27^{n}-2^{n}$ is a perfect square.",
    "options": [],
    "answer": "1",
    "solution": "For $n=1$ we have $27-2=25=5^{2}$. We will prove that there are no other positive integers with this property.\nIf $n$ is odd, $n=2k+1$, then\n$$\n\\begin{aligned}\n27^{n}-2^{n} &= 27^{2k+1}-2^{2k+1} = (28-1)^{2k+1}-2 \\cdot 4^{k} \\\\\n&= 4m+(-1)^{2k+1} = 4m-1\n\\end{aligned}\n$$\nIt is easy to see that integers of the form $4m-1$ are not perfect squares.\nIf $n$ is even, $n=2k$, then $27^{n}-2^{n} = (27^{k})^{2} - 2^{2k}$, and we have\n$$\n(27^{k}-1)^{2} < (27^{k})^{2} - 2^{2k} < 27^{k},\n$$\nhence $27^{2k}-2^{2k}$ is between two consecutive perfect squares, that is it cannot be a perfect square. The only solution is $n=1$.\n(Wael Hussain Al Saeed) Assume that $n \\geq 2$. For any integer $x$, we have $x^{2} \\equiv 0,1 \\pmod{3}$ and $x^{2} \\equiv 0,1 \\pmod{4}$. Considering the relation $27^{n}-2^{n}=x^{2}$ modulo 3 we get\n$$\n27^{n}-2^{n} \\equiv (-1)^{n+1} \\pmod{3}\n$$\nhence $(-1)^{n+1} \\equiv 1 \\pmod{3}$. It follows $n=2k+1$ for some positive integer $k$.\nConsidering the relation $27^{n}-2^{n}=x^{2}$ modulo 4 we obtain\n$$\n(-1)^{n} \\equiv x^{2} \\pmod{4}\n$$\nhence $(-1)^{n} \\equiv 1 \\pmod{4}$, that is $n=2m$ for some positive integer $m$, not possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40008,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $A$, $B$, $C$, $D$ lie on a line in this order. Draw parallel lines $a$ and $b$ through $A$ and $B$, respectively, and parallel lines $c$ and $d$ through $C$ and $D$, respectively, such that their points of intersection are vertices of a square. Prove that the side length of this square does not depend on the length of segment $B C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $x$ the side length of the square and construct $B B^{\\prime} \\perp a$, $C C^{\\prime} \\perp d$, where $B^{\\prime} \\in a$, $C^{\\prime} \\in d$. Let $\\alpha$ be the angle defined by lines $a$ and $A B$.\n\n![](attached_image_1.png)\n\nIn triangle $A B^{\\prime} B$ we have $x = A B \\sin \\alpha$, and in triangle $C C^{\\prime} D$ we have $x = C D \\cos \\alpha$. It follows\n$$\n\\left(\\frac{x}{A B}\\right)^{2} + \\left(\\frac{x}{C D}\\right)^{2} = \\sin^{2} \\alpha + \\cos^{2} \\alpha = 1,\n$$\nhence\n$$\nx = \\frac{A B \\cdot C D}{\\sqrt{A B^{2} + C D^{2}}}\n$$\nand the conclusion follows.\n\n![](attached_image_1.png)\n\nRemark. We have $\\tan \\alpha = \\frac{C D}{A B}$ and this formula shows how we can construct the lines $a$, $b$, $c$, $d$ through points $A$, $B$, $C$, $D$ in order to get the square. Let $B B^{\\prime\\prime} \\perp A B$ and $B B^{\\prime\\prime} = C D$. Then construct the line $a = A B^{\\prime\\prime}$ and $d \\perp a$. Finally, construct $b \\parallel a$ and $c \\parallel d$.\n(Saleh Saeed Al-Gamdi and Abrar Abdulmanem Al-Shaikh).\n\n![](attached_image_2.png)\n\nIn order to prove that the side length of the square does not depend on the length of segment $B C$ it suffices to draw parallel lines to line $A D$. Then the square is fixed; we get all possible configurations for the segment $B C$. The first position is $A^{\\prime} D^{\\prime}$, where points $B$, $C$ coincide with one vertex of the square.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40009,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $A B C$ let $O$ be the circumcenter, $H$ the orthocenter, and $M$ the midpoint of the segment $A H$. The perpendicular at $M$ onto $O M$ intersects lines $A B$ and $A C$ at $P$ and $Q$, respectively. Prove that $M P = M Q$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider $P'$ on $A B$, $Q'$ on $A C$ such that $A P' H Q'$ is a parallelogram. We shall prove that $P' Q' \\perp O M$. Let $T$ be the antipodal point of $A$ in the circumcircle of triangle $A B C$. We have $O M \\parallel T H$, hence it suffices to prove that $P' Q' \\perp T H$.\n\n![](attached_image_1.png)\n\nNotice that $B H \\perp A Q'$, and $C H \\perp A P'$. Since the diagonals $A H$ and $B C$ of parallelograms $B T C H$ and $A P' H Q'$ are perpendicular, it follows that diagonals $P' Q'$ and $H T$ are also perpendicular. Therefore $P = P'$ and $Q = Q'$, and we are done.\n\n![](attached_image_1.png)\nLet $S$ and $T$ be the midpoints of sides $A C$ and $A B$, respectively. Quadrilateral $M Q S O$ is cyclic, hence $\\widehat{M Q O} = \\widehat{M S O}$. Quadrilateral $M P T O$ is cyclic, hence we have $\\widehat{M P O} = \\widehat{M T O}$.\n\n![](attached_image_2.png)\n\nObserve that $M S O T$ is a parallelogram. Indeed, we have $M S \\parallel H C$ and $C H \\perp A B$ implies $S M \\perp A B$, therefore $A M \\parallel O T$. Also, we have $M T \\parallel H B$ and $B H \\perp A C$, implies $T M \\perp A C$, therefore $T M \\parallel O S$.\n\nFinally, from $\\widehat{M Q O} = \\widehat{M P O}$, it follows that triangle $O Q P$ is isosceles, that is $O Q = O P$. Since $O M \\perp P Q$, the conclusion follows.\n\n",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40010,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a non-zero real number $a$ such that $\\{a\\} + \\left\\{ \\frac{1}{a} \\right\\} = 1$, where $\\{x\\}$ denotes the fractional part of $x$. Prove that for any positive integer $n$, $\\left\\{ a^{n} \\right\\} + \\left\\{ \\frac{1}{a^{n}} \\right\\} = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have\n$$\na + \\frac{1}{a} = [a] + \\left[ \\frac{1}{a} \\right] + \\{ a \\} + \\left\\{ \\frac{1}{a} \\right\\} = [a] + \\left[ \\frac{1}{a} \\right] + 1\n$$\nis an integer and denote this integer by $k$. Let $S_{n} = a^{n} + \\frac{1}{a^{n}}$, $n = 0, 1, 2, \\ldots$ Since $a$ and $\\frac{1}{a}$ are the roots of quadratic equation $x^{2} - kx + 1 = 0$, it follows that\n$$\nS_{n+2} = k S_{n+1} - S_{n}, \\quad n = 0, 1, 2, \\ldots\n$$\nWe have $S_{0} = 2$, $S_{1} = k$ hence, by induction of step 2, we obtain that $S_{n} \\in \\mathbb{Z}$ for $n = 2, 3, \\ldots$\nIt follows\n$$\n\\left\\{ a^{n} \\right\\} + \\left\\{ \\frac{1}{a^{n}} \\right\\} = a^{n} + \\frac{1}{a^{n}} - \\left[ a^{n} \\right] - \\left[ \\frac{1}{a^{n}} \\right] = S_{n} - \\left[ a^{n} \\right] - \\left[ \\frac{1}{a^{n}} \\right] \\in \\mathbb{Z},\n$$\nthat is $\\left\\{ a^{n} \\right\\} + \\left\\{ \\frac{1}{a^{n}} \\right\\} \\in \\mathbb{Z}$. We get $\\left\\{ a^{n} \\right\\} + \\left\\{ \\frac{1}{a^{n}} \\right\\} \\in \\{ 0, 1 \\}$. If $\\left\\{ a^{n} \\right\\} + \\left\\{ \\frac{1}{a^{n}} \\right\\} = 0$, then $a^{n}$ and $\\frac{1}{a^{n}}$ are both integers, not possible. Therefore, $\\left\\{ a^{n} \\right\\} + \\left\\{ \\frac{1}{a^{n}} \\right\\} = 1$, and we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40011,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which the equation\n$$\nx^{3}+y^{3}=n!+4\n$$\nhas solutions in integers.",
    "options": [],
    "answer": "n = 4 and n = 5",
    "solution": "For any integer $a$ we have $a^{3} \\equiv 0,1,8 \\pmod{9}$, hence for any integers $x$ and $y$, we have\n$$\nx^{3}+y^{3} \\equiv 0,1,2,7,8 \\quad (\\bmod 9).\n$$\nFor $n \\geq 6$, we have $n!+4 \\equiv 4 \\pmod{9}$, that is there are no solutions in this case.\n\nIf $n=1$, then the equation becomes $x^{3}+y^{3}=5$, which has no solutions because of the possible residues above.\n\nIf $n=2$, then we get the equation $x^{3}+y^{3}=6$, which has no solutions because of the possible residues above.\n\nIf $n=3$, then the equation becomes $x^{3}+y^{3}=10$.\nFor any integer $a$ we have $a^{3} \\equiv 0, \\pm 1 \\pmod{7}$, hence for any integers $x$ and $y$, we have\n$$\nx^{3}+y^{3} \\equiv 0, \\pm 1, \\pm 2 \\quad (\\bmod 7).\n$$\nThe equation has no solutions because of the possible residues above.\n\nIf $n=4$, then the equation becomes $x^{3}+y^{3}=28$, with solution $(x, y)=(3,1)$.\n\nIf $n=5$, then we get the equation $x^{3}+y^{3}=124$, with solution $(x, y)=(5,-1)$.\n\nThe equation has solutions in integers if and only if $n=4$ and $n=5$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40012,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all quadruples $ (x, y, z, w) $ of integers satisfying the system of equations\n$$\nx + y + z + w = x y + y z + z x + w^{2} - w = x y z - w^{3} = -1.\n$$",
    "options": [],
    "answer": "(1, 1, -2, -1), (1, -2, 1, -1), (-2, 1, 1, -1)",
    "solution": "The system is equivalent to\n$$\n\\left\\{\n\\begin{array}{l}\nx + y + z = -(w + 1) \\\\\nx y + y z + z x = -\\left(w^{2} - w + 1\\right) \\\\\nx y z = w^{3} - 1\n\\end{array}\n\\right.\n$$\nMultiplying the first two equations of the system we get\n$$\n(x + y + z)(x y + y z + z x) = w^{3} + 1.\n$$\nEliminating $w$ between this relation and the last equation, it follows\n$$\n(x + y + z)(x y + y z + z x) - x y z = 2.\n$$\nThe equation (1) is equivalent to\n$$\n(x + y)(y + z)(z + x) = 2.\n$$\nTaking into account the symmetry of equation (3) we have to consider only the following situations:\n$$\n\\left\\{\n\\begin{array}{l}\nx + y = 2 \\\\\ny + z = -1 \\\\\nz + x = -1\n\\end{array}\n\\quad\n\\left\\{\n\\begin{array}{l}\nx + y = 2 \\\\\ny + z = 1 \\\\\nz + x = 1\n\\end{array}\n\\quad\n\\left\\{\n\\begin{array}{l}\nx + y = -2 \\\\\ny + z = 1 \\\\\nz + x = -1\n\\end{array}\n\\right.\n\\right.\n\\right.\n$$\nBecause $w^{2} - w + 1 > 0$, from the second equation of (1) it follows that the solutions to the above system must satisfy\n$$\nx y + y z + z x < 0.\n$$\nThe first system has solution $(x, y, z) = (1, 1, -2)$. Replacing in the first equation of (1) we obtain $w = -1$. In this case we get solutions $(1, 1, -2, -1)$, $(1, -2, 1, -1)$, $(-2, 1, 1, -1)$.\nThe solutions to the second and third system do not satisfy condition $x y + y z + z x < 0$, hence they will not give solutions to our problem.\n\n\nSolution 2:\n$x, y, z$ are the roots of the polynomial\n$$\nP(t) = t^{3} + (w + 1) t^{2} - \\left(w^{2} - w + 1\\right) t - w^{3} + 1\n$$\nWe have\n$$\nP(t) = (t + w + 1)\\left(t^{2} - w^{2} + w - 1\\right) + 2.\n$$\nSince $P(x) = 0$, it follows\n$$\n(x + w + 1)\\left(x^{2} - w^{2} + w - 1\\right) = -2\n$$\nThere are four possibilities\n$$\n\\left\\{\n\\begin{array}{l}\nx + w + 1 = 1 \\\\\nx^{2} - w^{2} + w - 1 = -2\n\\end{array}\n\\quad\n\\left\\{\n\\begin{array}{l}\nx + w + 1 = -1 \\\\\nx^{2} - w^{2} + w - 1 = 2\n\\end{array}\n\\right.\n$$\n$$\n\\left\\{\n\\begin{array}{l}\nx + w + 1 = 2 \\\\\nx^{2} - w^{2} + w - 1 = -1\n\\end{array}\n\\quad\n\\left\\{\n\\begin{array}{l}\nx + w + 1 = -2 \\\\\nx^{2} - w^{2} + w - 1 = 1\n\\end{array}\n\\right.\n$$\nThe second and the fourth system have no solutions in integers. From the first and from the second system we get $w = -1$ and $w = 1$.\nIf $w = -1$, the polynomial $P$ is $P(t) = t^{3} - 3 t + 2$ with roots $x = y = 1, z = -2$. We get solutions $(1, 1, -2, -1)$, $(1, -2, 1, -1)$, $(-2, 1, 1, -1)$.\nIf $w = 1$, the polynomial $P$ is $P(t) = t^{3} + 2 t^{2} - t$ and not all the roots are integers.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40013,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a rectangle of center $O$, such that $\\widehat{DAC}=60^{\\circ}$. The angle bisector of $\\widehat{DAC}$ 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": "Detailed solution",
    "solution": "We have $\\widehat{SAC}=\\widehat{SCA}=30^{\\circ}$, so $SA=SC$. Since $OA=OC$, we get $SO \\perp AC$, hence $LA=LC$. Moreover $\\widehat{LAC}=60^{\\circ}$, so triangle $ALC$ is equilateral.\n\n![](attached_image_1.png)\n\nPoint $S$ is the centroid of $\\triangle LAC$, thus $\\frac{LS}{SO}=2$. $DBCL$ is a parallelogram, hence $DQ=QC$. We get that $M$ is the centroid of $\\triangle DBC$, so $\\frac{CM}{MO}=2$.\n\nFinally, since $\\frac{LS}{SO}=\\frac{CM}{MO}$, we obtain that $SM \\parallel CL$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40014,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f_{n} = 2^{2^{n}} + 1$, $n = 1, 2, 3, \\ldots$, be the Fermat's numbers. Find the least real number $C$ such that\n$$\n\\frac{1}{f_{1}} + \\frac{2}{f_{2}} + \\frac{2^{2}}{f_{3}} + \\ldots + \\frac{2^{n-1}}{f_{n}} < C\n$$\nfor all positive integers $n$.",
    "options": [],
    "answer": "1/3",
    "solution": "We have for $k = 1, 2, \\ldots$\n$$\n2^{2^{k+1}} - 1 = \\left(2^{2^{k}}\\right)^{2} - 1 = \\left(2^{2^{k}} - 1\\right)\\left(2^{2^{k}} + 1\\right),\n$$\nhence we get\n$$\n\\frac{1}{2^{2^{k+1}} - 1} = \\frac{1}{2} \\left( \\frac{1}{2^{2^{k}} - 1} - \\frac{1}{2^{2^{k}} + 1} \\right)\n$$\nIt follows\n$$\n\\begin{equation*}\n\\frac{1}{f_{k}} = \\frac{1}{x_{k}} - \\frac{2}{x_{k+1}}, \\quad k = 1, 2, \\ldots \\tag{1}\n\\end{equation*}\n$$\nwhere $x_{k} = 2^{2^{k}} - 1$. From (1) we get for $n = 1, 2, \\ldots$\n$$\n\\begin{equation*}\n\\frac{1}{f_{1}} + \\frac{2}{f_{2}} + \\frac{2^{2}}{f_{3}} + \\ldots + \\frac{2^{n-1}}{f_{n}} = \\frac{1}{x_{1}} - \\frac{2^{n}}{x_{n+1}} = \\frac{1}{3} - \\frac{2^{n}}{x_{n+1}} \\tag{2}\n\\end{equation*}\n$$\nFrom (2) it follows\n$$\n\\frac{1}{f_{1}} + \\frac{2}{f_{2}} + \\frac{2^{2}}{f_{3}} + \\ldots + \\frac{2^{n-1}}{f_{n}} < \\frac{1}{3}\n$$\nfor all positive integers $n$. Since\n$$\n\\lim_{n \\rightarrow \\infty} \\frac{2^{n}}{x_{n+1}} = \\lim_{n \\rightarrow \\infty} \\frac{2^{n}}{2^{2^{n+1}} - 1} = \\lim_{t \\rightarrow \\infty} \\frac{t}{2^{2t} - 1} = 0\n$$\nwe obtain that the least real number $C$ with the desired property is $C = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40015,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be positive integers such that $a + b + c + d = 2011$. Prove that $2011$ is not a divisor of $a b - c d$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have\n$$\n(a + c)(b + c) = a b + a c + b c + c^2 = (a + b + c + d) c + a b - c d = 2011 c + a b - c d\n$$\nBecause $2011$ is a prime, if $2011 \\mid a b - c d$, then $2011 \\mid a + c$ or $2011 \\mid b + c$. This is not possible since $0 < a + c < 2011$, and $0 < b + c < 2011$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40016,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with medians $m_{a}, m_{b}, m_{c}$. Prove that:\n\na. There is a triangle with side lengths $m_{a}, m_{b}, m_{c}$.\n\nb. This triangle is similar to $ABC$ if and only if the squares of the side lengths of triangle $ABC$ form an arithmetical sequence.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A'$, $B'$, $C'$ be the midpoints of sides $BC$, $CA$, $AB$, respectively. Construct the parallelogram $AC'CC''$. The points $C'$, $B'$, $C''$ are collinear, hence $BA'C''B'$ is also a parallelogram, that is $A'C'' = BB'$.\n\n![](attached_image_1.png)\n\nThe desired triangle is $AA'C''$, and $AA' = m_{a}$, $A'C'' = m_{b}$, $AC'' = m_{c}$.\n\n![](attached_image_1.png)\n\nb. Recall the median formula\n$$\nm_{a}^{2} = \\frac{1}{4}\\left(2(b^{2} + c^{2}) - a^{2}\\right)\n$$\ntogether with the other two similar formulas for $m_{b}$ and $m_{c}$.\nAssume that the squares of the sides are in arithmetic progression, for example $2b^{2} = a^{2} + c^{2}$. Then $m_{a}^{2} = \\frac{3}{4}c^{2}$, $m_{b}^{2} = \\frac{3}{4}b^{2}$, $m_{c}^{2} = \\frac{3}{4}a^{2}$, implying that\n$$\n\\frac{m_{a}}{c} = \\frac{m_{b}}{b} = \\frac{m_{c}}{a} = \\frac{\\sqrt{3}}{2}\n$$\nObviously, the triangles are similar.\nConversely, assume the triangles are similar. Let $a \\leq b \\leq c$. Then $m_{a} \\geq m_{b} \\geq m_{c}$, so\n$$\n\\frac{m_{a}}{c} = \\frac{m_{b}}{b} = \\frac{m_{c}}{a} = k\n$$\nFrom the equality $m_{a}^{2} + m_{b}^{2} + m_{c}^{2} = \\frac{3}{4}(a^{2} + b^{2} + c^{2})$ we get $k = \\frac{\\sqrt{3}}{2}$. Hence $4m_{b}^{2} = 3b^{2}$, which yields $2b^{2} = a^{2} + c^{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40017,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\left(F_{n}\\right)_{n \\geq 0}$ be the sequence of Fibonacci numbers: $F_{0}=0$, $F_{1}=1$ and $F_{n+2}=F_{n+1}+F_{n}$, for every $n \\geq 0$. Prove that for any prime $p \\geq 3$, $p$ divides $F_{2p}-F_{p}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Observe that\n$$\n\\begin{gathered}\nF_{2p} = \\frac{1}{\\sqrt{5}} \\left[ \\left( \\frac{1+\\sqrt{5}}{2} \\right)^{2p} - \\left( \\frac{1-\\sqrt{5}}{2} \\right)^{2p} \\right] \\\\\n= \\frac{1}{\\sqrt{5}} \\left[ \\left( \\frac{1+\\sqrt{5}}{2} \\right)^p - \\left( \\frac{1-\\sqrt{5}}{2} \\right)^p \\right] \\left[ \\left( \\frac{1+\\sqrt{5}}{2} \\right)^p + \\left( \\frac{1-\\sqrt{5}}{2} \\right)^p \\right] \\\\\n= F_p L_p\n\\end{gathered}\n$$\nwhere $L_n$ is the Lucas number, that is\n$$\nL_n = \\left( \\frac{1+\\sqrt{5}}{2} \\right)^n + \\left( \\frac{1-\\sqrt{5}}{2} \\right)^n, \\quad n=0,1, \\ldots\n$$\nit follows $F_{2p} - F_p = F_p L_p - F_p = F_p (L_p - 1)$. Now\n$$\n\\begin{aligned}\n& L_p - 1 = \\left( \\frac{1+\\sqrt{5}}{2} \\right)^p + \\left( \\frac{1-\\sqrt{5}}{2} \\right)^p - 1 \\\\\n= & \\frac{1}{2^{p-1}} \\left[ \\binom{p}{0} + \\binom{p}{2} 5 + \\ldots + \\binom{p}{p-1} 5^{\\frac{p-1}{2}} \\right] - 1\n\\end{aligned}\n$$\n$$\n= \\frac{1}{2^{p-1}} \\left[ \\binom{p}{2} 5 + \\binom{p}{4} 5^{2} + \\ldots + \\binom{p}{p-1} 5^{\\frac{p-1}{2}} - \\left(2^{p-1} - 1\\right) \\right],\n$$\nhence $p \\mid L_p - 1$, since $p$ divides $\\binom{p}{2}, \\binom{p}{4}, \\ldots, \\binom{p}{p-1}$, and $2^{p-1} - 1$, and $\\gcd(p, 2) = 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40018,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p \\geq 3$ be a prime. For $j=1,2, \\ldots, p-1$, let $r_{j}$ be the remainder when the integer $\\frac{j^{p-1}-1}{p}$ is divided by $p$. Prove that\n$$\nr_{1}+2 r_{2}+\\ldots+(p-1) r_{p-1} \\equiv \\frac{p+1}{2}(\\bmod p)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $j=1,2, \\ldots, p-1$, we have\n$$\n\\frac{j^{p-1}-1}{p}=a_{j} p+r_{j}\n$$\nfor some integer $a_{j}$. It follows\n$$\n\\frac{j^{p}-j}{p}=j a_{j} p+j r_{j},\n$$\nhence\n$$\n\\frac{j^{p}-j+(p-j)^{p}-(p-j)}{p}=j a_{j} p+j r_{j}+(p-j) a_{p-j} p+(p-j) r_{p-j}.\n$$\nWe obtain\n$$\n\\frac{j^{p}+(p-j)^{p}}{p}=j a_{j} p+j r_{j}+(p-j) a_{p-j} p+(p-j) r_{p-j}+1.\n$$\nBecause\n$$\nj^{p}+(p-j)^{p}=\\binom{p}{0} p^{p}-\\binom{p}{1} p^{p-1} j+\\ldots+\\binom{p}{p-1} p j^{p-1}\n$$\nwe obtain that $p^{2} \\mid j^{p}+(p-j)^{p}$ and we get for all $j=1,2, \\ldots, p-1$,\n$$\nj r_{j}+(p-j) r_{p-j}+1 \\equiv 0 \\quad(\\bmod p)\n$$\nAdding up all these relations it follows\n$$\n2\\left(r_{1}+2 r_{2}+\\ldots+(p-1) r_{p-1}\\right) \\equiv-(p-1)(\\bmod p)\n$$\nhence\n$$\nr_{1}+2 r_{2}+\\ldots+(p-1) r_{p-1} \\equiv \\frac{p+1}{2}(\\bmod p)\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40019,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with circumcenter $O$. Points $P$ and $Q$ are interior to sides $CA$ and $AB$, respectively. Circle $\\mathcal{C}$ passes through the midpoints of segments $BP$, $CQ$, $PQ$. Prove that if line $PQ$ is tangent to circle $\\mathcal{C}$, then $OP = OQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K$, $L$, $M$, $B'$, $C'$ be the midpoints of $BP$, $CQ$, $PQ$, $CA$, and $AB$, respectively. Since $CA \\parallel LM$, we have $\\widehat{LMP} = \\widehat{QPA}$. Since $\\mathcal{C}$ touches the segment $PQ$ at $M$, we find $\\widehat{LMP} = \\widehat{LKM}$. It follows\n$$\n\\widehat{QPA} = \\widehat{LKM} \\tag{1}\n$$\n![](attached_image_1.png)\nSimilarly, from $AB \\parallel MK$ we get\n$$\n\\widehat{PQA} = \\widehat{KLM} \\tag{2}\n$$\nFrom (1) and (2) we obtain that triangles $APQ$ and $MKL$ are similar, hence\n$$\n\\frac{AP}{AQ} = \\frac{MK}{ML} = \\frac{\\frac{QB}{2}}{\\frac{PC}{2}} = \\frac{QB}{PC} \\tag{3}\n$$\nNow (3) is equivalent to $AP \\cdot PC = AQ \\cdot QB$ which means that the power of points $P$ and $Q$ with respect to the circumcircle of $\\triangle ABC$ are equal, hence $OP = OQ$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40020,
    "subject": "Mathematics (Multi-modal)",
    "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": "Detailed 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 therefore 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 $2012 < \\sum_{i=1}^{k} \\frac{1}{n_i}$.\n\nSuppose 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": 40021,
    "subject": "Mathematics (Multi-modal)",
    "question": "Divide a rectangle of dimension $m \\times n$ by lines parallel to the sides into $mn$ congruent squares. Find the number of squares in this configuration.",
    "options": [],
    "answer": "n(n+1)(3m - n + 1)/6",
    "solution": "There are $mn$ squares of length side $1$, $(m-1)(n-1)$ squares of length side $2$, $(m-2)(n-2)$ squares of side length $3$, etc.\nAssume that $m \\ge n$. The total number of squares in this configuration is\n$$\n\\sum_{k=0}^{n-1} (m-k)(n-k). \\quad (1)\n$$\nWe have\n$$\n\\begin{aligned}\n\\sum_{k=0}^{n-1} (m-k)(n-k) &= \\sum_{k=0}^{n-1} (mn - (m+n)k + k^2) \\\\\n&= mn^2 - (m+n) \\frac{(n-1)n}{2} + \\frac{(n-1)n(2n-1)}{6} \\\\\n&= mn^2 - m \\frac{(n-1)n}{2} - \\frac{(n-1)n^2}{2} + \\frac{(n-1)n(2n-1)}{6} \\\\\n&= \\frac{n(n+1)(3m-n+1)}{6}.\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40022,
    "subject": "Mathematics (Multi-modal)",
    "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 $ad = bc$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that\n$$\n\\begin{aligned}\n\\frac{\\gcd(acn + bc, acn + ad)}{ac} &\\le \\gcd(an + b, cn + d) \\\\\n&\\le \\gcd(acn + bc, acn + ad).\n\\end{aligned}\n$$\nSuppose $ad = bc$. We have\n$$\n\\begin{aligned}\n\\gcd(an + b, cn + d) &\\ge \\frac{\\gcd(acn + bc, acn + ad)}{ac} \\\\\n&= \\frac{\\gcd(acn + bc, acn + bc)}{ac} \\\\\n&= n + \\frac{b}{a}.\n\\end{aligned}\n$$\n$n + \\frac{b}{a}$ is unbounded, so $\\gcd(an+b, cn+d)$ is also unbounded as desired.\n\nOn the other hand, suppose $ad \\neq bc$. Then $|ad - bc| \\neq 0$, so therefore\n$$\n\\begin{aligned}\n\\gcd(an + b, cn + d) &\\le \\gcd(acn + bc, acn + ad) \\\\\n&= \\gcd(acn + bc, ad - bc) \\\\\n&\\le |ad - bc|.\n\\end{aligned}\n$$\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": 40023,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that\n$$\n\\sum_{n=0}^{1006} \\frac{2012!}{(n!(1006-n)!)^2}\n$$\nis a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that\n$$\n\\sum_{n=0}^{1006} \\frac{2012!}{(n!(1006-n)!)^2} = \\frac{2012!}{(1006!)^2} \\sum_{n=0}^{1006} \\left( \\frac{1006!}{n!(1006-n)!} \\right)^2 = \\left( \\frac{2012!}{1006!} \\right) \\sum_{n=0}^{1006} \\left( \\frac{1006!}{n!} \\right)^2.\n$$\nSince $\\binom{1006}{n} = \\binom{1006}{1006-n}$, we have\n$$\n\\sum_{n=0}^{1006} \\left(\\binom{1006}{n}\\right)^2 = \\sum_{n=0}^{1006} \\binom{1006}{n} \\binom{1006}{1006-n} = \\binom{2012}{1006},\n$$\nby the Vandermonde identity. Therefore,\n$$\n\\sum_{n=0}^{1006} \\frac{2012!}{(n!(1006-n)!)^2} = \\left(\\binom{2012}{1006}\\right)^2\n$$\nwhich is a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40024,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, points $D$ and $E$ lie on side $BC$ and $AC$ respectively such that $AD \\perp BC$ and $DE \\perp AC$. The circumcircle of triangle $ABD$ meets segment $BE$ at point $F$ (other than $B$). Ray $AF$ meets segment $DE$ at point $P$. Prove that\n$$\n\\frac{DP}{PE} = \\frac{CD}{DB}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $AED$ and $ADC$ are both right triangles, $\\widehat{ADP} = 90^\\circ - \\widehat{DAC} = \\widehat{ECB}$. Also, since $AFDB$ is a cyclic quad, $\\widehat{DAP} = \\widehat{EBC}$.\n\n![](attached_image_1.png)\n\nTherefore, $ADP$ and $BCE$ are similar and $\\frac{BC}{EC} = \\frac{AD}{DP}$. Also, $ADC$ and $DEC$ are similar since both are right angled triangles that share an acute angle. Therefore,\n$$\n\\frac{AD}{DC} = \\frac{DE}{EC}.\n$$\nThese two equations imply that\n$$\nBC \\cdot DP = AD \\cdot EC = DC \\cdot DE,\n$$\nso we have\n$$\n\\frac{BC}{DC} = \\frac{DE}{DP}.\n$$\nSince $BC = BD + DC$ and $DE = DP + PE$, subtracting one from both sides gives us $\\frac{BD}{DC} = \\frac{PE}{DP}$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40025,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three equal circles of radius $R$ are given such that each one passes through the centers of the other two. Find the area of the common region.",
    "options": [],
    "answer": "(1/2)(pi - sqrt(3)) R^2",
    "solution": "![](attached_image_1.png)\nLet $O_1$, $O_2$, $O_3$ be the centers of the three circles and $S$ the area of the common region. The three sectors with centers $O_1$, $O_2$, $O_3$ which subtend the arcs $O_2O_3$, $O_1O_3$, $O_2O_1$, respectively, cover the surface of area $S$ and twice more the surface of triangle $O_1O_2O_3$.\n\nThe area of triangle $O_1O_2O_3$ is $\\frac{R^2\\sqrt{3}}{4}$.\n\nOn the other hand, the area of each of these three circular sectors equals $\\frac{1}{3}$ the area of a semicircle of radius $R$, hence it is $\\frac{1}{6}\\pi R^2$.\n\nHence\n$$\n\\frac{1}{2}\\pi R^2 = S + 2 \\cdot \\frac{R^2\\sqrt{3}}{4}.\n$$\n$$\n\\text{Therefore } S = \\frac{1}{2}(\\pi - \\sqrt{3})R^2.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40026,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive integer $k$ define\n$$\nH_k = 1 + \\frac{1}{2} + \\dots + \\frac{1}{k}.\n$$\nProve the relation\n$$\n1 + \\frac{1}{n+1} \\sum_{k=1}^{n} H_k = H_{n+1}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have\n$$\n\\begin{align*}\n\\sum_{k=1}^{n} H_k &= 1 + \\left(1 + \\frac{1}{2}\\right) + \\left(1 + \\frac{1}{2} + \\frac{1}{3}\\right) + \\dots + \\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{n}\\right) \\\\\n&= n + \\frac{n-1}{2} + \\frac{n-2}{3} + \\dots + \\frac{1}{n} \\\\\n&= (n+1) - 1 + \\frac{n+1-2}{2} + \\frac{n+1-3}{3} + \\dots + \\frac{n+1-n}{n} \\\\\n&= (n+1) \\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{n}\\right) - n = (n+1)H_n - n.\n\\end{align*}\n$$\nUsing the above relation we get\n$$\n\\begin{align*}\n1 + \\frac{1}{n+1} \\sum_{k=1}^{n} H_k &= 1 + \\frac{1}{n+1} [(n+1)H_n - n] \\\\\n&= H_n + 1 - \\frac{n}{n+1} = H_n + \\frac{1}{n+1} = H_{n+1}.\n\\end{align*}\n$$\nFor $n=1$ we have $1+\\frac{1}{2}H_1 = 1+\\frac{1}{2}=H_2$. Assume that\n$$\n1 + \\frac{1}{n+1} \\sum_{k=1}^{n} H_k = H_{n+1}.\n$$\nWe get\n$$\n\\begin{align*}\n1 + \\frac{1}{n+2} \\sum_{k=1}^{n+1} H_k &= 1 + \\frac{1}{n+2} H_{n+1} + \\frac{n+1}{n+2} \\cdot \\frac{1}{n+1} \\sum_{k=1}^{n} H_k \\\\\n&= 1 + \\frac{1}{n+2} H_{n+1} - \\frac{n+1}{n+2} + \\frac{n+1}{n+2} \\cdot H_{n+1} \\\\\n&= H_{n+1} + \\frac{1}{n+2} = H_{n+2},\n\\end{align*}\n$$\nand the relation is proved by induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40027,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral with $\\hat{B} \\neq 90^\\circ$ and\n$$\nAB^2 + BC^2 + CD^2 + DA^2 = 2AC^2.\n$$\nProve that the midpoint of diagonal $BD$ is on $AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ be the intersection point of the diagonals $AC$ and $BD$.\n\n![](attached_image_1.png)\n\nSince $\\hat{B} \\ne 90^\\circ$ it follows that $\\hat{D} \\ne 90^\\circ$, and hence we have $\\cos \\hat{B} \\ne 0$ and $\\cos \\hat{D} \\ne 0$. The quadrilateral $ABCD$ is cyclic, so therefore $\\hat{B} + \\hat{D} = 180^\\circ$. It follows $\\cos \\hat{B} + \\cos \\hat{D} = 0$, and by the Cosine Law we obtain\n$$\n\\frac{AB^2 + BC^2 - AC^2}{2AB \\cdot BC} + \\frac{AD^2 + DC^2 - AC^2}{2AD \\cdot DC} = 0. \\quad (1)\n$$\nThe given relation is equivalent to\n$$\nAB^2 + BC^2 - AC^2 = -(AD^2 + DC^2 - AC^2) \\neq 0,\n$$\nand from (1) it follows $AB \\cdot BC = AD \\cdot DC$. This implies that $K[ABC] = K[ADC]$, and hence $BB' = DD'$, where $BB'$ and $DD'$ are the altitudes of triangles $ABC$ and $ADC$, respectively.\n\nThe triangles $BB'M$ and $DD'M$ are congruent, so $MB = MD$, and we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40028,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M = \\{1, 2, 4, 5, 7, 8, \\dots\\}$ be the set of all positive integers not divisible by $3$. The sum of $2n$ consecutive elements of $M$ is $300$. Determine the possible values of $n$.",
    "options": [],
    "answer": "{1, 2, 4, 5, 10}",
    "solution": "Let $S_i$ be the sum of the first $i$ numbers in the set $M$. We have\n$$\n\\begin{aligned}\nS_{2k} &= (1+2) + (4+5) + \\dots + (3k-2+3k-1) \\\\\n&= (6 \\cdot 1 - 3) + (6 \\cdot 2 - 3) + \\dots + (6k-3) \\\\\n&= 3k(k+1) - 3k = 3k^2.\n\\end{aligned}\n$$\nThen $S_{2k+1} = S_{2k} + 3k + 1 = 3k^2 + 3k + 1$.\n\n**Case 1.** The sum of $2n$ consecutive elements of $M$ is\n$$\na_{2k+1} + a_{2k+2} + \\dots + a_{2l} = S_{2l} - S_{2k},\n$$\nwhere $n = l - k$. We obtain $3(l^2 - k^2) = 300$, so therefore\n$$\n(l-k)(l+k) = 100,\n$$\ngiving the systems\n$$\n\\begin{cases} l - k = 1 \\\\ l + k = 100 \\end{cases} \\quad ; \\quad \\begin{cases} l - k = 2 \\\\ l + k = 50 \\end{cases} \\quad ; \\quad \\begin{cases} l - k = 4 \\\\ l + k = 25 \\end{cases} ; \\\\\n\\begin{cases} l - k = 5 \\\\ l + k = 20 \\end{cases} ; \\quad \\begin{cases} l - k = 10 \\\\ l + k = 10 \\end{cases}\n$$\nOnly the second and the last are solvable in integers, and we obtain $l = 26$, $k = 24$ and $l = 10$, $k = 0$. Thus we get $n = 2$, $n = 10$.\n\n**Case 2.** The sum of $2n$ consecutive elements of $M$ is\n$$\na_{2k} + a_{2k+1} + \\dots + a_{2l-1},\n$$\nwhere $n = l - k$, $k \\ge 1$. We obtain $3(l^2 + l - k^2 - k) = 300$, and so $(l-k)(l+k+1) = 100$.\nThe solutions that work are\n$$\n\\begin{cases} l - k = 1 \\\\ l + k = 99 \\end{cases} \\quad ; \\quad \\begin{cases} l - k = 4 \\\\ l + k = 24 \\end{cases} \\quad ; \\quad \\begin{cases} l - k = 5 \\\\ l + k = 19 \\end{cases}\n$$\nWe get\n$l = 50, k = 49$, $l = 14, k = 10$, $l = 12, k = 7$,\nmeaning that $n = 1, 4, 5$.\n\nThe solutions are $n \\in \\{1, 2, 4, 5, 10\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40029,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $G$ be the centroid of triangle $ABC$ with the side-lengths $a$, $b$, $c$. Prove that if $a + BG = b + AG$ and $b + CG = c + BG$, then triangle $ABC$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The relations are equivalent to\n$$\na + \\frac{2}{3}m_b = b + \\frac{2}{3}m_a \\quad \\text{and} \\quad b + \\frac{2}{3}m_c = c + \\frac{2}{3}m_b.\n$$\nWe will prove that if $a \\le b$, then $m_a \\ge m_b$. Indeed, we have\n$$\nm_a^2 - m_b^2 = \\frac{2(b^2 + c^2) - a^2}{4} - \\frac{2(a^2 + c^2) - b^2}{4} = \\frac{3}{4}(b^2 - a^2) \\ge 0, \\\\\n\\text{and hence } m_a \\ge m_b.\n$$\nIt follows that if $a \\le b \\le c$, then $m_a \\ge m_b \\ge m_c$. From the given relations we have\n$$\na - b = \\frac{2}{3}(m_a - m_b) \\ge 0 \\quad \\text{and} \\quad b - c = \\frac{2}{3}(m_b - m_c) \\ge 0.\n$$\nThat is, $a \\ge b \\ge c$, and thus $a = b = c$.\nWe shall use the notation in the following figure.\n![](attached_image_1.png)\nThe relations in the problem imply that the triangles $AGB$, $BGC$, $CGA$ have the same perimeter, and hence the same semiperimeter $s'$. Also, these triangles have the same areas. From Heron's formula it follows that\n$$\ns'(s' - a)(s' - b')(s' - c') = s'(s' - a')(s' - b)(s' - c'),\n$$\nso $ab' = a'b$. Similarly, $ac' = a'c$ and $bc' = b'c$. Using these relations and the equality of the areas of $AGB$, $BGC$, $CGA$, we get $\\widehat{BAG} = \\widehat{BCG}$, $\\widehat{CBG} = \\widehat{CAG}$, and $\\widehat{ABG} = \\widehat{ACG}$. Denote these angles by $\\alpha$, $\\beta$, $\\gamma$ respectively.\nWe have $2(\\alpha + \\beta + \\gamma) = 180^\\circ$, so $\\alpha + \\beta + \\gamma = 90^\\circ$. It follows that $\\widehat{BB'C} = 90^\\circ$, that is $BB'$ is altitude. Similarly, the other medians are altitudes, hence triangle $ABC$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40030,
    "subject": "Mathematics (Multi-modal)",
    "question": "A square of side $1$ is decomposed into $9$ equal squares of sides $\\frac{1}{3}$ and the one in the center is painted black. The remaining eight squares are analogously divided into nine squares each, and the squares in the centers are painted in black. Prove that after $1000$ steps the total area of the black region exceeds $0.999$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The first step gives rise to one black square of area $\\left(\\frac{1}{3}\\right)^2 = \\frac{1}{9}$. After the second step we obtain eight more squares of side $\\frac{1}{9}$, the black region increasing thus by $\\frac{8}{9^2}$. In the same manner, the third step increases the black area by $8^2 = 64$ black squares, each of area $\\frac{1}{27^2}$, that is at this stage the black area becomes\n$$\n\\frac{1}{9} + \\frac{8}{9^2} + \\frac{8^2}{9^3}.\n$$\nWe conclude that after $1000$ steps, the area of the black region is\n$$\n\\begin{aligned}\n\\frac{1}{9} + \\frac{8}{9^2} + \\frac{8^2}{9^3} + \\dots + \\frac{8^{999}}{9^{1000}} &= \\frac{1}{9} \\left( 1 + \\frac{8}{9} + \\left(\\frac{8}{9}\\right)^2 + \\dots + \\left(\\frac{8}{9}\\right)^{999} \\right) \\\\\n&= \\frac{1}{9} \\cdot \\frac{1 - \\left(\\frac{8}{9}\\right)^{1000}}{1 - \\frac{8}{9}} = 1 - \\left(\\frac{8}{9}\\right)^{1000}.\n\\end{aligned}\n$$\nIt remains to prove that the last number is greater than $0.999$, i.e., $1 - \\left(\\frac{8}{9}\\right)^{1000} > 0.999$, or equivalently $\\left(\\frac{8}{9}\\right)^{1000} < 0.001$.\n\nThis easily follows by using a binomial expansion:\n$$\n\\left(\\frac{9}{8}\\right)^{1000} = \\left(1 + \\frac{1}{8}\\right)^{1000} > \\left(\\frac{1000}{2}\\right) \\cdot \\frac{1}{8^2} = \\frac{1000 \\cdot 999}{2 \\cdot 64} > 1000.\n$$\nThis completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40031,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ such that the inequality\n$$\n\\sqrt{x-1} + \\sqrt{x-2} + \\cdots + \\sqrt{x-n} < x\n$$\nholds for every real number $x \\ge n$.",
    "options": [],
    "answer": "n = 1 and n = 2",
    "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$$\n\\sqrt{x-1} + \\sqrt{x-2} < 2\\sqrt{x-1} \\le x,\n$$\n$$\n\\text{since we have } x^2 - 4x + 4 = (x-2)^2 \\ge 0.\n$$\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": 40032,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that $5^{n^2} + 7$ is divisible by $6$.",
    "options": [],
    "answer": "All odd positive integers",
    "solution": "We have $5^{n^2} + 7 = (6 - 1)^{n^2} + 7 \\equiv (-1)^{n^2} + 1 \\pmod{6}$. It follows that $5^{n^2} + 7$ is divisible by $6$ if and only if $(-1)^{n^2} + 1 = 0$. This is equivalent to $n^2$ being odd. Therefore, the possible values of $n$ are all odd positive integers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40033,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square. Points $E$ and $F$ lie on sides $CD$ and $DA$, respectively, such that $\\overline{EBF} = 45^\\circ$. Given that $DE = 12^{34}$, determine the number of triples $(k, m, n)$ of positive integers with $AB = k$, $DF = m$, and $EF = n$.",
    "options": [],
    "answer": "4657",
    "solution": "Let $\\widehat{DBE} = x$, $\\widehat{DBF} = y$, and construct the point $E'$ on the ray $DA$ such that $\\widehat{E'BA} = y$.\n\n![](attached_image_1.png)\n\nThe triangles $BEC$ and $BE'A$ are congruent, so $BE' = BE$. Also, we have $\\triangle BEF \\cong \\triangle BE'F$ by SAS congruency.\n\nFrom $E'F = E'A + AF = EC + AF = (k - 12^{34}) + k - m$ and $E'F = EF = n$, it follows $n = (k - 12^{34}) + (k - m)$, and hence we get\n$$\nk = \\frac{1}{2}(m + n + 12^{34}). \\quad (1)\n$$\nThe triples $(k, m, n)$ are in bijection with the pairs $(m, n)$ satisfying $m^2 + 12^{68} = n^2$. Indeed, any integral solution $(m, n)$ to $m^2 + 12^{68} = n^2$ satisfies $m \\equiv n \\pmod 2$, meaning $k$ is an integer. We can write\n$$\n12^{68} = n^2 - m^2 = (n - m)(n + m) = 4 \\cdot \\frac{n - m}{2} \\cdot \\frac{n + m}{2}, \\quad (2)\n$$\nwhich is the same as\n$$\n2^{134} \\cdot 3^{68} = \\frac{n-m}{2} \\cdot \\frac{n+m}{2}. \\quad (3)\n$$\nFor every divisor $d$ of $2^{134} \\cdot 3^{68}$, with $d^2 < 2^{134} \\cdot 3^{68}$, we obtain a solution $\\frac{n-m}{2} = d$, $\\frac{n+m}{2} = \\frac{2^{134} \\cdot 3^{68}}{d} > d$ and conversely. We get $\\frac{135 \\cdot 69 - 1}{2}$ solutions since $2^{134} \\cdot 3^{68}$ has $135 \\cdot 69$ divisors.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40034,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define the sequence $x_1, x_2, \\dots$, by $x_1 = \\frac{1}{6}$ and\n$$\nx_{n+1} = \\frac{n+1}{n+3} \\left( x_n + \\frac{1}{2} \\right),\n$$\nfor every $n \\ge 1$. Find $x_{2011}$.",
    "options": [],
    "answer": "2011/6",
    "solution": "We have\n$$\nx_2 = \\frac{2}{4} \\left( \\frac{1}{6} + \\frac{1}{2} \\right) = \\frac{2}{6}, \\quad x_3 = \\frac{3}{5} \\left( \\frac{2}{6} + \\frac{1}{2} \\right) = \\frac{3}{6}, \\text{ etc.}\n$$\nWe will prove by induction that for every $n \\ge 1$, we have $x_n = \\frac{n}{6}$.\nIndeed, assuming $x_n = \\frac{n}{6}$, it follows\n$$\nx_{n+1} = \\frac{n+1}{n+3} \\left( \\frac{n}{6} + \\frac{1}{2} \\right) = \\frac{n+1}{n+3} \\cdot \\frac{n+3}{6} = \\frac{n+1}{6},\n$$\nand we are done.\nThe answer is $x_{2011} = \\frac{2011}{6}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40035,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, let $\\phi(n)$ be the number of positive integers less than $n$ and relatively prime to $n$ (by convention $\\phi(1) = 1$). Let $\\tau(n)$ be the number of positive integers that are divisors of $n$. Find all positive integers $n$ such that\n$$\n\\phi(n) + \\tau(n) > n.\n$$",
    "options": [],
    "answer": "n = 1, n = 4, or n is prime",
    "solution": "The answer is $n = 1$, $n = 4$, or $n$ is any prime number. Let $S$ be the set of integers less than $n$ and relatively prime to $n$, and let $T$ be the set of integers that are positive divisors of $n$. Then\n$$\n(i) \\quad |S| = \\phi(n) \\text{ and } |T| = \\tau(n),\n$$\n$$\n(ii) \\quad S, T \\subseteq \\{1, 2, \\dots, n\\}, \\text{ and}\n$$\n$$\n(iii) \\quad S \\cap T = \\{1\\}.\n$$\nWe require $|S| + |T| > n$. By the principle of inclusion-exclusion, $|S| + |T| - |S \\cap T| = |S \\cup T|$. We know $|S \\cap T| = 1$, and by $S, T \\subseteq \\{1, 2, \\dots, n\\}$, we have $|S \\cup T| \\le n$. Therefore $|S| + |T| > n$ if and only if $|S \\cup T| = n$, meaning that every positive integer in $\\{1, 2, \\dots, n\\}$ is either relatively prime to $n$ or a divisor of $n$. It is easy to check that this is true for $n = 1$, $n$ prime, and $n = 4$. Now suppose $n$ is a composite number.\n\nLet $d$ be the smallest divisor of $n$ greater than $1$. $d < n$ since $n$ is not prime. Consider $n-d$. We have $\\gcd(n-d, n) = d > 1$, so $n-d$ must be a divisor of $n$. But $n = k d$ for some $k$, so if $n-d$ divides $n$ then we have an $m$ such that $m(kd - d) = k d$. This gives $m(k-1) = k$. This is only possible for $k=2$, so $n = 2d$.\n\nThen since $n$ is even, $2$ is the smallest divisor of $n$ greater than $1$ and $d = 2$. Therefore $n = 4$, and this is the only composite number that works.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40036,
    "subject": "Mathematics (Multi-modal)",
    "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": "96",
    "solution": "Partition the set $\\{1, 2, 3, \\dots, 30\\}$ into 20 subsets as follows:\n$$\n\\begin{aligned}\n& \\{1, 3, 9, 27\\}, \\\\\n& \\{2, 6, 18\\}, \\\\\n& \\{4, 12\\}, \\{5, 15\\}, \\{7, 21\\}, \\{8, 24\\}, \\{10, 30\\}, \\\\\n& \\{11\\}, \\{13\\}, \\{14\\}, \\{16\\}, \\{17\\}, \\{19\\}, \\{20\\}, \\\\\n& \\{22\\}, \\{23\\}, \\{25\\}, \\{26\\}, \\{28\\}, \\{29\\}.\n\\end{aligned}\n$$\nA subset of $\\{1, 2, 3, \\dots, 30\\}$ is delicious if and only if it does not contain two consecutive elements from any one of the 20 sets listed above. 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$$\n3 \\cdot 1 \\cdot 2^5 \\cdot 1^{13} = 96\n$$\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": 40037,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)_{n \\ge 1}$ be a sequence with the property that for every prime $p$ and for every positive integer $k$ the following relation holds:\n$$\na_{kp+1} = pa_k - 3a_p + 13.\n$$\n\nFind $a_{2011}$.",
    "options": [],
    "answer": "2014",
    "solution": "For $k = q$, prime, we have\n$$\na_{qp+1} = pa_q - 3a_p + 13 = qa_p - 3a_q + 13,\n$$\nso $(p+3)a_q = (q+3)a_p$. It follows that for every primes $p$ and $q$,\n$$\n\\frac{a_p}{p+3} = \\frac{a_q}{q+3}. \\qquad (1)\n$$\nThen, since $2011$ is a prime, we obtain\n$$\n\\frac{a_{2011}}{2014} = \\frac{a_7}{10}. \\qquad (2)\n$$\nOn the other hand, $a_7 = a_{2 \\cdot 3 + 1} = 2a_3 - 3a_2 + 13$. Hence, using (1),\n$$\n\\frac{a_7}{10} = \\frac{a_3}{6} = \\frac{a_2}{5} = \\frac{a_7 - 2a_3 + 3a_2}{10 - 12 + 15} = \\frac{13}{13} = 1.\n$$\nFrom (2) we get $a_{2011} = 2014$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40038,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be the set of positive integers. Determine all functions $f: S \\to S$ such that $x^2 + f(y)$ divides $f(x)^2 + y$ for every pair of positive integers $x$ and $y$.",
    "options": [],
    "answer": "f(x) = x for all positive integers x",
    "solution": "Consider $x = y = 1$. Then we get that $\\frac{1 + f^2(1)}{1 + f(1)}$ is an integer, so $1 + f(1)$ divides $2f(1)$. It follows $1 + f(1) = 2$, that is $f(1) = 1$.\n\nFor $y = 1$, we obtain $x^2 + 1$ divides $f^2(x) + 1$. Therefore $x^2 \\leq f^2(x)$, which means that $x \\leq f(x)$ for every positive integer $x$.\n\nFor $x = 1$, we get $1 + f(y)$ divides $1 + y$, and hence $f(y) \\leq y$, for every positive integer $y$.\n\nFrom the above inequalities it follows that the unique function is $f(x) = x$, $x \\in S$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40039,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive real numbers $x$, $y$, $z$ if\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": "x = 1, y = 2, z = 3",
    "solution": "We have $x^2+1 \\ge 2x$, $y^2+4 \\ge 4y$, $z^2+9 \\ge 6z$, hence\n$$\n\\frac{1}{x^2+1} + \\frac{1}{y^2+4} + \\frac{1}{z^2+9} \\le \\frac{1}{2x} + \\frac{1}{4y} + \\frac{1}{6z}. \\quad (1)\n$$\nUsing Cauchy-Schwarz inequality it follows\n$$\n\\left(\\frac{1}{2x} + \\frac{1}{4y} + \\frac{1}{6z}\\right)^2 \\le \\left(\\frac{1}{2^2} + \\frac{1}{4^2} + \\frac{1}{6^2}\\right) \\left(\\frac{1}{x^2} + \\frac{1}{y^2} + \\frac{1}{z^2}\\right) = \\frac{1}{4} \\cdot \\frac{49}{36} \\left(\\frac{1}{x^2} + \\frac{1}{y^2} + \\frac{1}{z^2}\\right),\n$$\nhence\n$$\n\\frac{1}{2x} + \\frac{1}{4y} + \\frac{1}{6z} \\le \\frac{7}{12} \\sqrt{\\frac{1}{x^2} + \\frac{1}{y^2} + \\frac{1}{z^2}}. \\quad (2)\n$$\nFrom (1) and (2) we get\n$$\n\\frac{1}{x^2+1} + \\frac{1}{y^2+4} + \\frac{1}{z^2+9} \\le \\frac{7}{12} \\sqrt{\\frac{1}{x^2} + \\frac{1}{y^2} + \\frac{1}{z^2}}.\n$$\nwith equality if and only if $x = 1$, $y = 2$, $z = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40040,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every real number $x$ the following inequality holds:\n$$\nx^6 + x^4 - x^3 - x + \\frac{3}{4} > 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The inequality is equivalent to\n$$\nx^6 - x^3 + \\frac{1}{4} + x^4 - x^2 + \\frac{1}{4} + x^2 - x + \\frac{1}{4} > 0,\n$$\nwhich in turn is equivalent to\n$$\n\\left(x^3 - \\frac{1}{2}\\right)^2 + \\left(x^2 - \\frac{1}{2}\\right)^2 + \\left(x - \\frac{1}{2}\\right)^2 > 0.\n$$\n\nSince $\\frac{1}{\\sqrt{2}} \\neq \\frac{1}{2}$, equality cannot occur.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40041,
    "subject": "Mathematics (Multi-modal)",
    "question": "The median $AM$, $M \\in BC$, of the acute-angled triangle $ABC$ intersects the circumcircle of triangle at $D$. Let $E$ be the symmetric point of $A$ with respect to $M$. Prove that $BC$ is the common tangent of the circumcircles of triangles $BDE$ and $CDE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\n$ABEC$ is a parallelogram. We have $\\overline{CBD} \\equiv \\overline{CAD} \\equiv \\overline{AEB}$, so $BC$ is tangent to the circumcircle of triangle $BDE$. Similarly, from $\\overline{BCD} \\equiv \\overline{BAD} \\equiv \\overline{AEC}$, it follows that $BC$ is tangent to the circumcircle of triangle $CDE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40042,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. Point $D$ lies on side $BC$. Let $O$, $O_1$, and $O_2$ be the circumcenters of triangle $ABC$, $ABD$, and $ACD$, respectively. Prove that circumcircles of triangles $BOO_1$ and $COO_2$ meet on line $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be on $BC$ such that $OP$ is parallel to $AD$. If $O$ is on $BC$, then $P = O$ and the result is clear. We claim that the circumcircles of $BOO_1$ and $COO_2$ both pass through $P$. One of the angles $\\widehat{ADB}$ and $\\widehat{ADC}$ is not acute. Without loss of generality, assume that $\\widehat{ADB} \\ge 90^\\circ$. Then $O_1$ does not lie in the interior of triangle $ADB$. Note that\n$$\n\\widehat{OO_1B} = \\frac{\\widehat{AO_1B}}{2} = 180^\\circ - \\widehat{ADB} = 180^\\circ - \\widehat{OPB},\n$$\nimplying that $BO_1OP$ is cyclic.\n\n![](attached_image_1.png)\n\nSimilarly, we can show that $CO_2OP$ is cyclic. Therefore, the circumcircles of $BOO_1$ and $COO_2$ pass through $P$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40043,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be rational numbers such that\n$$\n\\frac{1}{a+bc} + \\frac{1}{b+ac} = \\frac{1}{a+b}.\n$$\nProve that $\\sqrt{\\frac{c-3}{c+1}}$ is rational.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The given relation is equivalent to\n$$\n(b + ac + a + bc)(a + b) = ab + c(a^2 + b^2) + abc^2,\n$$\nso therefore\n$$\n(a+b)^2c + (a+b)^2 = ab(c^2+1) + c(a^2+b^2).\n$$\nIt follows\n$$\n\\begin{aligned}\n(a+b)^2 &= ab(c^2+1) + c[a^2 + b^2 - (a+b)^2] \\\\\n&= ab(c^2+1) - 2abc = ab(c-1)^2.\n\\end{aligned} \\quad (1)\n$$\nIf $c=1$, from the given relation we get\n$$\n\\frac{1}{a+b} + \\frac{1}{b+a} = \\frac{1}{a+b},\n$$\nwhich is not possible. Therefore, we have $c \\ne 1$, and from (1) we obtain\n$$\nab = \\left(\\frac{a+b}{c-1}\\right)^2. \\quad (2)\n$$\nUsing (2), it follows\n$$\n\\begin{aligned}\n(c-3)(c+1) &= (c-1)^2 - 4 = \\frac{(a+b)^2}{ab} - 4 = \\frac{(a-b)^2}{ab} \\\\\n&= \\left[ \\frac{(a-b)(c-1)}{a+b} \\right]^2.\n\\end{aligned}\n$$\nFinally,\n$$\n\\sqrt{\\frac{c-3}{c+1}} = \\frac{\\sqrt{(c-3)(c+1)}}{|c+1|} = \\frac{|a-b| \\cdot |c-1|}{|c+1| \\cdot |a+b|} \\in \\mathbb{Q}.\n$$\nAssume $b=0$. Then we get\n$$\n\\frac{1}{a} + \\frac{1}{ac} = \\frac{1}{a},\n$$\nwhich is not possible. It follows that $b \\neq 0$, so we can divide by $b$ in the given relation and obtain\n$$\n\\frac{1}{\\frac{a}{b} + c} + \\frac{1}{1 + \\frac{a}{b}c} = \\frac{1}{\\frac{a}{b} + 1}, \\quad \\frac{a}{b} \\in \\mathbb{Q}. \\quad (1)\n$$\nLet $\\frac{a}{b} = r \\in \\mathbb{Q}$. Then (1) is equivalent to\n$$\n\\frac{1}{r+c} + \\frac{1}{1+rc} = \\frac{1}{r+1}. \\quad (2)\n$$\nFrom (2) we obtain $(1+r+rc+c)(r+1) = (r+c)(rc+1)$, hence $r^2 - [(c-1)^2 - 2]r + 1 = 0$. This is a quadratic equation with rational roots, hence the discriminant $\\Delta$ must be a perfect square of a rational number. We have\n$$\n\\begin{aligned}\n\\Delta &= [(c-1)^2 - 2]^2 - 4 = [(c-1)^2 - 4](c-1)^2 \\\\\n&= (c-3)(c+1)(c-1)^2 = \\frac{c-3}{c+1}(c+1)^2(c-1)^2,\n\\end{aligned}\n$$\nimplying that $\\frac{c-3}{c+1}$ is a perfect square of a rational number, and we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40044,
    "subject": "Mathematics (Multi-modal)",
    "question": "Sequence $(x_1, x_2, \\dots)$ is defined as $x_1 = 20$, $x_2 = 12$,\n$$\nx_{n+2} = x_n + x_{n+1} + 2\\sqrt{x_n x_{n+1} + 121},\n$$\nfor $n \\ge 1$.\n1) Compute $x_{10}$.\n2) Determine with justification if every term in the sequence is an integer?",
    "options": [],
    "answer": "x_10 = 49824; every term is an integer",
    "solution": "It is clear that $x_3 = 20 + 12 + 2 \\cdot 19 = 70$. We note that\n$$\n\\begin{aligned}\nx_{n+3} &= x_{n+1} + x_{n+2} + 2\\sqrt{x_{n+1}x_{n+2} + 121} \\\\\n&= x_{n+1} + x_{n+2} + 2\\sqrt{x_{n+1}(x_n + x_{n+1} + 2\\sqrt{x_n x_{n+1} + 121}) + 121} \\\\\n&= x_{n+1} + x_{n+2} + 2\\sqrt{x_{n+1}^2 + 2x_{n+1}\\sqrt{x_n x_{n+1} + 121} + x_n x_{n+1} + 121} \\\\\n&= x_{n+1} + x_{n+2} + 2(x_{n+1} + \\sqrt{x_n x_{n+1} + 121}) \\\\\n&= 3x_{n+1} + x_{n+2} + 2\\sqrt{x_n x_{n+1} + 121} \\\\\n&= 3x_{n+1} + x_{n+2} + x_{n+2} - x_n - x_{n+1} \\\\\n&= 2x_{n+2} + 2x_{n+1} - x_n.\n\\end{aligned}\n$$\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": 40045,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ and primes $p \\ge 5$ such that\n$$\n(2p)^n + 1\n$$\nis a perfect cube.",
    "options": [],
    "answer": "n = 1, p = 13",
    "solution": "If $(2p)^n + 1 = a^3$, then we have\n$$\n(2p)^n = a^3 - 1 = (a-1)(a^2+a+1) = (a-1)[(a-1)^2+3(a-1)+3].\n$$\nSince $a$ is odd, it follows that $a^2+a+1$ is also odd, hence $2^n \\nmid a-1$. This means that we have $a = 2^n p^k + 1$ for some integer $k \\ge 0$.\n\nWe obtain\n$$\n2^n p^n = 2^n p^k (2^{2n} p^{2k} + 3 \\cdot 2^n p^k + 3). \\quad (1)\n$$\n\n**Case 1.** $k > 0$. Clearly, we have $k < n$, hence from (1) it follows $p|2^{2n} p^{2k} + 3 \\cdot 2^n p^k + 3$. We get $p|3$, not possible since $p \\ge 5$.\n\n**Case 2.** $k = 0$. From (1) we obtain\n$$\np^n = 4^n + 3 \\cdot 2^n + 3. \\quad (2)\n$$\nFor $n = 1$ we have $p = 13$.\n\nFor $n = 2$ we have $p^2 = 31$, not possible.\n\nFor $n = 3$ we have $p^3 = 64 + 24 + 3 < 5^3$, not possible.\n\nFor $n \\ge 3$ we prove by induction that\n$$\n5^n > 4^n + 3 \\cdot 2^n + 3 = p^n,\n$$\nand so $p < 5$, which is not possible.\n\nThe unique solutions are $n = 1, p = 13$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40046,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $P$ lies inside quadrilateral $ABCD$ such that $\\widehat{APD} = \\widehat{BPC} = 90^\\circ$ and $AP \\cdot DP = BP \\cdot CP$. Let $O$ denote the circumcenter of triangle $CDP$. Prove that line $OP$ bisects segment $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ be the midpoint of $AB$, and let $E$ be the point on line $BP$ such that $AE \\parallel MP$. Then $P$ is the midpoint of $EB$. Since $\\frac{AP}{PB} = \\frac{CP}{PD}$, we have $\\frac{AP}{PE} = \\frac{CP}{PD}$. Also, note that\n\n$\\widehat{EPC} = \\widehat{APD} = 90^\\circ$, so $\\widehat{EPA} = \\widehat{DPC}$. These last two facts imply that triangles $APE$ and $CPD$ are similar.\n\n![](attached_image_1.png)\n\nFrom this we conclude that $\\widehat{PAE} = \\widehat{PCD}$. Since $AE \\parallel MP$, $\\widehat{MPA} = \\widehat{PAE} = \\widehat{PCD}$. Take a point $O'$ on ray $MP$ past $P$. Then\n\n$$\n\\widehat{O'PD} = 180^\\circ - \\widehat{APD} - \\widehat{MPA} = 90^\\circ - \\widehat{PCD} = \\widehat{OPD}.\n$$\n\nTherefore $M, P, O$ are collinear, which completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40047,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every positive real numbers $a$, $b$, $c$,\n$$\n\\frac{1+a^2}{1+b} + \\frac{1+b^2}{1+c} + \\frac{1+c^2}{1+a} \\ge 6(\\sqrt{2}-1).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Solution 1.** Using AM-GM inequality we have\n$$\n\\frac{1+a^2}{1+b} + \\frac{1+b^2}{1+c} + \\frac{1+c^2}{1+a} \\ge 3\\sqrt[3]{\\frac{1+a^2}{1+a} \\cdot \\frac{1+b^2}{1+b} \\cdot \\frac{1+c^2}{1+c}} \\quad (1)\n$$\nOn the other hand, for every positive real number $x$, the following inequality holds:\n$$\n\\frac{1+x^2}{1+x} \\ge 2(\\sqrt{2}-1). \\quad (2)\n$$\nIndeed, the inequality (2) is equivalent to\n$$\nx^2 - 2(\\sqrt{2}-1)x + 1 - (2\\sqrt{2}-2) \\ge 0,\n$$\ni.e.\n$$\nx^2 - 2(\\sqrt{2}-1)x + (\\sqrt{2}-1)^2 \\ge 0,\n$$\nand hence $(x - (\\sqrt{2} - 1))^2 \\ge 0$.\nFrom (1) and (2) we obtain\n$$\n\\frac{1+a^2}{1+b} + \\frac{1+b^2}{1+c} + \\frac{1+c^2}{1+a} \\ge 3\\sqrt[3]{8(\\sqrt{2}-1)^3} = 6(\\sqrt{2}-1).\n$$\nWe have equality if and only if $a = b = c = \\sqrt{2} - 1$.\n\n\n**Solution 2.** The triples\n$$\n(1+a^2, 1+b^2, 1+c^2) \\quad \\text{and} \\quad \\left( \\frac{1}{1+a}, \\frac{1}{1+b}, \\frac{1}{1+c} \\right)\n$$\nhave opposite monotony. From the Rearrangement Inequality, it follows\n$$\n\\sum_{cyc} \\frac{1+a^2}{1+b} \\ge \\sum \\frac{1+a^2}{1+a} \\ge 6(\\sqrt{2}-1),\n$$\nwhere we have used the inequality (2) in the previous solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40048,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every positive integer $n$ the number\n$$\nN = \\underbrace{44\\dots4}_{n} \\underbrace{88\\dots8}_{n} - 1 \\underbrace{33\\dots3}_{n-1} 2,\n$$\nis a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have\n$$\n\\begin{aligned}\nN &= \\underbrace{44\\dots4}_{n-1} \\underbrace{355\\dots56}_{n-1} = \\underbrace{44\\dots4}_{2n} - \\underbrace{88\\dots8}_{n} \\\\\n&= 4(10^{2n-1} + \\dots + 10 + 1) - 8(10^{n-1} + \\dots + 10 + 1) \\\\\n&= 4 \\cdot \\frac{10^{2n}-1}{9} - 8 \\cdot \\frac{10^n-1}{9} = \\frac{1}{9}(4 \\cdot 10^{2n} - 8 \\cdot 10^n + 4) \\\\\n&= \\left(2 \\cdot \\frac{10^n-1}{3}\\right)^2,\n\\end{aligned}\n$$\nand we are done, since $\\frac{10^n - 1}{3}$ is an integer. Notice that, in fact,\n$$\nN = \\underbrace{66\\dots6}_{n}^2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40049,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an $n \\times n$ board, the numbers $0$ through $n^2 - 1$ are written so that the number in row $i$ and column $j$ is equal to\n$$\n(i-1) + n(j-1) \\text{ where } 1 \\le i, j \\le n.\n$$\nSuppose we select $n$ different cells of the board, where no two cells are in the same row or column. Find the maximum possible product of the numbers in the $n$ cells.",
    "options": [],
    "answer": "n!·(n−1)^n",
    "solution": "The answer is $n! \\cdot (n-1)^n$. This is achievable by choosing the numbers $n-1, 2(n-1), 3(n-1), \\dots, n(n-1)$, which are at positions $(n, 1), (n-1, 2), (n-2, 3), \\dots, (1, n)$. To show this is the best possible, we begin with a lemma.\n\n**Lemma.** If $a < b < c < d$ are real numbers such that $a+d = b+c$, then $bc > ad$.\n\n**Proof.** We have $(b-a)(c-a) > 0$, which implies\n$$\nbc + (a - b - c)a > 0.\n$$\nSince $a - b - c = -d$, this simplifies to $bc > ad$ as desired. $\\square$\n\nLet $f(i, j) = (i - 1) + n(j - 1)$ be the number in row $i$ and column $j$. Suppose we have chosen the $n$ numbers $f(1, j_1), f(2, j_2), \\dots, f(n, j_n)$. If the $j_i$ sequence is strictly decreasing, then $j_i = n + 1 - i$ for all $i$ and we get a configuration in the first paragraph. Else there exists $k, m$ such that $k < m$ and $j_k < j_m$. Notice that\n$$\nf(k, j_k) < f(m, j_k) < f(k, j_m) < f(m, j_m).\n$$\nMoreover,\n$$\n\\begin{aligned}\nf(k, j_k) + f(m, j_m) &= f(m, j_k) + f(k, j_m) \\\\\n&= (k + m - 2) + n(j_k + j_m - 2).\n\\end{aligned}\n$$\nSo by the lemma, we have that\n$$\nf(k, j_k)f(m, j_m) < f(k, j_m)f(m, j_k).\n$$\nTherefore we increase the product of the $n$ numbers by changing $f(k, j_k)$ and $f(m, j_m)$ to $f(k, j_m)$ and $f(m, j_k)$, so our original arrangement could not have been a maximum.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40050,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a polynomial with real coefficients and odd degree. Suppose that the number of real solutions to\n$$\nP(P(x)) = P(x), \\quad P(x) \\neq x\n$$\nis finite and odd. Show that there exists a real $c$ such that $P(c) = c$ and the polynomial $P(x) - c$ has a real root of multiplicity at least two. (That is, it is divisible by $(x - r)^2$ for some real $r$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S$ be the set of all $a$ with $P(a) = a$. For each $a \\in S$, let $T_a$ be the set of solutions to $P(P(x)) = P(x) = a$ with $x \\neq a$. Since $P(P(x)) = P(x)$ and $P(x) \\neq x$ has an odd number of solutions, the union of all $T_a$ has an odd number of elements. Therefore one of the $T_a$ has an odd number of elements; choose this $a$ to be our $c$ and let $T_a = \\{b_1, b_2, \\dots, b_k\\}$ where $k$ is odd.\nWe know that $P(x) - a$ has an even number of distinct real roots: $a, b_1, b_2, \\dots, b_k$. Because $P(x) - a$ has odd degree and its nonreal roots come in conjugate pairs, the combined multiplicity of its real roots must be odd. Then if neither $(x-a)^2$ nor $(x-b_i)^2$ for any $i$ divide $P(x) - a$, the combined multiplicity of its real roots would be even, a contradiction. So either $(x-a)^2$ or $(x-b_i)^2$ for some $i$ divide $P(x) - a$. This shows that our choice of $c$ works, and the proof is complete.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40051,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, $I_a$ the excenter opposite to $A$, and $M$ its reflection across $BC$. Prove that $AM$ is parallel to the Euler line of triangle $BCI_a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $I$ be the incenter of $ABC$, $H_a$ the orthocenter of $I_aBC$, and $O_a$ the midpoint of the segment $II_a$.\n\n![](attached_image_1.png)\n\nWe have\n$$\nBO_a = IO_a = I_aO_a = CO_a,\n$$\nso $O_a$ is the circumcenter of triangle $I_aBC$.\n\nMoreover, $IB \\parallel CH_a$ (both are perpendicular to $BI_a$) and, similarly, $IC \\parallel BH_a$, hence $BICH_a$ is a parallelogram.\n\nLet $P$ denote the projection of $I_a$ onto $BC$ and $T$ the midpoint of segment $AI_a$. We have $H_aP = r$, hence\n$$\n\\frac{IA}{AI_a} = \\frac{r}{r_a} = \\frac{H_aP}{I_aP}.\n$$\nTherefore\n$$\n\\frac{H_a I_a}{I_a P} = \\frac{I I_a}{I_a A} = \\frac{2 I_a O_a}{2 I_a T} = \\frac{I_a O_a}{I_a T}. \\quad (1)\n$$\nThe relation (1) proves that $O_a H_a$ is parallel to $TP$. But $TP \\parallel AM$, hence $AM \\parallel O_a H_a$, and we are done.\nWe will use the notation in the previous solution. The quadrilateral $BICI_a$ is cyclic, inscribed in a circle of diameter $II_a$, which implies $O_a$ is the midpoint of segment $II_a$. Also, we have that $IBH_aC$ is a parallelogram.\n\n![](attached_image_2.png)\n\nWe want to prove that\n$$\n\\frac{I_a O_a}{I_a A} = \\frac{I_a H_a}{I_a M}. \\qquad (1)\n$$\nWe have\n$$\n\\frac{IA}{II_a} = \\frac{K[BIA]}{K[BII_a]} = \\frac{c \\sin \\frac{B}{2}}{BI_a}.\n$$\nHence\n$$\n\\frac{AO_a}{O_a I_a} = \\frac{IA + \\frac{1}{2}II_a}{\\frac{1}{2}II_a} = \\frac{c \\sin \\frac{B}{2} + \\frac{1}{2}BI_a}{\\frac{1}{2}BI_a} = 1 + \\frac{c \\sin \\frac{B}{2}}{\\frac{1}{2}BI_a}, \\quad (2)\n$$\nand\n$$\n\\frac{TH_a}{H_a I_a} = \\frac{TC \\sin \\frac{B}{2}}{CI_a \\sin \\frac{A}{2}}.\n$$\nThe relation (3) implies\n$$\n\\frac{MH_a}{H_a I_a} = 1 + \\frac{2TC \\sin \\frac{B}{2}}{CI_a \\sin \\frac{A}{2}}. \\qquad (4)\n$$\nWe have\n$$\n\\frac{c \\sin \\frac{B}{2}}{BI_a} = \\frac{TC \\sin \\frac{B}{2}}{CI_a \\sin \\frac{A}{2}} \\Leftrightarrow \\frac{c \\sin \\frac{A}{2}}{TC} = \\frac{BI_a}{CI_a} = \\frac{\\sin\\left(90^\\circ - \\frac{C}{2}\\right)}{\\sin\\left(90^\\circ - \\frac{B}{2}\\right)} \\Leftrightarrow\n$$\n$$\n\\frac{c \\sin \\frac{A}{2}}{s-b} = \\frac{\\cos \\frac{C}{2}}{\\cos \\frac{B}{2}} \\Leftrightarrow \\frac{c \\sqrt{\\frac{(s-b)(s-c)}{bc}}}{s-b} = \\frac{\\sqrt{\\frac{s(s-c)}{ab}}}{\\sqrt{\\frac{s(s-b)}{ac}}} \\Leftrightarrow\n$$\n$$\n\\frac{\\sqrt{c}}{\\sqrt{s-b}} = \\frac{\\sqrt{c}}{\\sqrt{s-b}},\n$$\na relation which is obviously true. It follows that\n$$\n\\frac{AO_a}{O_a I_a} = \\frac{MH_a}{H_a I_a},\n$$\nand therefore $AM \\parallel O_aH_a$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40052,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every positive real numbers $x, y, z$ the following inequality holds\n$$\n\\frac{9}{x+y+z} - \\frac{1}{xyz} \\le 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the AM-GM inequality we have\n$$\n\\frac{9}{x+y+z} - \\frac{1}{xyz} \\le \\frac{3}{\\sqrt[3]{xyz}} - \\frac{1}{xyz}. \\quad (1)\n$$\nLet $t = \\frac{1}{\\sqrt[3]{xyz}}$. According to inequality (1) it is sufficient to prove that $3t - t^3 \\le 2$ for $t > 0$. The last inequality is equivalent to $t^3 - 3t + 2 \\ge 0$, which is the same as $(t - 1)^2(t + 2) \\ge 0$.\n\nWe have equality if and only if $t = 1$ and we have equality in the AM-GM inequality. That is, $xyz = 1$ and $x = y = z$, so $x = y = z = 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40053,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle right-angled at $A$. A circle passing through $B$ and $C$ intersects the sides $AB$ and $AC$ at $M$, respectively $N$. Prove that if $BM \\cdot CN \\cdot BC = MN^3$, then the symmetric point of $A$ with respect to the midpoint of the segment $MN$ belongs to $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $BC = a$, $CA = b$, $AB = c$, $AM = x$.\n\n![](attached_image_1.png)\n\nTriangles $AMN$ and $ACB$ are similar, so\n$$\n\\frac{x}{b} = \\frac{AN}{c} = \\frac{MN}{a}.\n$$\nWe obtain\n$$\nAN = \\frac{cx}{b}, \\quad MN = \\frac{ax}{b}. \\qquad (1)\n$$\nThe relation $BM \\cdot CN \\cdot BC = MN^3$ is equivalent to\n$$\na(c-x)\\left(b-\\frac{cx}{b}\\right) = \\frac{a^3x^3}{b^3},\n$$\nand hence we get\n$$\na^2x^3 - b^2cx^2 + b^2(b^2 + c^2)x - cb^4 = 0. \\qquad (2)\n$$\nSince $b^2 + c^2 = a^2$, it follows that\n$$\na^2x^3 - b^2cx^2 + a^2b^2x - cb^4 = 0. \\qquad (3)\n$$\nThe relation (3) is equivalent to\n$$\n(a^2x - b^2c)x^2 + b^2(a^2x - b^2c) = 0,\n$$\nso we get\n$$\n(a^2x - b^2c)(x^2 + b^2) = 0. \\qquad (4)\n$$\nFrom (4) it follows\n$$\nx = \\frac{b^2c}{a^2} \\qquad (5)\n$$\nDraw $AD \\perp BC$, $DM' \\perp AB$, $DN' \\perp AC$, where $D \\in BC$, $M' \\in AB$, $N' \\in AC$. We have $AD^2 = AM' \\cdot AB$, so\n$$\nAM' = \\frac{AD^2}{AB} = \\frac{b^2c^2}{a^2c} = \\frac{b^2c}{a^2} = x = AM.\n$$\nThis implies $M = M'$. Also, from $AD^2 = AN' \\cdot AC$, we get\n$$\nAN' = \\frac{AD^2}{AC} = \\frac{b^2c^2}{a^2b} = \\frac{bc^2}{a^2} = \\frac{cx}{b} = AN, \\text{ hence } N = N'.\n$$\nTherefore $AMDN$ is a rectangle. It follows that the symmetric point of $A$ with respect to the midpoint of segment $MN$ is $D$, which completes the proof.\nConsider the coordinates system with the origin at $A$ and $AB$ and $AC$ are the coordinate axes. We have $A(0, 0)$, $B(b, 0)$, $C(0, c)$, $M(m, 0)$, $N(0, n)$.\n\n![](attached_image_2.png)\n\nThe relation $BM \\cdot CN \\cdot BC = MN^3$ is equivalent to\n$$\n(b-m)(c-n)\\sqrt{b^2+c^2} = (\\sqrt{m^2+n^2})^3 \\quad (1)\n$$\nThe quadrilateral $CNMB$ is cyclic. Hence, from the power of $A$ with respect to its circumcircle we get $nc = mb$. That is, $n = \\frac{b}{c}m$. Replacing in (1) we get\n$$\n(b-m) \\left(c - \\frac{b}{c}m\\right) \\sqrt{b^2+c^2} = \\left(\\sqrt{m^2\\left(1+\\frac{b^2}{c^2}\\right)}\\right)^3 \\quad (2)\n$$\nBut the relation (2) is equivalent to\n$$\n(b-m) \\left(c - \\frac{b}{c}m\\right) a = \\frac{a^3}{c^3}m^3. \\quad (3)\n$$\nHence\n$$\n(m^2 + c^2)(a^2m - bc^2) = 0,\n$$\nwhich gives $m = \\frac{bc^2}{a^2}$. The symmetric point $A_1$ of $A$ with respect to the midpoint of segment $MN$ has the coordinates $(m, n)$. But $A_1 \\in BC$ if and only if $\\frac{m}{b} + \\frac{n}{c} = 1$, which is equivalent to $mc + nb = bc$. The last relation is equivalent to\n$$\nmc + \\frac{b^2}{c}m = bc, \\text{ and hence } m = \\frac{bc^2}{b^2 + c^2} = \\frac{bc^2}{a^2},\n$$\nwhich is already proved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40054,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle $ABC$ is inscribed in circle $\\omega$. Line $\\ell$ is tangent to $\\omega$ at $A$. Points $B_1$ and $C_1$ lie on $\\ell$ such that rays $CA$ and $BA$ bisect $\\widehat{BCB_1}$ and $\\widehat{CBC_1}$, respectively. Segments $BB_1$ and $CC_1$ intersect at $P$. The line through $P$ parallel to segment $BC$ intersects sides $AC$ and $AB$ at $B_2$ and $C_2$, respectively. Prove that if $P$ is the midpoint of $B_2C_2$ then $ABC$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that $P$ is the midpoint of segment $B_2C_2$; that is, $B_2P = C_2P$. We will show that $AB = AC$.\n\nSet $\\widehat{ABC} = B$, $\\widehat{BCA} = C$, and $\\widehat{CAB} = A$. Extend segment $AP$ through $P$ to meet segment $BC$ at $A_3$. Then it is clear $A_3$ is the midpoint of side $BC$ or $BA_3 = A_3C$. Let $B_3, C_3$ denote the intersections of pairs of segments $BB_1$ and $AC$, $CC_1$ and $AB$, respectively. By Ceva's theorem, we have\n$$\n\\frac{AC_3 \\cdot BA_3 \\cdot CB_3}{C_3B \\cdot A_3C \\cdot B_3A} = 1 \\quad \\text{or} \\quad \\frac{AC_3}{C_3B} = \\frac{AB_3}{B_3C}\n$$\n![](attached_image_1.png)\n\nNote that $AB_3/CB_3$ is equal to the ratio between the areas of triangle $ABB_1$ and $CBB_1$; that is,\n$$\n\\begin{aligned} \\frac{AB_3}{CB_3} &= \\frac{[ABB_1]}{[CBB_1]} = \\frac{AB \\cdot AB_1 \\cdot \\sin \\widehat{BAB_1}}{BC \\cdot CB_1 \\cdot \\sin \\widehat{BCB_1}} \\\\ &= \\frac{AB}{BC} \\cdot \\frac{AB_1}{CB_1} \\cdot \\frac{\\sin \\widehat{BAB_1}}{\\sin \\widehat{BCB_1}}. \\end{aligned}\n$$\nBecause line $AB_1$ is tangent to $\\omega$, $\\widehat{C_1AB} = \\widehat{ACB} = C$. Hence $\\sin \\widehat{BAB_1} = \\sin \\widehat{C_1AB} = \\sin C$ and\n$$\n\\frac{\\sin \\widehat{BAB_1}}{\\sin \\widehat{BCB_1}} = \\frac{\\sin C}{\\sin 2C} = \\frac{1}{\\cos C}.\n$$\nBecause line $AB_1$ is tangent to $\\omega$, we also have $\\widehat{B_1AC} = \\widehat{ABC} = B$. Thus, triangle $AB_1C$ is similar to triangle $BAC$. By the Law of sines, we have\n$$\n\\frac{AB}{BC} = \\frac{\\sin C}{\\sin A} \\quad \\text{and} \\quad \\frac{AB_1}{CB_1} = \\frac{\\sin C}{\\sin B}.\n$$\nIt follows that\n$$\n\\frac{AB_3}{CB_3} = \\frac{AB}{BC} \\cdot \\frac{AB_1}{CB_1} \\cdot \\frac{\\sin \\widehat{BAB_1}}{\\sin \\widehat{BCB_1}} = \\frac{\\sin^2 C}{\\sin A \\sin B \\cos C}.\n$$\nIn exactly the same way, we can show that\n$$\n\\frac{AC_3}{BC_3} = \\frac{\\sin^2 B}{\\sin A \\sin C \\cos B}.\n$$\nIt follows that\n$$\n\\frac{\\sin^2 C}{\\sin A \\sin B \\cos C} = \\frac{AB_3}{CB_3} = \\frac{AC_3}{BC_3} = \\frac{\\sin^2 B}{\\sin A \\sin C \\cos B}.\n$$\nor\n$$\n\\sin^2 C \\tan C = \\sin^2 B \\tan B.\n$$\nBecause we can have at most one of $\\tan B$ and $\\tan C$ being negative, we must have both of them positive, from which it follows that $B$ and $C$ are acute angles. For $0^\\circ < \\alpha < 90^\\circ$, both $\\sin \\alpha$ and $\\tan \\alpha$ are monotonically increasing, thus we must have $B = C$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40055,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive integers. Prove that if the numbers $\\frac{a^2}{a+b}$, $\\frac{b^2}{b+c}$, $\\frac{c^2}{c+a}$ are integers and primes, then $a = b = c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will use the following result:\n\n**Lemma.** If $x$ and $y$ are positive integers such that $\\frac{x^2}{x+y}$ is an integer and prime, then $y \\ge x$.\n\n**Proof of Lemma.** Assume that $\\frac{x^2}{x+y} = p$, where $p$ is a prime.\nWe have $py = x(x - p)$, so $p \\mid x(x - p)$, i.e. $p \\mid x$. Let $x = up$, for some positive integer $u$. We get\n$$\ny = x(up - p) = x(u - 1)p \\ge x,\n$$\nand we are done, since $u \\ge 2$.\n\nApplying Lemma in our situation, it follows $b \\ge a$, $c \\ge b$, $a \\ge c$, meaning that $a = b = c$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40056,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the greatest real number $a$ such that for every positive real numbers $x, y, z$ we have\n$$\n\\frac{x+1}{y} + \\frac{2y+1}{z} + \\frac{3z+1}{x} > a.\n$$",
    "options": [],
    "answer": "3√6",
    "solution": "For any positive real numbers $x, y, z$ we have\n$$\n\\begin{aligned}\n\\frac{x+1}{y} + \\frac{2y+1}{z} + \\frac{3z+1}{x} &= \\frac{x}{y} + \\frac{2y}{z} + \\frac{3z}{x} + \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} \\\\\n&\\geq 3\\sqrt{6} + \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} > 3\\sqrt{6}.\n\\end{aligned}\n$$\nHence $a \\geq 3\\sqrt{6}$.\n\nConsider the triples $(x, y, z)$ satisfying the equality case in the previous AM-GM inequality. We have\n$$\n\\frac{x}{y} = \\frac{2y}{z} = \\frac{3z}{x} = \\sqrt{6}.\n$$\nIt follows that $x = \\frac{3}{\\sqrt{6}}z$, $y = \\frac{\\sqrt{6}}{2}z$. So all the desired triples are given by $\\left(\\frac{3}{\\sqrt{6}}z, \\frac{\\sqrt{6}}{2}z, z\\right)$, and we obtain\n$$\n3\\sqrt{6} + \\left( \\frac{\\sqrt{6}}{3} + \\frac{2}{\\sqrt{6}} + 1 \\right) \\cdot \\frac{1}{z} > a,\n$$\nFor $z \\to +\\infty$, we deduce that $3\\sqrt{6} \\geq a$.\n\nThus, the answer is $a = 3\\sqrt{6}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40057,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with incenter $I$ and circumcenter $O$ and let $M$ be the midpoint of $BC$. The bisector of angle $A$ intersects lines $BC$ and $OM$ at $L$ and $Q$, respectively. Prove that\n$$\nAI \\cdot LQ = IL \\cdot IQ.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The bisector of $\\widehat{BAC}$ and the perpendicular bisector of side $BC$ intersect at $Q$, the midpoint of arc $\\widehat{BC}$ not containing $A$.\nWe have $BQ = IQ$, since $\\widehat{QBI} = \\widehat{QIB} = 90^\\circ - \\frac{1}{2}\\widehat{C}$.\nOn the other hand, the triangles $BLQ$ and $ALC$ are similar, meaning that\n$$\n\\frac{BQ}{LQ} = \\frac{AC}{LC}.\n$$\nUsing the angle bisector theorem in the triangles $ABC$ and $ABL$, we get\n$$\n\\frac{IQ}{LQ} = \\frac{BQ}{LQ} = \\frac{AC}{LC} = \\frac{AB}{BL} = \\frac{AI}{IL}.\n$$\nso $AI \\cdot LQ = IQ \\cdot IL$, as desired.\nLet us use the diagram in the previous solution. We have\n$$\n\\frac{AI}{IQ} = \\frac{K[ABI]}{K[QBI]} = \\frac{c \\sin \\frac{B}{2}}{BQ \\sin \\frac{A+B}{2}} = \\frac{c \\sin \\frac{B}{2}}{BQ \\cos \\frac{C}{2}}\n$$\n$$\n= \\frac{2R \\sin C \\sin \\frac{B}{2}}{2R \\sin \\frac{A}{2} \\cos \\frac{C}{2}} = \\frac{2 \\sin \\frac{B}{2} \\sin \\frac{C}{2}}{\\sin \\frac{A}{2}}, \\quad (1)\n$$\nand\n$$\n\\begin{aligned}\n\\frac{IL}{LQ} &= \\frac{K[BIL]}{K[BLQ]} = \\frac{BI \\sin \\frac{B}{2}}{BQ \\sin \\frac{A}{2}} = \\frac{\\sin C}{\\sin \\widehat{BIQ}} \\cdot \\frac{\\sin \\frac{B}{2}}{\\sin \\frac{A}{2}} \\\\\n&= \\frac{\\sin C \\sin \\frac{B}{2}}{\\sin \\frac{A+B}{2} \\sin \\frac{A}{2}} = \\frac{\\sin C \\sin \\frac{B}{2}}{\\cos \\frac{C}{2} \\sin \\frac{A}{2}} = \\frac{2 \\sin \\frac{C}{2} \\sin \\frac{B}{2}}{\\sin \\frac{A}{2}}, \\quad (2)\n\\end{aligned}\n$$\nsince\n$$\n\\widehat{BIQ} = 180^\\circ - \\frac{\\widehat{B}}{2} - \\widehat{ALB} = 180^\\circ - \\frac{\\widehat{B}}{2} - \\left( \\widehat{C} + \\frac{\\widehat{A}}{2} \\right) = \\frac{\\widehat{A} + \\widehat{B}}{2}.\n$$\nFrom (1) and (2) it follows $\\frac{AI}{IQ} = \\frac{IL}{LQ}$, hence $AI \\cdot LQ = IL \\cdot IQ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40058,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Q}$ be the set of rational numbers. Find all functions $f: \\mathbb{Q} \\to \\mathbb{Q}$ such that for all rational numbers $x, y$,\n$$\nf(f(x) + x f(y)) = x + f(x)y.\n$$",
    "options": [],
    "answer": "f(x) = x for all rational x",
    "solution": "**Solution 1.** We show $f(x) = x$ is the only solution. It is easy to check that it works.\n\nPut in $y = 0$ to obtain that $f$ is surjective. Let $c$ be a real with $f(c) \\neq 0$ and suppose $f(a) = f(b)$. Then\n$$\na = \\frac{f(f(c) + c f(a)) - c}{f(c)} = \\frac{f(f(c) + c f(b)) - c}{f(c)} = b,\n$$\nso $f$ is also injective.\n\nLet $z$ be the real number such that $f(z) = 0$. Then $x = z$ gives $f(z f(y)) = z$. Choosing $y$ such that $f(y) = 1$ or $f(y) = 0$, possible by surjectivity, shows that $f(z) = f(0) = z$. By injectivity $z = 0$, so $f(0) = 0$.\n\nPut in $x = y = -1$ to obtain $f(-1) = -1$. Put in $y = -1$ to get $f(f(x) - x) = x - f(x)$. Let $d = f(1) - 1$; with $x = 1$ our previous equation this gives us $f(d) = -d$. Put in $x = d, y = 1$ to get $f(d f(1) - d) = f(d^2) = 0$. By injectivity, $d = 0$ and so $f(1) = 1$.\n\nPut in $y = 0$ to obtain $f(f(x)) = x$. Now put in $x = 1$ and change $y$ to $f(y)$ to get $f(1 + y) = 1 + f(y)$. This implies $f(n) = n$ for all integers $n$. Finally, for arbitrary integers $m, n$ with $n \\neq 0$, put in $x = n, y = \\frac{m}{n}$ to get $f(n + n f(\\frac{m}{n})) = n + m = f(n + m)$. By injectivity, $n f(\\frac{m}{n}) = m$, so $f(\\frac{m}{n}) = \\frac{m}{n}$. Since $\\frac{m}{n}$ is an arbitrary rational number, $f(x) = x$ for all $x$ in the domain.\n**Solution 2.** Put in $x = 0$ to obtain $f(f(0)) = f(0)y$ for all rational $y$. Thus $f(0) = 0$ since $f(0)y$ is constant. Now put in $y = 0$ to get $f(f(x)) = x$.\n\nLet $y = f(1)$. We get $f(f(x) + x) = x + f(x)f(1)$. If we replace $x$ with $f(x)$, then by $f(f(x)) = x$ we get $f(f(x) + x) = f(x) + x f(1)$. Thus $x + f(x)f(1) = f(x) + x f(1)$, or $(f(1) - 1)(x - f(x)) = 0$. Putting in $x = 1$ gives us $f(1) = 1$.\n\nPut $x = 1$ into the given equation to get $f(y) + 1 = f(y + 1)$. This implies $f(y + n) = f(y) + n$ for all integers $n$. Since $f(0) = 0$, $f(n) = n$ for all integers $n$. Finally, let $x = \\frac{2}{q}$ and $y = q$ where $\\frac{2}{q}$ is an arbitrary rational. This gives us $f(f(\\frac{2}{q}) + p) = \\frac{2}{q} + f(\\frac{2}{q})q$. But $f(f(\\frac{2}{q}) + p) = f(f(\\frac{2}{q})) + p = \\frac{2}{q} + p$, so $\\frac{2}{q} + p = \\frac{2}{q} + f(\\frac{2}{q})q$ and $f(\\frac{2}{q}) = \\frac{2}{q}$. Since $\\frac{2}{q}$ was arbitrary, $f(x) = x$ for all $x$, and this clearly satisfies the equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40059,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the isosceles triangle $ABC$ with $AB = AC$. A semicircle of diameter $EF$ situated on the side $BC$, is tangent to the sides $AB$ and $AC$ at $M$ and $N$, respectively. The line $AE$ intersects the semicircle at $P$. Prove that the line $PF$ passes through the midpoint of the chord $MN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the center of the semicircle and let $R$ be the midpoint of segment $MN$.\n\n![](attached_image_1.png)\n\nIn triangle $ANO$ we have $AN^2 = AR \\cdot AO$. Using the power of the point $A$ with respect to the circle we get\n$$\nAM^2 = AP \\cdot AE = AN^2 = AR \\cdot AO. \\quad (1)\n$$\nFrom (1) it follows that the quadrilateral $PROE$ is cyclic, hence $RP \\perp AE$. Since $FP \\perp AE$, we get that the points $F$, $R$, $P$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40060,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n \\ge 2$ for which the following statement is true:\nGiven any $n$ distinct points on the plane such that the distance between each pair of points is distinct, there exists a pair of points $A$, $B$ for which the difference between the number of points lying on either side of the perpendicular bisector of segment $AB$ is not greater than $1$.",
    "options": [],
    "answer": "n = 2 and n = 3",
    "solution": "The statement trivially holds for $n = 2$ and $n = 3$. We will show that it does not hold for any $n \\ge 4$.\n\nFirst suppose that $n$ is even. Setup a coordinate, and place a point at $(0,0)$. Place $n/2 - 1$ points on the negative $x$-axis and $n/2 - 1$ points on the positive $x$-axis at $(-1/2^i, 0)$ and $(2^i, 0)$ for $i = 1, 2, \\dots, n-1$. Place the last point on the positive $y$-axis, so that the perpendicular bisector of the segment joining that point and any of the previously placed point divides the plane into two parts, one of which contains only the last point. (It is clear that we can do this, and that this construction serves as a counterexample to the given statement.)\n\nIf $n$ is odd, we simply add another point on the negative $y$-axis, so that perpendicular bisector of the segment joining that point and any of the previously placed point of the $x$-axis divides the plane into two parts, one of which contains only the last point. Also, the distance from the $x$-axis of the point on the positive $y$-axis and the point on the negative $y$-axis should be distinct.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40061,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an arbitrary triangle. A circle passes through $B$ and $C$ and intersects the lines $AB$ and $AC$ in $D$ and $E$, respectively. The projections of the points $B$ and $E$ on $CD$ are denoted by $B'$ and $E'$, respectively. The projections of the points $D$ and $C$ on $BE$ are denoted by $D'$ and $C'$, respectively.\n\nProve that the points $B'$, $D'$, $E'$ and $C'$ lie on the same circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $I$ be the intersection point of the lines $BE$ and $CD$. The quadrilaterals $BD'B'D$ and $CE'C'E$ are cyclic, so $\\overline{BDB'} = \\overline{B'D'I}$ and $\\overline{CEC'} = \\overline{IE'C'}$. Since $BDEC$ is also cyclic, $\\overline{BDB'} = \\overline{CEC'}$. It follows that $\\overline{B'D'I} = \\overline{IE'C'}$, so $B'D'E'C'$ is a cyclic quadrilateral.\nUsing the power of a point theorem, one has:\n$$\nIB' \\cdot ID = ID' \\cdot IB\n$$\n$$\nIC' \\cdot IE = IE' \\cdot IC\n$$\n$$\nIE \\cdot IB = ID \\cdot IC\n$$\nFrom these one easily obtains\n$$\nIB' \\cdot IE' = ID' \\cdot IC'\n$$\nwhich proves that the quadrilateral $B'D'E'C'$ is cyclic, using the reciprocal of the power of a point theorem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40062,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that for every positive integers $n \\ge 3$ there are distinct positive integers $a_1, a_2, \\dots, a_n$ with\n$$\na_1! a_2! \\dots a_{n-1}! = a_n!\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $n = 3$ we have $3! \\cdot 5! = 6!$.\n\nAssume that $a_1! a_2! \\dots a_{k-1}! = a_k!$. Since $a_k!(a_k! - 1)! = (a_k!)!$, it follows that\n$$\na_1! a_2! \\dots a_{k-1}! (a_k! - 1)! = a_k! (a_k! - 1)! = (a_k!)!\n$$\nand obviously $a_k! > a_k! - 1 > a_{k-1}!$. We are done by induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40063,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider $S = \\{(x, y, z) \\mid x, y, z \\in \\{1, 2, \\dots, 2012\\}\\}$ as a set of $2012^3$ points in three-dimensional space. For any segment joining two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the space, we define its *distance triplet* to be the ordered triple\n$$\n(|x_1 - x_2|, |y_1 - y_2|, |z_1 - z_2|).\n$$\nAlice wants to draw segments in such a way that\n\na. Each segment joins two distinct points in $S$;\n\nb. Each point in $S$ is an endpoint of at most one segment;\n\nc. For any two segments, their distance triplets are different.\n\nFind the greatest number of segments that Alice can draw.",
    "options": [],
    "answer": "2012^3/2",
    "solution": "We claim that Alice can draw up to $K = \\frac{2012^3}{2}$ segments. Since there are $2012^3$ points, conditions (a) and (b) guarantee that Alice can draw at most $K$ segments. We will prove that she can do so.\n\nLet $T = \\{1, 2, \\dots, 2012\\}$. We will define a bijection $f: T \\to T$ such that for all distinct $i, j \\in T$, the inequality\n$$\n|f(i) - i| \\neq |f(j) - j|\n$$\nholds. We let\n$$\n\\begin{aligned}\n& (f(1), f(2), \\dots, f(2012)) \\\\\n= & (2012, 2011, \\dots, 1510, 504, 1509, 1508, \\dots, 1008, 1006, \\\\\n& \\qquad 1005, \\dots, 505, 503, 502, \\dots, 1, 1007).\n\\end{aligned}\n$$\nIt is not hard to verify that $f$ satisfies the desired inequality condition.\n\nFor each point $(x, y, z) \\in S$ such that $z \\le 1006$, Alice draws a segment between it and the point $(f(x), f(y), 2013 - z)$. The $K$ segments she has drawn are easily seen to satisfy (a) and (b). To verify that they satisfy (c), suppose that two segments have the same distance triplets. Assume that the first segment has $(x_1, y_1, z_1)$ where $z_1 \\le 1006$ as an endpoint, and the second segment has $(x_2, y_2, z_2)$ where $z_2 \\le 1006$ as an endpoint. The distance triplets of the two segments are\n$$\n(|f(x_1) - x_1|, |f(y_1) - y_1|, |2013 - 2z_1|)\n$$\nand\n$$\n(|f(x_2) - x_2|, |f(y_2) - y_2|, |2013 - 2z_2|)\n$$\nrespectively. For them to coincide, we must have that\n$$\n(x_1, y_1, z_1) = (x_2, y_2, z_2),\n$$\na contradiction.\nWe will use the following auxiliary result:\n\n**Lemma.** If $n \\equiv 0 \\pmod{4}$, then there exists a permutation $\\sigma \\in S_n$ such that\n$$\n\\{|\\sigma(i) - i| : i = 1, \\dots, n\\} = \\{0, 1, \\dots, n-1\\}. \\quad (1)\n$$\n**Proof of Lemma.** Consider the cycle defined by\n$$\n\\sigma = (1, n, 2, n-1, \\dots, \\frac{n}{4}, \\frac{3n}{4}+1, \\frac{n}{4}+1, \\frac{3n}{4}-1, \\dots, \\frac{n}{2}-1, \\frac{n}{2}+1, \\frac{n}{2}).\n$$\nIf $n = 4k$, then we have\n$$\n|\\sigma(1) - 1| = 4k - 1 \\qquad |\\sigma(2k + 1) - (2k + 1)| = 1\n$$\n$$\n|\\sigma(2) - 2| = 4k - 3 \\qquad |\\sigma(2k + 2) - (2k + 2)| = 3\n$$\n$$\n|\\sigma(k) - k| = 2k + 1 \\qquad |\\sigma(3k - 1) - (3k - 1)| = 2k - 3\n$$\n$$\n|\\sigma(k + 1) - (k + 1)| = 2k - 2 \\qquad |\\sigma(3k) - 3k| = 0\n$$\n$$\n|\\sigma(k + 2) - (k + 2)| = 2k - 4 \\qquad |\\sigma(3k + 1) - (3k + 1)| = 2k + 2\n$$\n$$\n|\\sigma(2k - 1) - (2k - 1)| = 2 \\qquad |\\sigma(4k - 1) - (4k - 1)| = 4k - 4\n$$\n$$\n|\\sigma(2k) - 2k| = 2k - 1 \\qquad |\\sigma(4k) - 4k| = 4k - 2.\n$$\n**Remark.** Such a permutation exists if and only if\n$$\nn \\equiv 0 \\pmod{4} \\text{ or } n \\equiv 1 \\pmod{4}.\n$$\nTo see the condition is necessary, notice that those $n$ distinct differences must be, in some order, the numbers $0, 1, \\dots, n-1$, as $0 \\le |\\sigma(k) - k| \\le n-1, k = 1, \\dots, n$. We must have\n$$\n\\frac{n(n-1)}{2} = \\sum_{k=1}^{n} |\\sigma(k) - k| \\equiv \\sum_{k=1}^{n} (\\sigma(k) - k) \\equiv 0 \\pmod{2}.\n$$\nIn our problem, clearly $2012 \\equiv 0 \\pmod{4}$. We connect the point $(x, y, z)$ with $(\\sigma(x), \\sigma(y), 2013 - z)$, where $z \\le 1006$, $x, y \\le 2012$, and $\\sigma \\in S_{2012}$ is the permutation in Lemma. The \"distance\" of a such segment is $(|x - \\sigma(x)|, |y - \\sigma(y)|, 2013 - 2z)$. Moreover, for two pairs $(x_1, y_1, z_1), (x_2, y_2, z_2), z_1, z_2 \\le 1006$, we have\n$$\n\\left\\{ \\begin{array}{l}\n|x_1 - \\sigma(x_1)| = |x_2 - \\sigma(x_2)| \\\\\n|y_1 - \\sigma(y_1)| = |y_2 - \\sigma(y_2)| \\\\\n2013 - 2z_1 = 2013 - 2z_2\n\\end{array} \\right.\n$$\nso all the \"distances\" are different and every point lies in some segment. It follows that we have $\\frac{2012^3}{2}$ segments.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40064,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $ABC$, the internal bisector of $\\hat{A}$ intersects $BC$ at $D$ and the external bisector of $\\hat{A}$ intersects $BC$ at $E$. Let $C_1$ be the circumcircle of $ADE$. $AC$ intersects $C_1$ at $F$. Let $C_2$ be the circumcircle of $ABF$ and the tangent of $C_2$ at $A$ intersects $C_1$ at $G$. Prove that $AF = AG$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\widehat{ABC} = 2\\beta$, $\\widehat{BAC} = 2\\alpha$, $\\widehat{ACB} = 2\\gamma$. Thus, the law of sines on triangle $AEF$ gives us\n\n$$\n\\frac{AE}{AF} = \\frac{\\sin(\\alpha + 2\\beta)}{\\sin(\\alpha - \\beta + \\gamma)}.\n$$\n![](attached_image_1.png)\n\nOn the other hand, using the law of sines on $ABD$, $ADF$ and the fact that $\\widehat{BAD} = 180^\\circ - \\widehat{DAF}$, we get that\n\n$$\n\\frac{AB}{AF} = \\frac{BD \\sin(\\widehat{BDA})}{DF \\sin(\\widehat{FDA})} = \\frac{BD \\sin(\\alpha + 2\\beta)}{DF \\sin(\\alpha - \\beta + \\gamma)}.\n$$\n\nCombining these two identities, we obtain that $\\frac{AE}{AB} = \\frac{DF}{DB}$. Then since we also have $\\widehat{FDB} = \\widehat{FAE} = \\widehat{EAB}$, triangle $EAB$ is similar to triangle $FDB$. Thus we find that $\\widehat{FBE} = 2\\beta$. Using this and the fact that $A, D, F, E, G$ are on the same circle, we are able to compute that $\\widehat{GEA} = \\alpha - \\beta + \\gamma$. But this was also computed to be equal to $\\widehat{AEF}$. Then $\\widehat{GFA} = \\widehat{GEA} = \\widehat{AEF} = \\widehat{AGF}$, so $AF = AG$ as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40065,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime. At any vertex of a regular polygon with $p$ sides it is written an integer. For any vertex of the polygon we compute the difference between the sum of the integers written at his neighbors and his number. After that we delete all the initial integers and replace them by the new obtained integers. Prove that the integers obtained after $p$ such steps are the same modulo $p$ with the initial integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_0A_1...A_{p-1}$ be the regular polygon and let $(a_0, a_1, ..., a_{p-1})$ be the integers written at its vertices, $a_k$ at $A_k$ for $k = 0, 1, ..., p-1$. Consider the polynomial with integer coefficients\n$$\nH(x) = a_0 + a_1x + ... + a_{p-1}x^{p-1}.\n$$\nApplying the transformation\n$$\na_0 \\to a_{p-1} + a_1 - a_0, \\quad a_1 \\to a_0 + a_2 - a_1, \\dots, a_{p-1} \\to a_{p-2} + a_0 - a_{p-1}\n$$\non the coefficients of the polynomial $H(x)$ we obtain the polynomial\n$$\nH_1(x) = H(x)(x^{p-1} + x - 1) \\pmod{(x^p - 1)}.\n$$\nAfter $s$ steps we obtain the polynomial\n$$\n\\begin{aligned}\nH_s(x) &= H_{s-1}(x)(x^{p-1} + x - 1) \\pmod{(x^p - 1)} \\\\\n&= H(x)(x^{p-1} + x - 1)^s \\pmod{(x^p - 1)}.\n\\end{aligned}\n$$\nFinally, after $p$ steps we get\n$$\nH_p(x) = (x^{p-1} + x - 1)^p H(x) \\pmod{(x^p - 1)} \\quad (1)\n$$\nBut\n$$\n(x^{p-1} + x - 1)^p = p \\cdot Q(x) + x^{p(p-1)} + x^p - 1\n$$\n$= p \\cdot Q(x) + (x^p)^{p-1} - 1 + x^p - 1 + 1 = p \\cdot Q(x) + 1 \\pmod{(x^p - 1)}$,\nwhere $Q$ is a polynomial with integer coefficients. From (1) we get\n$$\nH_p(x) = p \\cdot Q(x)H(x) + H(x) \\pmod{(x^{p-1})},\n$$\nTherefore $H_p$ and $H$ have the same coefficients modulo $p$.\nThe transformation in the problem is described by the following matrix relation\n$$\n\\begin{pmatrix} a_0^{(1)} \\\\ \\vdots \\\\ a_{p-1}^{(1)} \\end{pmatrix} = \\begin{pmatrix} -1 & 1 & 0 & \\cdots & 0 & 1 \\\\ 1 & -1 & 1 & \\cdots & 0 & 0 \\\\ 0 & 1 & -1 & \\cdots & 0 & 0 \\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots & \\vdots \\\\ 1 & 0 & 0 & \\cdots & 1 & -1 \\end{pmatrix} \\begin{pmatrix} a_0 \\\\ \\vdots \\\\ a_{p-1} \\end{pmatrix} \\quad (1)\n$$\nAfter $p$ steps we get\n$$\n\\begin{pmatrix} a_0^{(p)} \\\\ \\vdots \\\\ a_{p-1}^{(p)} \\end{pmatrix} = A^p \\cdot \\begin{pmatrix} a_0 \\\\ \\vdots \\\\ a_{p-1} \\end{pmatrix}, \\quad (2)\n$$\nwhere $A$ is the square matrix in relation (1).\nWe can write $A = -I_p + X + X^{p-1}$, where $X$ is the permutation matrix\n$$\nX = \\begin{pmatrix} 0 & 0 & \\dots & 0 & 1 \\\\ 1 & 0 & \\dots & 0 & 0 \\\\ 0 & 1 & \\dots & 0 & 0 \\\\ \\dots & \\dots & \\dots & \\dots & \\dots \\\\ 0 & 0 & \\dots & 1 & 0 \\end{pmatrix} .\n$$\nIn $M_p(\\mathbb{Z}_p)$ we have\n$$\n\\begin{align*}\nA^p &= (-I_p + X + X^{p-1})^p = (-I_p)^p + X^p + X^{p(p-1)} \\\\\n&= -I_p + I_p + I_p = I_p,\n\\end{align*}\n$$\nsince $X^p = I_p$. From (2) it follows that in $M_{p,1}(\\mathbb{Z}_p)$ we have\n$$\n\\begin{pmatrix} a_0^{(p)} \\\\ \\vdots \\\\ a_{p-1}^{(p)} \\end{pmatrix} = \\begin{pmatrix} a_0 \\\\ \\vdots \\\\ a_{p-1} \\end{pmatrix},\n$$\nand we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40066,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be a positive integer. Suppose we have a circle with $n$ positions labeled $1, 2, \\dots, n$ in clockwise order. $n$ counters, each with one side white and the other side black, are placed on the circle with one counter in each of the numbered positions. Initially all counters have the white side facing up, except the counter at position $1$ which shows black. We are allowed to perform the following operation:\n(i) Choose a counter $X$ whose black side is facing up, and let $Y, Z$ be the next two counters in clockwise order.\n(ii) Flip counter $Y$ so that it is showing the other color.\n(iii) Move $X$ two spaces clockwise, and move $Y, Z$ one space counterclockwise. (So after the operation we still have one counter in each position.)\nLet $S$ be any nonempty subset of $\\{1, 2, \\dots, n\\}$. Show that we can perform a finite sequence of moves after which the numbers in $S$ correspond exactly to the positions of the black counters.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The operation is invertible; its inverse is to take consecutive counters $Y, Z, X$ with $X$ black, move $X$ to the front, and flip $Y$. We will work with this operation instead, and show that from any initial position with at least one black counter, we can reach the configuration with one black counter on position $1$.\n\nLet $B$ denote a black counter and $W$ denote a white counter. Suppose we do not have two consecutive counters of the same color anywhere. Take a configuration $BWB$ and perform the move changing it to $BWW$. Given that we have two consecutive white counters, we can find a configuration $WWB$ and perform the move changing it to $BBW$. So no matter what we start with, we will be able to obtain two consecutive black counters somewhere.\n\nNow we use our two consecutive black counters to turn everything black. Suppose there is a white counter; by taking the first white counter preceding a run of at least two consecutive black counters we have a configuration of the form $WBB$. Perform the move to obtain $BBB$, eliminating this white counter. By repeating this process we change all the counters to black.\n\nIt now suffices to show that we can get to just one black counter from this, since we can cyclically shift any moves from this point to ensure it ends up on space $1$. Given the configuration $BBB$, two moves will change it to $BWW$. So we can change the black counters to white in pairs. Repeat this process in a way that keeps all the white counters and black counters consecutive. If $n$ is odd, this gives us one black counter at the end. If $n$ is even, we get two consecutive black counters with the rest white. Then perform three moves on $WBB$, which changes it to $BWW$. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40067,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the finite sets $A$ of real positive numbers having at least two elements, with the property that $a^2 + b^2 \\in A$ for every $a, b \\in A$ with $a \\neq b$.",
    "options": [],
    "answer": "All such sets are exactly the two-element sets {t, sqrt(t(1 − t))} with t in (0, 1) and t ≠ 1/2; equivalently, two distinct positive reals x, y with x^2 + y^2 equal to one of x or y. No set with three or more elements exists.",
    "solution": "If $a_1 < a_2 < \\dots < a_n$ are the elements of $A$, then $a_1^2 + a_2^2, a_1^2 + a_3^2, \\dots, a_1^2 + a_n^2, a_2^2 + a_n^2, \\dots, a_{n-1}^2 + a_n^2$ belong to $A$, and we have\n$a_1^2 + a_2^2 < a_1^2 + a_3^2 < \\dots < a_1^2 + a_n^2 < a_2^2 + a_n^2 < \\dots < a_{n-1}^2 + a_n^2$,\nwhich implies $2n - 3 \\le n$. It follows that $A$ has at most 3 elements.\nIf $n=3$ and $A = \\{a, b, c\\}$ with $a < b < c$, then $a^2 < b^2 < c^2$, hence $a^2 + b^2 < a^2 + c^2 < b^2 + c^2$.\nSince $A = \\{a, b, c\\} = \\{a^2 + b^2, a^2 + c^2, b^2 + c^2\\}$, it follows\n$$\na^2 + b^2 = a, \\quad a^2 + c^2 = b, \\quad b^2 + c^2 = c. \\quad (1)\n$$\nFrom the first and the last relation in (1) we get $a^2 - c^2 = a - c$, hence $a + c = 1$. Substituting in the second relation of (1) it follows\n$$\nb = a^2 + (1 - a)^2 = 2a^2 - 2a + 1 = 2(a^2 - a) + 1 = -2b^2 + 1,\n$$\ntherefore $(2b - 1)(b + 1) = 0$, hence $b = \\frac{1}{2}$. Then $a^2 - a + \\frac{1}{4} = 0$\nleads to $a = \\frac{1}{2}$, which contradicts $a < b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40068,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDE$ be a pentagon with $\\widehat{A} = \\widehat{B} = \\widehat{C} = \\widehat{D} = 120^\\circ$. Prove that\n$$\n4AC \\cdot BD \\ge 3AE \\cdot ED.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\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$$\n\\frac{AC}{AF} \\cdot \\frac{BD}{FD} \\ge \\frac{3}{4}. \\quad (1)\n$$\nUsing the Law of Sines in triangles $ACF$ and $BDF$, we get that (1) is equivalent to\n$$\n\\frac{\\sqrt{3}}{2} \\cdot \\frac{\\sqrt{3}}{\\sin \\widehat{ACF}} \\cdot \\frac{\\sqrt{3}}{\\sin \\widehat{DBF}} \\ge \\frac{3}{4},\n$$\nso $\\sin \\widehat{ACF} \\cdot \\sin \\widehat{DBF} \\le 1$, which is clearly true.\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": 40069,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of prime numbers $p, q$ such that $p^2 - p - 1 = q^3$.",
    "options": [],
    "answer": "(37, 11)",
    "solution": "The only such pair is $(p, q) = (37, 11)$.\nWe have $p(p-1) = (q+1)(q^2-q+1)$. Since $p$ is prime and $p > p-1$ and $q^2-q+1 > q+1$, there exists an integer $m > 0$ such that\n$$\nq^2 - q + 1 = mp, \\quad (1)\n$$\n$$\np-1 = m(q+1). \\quad (2)\n$$\nSince $q^3 = p^2 - p + 1 > (p-1)^2 = m^2(q+1)^2 > m^2q^2$ we deduce that $q > m^2$. We have from (1) and (2) that\n$$\nmp \\equiv 1 \\pmod{q} \\qquad p \\equiv m+1 \\pmod{q}\n$$\nand so we obtain that $m^2 + m - 1$ is divisible by $q$.\nSince $m^2 + m - 1 < 2m^2 < 2q$ this means that\n$$\nq = m^2 + m - 1. \\quad (3)\n$$\nApplying the substitution from (3) to (1) and (2), we obtain\n$$\nq^2 - q + 1 = mp = m^2(q+1) + m = (q - m + 1)(q + 1) + m\n$$\nwhich after cancellations gives $(m-3)q = 0$ and therefore $m = 3$. Replacing in (3) and then in (2) we obtain $(p, q) = (37, 11)$, the only solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40070,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive integer $n$ denote by $a_n$ the number of quadratic functions $f(x) = ax^2 + bx + c$, $a, b, c \\in \\{1, 2, \\dots, n\\}$, having only integer roots. Prove that for every $n \\ge 4$,\n$$\nn < a_n < n^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The equations $x^2 + kx + k - 1 = 0$, $k = 2, \\dots, n$, and $2x^2 + 4x + 2 = 0$, $x^2 + 4x + 4 = 0$, satisfy the property, hence $n+1 \\le a_n$.\nIf $f$ is a quadratic function satisfying our property, then $f(x) = a(x + x_1)(x + x_2)$, where $x_1, x_2 \\in \\mathbb{Z}_+$, and $a, a(x_1 + x_2)$, $a x_1 x_2 \\in \\{1, 2, \\dots, n\\}$. From the last condition it follows that\n$$\nx_2 \\le \\frac{n}{a x_1}.\n$$\nWe obtain\n$$\na_n \\le \\sum_{\\substack{1 \\le x_1 \\le n \\\\ 1 \\le a \\le n}} \\frac{n}{a x_1} = n \\left( 1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n} \\right)^2. \\quad (1)\n$$\nIt is easy to prove by induction that for every $n \\ge 5$ the following inequality holds:\n$$\n1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n} < \\sqrt{n}.\n$$\nMoreover, $a_4 = 5$. Replacing in (1), we get $a_n < n^2$, $n \\ge 5$, and the conclusion follows.\nWrite $f(x) = a(x + x_1)(x + x_2)$, with $x_1, x_2 \\in \\mathbb{Z}_+^*$. We have $b = a(x_1 + x_2)$, $c = a x_1 x_2$, and $c = a x_1 x_2 \\le n$ implies $x_1, x_2 \\in \\{1, 2, \\dots, n\\}$.\n\nFirst we count the quadratics with root $x_1 = 1$. If $x_2 = 1$, then $f(x) = a(x^2 + 2x + 1)$, so we have $\\left[ \\frac{n}{2} \\right]$ solutions.\nIf $x_2 = 2$, then $f(x) = a(x^2 + 3x + 2)$, so we have $\\left[ \\frac{n}{3} \\right]$ solutions.\n......\nIf $x_2 = n - 1$, then $f(x) = a(x^2 + n x + n - 1)$, so we have $\\left[ \\frac{n}{n} \\right]$ solutions.\nAltogether, we have\n$$\nb_n = \\sum_{k=2}^n \\left\\lfloor \\frac{n}{k} \\right\\rfloor \\quad (1)\n$$\nquadratics that have a root 1.\n\nLet us now count the quadratics with both roots $x_1, x_2 \\ge 2$. We have $b = a(x_1 + x_2) \\le a x_1 x_1 = c \\le n$. For any pair $(a, c)$, we want to count how many choices of $b$ there are. The only condition is that $b = a(x_1 + x_2)$, where $x_1 x_2 = \\frac{c}{a} > 1$, and $x_1, x_2 \\ge 2$ (since we have already proved that $0 < b \\le c \\le n$). Write the set of divisors of $\\frac{c}{a}$ as\n$$\n1 = d_1 < d_2 < \\dots < d_{\\tau(\\frac{c}{a})} = \\frac{c}{a}.\n$$\nThe pairs $d_1 d_{\\tau(\\frac{c}{a})} = d_2 d_{\\tau(\\frac{c}{a})-1} = \\dots = d_{\\tau(\\frac{c}{a})} d_1$ are the solutions of $x_1 x_2 = \\frac{c}{a}$. There are $\\tau(\\frac{c}{a})$ such pairs. Since\n$$\nd_1 d_{\\tau(\\frac{c}{a})} = 1 \\cdot \\frac{c}{a} = \\frac{c}{a} \\cdot 1 = d_{\\tau(\\frac{c}{a})} d_1,\n$$\nthese two pairs are not acceptable (we assumed $x_1, x_2 \\ne 1$). Now, any two symmetric pairs $d_m d_{\\tau(\\frac{c}{a})-m} = d_{\\tau(\\frac{c}{a})-m} d_m$ yield a unique value of the sum $d_m + d_{\\tau(\\frac{c}{a})-m}$, which gives a unique value of $b$.\n\nThere is an even number of divisors of $\\frac{c}{a}$, unless $\\frac{c}{a}$ is a square. So, if $\\frac{c}{a}$ is not a square, we get $\\frac{1}{2}(\\tau(\\frac{c}{a}) - 2)$ different sum $d_{\\tau(\\frac{c}{a})-m} + d_m$, with $m \\in \\{2, \\dots, \\tau(\\frac{c}{a}) - 1\\}$. If $\\frac{c}{a}$ is a square, its pairs of divisors have $\\frac{1}{2}(\\tau(\\frac{c}{a}) - 1)$ different sums (disconsidering pairs that contain 1). A unified formula for these results is\n$$\n\\left\\lfloor \\frac{\\tau\\left(\\frac{c}{a}\\right) - 1}{2} \\right\\rfloor.\n$$\nNow we have established that for each pair\n$$\n(a, c) \\in \\{1, \\dots, n\\} \\times \\{1, \\dots, n\\}\n$$\nwith $a|c$, there exist $\\left[ \\frac{\\tau\\left(\\frac{c}{a}\\right) - 1}{2} \\right]$ quadratics which do not have a root 1, and for which $b \\le n$.\n\nNow we count the quadratics by taking the values of $\\frac{c}{a}$ into account.\nFor $\\frac{c}{a} = 2$, we can choose\n$$\n(a, c) \\in \\left\\{ (1, 2), (2, 4), \\dots, \\left( \\left[ \\frac{n}{2} \\right], 2 \\left[ \\frac{n}{2} \\right] \\right) \\right\\},\n$$\nhence we have $\\left[ \\frac{n}{2} \\right] \\left[ \\frac{\\tau(2) - 1}{2} \\right]$ quadratics.\nGenerally, for $\\frac{c}{a} = k$, we can choose $(a, c)$ from the set\n$$\n\\left\\{ (1, k), (2, 2k), \\dots, \\left( \\left[ \\frac{n}{k} \\right], k \\left[ \\frac{n}{k} \\right] \\right) \\right\\}.\n$$\nhence we have $\\left[ \\frac{n}{k} \\right] \\left[ \\frac{\\tau(k) - 1}{2} \\right]$ quadratics.\nAltogether, we obtain the following exact formula for $a_n$\n$$\n\\begin{aligned}\na_n = b_n + \\sum_{k=2}^n \\left[ \\frac{n}{k} \\right] \\left[ \\frac{\\tau(k) - 1}{2} \\right] &= \\sum_{k=2}^n \\left[ \\frac{n}{k} \\right] \\left( \\left[ \\frac{\\tau(k) - 1}{2} \\right] + 1 \\right) \\\\\n&= \\sum_{k=2}^n \\left[ \\frac{n}{k} \\right] \\left[ \\frac{\\tau(k) + 1}{2} \\right].\n\\end{aligned} \\quad (2)\n$$\nFrom (2) we obtain\n$$\na_n \\le n \\sum_{k=2}^n \\frac{\\tau(k) + 1}{2k} \\le n \\sum_{k=2}^n \\frac{2\\sqrt{k} + 1}{2k}\n$$\n$$\n= n \\sum_{k=2}^n \\frac{1}{\\sqrt{k}} + \\frac{n}{2} \\sum_{k=2}^n \\frac{1}{k}\n$$\n$$\n\\le n(2\\sqrt{n+1} - 1) + \\frac{n}{2}(\\ln(n+1) + C - 1) < n^2,\n$$\nwhere $C$ is the well-known Euler's constant.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40071,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle $ABC$ is inscribed in circle $\\omega$. Point $D$ is midpoint of side $AC$, and point $M$ lies on segment $BD$ with $DM = 2BM$. Ray $AM$ meets side $BC$ at $E$, and ray $CM$ meets side $BA$ at $F$. Ray $FE$ intersects $\\omega$ at $N$. Suppose that $AM \\perp CM$. Prove that $ADEF$ is cyclic if and only if line $AN$ bisects segment $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $a$, $b$, $c$ the side lengths, and by $m_a$, $m_b$, $m_c$ the lengths of the medians of the triangle $ABC$. Since $MD$ is median in the right-angled triangle $AMC$, it follows that\n$$\n2m_b/3 = MD = AD = CD = b/2,\n$$\nso $m_b = 3b/4$, which means that\n$$\n(3b/4)^2 = m_b^2 = (a^2 + c^2)/2 - b^2/4.\n$$\nThis is equivalent to $13b^2 = 8(a^2 + c^2)$.\n\nNext, apply the Menelaus theorem to get\n$$\nEC/EB = 4 = FA/FB\n$$\nand deduce thereby that the lines $AC$ and $EF$ are parallel. The quadrilateral $AFED$ is therefore a trapezoid; it is cyclic if and only if $AF = DE$.\n\nWe now express the two lengths in terms of $a$, $b$ and $c$.\nRecall that $FA/FB = 4$ to obtain $AF = 4c/5$. Next, apply Stewart's theorem in triangle $BCD$ to get $DE^2 = b^2/2 - 4a^2/25$. 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 $a$:\n$$\nb = 2a\\sqrt{2}/3 \\quad \\text{and} \\quad c = 2a/3.\n$$\nFinally, let $N$ be the midpoint of the side $BC$ and let the lines $AN$ and $EF$ meet at $P$. Notice that\n$$\nEN = a/2 - a/5 = 3a/10,\n$$\nand that the triangles $ANC$ and $PNE$ are similar. Then we obtain\n$$\nNP = 3m_a/5,\n$$\nso\n$$\nNA \\cdot NP = 3m_a^2/5 = 3(2(b^2 + c^2) - a^2)/20 = a^2/4 = NB \\cdot NC.\n$$\nThe conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40072,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ with the following property: there are two divisors $a$ and $b$ of the number $n$ such that $a^2 + b^2 + 1$ is a multiple of $n$.",
    "options": [],
    "answer": "All n equal to 1 or 3, or of the form F_{2k-1} F_{2k+1} or 3 F_{2k-1} F_{2k+1} for k ≥ 1, where F_m denotes the mth Fibonacci number with F_0 = 0 and F_1 = 1.",
    "solution": "Since $a$ divides $n$ divides $a^2 + b^2 + 1$, it follows that $a \\mid b^2 + 1$. Similarly $b \\mid a^2 + 1$. Thus we have $\\gcd(a, b) = 1$.\nWe have $ab \\mid a^2 + b^2 + 1$. By symmetry, we can assume $a \\le b$. It follows that\n$$\na^2 + b^2 + 1 = kab \\quad (1)\n$$\nfor some positive integer $k$. In the case $a = b$ we have $2a^2 + 1$ is a multiple of $a$, meaning $a = 1$. Consequently, $(a, b) = (1, 1)$ and $k = 3$.\n\nAssume now $a > b$. The equation (1) considered as a quadratic equation in $a$ has a positive integer solution. Its second solution is\n$$\na' = kb - a = \\frac{b^2 + 1}{a},\n$$\nand it is also a positive integer. Moreover,\n$$\na' = \\frac{b^2 + 1}{a} \\le \\frac{b^2 + 1}{b+1} \\le b < a.\n$$\nThus, if equation (1) has a solution $(a, b)$ with $a > b$, then it also has another solution $(b, a')$ with a strictly smaller sum of numbers. Applying the same argument to this new solution, then to the next solution, etc., we eventually arrive at a pair $(a_0, b_0)$ that cannot be further reduced. Hence,\n$$\na_0 = b_0 = 1,\n$$\nand thus $k = 3$.\nThus, starting with an arbitrary solution one can descend to the pair $(1, 1)$. Therefore, all possible solution $(a, b)$ of (1) are constructed in the infinite chain\n$$\n(1, 1) \\rightarrow (2, 1) \\rightarrow (5, 2) \\rightarrow (13, 5) \\rightarrow (34, 13) \\rightarrow \\dots\n$$\nThe transfer to the next pair is by the rule $(x, y) \\rightarrow (3x - y, x)$.\nIt can easily be observed and then proved by induction that the $k^{\\text{th}}$ pair, if the pair $(1, 1)$ has number 0, consists of numbers $(F_{2k+1}, F_{2k-1})$, where $(F_i)_{i \\ge 0}$ is the Fibonacci sequence, $F_0 = 0$, $F_1 = 1$, and $F_{i+2} = F_{i+1} + F_i$, $i = 0, 1, \\dots$.\nFinally, since $n$ is a multiple of $ab$ and is a divisor of $3ab$, it follows that either\n$$\nn = ab \\text{ or } n = 3ab.\n$$\nThus, the solutions are\n$$\nn = 1, 3, \\quad n = F_{2k-1}F_{2k+1}, \\quad n = 3F_{2k-1}F_{2k+1}, \\quad k = 1, 2, \\dots\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40073,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine if there are polynomials $p(x)$ and $q(x)$ with real coefficients such that\n$$\n\\frac{p(n)}{q(n)} = 1 + \\frac{1}{2!} + \\frac{1}{3!} + \\dots + \\frac{1}{n!}\n$$\nfor every positive integer $n$.",
    "options": [],
    "answer": "No",
    "solution": "Assume that there are polynomials $p, q \\in \\mathbb{R}[X]$ such that\n$$\n\\frac{p(n)}{q(n)} = 1 + \\frac{1}{2!} + \\dots + \\frac{1}{n!}, \\quad n \\ge 1. \\quad (1)\n$$\nThen\n$$\n\\frac{p(n+1)}{q(n+1)} - \\frac{p(n)}{q(n)} = \\frac{1}{(n+1)!}, \\quad n \\ge 1,\n$$\nso\n$$\n\\frac{p(n+1)q(n) - p(n)q(n+1)}{q(n)q(n+1)} = \\frac{1}{(n+1)!}, \\quad n \\ge 1. \\quad (2)\n$$\nDefine the polynomials $u, v \\in \\mathbb{R}[X]$ by\n$$\nu(x) = p(x+1)q(x) - p(x)q(x+1), \\quad v(x) = q(x)q(x+1).\n$$\nFrom (2) it follows that $u$ is not the 0-constant polynomial and we have\n$$\n\\frac{u(n)}{v(n)} = \\frac{1}{(n+1)!}, \\quad n \\ge 1 \\quad (3)\n$$\nand\n$$\n\\frac{u(n+1)}{v(n+1)} = \\frac{1}{(n+2)!}, \\quad n \\ge 1.\n$$\nIt follows\n$$\n\\frac{u(n+1)}{u(n)} \\cdot \\frac{v(n)}{v(n+1)} = \\frac{1}{n+2}, \\quad n \\ge 1. \\quad (4)\n$$\nWe have\n$$\n\\lim_{n \\to \\infty} \\frac{u(n+1)}{u(n)} = \\lim_{n \\to \\infty} \\frac{v(n+1)}{v(n)} = 1,\n$$\nand from (4) we obtain the contradiction $1 = 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40074,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every real numbers $x_1, \\dots, x_n$, prove that\n$$\n\\frac{x_1}{1+x_1^2} + \\frac{x_2}{1+x_1^2+x_2^2} + \\dots + \\frac{x_n}{1+x_1^2+\\dots+x_n^2} < \\sqrt{n}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using Cauchy-Schwarz inequality we have\n$$\n\\begin{aligned}\n& \\left( \\frac{x_1}{1+x_1^2} + \\dots + \\frac{x_n}{1+x_1^2 + \\dots + x_n^2} \\right) \\\\\n& \\le n \\left[ \\frac{x_1^2}{(1+x_1^2)^2} + \\dots + \\frac{x_n^2}{(1+x_1^2 + \\dots + x_n^2)^2} \\right] \\\\\n& \\le n \\left[ \\frac{x_1^2}{1 \\cdot (1+x_1^2)} + \\frac{x_2^2}{(1+x_1^2)(1+x_1^2+x_2^2)} + \\dots + \\frac{x_n^2}{(1+x_1^2+\\dots+x_{n-1}^2)(1+x_1^2+\\dots+x_n^2)} \\right] \\\\\n& = n \\left[ 1 - \\frac{1}{1+x_1^2} + \\frac{1}{1+x_1^2} - \\frac{1}{1+x_1^2+x_2^2} + \\dots + \\frac{1}{1+x_1^2+\\dots+x_{n-1}^2} - \\frac{1}{1+x_1^2+\\dots+x_n^2} \\right] \\\\\n& = n \\left( 1 - \\frac{1}{1+x_1^2+\\dots+x_n^2} \\right) < n,\n\\end{aligned}\n$$\nand the inequality follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40075,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, find the first decimal of the number:\n$$\na_n = \\frac{1}{n+1} + \\frac{1}{n+2} + \\dots + \\frac{1}{2n}.\n$$",
    "options": [],
    "answer": "The first decimal digit is 5 for n = 1, 2, and 6 for all n ≥ 3.",
    "solution": "The numbers $a_1 = \\frac{1}{2}$ and $a_2 = \\frac{7}{12}$ have the first decimal 5. The number\n$$\na_3 = \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} = \\frac{37}{60} > 0.6\n$$\nhas the first decimal 6.\n\nWe will prove that for every $n \\ge 3$, we have $0.6 < a_n < 0.7$. Notice that\n$$\na_{n+1} - a_n = \\frac{1}{2n+1} + \\frac{1}{2n+2} - \\frac{1}{n+1} = \\frac{1}{(2n+1)(2n+2)} > 0,\n$$\nso $a_n \\ge a_3 > 0.6$, for every $n \\ge 3$.\n\nWe will prove by induction that\n$$\na_n \\le 0.7 - \\frac{1}{4n}, \\quad n \\ge 3. \\quad (1)\n$$\nFor $n=3$, we have $a_3 = \\frac{37}{60} = 0.7 - \\frac{1}{12}$. Assume that\n$$\na_n \\le 0.7 - \\frac{1}{4n}.\n$$\nThen we get\n$$\na_{n+1} = a_n + \\frac{1}{(2n+1)(2n+2)} \\le 0.7 - \\frac{1}{4n} + \\frac{1}{(2n+1)(2n+2)} < 0.7 - \\frac{1}{4(n+1)},\n$$\nsince we have\n$$\n\\frac{1}{2(2n+1)(n+1)} \\le \\frac{1}{4n} - \\frac{1}{4(n+1)} \\Leftrightarrow \\\\\n\\frac{1}{(n+1)(2n+1)} < \\frac{1}{2n(n+1)} \\Leftrightarrow \\\\\n2n < 2n+1,\n$$\nand we are done.\n\nFor $n \\ge 3$, the first decimal of $a_n$ is 6.\nWe have\n$$\na_{n+1} - a_n = \\frac{1}{2n+1} + \\frac{1}{2n+2} - \\frac{1}{n+1} = \\frac{1}{(2n+1)(2n+2)} > 0,\n$$\nso the sequence $(a_n)_{n \\ge 1}$ is strictly increasing.\n\nConsider the sequence $(b_n)_{n \\ge 1}$, defined by\n$$\nb_n = 1 + \\frac{1}{2} + \\dots + \\frac{1}{n} - \\ln n.\n$$\nIt is clear that\n$$\nb_{2n} - b_n = \\frac{1}{n+1} + \\dots + \\frac{1}{2n} - \\ln 2 = a_n - \\ln 2. \\quad (1)\n$$\nWe will show that $(b_n)_{n \\ge 1}$ is strictly decreasing. Indeed, we have\n$$\nb_{n+1} - b_n = \\frac{1}{n+1} - \\ln \\frac{n+1}{n} = \\frac{1}{n+1} \\left[ 1 - \\ln \\left( 1 + \\frac{1}{n} \\right)^{n+1} \\right] < 0, \\\\\n\\text{since } \\left(1 + \\frac{1}{n}\\right)^{n+1} > e, n = 1, 2, \\dots\n$$\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 hence the first decimal in this case is 6.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40076,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots$ be a sequence of positive integers such that\n$$\na_{n+1} = \\begin{cases} \\frac{a_n}{2} & \\text{if } a_n \\text{ is even,} \\\\ 2^r + \\frac{a_n+1}{2} & \\text{if } a_n \\text{ is odd and } 2^{r-1} \\le a_n < 2^r. \\end{cases}\n$$\nProve that no matter what the value of $a_1$ is, there exists an $N$ such that for all $n > N$, $a_n = a_{n+2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose $a_i$, when expressed in binary, is a number formed by appending some number of copies of $10$ at the beginning followed by a generic $k$-digit binary number. We show that continuing the recurrence from $a_i$ will eventually result in an $N$ for which $n > N$ implies $a_n = a_{n+2}$. We do so by strong induction on $k$.\n\nOur base cases will be $k = 0, 1, 2$. First, notice that if $a_i = \\overline{1010...1011}$, then $a_{i+1} = \\overline{101010...110}$ (with one extra $10$ at the beginning) and $a_{i+2} = \\overline{101010...11} = a_i$, so $N = i$ works. Now we manually check the other cases:\n\n$a_i = \\overline{1010...10101}$ : $a_{i+1} = \\overline{101010...1011}$, already checked.\n\n$a_i = \\overline{1010...1010}$ : $a_{i+1} = \\overline{1010...101}$, already checked.\n\n$a_i = \\overline{1010...10100}$ : $a_{i+1} = \\overline{1010...1010}$, already checked.\n\n$a_i = \\overline{1010...101000}$ : $a_{i+1} = \\overline{1010...10100}$, already checked.\n\n$a_i = \\overline{1010...101001}$ : $a_{i+1} = \\overline{101010...10101}$, already checked.\n\nNow for the inductive step. Assume $a_i$ has some copies of $10$ in its binary representation followed by an arbitrary $k$-digit number, and that we have shown our inductive hypothesis for all numbers less than $k$.\n\n**Case 1:** $a_i$ ends with a $0$. Then $a_{i+1}$ is the same number without that $0$, so we have $k-1$ digits following the copies of $10$ and we can apply the inductive hypothesis.\n\n**Case 2:** $a_i$ ends with a $1$ and the $k$-digit number is not all $1$s. Then $a_{i+1}$ is obtained by removing the $1$ at the end, adding $1$ to the number, and adding a $10$ to the beginning. Since the $k$-digit number is not all $1$s, adding $1$ to it will not mess up the last copy of $10$, so the result is a $(k-1)$-digit binary number preceded by copies of $10$, and we can apply the inductive hypothesis.\n\n**Case 3:** $a_i$ ends with $k$ $1$s. Then $a_{i+1}$ ends with $1100...0$, where there are $k-1$ $0$s, preceded by some copies of $10$. It can be easily computed that $a_{i+k}$ will be the same number without the last $k-1$ $0$s, so $a_{i+k}$ is a $2$-digit binary number preceded by copies of $10$, and our base case finishes this case.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40077,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an odd number $n \\ge 3$, let\n$$\nS = \\{k : 1 \\le k \\le n, \\gcd(k, n) = 1\\}\n$$\nand let\n$$\nT = \\{k : k \\in S, \\gcd(k + 1, n) = 1\\}.\n$$\nFor each $k \\in S$, let $r_k$ be the remainder left by $\\frac{k|S| - 1}{n}$ upon division by $n$. Show that\n$$\n\\prod_{k \\in T} (r_k - r_{n-k}) \\equiv |S|^{|T|} \\pmod{n}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $n$ is odd,\n$$\n|S| = \\phi(n) = n \\prod_{p|n} (1 - 1/p)\n$$\nis even. Given an element $k$ of $S$, write\n$k^{|S|} \\equiv 1+nr_k \\pmod{n^2}$ and $(n-k)^{|S|} \\equiv 1+nr_{n-k} \\pmod{n^2}$, and notice that\n$$\n(n-k)^{|S|} \\equiv k^{|S|} \\cdot n \\cdot k^{|S|-1} \\pmod{n^2},\n$$\nto get\n$$\nr_k - r_{n-k} \\equiv |S| \\cdot k^{|S|-1} \\pmod{n}.\n$$\nHence\n$$\n\\prod_{k \\in T} (r_k - r_{n-k}) \\equiv |S|^{|T|} \\left(\\prod_{k \\in T} k\\right)^{|S|-1} \\pmod{n}.\n$$\nFinally, notice that the product in the right-hand member is congruent to 1 modulo $n$. To see this, let $k'$ denote the modulo $n$ multiplicative inverse of an element $k$ of $S$, and notice that\n$$\nk'+1 \\equiv k'(k+1) \\pmod{n} \\quad \\text{and} \\quad (k-1)(k'+1) \\equiv k-k' \\pmod{n}.\n$$\nThe first congruence shows that if $k$ belongs to $T$, then so does $k'$, and the second shows that $k = k'$ if and only if $k = 1$. Consequently, if $k \\neq 1$, the factors $k$ and $k'$ in the product can be paired off and the conclusion follows.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40078,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral such that $AB = AC = BD$. The lines $AC$ and $BD$ meet at point $O$, the circles $ABC$ and $ADO$ meet again at point $P$, and the lines $AP$ and $BC$ meet at point $Q$. Show that $\\overline{COQ} = \\overline{DOQ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We shall prove that the circles $ADO$ and $BCO$ meet again at the incenter $I$ of the triangle $ABO$, so the line $IO$ is the radical axis of the circles $ADO$ and $BCO$. Noticing further that the lines $AP$ and $BC$ are the radical axes of the pairs of circles $(ABC, ADO)$ and $(ABC, BCO)$, respectively, it follows that the lines $AP$, $BC$ and $IO$ are concurrent (at point $Q$), and the conclusion follows.\n\nTo show that the point $I$ lies on the circle $ADO$, notice that\n$$\n\\begin{aligned}\n\\overline{AIO} &= 90^\\circ + \\frac{1}{2} \\overline{ABO} = 90^\\circ + \\frac{1}{2} \\overline{ABD} = 90^\\circ + \\frac{1}{2} (180^\\circ - 2\\overline{ADB}) \\\\\n&= 180^\\circ - \\overline{ADB} = 180^\\circ - \\overline{ADO}.\n\\end{aligned}\n$$\nSimilarly, the point $I$ lies on the circle $BCO$, for\n$$\n\\begin{aligned}\n\\overline{BIO} &= 90^\\circ + \\frac{1}{2} \\overline{BAO} = 90^\\circ + \\frac{1}{2} \\overline{BAC} = 90^\\circ + \\frac{1}{2} (180^\\circ - 2\\overline{ACB}) \\\\\n&= 180^\\circ - \\overline{ACB} = 180^\\circ - \\overline{BCO}.\n\\end{aligned}\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40079,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $k$ such that there is a function $f: \\mathbb{N} \\to \\mathbb{N}$ satisfying the following condition: for every positive integer $n$, we have both $f(f(n)) = kn$ and $f(n) < f(n+1)$.",
    "options": [],
    "answer": "all positive integers except 2",
    "solution": "The answer is all $k \\neq 2$. If $k = 2$, note that $f(f(n)) = 2n$ implies that $f(n) \\neq n$ for all $n$. Therefore $f(1) \\neq 1$. We also have $f(1) \\neq 2$ since then $f(f(1)) = f(2) = 2$, so $f(1) > 2$. But then $2 = f(f(1)) < f(1)$ while $f(1) > 1$, so we have a contradiction of the second condition.\n\nFor $k = 1$, we simply take $f(n) = n$. We give an algorithm to construct a function that works for any $k > 2$. Note that each time we set $f(a) = b$, this implicitly means we also set $f(k^m a) = k^m b$ and $f(k^m b) = k^{m+1} a$. Begin by setting $f(1) = 2$. Then repeat the following: for the smallest number $m$ for which $f(m)$ has yet to be defined, set $f(m) = m'$ where $m'$ is the smallest nonmultiple of $k$ greater than $f(m-1)$. It is clear we will obtain $f(f(n)) = kn$ this way for all $n$; we now need to show $f(n) < f(n+1)$. We go by strong induction, with $n = 1$ obvious. Given it is true up to $n$, we show $f(n) < f(n+1)$ with two cases:\n\n(i) $f(n+1)$ is not a multiple of $k$: This means $n+1$ was selected as an $m$ in the algorithm, and $f(n+1)$ was deliberately set to be greater than $f(n)$.\n\n(ii) $f(n + 1)$ is a multiple of $k$: Let $x$ be the largest integer less than or equal to $n$ such that $f(x)$ is a multiple of $k$. Because all multiples of $k$ map to multiples of $k$ under $f$, we have $n - k < x \\le n$. Let $f(x) = ka$ and $f(n+1) = kb$. Then $f(a) = x$ and $f(b) = n + 1$. Since $x < n + 1$ and $a, b < n$, the inductive hypothesis implies $a < b$. Note that for each $m \\in \\{n - x + 1, n - x + 2, \\dots, n\\}$, $f(m)$ was set equal to $f(m - 1) + 1$. So this means that the algorithm would have set $f(n) = f(x) + (n - x) = ka + (n - x)$. Since $n - x < k$, $ka + (n - x) < k(a + 1) \\le kb$. Therefore $f(n) < f(n + 1)$ as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40080,
    "subject": "Mathematics (Multi-modal)",
    "question": "The set $G$ is defined by the points $(x, y)$ with integer coordinates, $1 \\leq x \\leq 5$ and $1 \\leq y \\leq 5$. Determine the number of five-point sequences $\\left(P_{1}, P_{2}, P_{3}, P_{4}, P_{5}\\right)$ such that for $1 \\leq i \\leq 5, P_{i}=\\left(x_{i}, i\\right)$ is in $G$ and $\\left|x_{1}-x_{2}\\right|= \\left|x_{2}-x_{3}\\right|=\\left|x_{3}-x_{4}\\right|=\\left|x_{4}-x_{5}\\right|=1$.",
    "options": [],
    "answer": "42",
    "solution": "Let us count the number of five-point sequences according to the position of point $P_{3}$.\n\n![](attached_image_1.png)\n\nBecause the horizontal line of equation $y=3$ is a symmetry axis for this figure, the number of sequences with $x_{3}=i$ is equal to the number of sequences with $x_{3}=6-i$, for $i=1,2$.\n\n![](attached_image_2.png)\n\nSimilarly, the vertical line of equation $x=3$ is a symmetry axis for this figure, so that the number $N_{i}$ of three-point sequences $\\left(P_{1}, P_{2}, P_{3}\\right)$ ending at $P_{3}$ with $x_{3}=i$ is equal to the number of three-point sequences $\\left(P_{3}, P_{4}, P_{5}\\right)$ starting at $P_{3}$, with $x_{3}=i$, for $i=1,2,3$. Therefore, the number of five-point sequences $\\left(P_{1}, P_{2}, P_{3}, P_{4}, P_{5}\\right)$ with $x_{3}=i$ is $N_{i}^{2}$. Hence, the total number of five-point sequences $\\left(P_{1}, P_{2}, P_{3}, P_{4}, P_{5}\\right)$ is\n$$\n2\\left(N_{1}^{2}+N_{2}^{2}\\right)+N_{3}^{2} .\n$$\nIt is easy to see that $N_{1}=2$, since the only possibilites for $x_{4}, x_{5}$ are $x_{4}=2$ and $x_{5}=1$ or $3$.\n\n![](attached_image_3.png)\n\nIt is also easy to see that $N_{2}=3$, since $x_{4}=1$ or $3$ and $x_{5}=2$ in the first case and $2$ or $4$ in the second case.\n\n![](attached_image_4.png)\n\nFinally, $N_{3}=4$, since $x_{4}=2$ or $4$, and $x_{5}=1$ or $3$ in the first case, and $3$ or $5$ in the second case.\n\n![](attached_image_5.png)\n\nHence, the number of sequences is $2\\left(2^{2}+3^{2}\\right)+4^{2}=42$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40081,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ which satisfy\n$$\n\\begin{aligned}\n& f\\left(\\frac{\\sqrt{3}}{3} x\\right)=\\sqrt{3} f(x)-\\frac{2 \\sqrt{3}}{3} x \\\\\n& f(x) f(y)=f(x y)+f\\left(\\frac{x}{y}\\right)\n\\end{aligned}\n$$\nfor all $x, y \\in \\mathbb{R}$, with $y \\neq 0$.",
    "options": [],
    "answer": "f(x) = x + 1/x for x ≠ 0, and f(0) = 0",
    "solution": "Let $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ be a function which satisfies the two functional equations.\nBecause $f\\left(\\frac{\\sqrt{3}}{3} x\\right)-\\sqrt{3} f(1)=-\\frac{2 \\sqrt{3}}{3} \\neq 0$, either $f(1) \\neq 0$ or $f\\left(\\frac{\\sqrt{3}}{3} x\\right) \\neq 0$.\nFix $x_{0} \\in \\mathbb{R}$ with $f\\left(x_{0}\\right) \\neq 0$ and let $y \\in \\mathbb{R}$ with $y \\neq 0$. We have\n$$\nf\\left(x_{0}\\right) f\\left(\\frac{1}{y}\\right)=f\\left(\\frac{x_{0}}{y}\\right)+f\\left(x_{0} y\\right)=f\\left(x_{0}\\right) f(y) .\n$$\nTherefore, $f\\left(\\frac{1}{y}\\right)=f(y)$ for all $y \\neq 0$.\nLet $x \\neq 0$. We have\n$$\n\\begin{aligned}\nf\\left(\\frac{\\sqrt{3}}{3} x\\right) & =\\sqrt{3} f(x)-\\frac{2 \\sqrt{3}}{3} x \\\\\n& =\\sqrt{3} f\\left(\\frac{\\frac{\\sqrt{3}}{3}}{\\frac{\\sqrt{3}}{3} x}\\right)-\\frac{2 \\sqrt{3}}{3} x \\\\\n& =\\sqrt{3}\\left(\\sqrt{3} f\\left(\\frac{1}{\\frac{\\sqrt{3}}{3} x}\\right)-\\frac{2}{x}\\right)-\\frac{2 \\sqrt{3}}{3} x \\\\\n& =3 f\\left(\\frac{\\sqrt{3}}{3} x\\right)-\\frac{2 \\sqrt{3}}{x}-\\frac{2 \\sqrt{3}}{3} x\n\\end{aligned}\n$$\nWe deduce that\n$$\nf\\left(\\frac{\\sqrt{3}}{3} x\\right)=\\frac{\\sqrt{3}}{x}+\\frac{\\sqrt{3}}{3} x\n$$\nand therefore\n$$\nf(x)=x+\\frac{1}{x}\n$$\nfor all $x \\neq 0$.\nFor $x=0$ and $y=2$, we have from the second functional equation\n$$\n\\frac{5}{2} f(0)=f(0)+f(0)\n$$\nwhich implies that $f(0)=0$. Hence\n$$\nf(x)=\\left\\{\\begin{array}{cr}\nx+\\frac{1}{x} & \\text{ if } x \\neq 0 \\\\\n0 & \\text{ otherwise }\n\\end{array}\\right.\n$$\nConversely, it is easy to check that this function satisfies the two functional equations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40082,
    "subject": "Mathematics (Multi-modal)",
    "question": "Adel draws an $m \\times n$ grid of dots on the coordinate plane, at the points of integer coordinates $(a, b)$ where $1 \\leq a \\leq m$ and $1 \\leq b \\leq n$. He proceeds to draw a closed path along $k$ of these dots, $(a_{1}, b_{1}),(a_{2}, b_{2}), \\ldots,(a_{k}, b_{k})$, such that $(a_{i}, b_{i})$ and $(a_{i+1}, b_{i+1})$ (where $(a_{k+1}, b_{k+1})=(a_{1}, b_{1})$ ) are 1 unit apart for each $1 \\leq i \\leq k$. Adel makes sure his path does not cross itself, that is, the $k$ dots are distinct. Find, with proof, the maximum possible value of $k$ in terms of $m$ and $n$.",
    "options": [],
    "answer": "If both m and n are odd, the maximum k is mn − 1; otherwise, the maximum k is mn.",
    "solution": "If $m$ is even, Adel can draw the following closed path which passes through all the dots of his grid. Therefore, the maximum possible value of $k$ is $mn$.\n\n![](attached_image_1.png)\n\nIf $n$ is even, Adel can draw a similar closed path obtained by symmetry with respect to the first diagonal. Therefore, the maximum possible value of $k$ is $mn$.\n\nIf $mn$ is odd, Adel can draw the following closed path which passes through $mn-1$ dots of his grid:\n\n![](attached_image_2.png)\n\nIt remains to prove that this is the maximal possible value of $k$. For this, assign black color to all dots of the grid of coordinates $(a, b)$ with $a+b$ odd and white color to all dots with $a+b$ even. Any consecutive dots in a path $(a_{1}, b_{1}),(a_{2}, b_{2}), \\ldots,(a_{k}, b_{k})$ that Adel can draw have different colors. Therefore, the colors of the first and last dots in Adel's path describe the parity of the length of the path. Since the path of Adel is closed, it will start and end with the same color and therefore, its length is even. Therefore $k$ is even. This proves that the maximum possible value of $k$ is $mn-1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40083,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $-1 \\leq x, y \\leq 1$. Prove the inequality\n$$\n2 \\sqrt{(1-x^{2})(1-y^{2})} \\leq 2(1-x)(1-y)+1 .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First solution. Because $-1 \\leq x, y \\leq 1$, we have $(1-x^{2}),(1-y^{2}) \\geq 0$. Therefore, by applying AM-GM, we obtain\n$$\n2 \\sqrt{(1-x^{2})(1-y^{2})} \\leq (1-x^{2})+(1-y^{2}) .\n$$\nSo, it remains to prove that\n$$\n(1-x^{2})+(1-y^{2}) \\leq 2(1-x)(1-y)+1\n$$\nwhich is equivalent to\n$$\nx^{2}+y^{2}-2x-2y+2xy+1 \\geq 0 .\n$$\nBut\n$$\nx^{2}+y^{2}-2x-2y+2xy+1=(x+y-1)^{2}\n$$\nThis ends the proof.\nThe equality holds when $1-x^{2}=1-y^{2}$ and $x+y-1=0$. This is equivalent to $x=y=\\frac{1}{2}$.\n\n\nSecond solution. Because $-1 \\leq x, y \\leq 1$, there exist $-\\frac{\\pi}{2} \\leq \\theta, \\phi \\leq \\frac{\\pi}{2}$ such that $x=\\sin \\theta$ and $y=\\sin \\phi$. The inequality to prove becomes\n$$\n2 \\cos \\theta \\cos \\phi \\leq 2(1-\\sin \\theta)(1-\\sin \\phi)+1,\n$$\nor, equivalently\n$$\n2[\\cos (\\theta+\\phi)+\\sin \\theta+\\sin \\phi] \\leq 3 .\n$$\nUsing the relations\n$\\cos (\\theta+\\phi)=1-2 \\sin^{2} \\frac{\\theta+\\phi}{2}$ and $\\sin \\theta+\\sin \\phi=2 \\sin \\frac{\\theta+\\phi}{2} \\cos \\frac{\\theta-\\phi}{2}$,\nthe inequality simplifies to\n$$\n\\cos^{2} \\frac{\\theta-\\phi}{2} \\leq 1+\\left(2 \\sin \\frac{\\theta+\\phi}{2}-\\cos \\frac{\\theta-\\phi}{2}\\right)^{2},\n$$\nwhich is satisfied.\nThe equality holds when $\\theta=\\phi=\\frac{\\pi}{6}$, which means $x=y=\\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40084,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is a cyclic quadrilateral and $\\omega$ its circumcircle. The perpendicular line to $AC$ at $D$ intersects $AC$ at $E$ and $\\omega$ at $F$. Denote by $\\ell$ the perpendicular line to $BC$ at $F$. The perpendicular line to $\\ell$ at $A$ intersects $\\ell$ at $G$ and $\\omega$ at $H$. Line $GE$ intersects $FH$ at $I$ and $CD$ at $J$. Prove that points $C$, $F$, $I$, and $J$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "To prove that $C$, $F$, $I$, and $J$ are concyclic, it is equivalent to prove that $\\angle CFI = \\angle CJI$ or $\\angle CFI + \\angle CJI = 180^{\\circ}$, depending on the configuration. We will present here the proof for one configuration. The proof for the other configuration is similar.\n\n![](attached_image_1.png)\n\nBecause $AFCH$ is cyclic, we have $\\angle CFI = \\angle CAH$.\n\nBecause $\\angle FEA = \\angle FGA = 90^{\\circ}$, the quadrilateral $AFEG$ is cyclic, and therefore $\\angle CAH = \\angle EFG$.\n\nIt remains to prove that $\\angle EFG = \\angle DJG$, which is equivalent to proving that quadrilateral $DGFJ$ is cyclic.\n\nBut $\\angle EGF = \\angle EAF$ since $AFEG$ is cyclic. On the other hand, because $AFCD$ is cyclic, we deduce that $\\angle EAF = \\angle CDF$. Therefore, $\\angle EGF = \\angle CDF$, which proves that $DGFJ$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40085,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{1}, a_{2}, a_{3}, \\ldots$ be a sequence of real numbers which satisfy the relation\n$$\na_{n+1}=\\sqrt{a_{n}^{2}+1}\n$$\nSuppose that there exists a positive integer $n_{0}$ such that $a_{2 n_{0}}=3 a_{n_{0}}$. Find the value of $a_{46}$.",
    "options": [],
    "answer": "19*sqrt(2)/4",
    "solution": "We have $a_{n+1}^{2}-a_{n}^{2}=1$ for all natural number $n$. Using a telescopic sum we obtain\n$$\nn_{0}=\\sum_{k=n_{0}}^{2 n_{0}-1}\\left(a_{k+1}^{2}-a_{k}^{2}\\right)=a_{2 n_{0}}^{2}-a_{n_{0}}^{2}=9 a_{n_{0}}^{2}-a_{n_{0}}^{2}=8 a_{n_{0}}^{2} .\n$$\nHence\n$$\na_{n_{0}}^{2}=\\frac{n_{0}}{8}\n$$\nOn the other hand,\n$$\nn_{0}-1=\\sum_{k=1}^{n_{0}-1}\\left(a_{k+1}^{2}-a_{k}^{2}\\right)=a_{n_{0}}^{2}-a_{1}^{2}=\\frac{n_{0}}{8}-a_{1}^{2},\n$$\nor equivalently\n$$\na_{1}^{2}=\\frac{8-7 n_{0}}{8}\n$$\nSince $a_{1}^{2} \\geq 0$, we deduce that $n_{0}=1$ and $a_{1}^{2}=\\frac{1}{8}$.\nTherefore\n$$\n45=\\sum_{k=1}^{45}\\left(a_{k+1}^{2}-a_{k}^{2}\\right)=a_{46}^{2}-a_{1}^{2}=a_{46}^{2}-\\frac{1}{8},\n$$\nwhich leads to\n$$\na_{46}=\\frac{19 \\sqrt{2}}{4} .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40086,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the maximum and the minimum values of\n$$\nS = (1 - x_{1})(1 - y_{1}) + (1 - x_{2})(1 - y_{2})\n$$\nfor real numbers $x_{1}, x_{2}, y_{1}, y_{2}$ with $x_{1}^{2} + x_{2}^{2} = y_{1}^{2} + y_{2}^{2} = 2013$.",
    "options": [],
    "answer": "maximum = 2015 + 2 sqrt(4026), minimum = -2012",
    "solution": "We present for this problem two solutions, one using trigonometric functions and the other using classical inequalities.\n\nFirst solution. The condition $x_{1}^{2} + x_{2}^{2} = y_{1}^{2} + y_{2}^{2} = c^{2}$, here $c^{2} = 2013$, is equivalent to saying that there exist $\\alpha, \\beta \\in \\mathbb{R}$ such that\n$$\nx_{1} = c \\cos \\alpha, \\quad x_{2} = c \\sin \\alpha, \\quad y_{1} = c \\cos \\beta, \\quad \\text{ and } \\quad y_{2} = c \\sin \\beta.\n$$\nTherefore\n$$\n\\begin{aligned}\nS & = (1 - x_{1})(1 - y_{1}) + (1 - x_{2})(1 - y_{2}) = 2 - (x_{1} + x_{2} + y_{1} + y_{2}) + x_{1} y_{1} + x_{2} y_{2} \\\\\n& = 2 - c(\\cos \\alpha + \\sin \\alpha + \\cos \\beta + \\sin \\beta) + c^{2}(\\cos \\alpha \\cos \\beta + \\sin \\alpha \\sin \\beta) \\\\\n& = 2 - \\sqrt{2} c \\left( \\sin \\left( \\alpha + \\frac{\\pi}{4} \\right ) + \\sin \\left( \\beta + \\frac{\\pi}{4} \\right ) \\right ) + c^{2} \\cos (\\alpha - \\beta) \\\\\n& = (2 - c^{2}) - 2 \\sqrt{2} c \\sin \\left( \\frac{\\alpha + \\beta}{2} + \\frac{\\pi}{4} \\right ) \\cos \\left( \\frac{\\alpha - \\beta}{2} \\right ) + 2 c^{2} \\cos^{2} \\left( \\frac{\\alpha - \\beta}{2} \\right ) \\\\\n& = (2 - c^{2}) - 2 \\sqrt{2} c s t + 2 c^{2} t^{2}\n\\end{aligned}\n$$\nwhere\n$$\ns = \\sin \\left( \\frac{\\alpha + \\beta}{2} + \\frac{\\pi}{4} \\right ), \\quad \\text{ and } \\quad t = \\cos \\left( \\frac{\\alpha - \\beta}{2} \\right )\n$$\nare two independent variables, since $\\alpha + \\beta$ and $\\alpha - \\beta$ are independent, taking all the real values between $-1$ and $1$ included.\n\nHence, the maximum of $S$ is equal to $2 + c^{2} + 2 \\sqrt{2} c = 2015 + 2 \\sqrt{4026}$, and is reached when $s = -t = \\pm 1$. That is precisely when $x_{1} = x_{2} = y_{1} = y_{2} = -\\frac{\\sqrt{4026}}{2}$.\n\nFor the minimum, notice that\n$$\nS = (2 - c^{2}) - s^{2} + (\\sqrt{2} c t - s)^{2}\n$$\nTherefore, the minimum of $S$ is equal to $1 - c^{2} = -2012$, and is reached when $s = \\pm 1$ and $t = \\frac{\\sqrt{2} s}{2 c} = \\frac{\\pm \\sqrt{4026}}{4026}$. That is precisely when\n$$\nx_{1} = y_{2} = \\frac{1 + \\sqrt{4025}}{2} \\quad \\text{ and } \\quad x_{2} = y_{1} = \\frac{1 - \\sqrt{4025}}{2},\n$$\nor vice-versa.\n\n\nSecond solution. For the maximum, we have by Cauchy-Schwartz inequality\n$$\n\\begin{aligned}\nS & = 2 - (x_{1} + x_{2} + y_{1} + y_{2}) + x_{1} y_{1} + x_{2} y_{2} \\\\\n& \\leq 2 + \\sqrt{4 \\cdot (x_{1}^{2} + x_{2}^{2} + y_{1}^{2} + y_{2}^{2})} + \\sqrt{(x_{1}^{2} + x_{2}^{2})(y_{1}^{2} + y_{2}^{2})} \\\\\n& \\leq 2015 + 2 \\sqrt{4026}\n\\end{aligned}\n$$\nand the equality holds when $x_{1} = x_{2} = y_{1} = y_{2} = -\\frac{\\sqrt{4026}}{2}$.\n\nFor the minimum, we have\n$$\n\\begin{aligned}\nS & = 2 - (x_{1} + y_{1}) - (x_{2} + y_{2}) + \\frac{(x_{1} + y_{1})^{2}}{2} + \\frac{(x_{2} + y_{2})^{2}}{2} - c^{2} \\\\\n& = (1 - c^{2}) + \\frac{(x_{1} + y_{1} - 1)^{2}}{2} + \\frac{(x_{2} + y_{2} - 1)^{2}}{2} \\\\\n& \\geq 1 - c^{2} = -2012\n\\end{aligned}\n$$\nwhere $c^{2} = 2013$.\n\nThus the minimum of $S$ is $-2012$ and is reached when $y_{1} = 1 - x_{1}, y_{2} = 1 - x_{2}$ and $x_{1}^{2} + x_{2}^{2} = y_{1}^{2} + y_{2}^{2} = 2013$. This is equivalent to\n$$\nx_{1} = y_{2} = \\frac{1 + \\sqrt{4025}}{2} \\quad \\text{ and } \\quad x_{2} = y_{1} = \\frac{1 - \\sqrt{4025}}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40087,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ be a function satisfying $f(f(x)) = 4x + 1$ for all real number $x$. Prove that the equation $f(x) = x$ has a unique solution.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $x_0$ be a solution of the equation $f(x) = x$. We have\n$$\nx_0 = f(x_0) = f(f(x_0)) = 4x_0 + 1.\n$$\nTherefore, $x_0 = -\\frac{1}{3}$. This proves the uniqueness of the solution.\n\nOn the other hand,\n$$\nf\\left(f\\left(-\\frac{1}{3}\\right)\\right) = 4\\left(-\\frac{1}{3}\\right) + 1 = -\\frac{1}{3}\n$$\nWe deduce that\n$$\nf\\left(-\\frac{1}{3}\\right) = f\\left(f\\left(f\\left(-\\frac{1}{3}\\right)\\right)\\right) = 4 f\\left(-\\frac{1}{3}\\right) + 1\n$$\nand therefore,\n$$\nf\\left(-\\frac{1}{3}\\right) = -\\frac{1}{3}\n$$\nThis proves the existence of the solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40088,
    "subject": "Mathematics (Multi-modal)",
    "question": "Tarik wants to choose some distinct numbers from the set $S = \\{2, \\ldots, 111\\}$ in such a way that each of the chosen numbers cannot be written as the product of two other distinct chosen numbers. What is the maximum number of numbers Tarik can choose?",
    "options": [],
    "answer": "101",
    "solution": "First, we see that it is possible for Tarik to choose the 101 numbers $11, 12, \\ldots, 111$, since the product $11 \\times 12 > 111$.\n\nAssume that Tarik has chosen $k$ numbers and let $d$ be the smallest among these numbers. If $d \\geq 11$, then clearly, $k \\leq 101$.\n\nIf $2 \\leq d \\leq 6$, from each of the 9 sets $\\{9, 9d\\}; \\{10, 10d\\}; \\ldots; \\{17, 17d\\}$, Tarik can choose at most one number. Because $9d > 17$, these sets are pairwise disjoint. Because $17d \\leq 102$, there are at least 9 numbers between 9 and 102 that Tarik could not choose. Therefore $k \\leq 101$.\n\nIf $3 \\leq d \\leq 10$, from each of the sets $\\{d+1, d(d+1)\\}; \\{d+2, d(d+2)\\}; \\ldots; \\{11, 11d\\}$, Tarik can choose at most one element. Because $d(d+1) \\geq 12$, these sets are pairwise disjoint. Because $11d < 111$, there are at least $11-d$ numbers from these sets that Tarik could not choose. But Tarik didn't choose the numbers $2, \\ldots, d-1$. Therefore, Tarik didn't choose at least $11-d + d-2 = 9$ numbers. Hence $k \\leq 101$.\n\nTherefore, the maximum number of numbers Tarik can choose is $101$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40089,
    "subject": "Mathematics (Multi-modal)",
    "question": "An acute triangle $ABC$ is inscribed in circle $\\omega$ centered at $O$. Line $BO$ and side $AC$ meet at $B_1$. Line $CO$ and side $AB$ meet at $C_1$. Line $B_1C_1$ meets circle $\\omega$ at $P$ and $Q$. If $AP = AQ$, prove that $AB = AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume $AP = AQ$. Because $OP = OQ$, the line $AO$ is perpendicular to $PQ$. But since $\\angle C_1AO = 90^{\\circ} - \\angle ACB$ then $\\angle B_1C_1A = \\angle ACB$, and therefore, quadrilateral $BCB_1C_1$ is cyclic.\n\nUsing the power of the point $O$ with respect to the circumcircle of $BCB_1C_1$ we get\n$$\nOB_1 = \\frac{OB_1 \\cdot OB}{R} = \\frac{OC_1 \\cdot OC}{R} = OC_1,\n$$\nwhere $R$ is the circumradius of triangle $ABC$. Therefore, $BB_1 = CC_1$, that is the minors $\\widehat{BB_1}$ and $\\widehat{CC_1}$ of the circumcircle of $BCB_1C_1$ have the same length. This is equivalent to saying that $\\angle CBC_1 = \\angle B_1CB$, and therefore $AB = AC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40090,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is a cyclic quadrilateral such that $AB = BC = CA$. Diagonals $AC$ and $BD$ intersect at $E$. Given that $BE = 19$ and $ED = 6$, find the possible values of $AD$.",
    "options": [],
    "answer": "10 or 15",
    "solution": "Applying Ptolemy relation to the cyclic quadrilateral $ABCD$, we get\n$$\nAB \\cdot CD + BC \\cdot DA = AC \\cdot BD\n$$\nwhich simplifies to\n$$\nCD + DA = 25\n$$\n![](attached_image_1.png)\nLet $a = AB = BC = CA$ and $x = AE$.\nWe have, from similarity of triangles $AED$ and $BEC$, that\n$$\n\\frac{DA}{a} = \\frac{x}{19}\n$$\nWe have, from similarity of triangles $CED$ and $BEA$, that\n$$\n\\frac{CD}{a} = \\frac{a - x}{19}.\n$$\nWe deduce that\n$$\n\\frac{a}{19} = \\frac{x}{19} + \\frac{a - x}{19} = \\frac{DA}{a} + \\frac{CD}{a} = \\frac{25}{a},\n$$\nand therefore $a = 5\\sqrt{19}$.\nApplying sine law on the circumcircle of $ABCD$, we obtain\n$$\n\\frac{\\sin \\angle BAD}{25} = \\frac{\\sin \\angle BAC}{5\\sqrt{19}} = \\frac{\\sqrt{57}}{190}.\n$$\nHence $\\sin \\angle BAD = \\frac{5\\sqrt{57}}{38}$, that is\n$$\n\\cos \\angle BAD = \\pm \\frac{\\sqrt{38^2 - 25 \\times 57}}{38} = \\pm \\frac{\\sqrt{19}}{38}.\n$$\nApplying cosine law to triangle $ABD$, we obtain\n$$\n25^2 = 25 \\times 19 + AD^2 \\pm 2 \\times 5\\sqrt{19} \\times AD \\times \\frac{\\sqrt{19}}{38},\n$$\nand therefore, $AD = 10$ or $15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40091,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if $a$ is an integer relatively prime with $35$ then\n$$\n\\left(a^{4}-1\\right)\\left(a^{4}+15 a^{2}+1\\right) \\equiv 0 \\quad \\bmod 35\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $a$ is relatively prime with $35$ then it is relatively prime with both $5$, $7$.\nSince $a$ is relatively prime with $5$ then, by Fermat, $a^{4} \\equiv 1 \\bmod 5$ which implies\n$$\n\\left(a^{4}-1\\right)\\left(a^{4}+15 a^{2}+1\\right) \\equiv 0 \\quad \\bmod 5\n$$\nSince $a$ is relatively prime with $7$ then $a^{6} \\equiv 1 \\bmod 7$. Hence\n$$\n\\begin{aligned}\n\\left(a^{4}-1\\right)\\left(a^{4}+15 a^{2}+1\\right) & \\equiv \\left(a^{2}+1\\right)\\left(a^{2}-1\\right)\\left(a^{4}+a^{2}+1\\right) \\\\\n& \\equiv \\left(a^{2}+1\\right)\\left(a^{6}-1\\right) \\equiv 0 \\quad \\bmod 7\n\\end{aligned}\n$$\nBut $5$, $7$ are relatively prime and therefore\n$$\n\\left(a^{4}-1\\right)\\left(a^{4}+15 a^{2}+1\\right) \\equiv 0 \\quad \\bmod 35.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40092,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, $AB = AC = 3$ and $\\angle A = 90^{\\circ}$. Let $M$ be the midpoint of side $BC$. Points $D$ and $E$ lie on sides $AC$ and $AB$ respectively such that $AD > AE$ and $ADME$ is a cyclic quadrilateral. Given that triangle $EMD$ has area $2$, find the length of segment $CD$.",
    "options": [],
    "answer": "(3 - sqrt(7))/2",
    "solution": "Because $AEMD$ is cyclic, we have $\\angle BEM = \\angle ADM$. But $\\angle MBE = \\angle MAD = 45^{\\circ}$ and $BM = AM$. We deduce that triangles $BME$ and $AMD$ are congruent and therefore $ME = MD$.\n\nBecause $AEMD$ is cyclic, $\\angle DME = 90^{\\circ}$. Therefore, the area of triangle $EMD$ is\n$$\n2 = \\frac{1}{2} ME \\cdot MD = \\frac{1}{2} MD^2\n$$\nand hence, $MD = ME = 2$.\n\n![](attached_image_1.png)\n\nWe have from the cosine law in triangle $AMD$\n$$\n4 = MD^2 = AD^2 + AM^2 - 2 AD \\cdot AM \\cos 45^{\\circ} = AD^2 + \\frac{9}{2} - 3 AD\n$$\nBecause $AE$ satisfies the same equation as $AD$ above, and $AE < AD$, we deduce that $AD = \\frac{3 + \\sqrt{7}}{2}$ and therefore\n$$\nCD = \\frac{3 - \\sqrt{7}}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40093,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle $ABC$ is inscribed in circle $\\omega$. Point $P$ lies inside triangle $ABC$. Lines $AP$, $BP$ and $CP$ intersect $\\omega$ again at points $A_1$, $B_1$ and $C_1$ (other than $A$, $B$, $C$), respectively. The tangent lines to $\\omega$ at $A_1$ and $B_1$ intersect at $C_2$. The tangent lines to $\\omega$ at $B_1$ and $C_1$ intersect at $A_2$. The tangent lines to $\\omega$ at $C_1$ and $A_1$ intersect at $B_2$. Prove that the lines $AA_2$, $BB_2$ and $CC_2$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $R$ be the circumradius of triangle $ABC$. We have, using sine law,\n$$\nC_1A = 2R \\sin \\angle ACP \\quad \\text{ and } \\quad AB_1 = 2R \\sin \\angle PBA.\n$$\n![](attached_image_1.png)\nOn the other hand, because lines $A_2C_1$ and $A_2B_1$ are tangent to $\\omega$, we have\n$$\n\\angle AC_1A_2 = \\angle ACP \\quad \\text{ and } \\quad \\angle A_2B_1A = \\angle PBA.\n$$\nApplying sine law to triangles $AA_2C_1$ and $AB_1A_2$, we obtain\n$$\n\\frac{\\sin \\angle AA_2C_1}{AC_1} = \\frac{\\sin \\angle AC_1A_2}{AA_2} \\quad \\text{ and } \\quad \\frac{\\sin \\angle AA_2B_1}{AB_1} = \\frac{\\sin \\angle A_2B_1A}{AA_2}.\n$$\nTherefore, we have\n$$\n\\frac{\\sin \\angle C_1A_2A}{\\sin \\angle AA_2B_1} = \\frac{AC_1 \\cdot \\sin \\angle AC_1A_2}{AB_1 \\cdot \\sin \\angle A_2B_1A} = \\frac{\\sin^2 \\angle ACP}{\\sin^2 \\angle PBA}\n$$\nWe have, in a similar way\n$$\n\\frac{\\sin \\angle A_1B_2B}{\\sin \\angle BB_2C_1} = \\frac{\\sin^2 \\angle BAP}{\\sin^2 \\angle PCB} \\quad \\text{ and } \\quad \\frac{\\sin \\angle B_1C_2C}{\\sin \\angle CC_2A_1} = \\frac{\\sin^2 \\angle CBP}{\\sin^2 \\angle PAC}\n$$\nBut, because $AP$, $BP$ and $CP$ are concurrent, we have by trigonometric Ceva\n$$\n\\frac{\\sin \\angle BAP}{\\sin \\angle PAC} \\cdot \\frac{\\sin \\angle CBP}{\\sin \\angle PBA} \\cdot \\frac{\\sin \\angle ACP}{\\sin \\angle PCB} = 1.\n$$\nWe deduce that\n$$\n\\frac{\\sin \\angle C_1A_2A}{\\sin \\angle AA_2B_1} \\cdot \\frac{\\sin \\angle A_1B_2B}{\\sin \\angle BB_2C_1} \\cdot \\frac{\\sin \\angle B_1C_2C}{\\sin \\angle CC_2A_1} = 1.\n$$\nThis proves, by using the reciprocal of trigonometric Ceva, that lines $AA_2$, $BB_2$ and $CC_2$ are concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40094,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all values of $n$ for which there exists a convex cyclic non-regular polygon with $n$ vertices such that the measures of all its internal angles are equal.",
    "options": [],
    "answer": "All even integers at least 4",
    "solution": "Let $P_{1} P_{2} \\cdots P_{n}$ be a convex cyclic non-regular polygon with the measures of all its internal angles equal and let $O$ be its circumcenter. Because all the angles are equal, all the arcs $\\widehat{P_{i} P_{i+2}}$, for $i=1, \\ldots, n$, have the same length. Hence, $\\angle P_{i} O P_{i+2} = \\frac{4\\pi}{n}$, for all $i=1, \\ldots, n$, since the polygon is convex.\n\n![](attached_image_1.png)\n\nLet $\\theta = \\angle P_{1} O P_{2}$. We have\n$$\n\\angle P_{2i-1} O P_{2i} = \\angle P_{1} O P_{2} + \\angle P_{2} O P_{2i} - \\angle P_{1} O P_{2i-1} = \\angle P_{1} O P_{2} = \\theta.\n$$\nIf $n$ is an odd integer,\n$$\n\\theta = \\angle P_{n} O P_{1} = \\frac{n+1}{2} \\angle P_{n} O P_{2} = \\frac{n+1}{2} \\cdot \\frac{4\\pi}{n} = \\frac{2\\pi}{n} = \\angle P_{1} O P_{2} \\quad \\bmod 2\\pi.\n$$\nThis means that the polygon is regular, which contradicts the hypothesis.\n\nIf $n$ is even, any value of $\\theta$ with $0 < \\theta < \\frac{4\\pi}{n}$ and $\\theta \\neq \\frac{2\\pi}{n}$ defines a unique non-regular such polygon.\n\nTherefore, the possible values of $n$ are all even positive integers $n \\geq 4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40095,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(X) = a_{n} X^{n} + a_{n-1} X^{n-1} + \\cdots + a_{1} X + p$ be a polynomial of integer coefficients where $p$ is a prime number. Assume that\n$$\np > \\sum_{i=1}^{n} \\left| a_{i} \\right| .\n$$\n\nProve that $f(X)$ is irreducible.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that there exist two non-constant polynomials $g(X)$ and $h(X)$ with integer coefficients such that $f(X) = g(X) h(X)$. Because $p = g(0) h(0)$ is prime, we can assume that $|g(0)| = 1$.\n\nBecause the modulus of the product of the complex roots of $g(X)$ is equal to $1$, at least one of these roots, say $\\omega_{0}$, has modulus less than or equal to $1$. But $f\\left(\\omega_{0}\\right) = 0$. We deduce that\n$$\n\\begin{aligned}\np & = \\left| a_{n} \\omega_{0}^{n} + a_{n-1} \\omega_{0}^{n-1} + \\cdots + a_{1} \\omega_{0} \\right| \\\\\n& \\leq \\left| a_{n} \\right| \\cdot \\left| \\omega_{0} \\right|^{n} + \\left| a_{n-1} \\right| \\cdot \\left| \\omega_{0} \\right|^{n-1} + \\cdots + \\left| a_{1} \\right| \\cdot \\left| \\omega_{0} \\right| \\\\\n& \\leq \\left| a_{n} \\right| + \\left| a_{n-1} \\right| + \\cdots + \\left| a_{1} \\right|\n\\end{aligned}\n$$\nwhich is a contradiction. Therefore, $f(X)$ is irreducible.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40096,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ which satisfy for all $x, y \\in \\mathbb{R}$ the relation\n$$\nf(f(f(x)+y)+y)=x+y+f(y) .\n$$",
    "options": [],
    "answer": "f(x) = x",
    "solution": "Plug in $y=0$. The functional equation becomes\n$$\nf(f(f(x)))=x+f(0),\n$$\nfor all $x \\in \\mathbb{R}$. Since the map $x \\mapsto x+f(0)$ is bijective, then so is $f$.\n\nPlug in $y=-x$. The functional equation becomes\n$$\nf(f(f(x)-x)-x)=f(-x),\n$$\nfor all $x \\in \\mathbb{R}$. By injectivity of $f$, we can cancel $f$ in both sides and obtain\n$$\nf(f(x)-x)=0,\n$$\nfor all $x \\in \\mathbb{R}$. By surjectivity of $f$ there exists a real number $a$ such that $f(a)=0$. Again by cancelling $f$ from both sides we obtain\n$$\nf(x)=x+a\n$$\nfor all $x \\in \\mathbb{R}$. But $0=f(a)=2a$. We deduce that\n$$\nf(x)=x\n$$\nfor all $x \\in \\mathbb{R}$.\nConversely, we check easily that this function is a solution to the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40097,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle, and let $AA_{1}$, $BB_{1}$, and $CC_{1}$ be its altitudes. Segments $AA_{1}$ and $B_{1}C_{1}$ meet at point $K$. The perpendicular bisector of segment $A_{1}K$ intersects sides $AB$ and $AC$ at $L$ and $M$, respectively. Prove that points $A$, $A_{1}$, $L$, and $M$ lie on a circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because $LM$ is parallel to $BC$, the problem is equivalent to proving that $\\angle AA_{1}L = \\angle AML = \\angle ACB$. We present two solutions:\n\n![](attached_image_1.png)\n\nFirst solution. Let $L_{1}$ be the point on $AB$ such that $\\angle AA_{1}L_{1} = \\angle ACB$ and let us prove that $L_{1} = L$.\n\nQuadrilateral $BCB_{1}C_{1}$ is cyclic since $\\angle BC_{1}C = \\angle BB_{1}C = 90^{\\circ}$. Therefore, $\\angle B_{1}C_{1}A = \\angle ACB = \\angle AA_{1}L_{1}$. We deduce that the quadrilateral $A_{1}KC_{1}L_{1}$ is cyclic. Hence, $\\angle L_{1}KA_{1} = \\angle L_{1}C_{1}A_{1}$.\n\nQuadrilateral $AC_{1}A_{1}C$ is cyclic since $\\angle CA_{1}A = \\angle CC_{1}A = 90^{\\circ}$. We deduce that $\\angle L_{1}C_{1}A_{1} = \\angle ACB$. Thus, $\\angle L_{1}KA_{1} = \\angle AA_{1}L_{1}$. This proves that $L_{1}A_{1} = L_{1}K$, that is $L_{1} = L$, the intersection point of the perpendicular bisector of $KA_{1}$ with $AB$.\n\n\nSecond solution. Since $LK = LA_{1}$, the problem is equivalent to proving that $\\angle LKA_{1} = \\angle ACB$. But $\\angle BHA_{1} = 90^{\\circ} - \\angle A_{1}BH = \\angle ACB$, where $H$ is the orthocenter of triangle $ABC$. Therefore, the problem is equivalent to proving that $LK$ and $BH$ are parallel.\n\nWe know that $\\frac{AL}{LB} = \\frac{AN}{NA_{1}}$, where $N$ is the midpoint of $KA_{1}$, since $LM$ and $BC$ are parallel. It remains to prove that $\\frac{AK}{KH} = \\frac{AN}{NA_{1}}$.\n\nTo simplify the notations let us write $\\alpha = \\angle BAC$, $\\beta = \\angle CBA$ and $\\gamma = \\angle ACB$.\n\nWe know that $\\angle KC_{1}A = \\gamma$, $\\angle HC_{1}K = 90^{\\circ} - \\gamma$, $\\angle C_{1}AK = 90^{\\circ} - \\beta$ and $\\angle KHC_{1} = \\beta$. We deduce by applying sine laws on triangles $AC_{1}K$ and $C_{1}HK$ that\n$$\n\\frac{AK}{KH} = \\frac{\\sin \\angle KC_{1}A \\cdot \\sin \\angle KHC_{1}}{\\sin \\angle HC_{1}K \\cdot \\sin \\angle C_{1}AK} = \\tan \\beta \\cdot \\tan \\gamma\n$$\nOn the other hand, we have $\\frac{AN}{NA_{1}} = \\frac{AK}{NA_{1}} + 1 = 2 \\frac{AK}{KA_{1}} + 1$. We also know that $\\angle A_{1}C_{1}K = 180^{\\circ} - 2\\gamma$ and $\\angle KA_{1}C_{1} = 90^{\\circ} - \\alpha$. We deduce by applying sine laws on triangles $AC_{1}K$ and $C_{1}A_{1}K$ that\n$$\n\\frac{AK}{KA_{1}} = \\frac{\\sin \\gamma \\cdot \\cos \\alpha}{\\sin 2\\gamma \\cdot \\cos \\beta} = \\frac{\\cos \\alpha}{2 \\cos \\beta \\cdot \\cos \\gamma}\n$$\nWe deduce that\n$$\n\\frac{AN}{NA_{1}} = \\frac{\\cos \\alpha}{\\cos \\beta \\cdot \\cos \\gamma} + 1 = \\frac{\\cos \\beta \\cdot \\cos \\gamma - \\cos (\\beta + \\gamma)}{\\cos \\beta \\cdot \\cos \\gamma} = \\tan \\beta \\cdot \\tan \\gamma = \\frac{AK}{KH}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40098,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive real numbers $a$, $b$ and $c$, prove that\n$$\n\\frac{a^{3}}{a^{2}+a b+b^{2}}+\\frac{b^{3}}{b^{2}+b c+c^{2}}+\\frac{c^{3}}{c^{2}+c a+a^{2}} \\geq \\frac{a+b+c}{3}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have\n$$\n\\begin{aligned}\n& \\frac{a^{3}}{a^{2}+a b+b^{2}}+\\frac{b^{3}}{b^{2}+b c+c^{2}}+\\frac{c^{3}}{c^{2}+c a+a^{2}} \\\\\n& \\quad=\\frac{a^{4}}{a^{3}+a^{2} b+a b^{2}}+\\frac{b^{4}}{b^{3}+b^{2} c+b c^{2}}+\\frac{c^{4}}{c^{3}+c^{2} a+c a^{2}} \\\\\n& \\quad \\geq \\frac{\\left(a^{2}+b^{2}+c^{2}\\right)^{2}}{a^{3}+a b^{2}+a c^{2}+b a^{2}+b^{3}+b c^{2}+c a^{2}+c b^{2}+c^{3}}\n\\end{aligned}\n$$\nby applying Cauchy-Schwarz inequality.\n\nOn the other hand\n$$\n\\begin{aligned}\n\\frac{\\left(a^{2}+b^{2}+c^{2}\\right)^{2}}{a^{3}+a b^{2}+a c^{2}+b a^{2}+b^{3}+b c^{2}+c a^{2}+c b^{2}+c^{3}} & =\\frac{a^{2}+b^{2}+c^{2}}{a+b+c} \\\\\n& \\geq \\frac{a+b+c}{3}\n\\end{aligned}\n$$\nby applying again Cauchy-Schwarz inequality.\n\nThe equality holds when $a=b=c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40099,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x, y$ be two integers. Prove that if $2013$ divides $x^{1433} + y^{1433}$ then $2013$ divides $x^{7} + y^{7}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $2013 = 3 \\times 11 \\times 61$, we will prove that for $p = 3, 11, 61$, if $p$ divides $x^{1433} + y^{1433}$ then $p$ divides $x^{7} + y^{7}$.\n\nLet $p = 3, 11, 61$, and assume that $p$ divides $x^{1433} + y^{1433}$.\n\nIf $p$ divides $x$, then it divides $x^{7}$ and $x^{1433}$. But $p$ divides $x^{1433} + y^{1433}$. Then it divides $y^{1433}$. Since $p$ is a prime number, we deduce that it divides $y$ and $y^{7}$. Therefore, it divides $x^{7} + y^{7}$. In a similar way we prove that if $p$ divides $y$, then it divides $x^{7} + y^{7}$.\n\nAssume now that $p$ is relatively prime with $x, y$. Then, using Fermat,\n$$\nx^{p-1} \\equiv y^{p-1} \\equiv 1 \\quad \\bmod p\n$$\nBut $p-1$ divides $1440$ for $p = 3, 11, 61$. We deduce that\n$$\nx^{1440} \\equiv y^{1440} \\equiv 1 \\quad \\bmod p .\n$$\nHence\n$$\nx^{7} + y^{7} \\equiv x^{7} y^{1440} + y^{7} x^{1440} \\equiv x^{7} y^{7} (y^{1433} + x^{1433}) \\equiv 0 \\quad \\bmod p .\n$$\nThis proves that $p$ divides $x^{7} + y^{7}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40100,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x, y$ be two non-negative integers. Prove that $47$ divides $3^{x} - 2^{y}$ if and only if $23$ divides $4x + y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because $2$ and $47$ are relatively prime numbers, $47$ divides $3^{x} - 2^{y}$ if and only if $47$ divides $2^{4x}(3^{x} - 2^{y})$. But\n$$\n2^{4x}(3^{x} - 2^{y}) = 48^{x} - 2^{4x + y} \\equiv 1 - 2^{4x + y} \\pmod{47}.\n$$\nTherefore $47$ divides $3^{x} - 2^{y}$ if and only if $2^{4x + y} \\equiv 1 \\pmod{47}$.\n\nOn the other hand, we have $2^{23} \\equiv 49^{23} \\equiv 7^{46} \\equiv 1 \\pmod{47}$. We deduce that the prime number $23$ is the order of $2$ modulo $47$. This implies that $2^{4x + y} \\equiv 1$ mod $47$ if and only if $23$ divides $4x + y$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40101,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S = \\{0, 1, 2, 3, \\ldots\\}$ be the set of the non-negative integers. Find all strictly increasing functions $f: S \\rightarrow S$ such that $n + f(f(n)) \\leq 2 f(n)$ for every $n$ in $S$.",
    "options": [],
    "answer": "All functions of the form f(n) = n for n < n0 and f(n) = n + k0 for n ≥ n0, for some n0 ∈ S and some integer k0 ≥ 0.",
    "solution": "Since $f$ is strictly increasing, we have $f(n) \\geq n$ for all $n$ in $S$.\n\nAssume that there exists an integer $n$ in $S$ such that $f(n) > n$. Let $n_0$ be the smallest such $n$ and write $f(n_0) = n_0 + k_0$, for some positive integer $k_0 \\geq 1$. Again, since $f$ is strictly increasing, we have $f(n) \\geq n + k_0$, for all integers $n \\geq n_0$.\n\nBecause $k_0 \\geq 1$, we have\n$$\nf(n_0 + k_0) \\geq n_0 + 2k_0.\n$$\nOn the other hand, we have\n$$\nf(n_0 + k_0) = f(f(n_0)) \\leq 2 f(n_0) - n_0 = n_0 + 2k_0\n$$\nWe deduce that\n$$\nf(n_0 + k_0) = n_0 + 2k_0\n$$\nBecause $f(n_0) = n_0 + k_0$, $f(n_0 + k_0) = (n_0 + k_0) + k_0$ and $f$ is strictly increasing, we deduce that $f(n) = n + k_0$, for all $n_0 \\leq n \\leq n_0 + k_0$.\n\nWe prove by induction on $m$ that for all integer $n$ such that\n$$\nn_0 + m k_0 \\leq n \\leq n_0 + (m+1) k_0\n$$\nwe have $f(n) = n + k_0$. Hence\n$$\nf(n) = \\begin{cases} n + k_0 & \\text{ if } n \\geq n_0 \\\\ n & \\text{ otherwise } \\end{cases}\n$$\nConversely, if $f$ is such a function and $n \\geq n_0$ then\n$$\nn + f(f(n)) = n + f(n + k_0) = 2n + 2k_0 = 2 f(n)\n$$\nAnd if $f(n) = n$, the inequality is clearly satisfied.\n\nTherefore, the solutions to this functional inequality are the functions that can be written as\n$$\nf(n) = \\begin{cases} n + k_0 & \\text{ if } n \\geq n_0 \\\\ n & \\text{ otherwise } \\end{cases}\n$$\nfor some $n_0 \\in S$ and some nonnegative integer $k_0 \\geq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40102,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\geq 2$, determine the number of ordered $n$-tuples of integers $(a_{1}, a_{2}, \\ldots, a_{n})$ such that\n\na) $a_{1} + a_{2} + \\cdots + a_{n} \\geq n^{2}$; and\n\nb) $a_{1}^{2} + a_{2}^{2} + \\cdots + a_{n}^{2} \\leq n^{3} + 1$.",
    "options": [],
    "answer": "1",
    "solution": "First solution. Notice that we have\n$$\n\\begin{aligned}\n\\left(a_{1}-n\\right)^{2}+\\cdots+\\left(a_{n}-n\\right)^{2} & =\\left(a_{1}^{2}+\\cdots+a_{n}^{2}\\right)-2 n\\left(a_{1}+\\cdots+a_{n}\\right)+n^{3} \\\\\n& \\leq n^{3}+1-2 n^{3}+n^{3}=1 .\n\\end{aligned}\n$$\nTherefore, there are two cases:\n\nThe first case is when $a_{1}=a_{2}=\\cdots=a_{n}=n$. This is a solution since it satisfies both inequalities.\n\nThe second case is when there exists $1 \\leq i_{0} \\leq n$ such that $|a_{i_{0}}-n|=1$ and $a_{i}=n$ for all $1 \\leq i \\leq n$ with $i \\neq i_{0}$.\n\nIf $a_{i_{0}}=n-1$, the first inequality becomes $n^{2}-1=a_{1}+\\cdots+a_{n} \\geq n^{2}$ which is impossible.\n\nIf $a_{i_{0}}=n+1$, the second inequality becomes $n^{3}+2 n+1=a_{1}^{2}+\\cdots+a_{n}^{2} \\leq n^{3}+1$ which is also impossible.\n\nHence, the only solution to both inequalities is given by $a_{1}=a_{2}=\\cdots= a_{n}=n$.\n\n\nSecond solution. We have by Cauchy-Schwartz inequality\n$$\n\\left(n^{2}+1\\right)^{2}>n^{4}+n \\geq n\\left(a_{1}^{2}+\\cdots+a_{n}^{2}\\right) \\geq\\left(a_{1}+\\cdots+a_{n}\\right)^{2} \\geq n^{4} .\n$$\nWe deduce that\n$$\na_{1}+\\cdots+a_{n}=n^{2}\n$$\nand\n$$\nn^{3}+1 \\geq a_{1}^{2}+\\cdots+a_{n}^{2} \\geq n^{3}\n$$\nand the right hand side equality occurs if and only if $a_{1}=\\cdots=a_{n}=n$.\n\nNow, assume, looking for a contradiction, that $a_{1}^{2}+\\cdots+a_{n}^{2}=n^{3}+1$. Since a number and all his powers have the same parity, we deduce that\n$$\nn \\equiv n^{2} \\equiv a_{1}+\\cdots+a_{n} \\equiv a_{1}^{2}+\\cdots+a_{n}^{2} \\equiv n^{3}+1 \\equiv n+1 \\quad \\bmod 2\n$$\nwhich is impossible.\n\nHence, the only solution to both inequalities is given by $a_{1}=a_{2}=\\cdots= a_{n}=n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40103,
    "subject": "Mathematics (Multi-modal)",
    "question": "The quadratic equation $a x^{2} + b x + c = 0$ has its roots in the interval $[0, 1]$. Find the maximum of\n$$\n\\frac{(a-b)(2a-b)}{a(a-b+c)}\n$$",
    "options": [],
    "answer": "3",
    "solution": "Let $u, v$ be the roots of the quadratic equation $a x^{2} + b x + c = 0$ such that $0 \\leq u \\leq v \\leq 1$. We have the relations $b = -a(u + v)$ and $c = u v$. Therefore\n$$\n\\begin{aligned}\n\\frac{(a-b)(2a-b)}{a(a-b+c)} &= \\frac{(1 + u + v)(2 + u + v)}{1 + u + v + u v} \\\\\n&= 2 + \\frac{u}{1 + v} + \\frac{v}{1 + u} \\\\\n&\\leq 2 + \\frac{u}{1 + u} + \\frac{1}{1 + u} = 3\n\\end{aligned}\n$$\nClearly, when $u = v = 1$, the equality holds. Thus, $3$ is the maximum.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40104,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive integers $a$ and $b$, $\\operatorname{gcd}(a, b)$ denotes their greatest common divisor and $\\operatorname{lcm}(a, b)$ their least common multiple. Determine the number of ordered pairs $(a, b)$ of positive integers satisfying the equation\n$$\na b+63=20 \\operatorname{lcm}(a, b)+12 \\operatorname{gcd}(a, b).\n$$",
    "options": [],
    "answer": "2",
    "solution": "Let $d=\\operatorname{gcd}(a, b)$ and $a=d a'$, $b=d b'$ with $a', b'$ two relatively prime positive integers. The equation becomes\n$$\na' b' d^2 + 63 = 20 a' b' d + 12 d\n$$\nTherefore, $d$ divides $63$ and we have\n$$\na' b' d + \\frac{63}{d} = 20 a' b' + 12\n$$\nIf $5 < d < 20$, then $a' b' d + \\frac{63}{d} < 20 a' b' + 12$, and this is impossible. Hence, $d = 1, 3, 21$, or $63$.\n\n1. If $d = 1$, the equation is equivalent to $51 = 19 a' b'$, which is impossible since $19$ does not divide $51$.\n\n2. If $d = 3$, the equation is equivalent to $9 = 17 a' b'$, which is impossible since $17$ does not divide $9$.\n\n3. If $d = 21$, the equation is equivalent to $a' b' = 9$. Because $a', b'$ are relatively prime positive integers, either $a' = 1, b' = 9$ or $a' = 9, b' = 1$. This leads to two solutions $(a, b) = (21, 189)$ and $(a, b) = (189, 21)$.\n\n4. If $d = 63$, the equation is equivalent to $43 a' b' = 11$, which is impossible since $43$ does not divide $11$.\n\nTherefore, the equation has two solutions $(a, b) = (21, 189)$ and $(a, b) = (189, 21)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40105,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{1}, a_{2}, \\ldots, a_{9}$ be integers. Prove that if $19$ divides $a_{1}^{9} + a_{2}^{9} + \\cdots + a_{9}^{9}$ then $19$ divides the product $a_{1} a_{2} \\cdots a_{9}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that $19$ does not divide the product $a_{1} a_{2} \\cdots a_{9}$. This means that $a_{1}, a_{2}, \\ldots, a_{9}$ are relatively prime with $19$. Using Fermat,\n$$\na_{1}^{18} \\equiv a_{2}^{18} \\equiv \\cdots \\equiv a_{9}^{18} \\equiv 1 \\pmod{19}.\n$$\nBut $a_{i}^{18} \\equiv 1 \\pmod{19}$ is equivalent to $(a_{i}^{9} - 1)(a_{i}^{9} + 1) \\equiv 0 \\pmod{19}$. Since $19$ is a prime number, this implies that $a_{i}^{9} \\equiv \\pm 1 \\pmod{19}$, for $i = 1, \\ldots, 9$, and therefore $a_{1}^{9} + a_{2}^{9} + \\cdots + a_{9}^{9} \\equiv \\pm k \\pmod{19}$ for some odd number $k$ between $1$ and $9$. This is a contradiction. Thus $19$ divides the product $a_{1} a_{2} \\cdots a_{9}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40106,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define Fibonacci sequence $\\{F\\}_{n=0}^{\\infty}$ as $F_{0}=0$, $F_{1}=1$ and $F_{n+1}=F_{n}+ F_{n-1}$ for every integer $n>1$. Determine all quadruples $(a, b, c, n)$ of positive integers with $a<b<c$ such that each of $a, b, c, a+n, b+n, c+2n$ is a term of the Fibonacci sequence.",
    "options": [],
    "answer": "(1, 2, 3, 1)",
    "solution": "Let $(a, b, c, n)$ be a quadruplet of positive integers with $a<b<c$ such that each of $a, b, c, a+n, b+n, c+2n$ is a term of the Fibonacci sequence, and let\n$$\nb+n=F_{k}\n$$\nfor some positive integer $k$. Because $b<b+n$ and $a+n<b+n$, we have $\\max \\{b, a+n\\} \\leq F_{k-1}$.\nAssume $\\min \\{b, a+n\\} \\leq F_{k-2}$. We have $a+n+b \\leq F_{k-1}+F_{k-2}=F_{k}=b+n$, which is impossible since $a>0$. Therefore,\n$$\n\\begin{gathered}\na+n=b=F_{k-1}, \\\\\nn=(b+n)-b=F_{k}-F_{k-1}=F_{k-2},\n\\end{gathered}\n$$\nand\n$$\na=(a+n)-n=F_{k-1}-F_{k-2}=F_{k-3} .\n$$\nLet $c+2n=F_{m}$. We have $F_{k} \\leq c \\leq F_{m-1}$ and therefore $F_{m-2} \\leq 2n= 2F_{k-2} \\leq F_{k}$.\nIf $F_{m-2}=F_{k}$ then $F_{k-2}=F_{k-1}=1$ and $a=F_{k-3}=0$ which is impossible. Therefore\n$$\nc=b+n=F_{m-1}=F_{k},\n$$\nand\n$$\nc+2n=F_{m}=F_{k+1}\n$$\nBut\n$$\n2n=(c+2n)-c=F_{k+1}-F_{k}=F_{k-1}=a+n .\n$$\nWe deduce that\n$$\nF_{k-3}=a=n=F_{k-2}=1 .\n$$\nHence, $k=4$, $(a, b, c, n)=(1,2,3,1)$ and we check easily that\n$$\na=F_{1},\\ b=a+n=F_{3},\\ c=b+n=F_{4}\\ \\text{ and }\\ c+2n=F_{5} .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40107,
    "subject": "Mathematics (Multi-modal)",
    "question": "The base-7 representation of number $n$ is $\\overline{a b c}_{(7)}$, and the base-9 representation of number $n$ is $\\overline{c b a}_{(9)}$. What is the decimal (base-10) representation of $n$?",
    "options": [],
    "answer": "0 or 248",
    "solution": "We have $n = c + 7b + 49a = a + 9b + 81c$. This implies that $20(2c - a) = (4a - b)$. Because $0 \\leq a, b, c \\leq 6$, either $4a - b = 2c - a = 0$ or $4a - b = 20$ and $2c - a = 1$.\n\n1. If $4a - b = 2c - a = 0$, then $b = 8c \\leq 6$ and therefore $a = b = c = 0$, that is $n = 0$.\n\n2. If $4a - b = 20$ and $2c - a = 1$, then $a = 5$, or $6$. If $a = 5$ then $b = 0$ and $c = 3$ and $n = 248$. If $a = 6$, then $2c = 7$, which is impossible.\n\nTherefore, $n = 0$ or $248$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40108,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the area of the set of points of the plane whose coordinates $(x, y)$ satisfy\n$$\nx^{2}+y^{2} \\leq 4|x|+4|y|\n$$",
    "options": [],
    "answer": "32 + 16π",
    "solution": "Notice that, if $(x, y)$ is a solution of the inequality, then $(-x, y)$, $(x,-y)$, and $(-x,-y)$ are all solutions of the inequality. Therefore, we can assume $x, y \\geq 0$ and deduce the other solutions by symmetries with respect to $x$-axis and $y$-axis.\n\nAssume $x, y \\geq 0$. The inequality is equivalent to\n$$\n(x-2)^{2}+(y-2)^{2} \\leq 8 .\n$$\nThe points whose coordinates satisfy this inequality are precisely the points in the intersection of the first quadrant with the disk of center $(2,2)$ and radius $\\sqrt{8}$. By applying the above symmetries we obtain the following surface of points whose coordinates satisfy the inequality.\n\n![](attached_image_1.png)\n\nIts area, is the area of a square of side length $4 \\sqrt{2}$ and four half disks of radius $2 \\sqrt{2}$, that is $32+16 \\pi$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40109,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many permutations $\\left(s_{1}, s_{2}, \\cdots, s_{n}\\right)$ of $(1,2, \\ldots, n)$ are there satisfying the condition $s_{i}>s_{j}$ for all $i \\geq j+3$ when $n=5$ and when $n=7$?",
    "options": [],
    "answer": "25 for n=5, 124 for n=7",
    "solution": "Let $N_{n}$ be the number of permutations satisfying these conditions.\n\nFor $n=5$, the conditions are $s_{4}>s_{1}$ and $s_{5}>s_{2}, s_{1}$. Among the $5!$ permutations of $(1, \\cdots, 5)$, half of them satisfy the condition $s_{4}>s_{1}$. Among these permutations, half of them satisfy also the condition $s_{5}>s_{2}$. Therefore, there are $30$ permutations satisfying both conditions $s_{4}>s_{1}$ and $s_{5}>s_{2}$.\n\nTo compute $N_{5}$, it is easier to subtract from $30$ the number of permutations which do not satisfy the condition $s_{5}>s_{1}$. These permutations satisfy the condition $s_{4}>s_{1}>s_{5}>s_{2}$ with no condition on $s_{3}$. Since there are $5$ possibilities for $s_{3}$, there are $5$ such permutations and\n$$\nN_{5}=30-5=25\n$$\nFor $n=7$. There are more conditions and the method used for $n=5$ becomes complicated. That is why, we will use a different method based on an inductive relation for $N_{n}$.\n\nFor $n=1,2,3$ there are no conditions on the permutations. So\n$$\nN_{1}=1!=1, \\quad N_{2}=2!=2, \\quad N_{3}=3!=6 .\n$$\nFor $n=4$, there is only one condition: $s_{4}>s_{1}$. This gives,\n$$\nN_{4}=\\frac{1}{2} 4!=12\n$$\nFor $n \\geq 5$, we have the conditions $s_{n} \\geq s_{1}, s_{2}, \\ldots, s_{n-3}$. This means that at most there are only $s_{n-1}, s_{n-2}$ at most which can be greater than $s_{n}$. Therefore $s_{n} \\in\\{n, n-1, n-2\\}$.\n\n(a) When $s_{n}=n$, there are as many permutations as for $n-1$, that is $N_{n-1}$.\n\n(b) When $s_{n}=n-1$, because of the conditions $s_{n} \\geq s_{1}, s_{2}, \\ldots, s_{n-3}$, we have $n \\in\\left\\{s_{n-1}, s_{n-2}\\right\\}$.\n- If $s_{n-1}=n$, there are as many permutations as for $n-2$, that is $N_{n-2}$.\n- If $s_{n-2}=n$, because of the conditions $s_{n-1} \\geq s_{1}, s_{2}, \\ldots, s_{n-4}$ we have $s_{n-1} \\in\\{n-2, n-3\\}$.\n  i. If $s_{n-1}=n-2$, there are as many permutations as for $n-3$, that is $N_{n-3}$.\n  ii. If $s_{n-1}=n-3$ then $s_{n-3}=n-2$ and there are as many permutations as for $n-4$, that is $N_{n-4}$.\n\n(c) When $s_{n}=n-2$, because of the conditions $s_{n} \\geq s_{1}, s_{2}, \\ldots, s_{n-3}$, we have either $s_{n-1}=n, s_{n-2}=n-1$ or $s_{n-1}=n-1, s_{n-2}=n$. In each case there are as many permutations as for $n-3$, that is $N_{n-3}$.\n\nHence, we obtain the inductive relation\n$$\nN_{n}=N_{n-1}+N_{n-2}+3 N_{n-3}+N_{n-4}\n$$\nTherefore\n$$\n\\begin{gathered}\nN_{5}=12+6+3 \\times 2+1=25 \\\\\nN_{6}=25+12+3 \\times 6+2=57 \\\\\nN_{7}=57+25+3 \\times 12+6=124\n\\end{gathered}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 40110,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define a regular $n$-pointed star to be a union of $n$ line segments $P_{1}P_{2}$, $P_{2}P_{3}$, $\\ldots$, $P_{n}P_{1}$ such that\n- the points $P_{1}, P_{2}, \\ldots, P_{n}$ are coplanar and no three of them are collinear;\n- each of the $n$ line segments intersects at least one of the other line segments at a point other than an endpoint;\n- all of the angles at $P_{1}, P_{2}, \\ldots, P_{n}$ are congruent;\n- all of the $n$ line segments $P_{1}P_{2}, P_{2}P_{3}, \\ldots, P_{n}P_{1}$ are congruent; and\n- the path $P_{1}P_{2}\\ldots P_{n}P_{1}$ turns counterclockwise at an angle less than $180^{\\circ}$ at each vertex.\nThere are no regular 3-pointed, 4-pointed, or 6-pointed stars. All regular 5-pointed stars are similar, but there are two non-similar regular 7-pointed stars. Find all possible values of $n$ such that there are exactly 29 non-similar regular $n$-pointed stars.",
    "options": [],
    "answer": "61, 77, 93, 99, 122, 124, 154, 186, 198",
    "solution": "Because all angles are congruent and all line segments are congruent, the regular $n$-pointed star is cyclic. Indeed, consider any four consecutive vertices $P_{i}, P_{i+1}, P_{i+2}, P_{i+3}$, for some $i \\in \\{1,2, \\ldots, n\\}$. They form an isosceles trapezoid. So they are cocyclic. By induction, all vertices $P_{1}, P_{2}, \\ldots, P_{n}$ are on the same circle. Let $O$ be the center of this circle.\n\nBecause all the line segments are congruent and the path turns counterclockwise, there exists a positive real number $0^{\\circ} < \\theta < 180^{\\circ}$ such that $\\angle P_{i} O P_{i+1} = \\theta$ for all $i = 1,2, \\ldots, n$. Because $P_{n+1} = P_{1}$, there exists a positive integer $k$ such that $n \\theta = 360k^{\\circ}$. But $\\theta < 180^{\\circ}$. Therefore $k < \\frac{n}{2}$. Because each of the $n$ line segments intersects at least one of the other line segments at a point other than an endpoint, $k > 1$. Because all the vertices $P_{1}, P_{2}, \\ldots, P_{n}$ are different, the integers $k$ and $n$ have no common divisors.\n\nConversely, for a given circumradius, if $1 < k < \\frac{n}{2}$ is an integer relatively prime to $n$, there exists a unique regular $n$-pointed star of angle $\\theta = \\frac{360k^{\\circ}}{n}$. Hence, the number of non-similar regular $n$-pointed stars is the number of such integers $k$, that is $\\frac{\\phi(n)}{2} - 1$, where $\\phi$ is the Euler totient function. Therefore, it remains to solve the equation $\\phi(n) = 60$.\n\nThere are three cases. The first case is when there exists a prime $p$ divisor of $n$ such that $3 \\times 5$ divides $p-1$. Here, the only possibilities are $p = 61$ or $31$.\n\nIf $p = 61$, there are 2 solutions for $n$. Either $n = 61$ or $n = 2 \\times 61 = 122$.\nIf $p = 31$, there are 3 solutions for $n$. Either $n = 3 \\times 31 = 93$, or $n = 2 \\times 3 \\times 31 = 186$, or $n = 2^{2} \\times 31 = 124$.\n\nThe second case is when there exist two prime numbers $p$ and $q$ such that $3$ divides $p-1$ and $5$ divides $q-1$. In this case, the only possibility is $p = 7$ and $q = 11$. This gives only two solutions for $n$. Either $n = 7 \\times 11 = 77$ or $n = 2 \\times 7 \\times 11 = 154$.\n\nThe third case is when there exists an odd prime number $p$ such that $p^{2}$ divides $n$. Then $p = 3$ or $5$. If $n = 5^{2} m$ and $m$ is relatively prime to $5$, we obtain $\\phi(m) = 3$ which is impossible since $3$ is an odd number. If $n = 3^{2} m$ and $m$ is relatively prime to $3$, we obtain $\\phi(m) = 10$. Therefore, $m = 11$ or $22$. Hence $n = 99$ or $198$.\n\nIn conclusion, all possible values for $n$ are $61, 77, 93, 99, 122, 124, 154, 186$ and $198$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 40111,
    "subject": "Mathematics (Multi-modal)",
    "question": "A Saudi company has two offices. One office is located in Riyadh and the other in Jeddah. To insure the connection between the two offices, the company has designated from each office a number of correspondents so that\n(a) each pair of correspondents from the same office share exactly one common correspondent from the other office.\n(b) there are at least 10 correspondents from Riyadh.\n(c) Zayd, one of the correspondents from Jeddah, is in contact with exactly 8 correspondents from Riyadh.\nWhat is the minimum number of correspondents from Jeddah who are in contact with the correspondent Amr from Riyadh?",
    "options": [],
    "answer": "8",
    "solution": "Assume that we have two correspondents, $R$ from Riyadh and $J$ from Jeddah, who are not connected to each other. Let $\\mathcal{J}_R$ be the set of correspondents from Jeddah connected to $R$ and $\\mathcal{R}_J$ the set of correspondents from Riyadh connected to $J$. Define the function\n$$\nf: \\mathcal{J}_R \\longrightarrow \\mathcal{R}_J\n$$\nin the following way:\nFor each correspondent $J'$ $\\in \\mathcal{J}_R$, since the correspondents $J$ and $J'$ are different, they share a unique correspondent in $\\mathcal{R}_J$. Let $f\\left(J'\\right)$ be this unique correspondent. So $f$ is well-defined.\nAssume that there exist two correspondents $J_1, J_2 \\in \\mathcal{J}_R$ such that $f\\left(J_1\\right)= f\\left(J_2\\right)=R' \\in \\mathcal{R}_J$. This means that the correspondent $R'$ from Riyadh shares with the correspondent $R$ two correspondents $J_1, J_2$ from Jeddah. Hence $J_1=J_2$ and the map $f$ is injective.\nConsider a correspondent $R' \\in \\mathcal{R}_J$ and let $J'$ be the unique correspondent in $\\mathcal{J}_R$ that $R'$ shares with $R$. Then $R'$ is the unique correspondent in $\\mathcal{R}_J$ that $J'$ shares with $J$. Thus $R'=f\\left(J'\\right)$ and $f$ is surjective. This proves that $R$ and $J$ have the same number of correspondents from the other city.\nWe deduce from this that if the correspondent Amr from Riyadh is not connected to the correspondent Zayd from Jeddah, the correspondent Amr has exactly eight correspondents from Jeddah.\nAssume now that the correspondent Amr is connected to the correspondent Zayd from Jeddah. Since there are at least ten correspondents in Riyadh and Zayd is connected only with eight, let $R_1, R_2$ be two correspondents from Riyadh who are not connected to Zayd. Let $J_1$ be the unique correspondent that Amr and $R_1$ share. Let $J_2$ be the unique correspondent that $R_1$ shares with $R_2$. Since $R_1$ is not in contact with Zayd, he has eight correspondents from Jeddah. So there exists at least six correspondents in contact with $R_1$ who are not in contact neither with Amr nor with $R_2$. Let $J$ be one of these correspondents.\nThe correspondent $R_2$ from Riyadh has exactly eight correspondents from Jeddah since he is not connected to the correspondent Zayd. The correspondent $J$ from Jeddah has exactly eight correspondents from Riyadh since he is not connected to the correspondent $R_2$ from Riyadh. The correspondent Amr from Riyadh has exactly eight correspondents from Jeddah since he is not connected to the correspondent $J$ from Jeddah.\nThis proves that in all cases the correspondent Amr from Riyadh has exactly eight correspondents from Jeddah.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40112,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABC$ is a triangle, $H$ its orthocenter, $I$ its incenter, $O$ its circumcenter and $\\omega$ its circumcircle. Line $CI$ intersects circle $\\omega$ at point $D$ different from $C$. Assume that $AB = ID$ and $AH = OH$. Find the angles of triangle $ABC$.",
    "options": [],
    "answer": "∠A = 15°, ∠B = 45°, ∠C = 120°",
    "solution": "We already know that $DA = DB = DI$. Because $DI = AB$, triangle $ADB$ is equilateral. But $\\angle ACB + \\angle BDA = 180^\\circ$, we deduce that $\\angle ACB = 120^\\circ$.\n\nBecause $\\angle BOA = 2 \\angle BDA = 120^\\circ$, we deduce that $\\angle OAB = 30^\\circ$. On the other hand, we have $\\angle CAH = 90^\\circ - \\angle CBA - \\angle BAC = 30^\\circ$. By applying cosine law in the triangle $AOH$ we obtain\n$$\nOH^2 = AH^2 + R^2 - 2R \\cdot AH \\cos(\\angle BAC + 60^\\circ),\n$$\nwhere $R$ is the circumradius of triangle $ABC$. This is equivalent to\n$$\n2AH \\cos(\\angle BAC + 60^\\circ) = R.\n$$\n![](attached_image_1.png)\nBecause $HC$ is perpendicular to $AB$ and $BC$ is perpendicular to $AH$, we have $\\angle AHC = \\angle CBA$. On the other hand, we have $\\angle HCA = 180^\\circ - \\angle AHC - 30^\\circ = 90^\\circ + \\angle BAC$. We deduce, by applying sine law to the triangle $ACH$ that\n$$\n\\frac{AH}{\\sin(90^\\circ + \\angle BAC)} = \\frac{AC}{\\sin \\angle CBA} = 2R.\n$$\nTherefore, $AH = 2R \\cos \\angle BAC$. By plugging this into our previous relation we obtain\n$$\n2 \\cos \\angle BAC \\cdot \\cos(\\angle BAC + 60^\\circ) = \\frac{1}{2}.\n$$\nThis is equivalent to\n$$\n\\cos(2\\angle BAC + 60^\\circ) + \\cos 60^\\circ = \\frac{1}{2},\n$$\nand therefore, $\\angle BAC = 15^\\circ$. Thus\n$$\n\\angle BAC = 15^\\circ, \\quad \\angle CBA = 45^\\circ, \\quad \\text{and} \\quad \\angle ACB = 120^\\circ.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40113,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the following equation where $x$ is a real number:\n$$\n\\left\\lfloor x^{2}\\right\\rfloor - 10 \\lfloor x \\rfloor + 24 = 0\n$$",
    "options": [],
    "answer": "[4, \\sqrt{17}) \\cup [\\sqrt{26}, \\sqrt{27}) \\cup [6, \\sqrt{37})",
    "solution": "Notice first that if $\\lfloor x \\rfloor \\leq 0$ then\n$$\n\\left\\lfloor x^{2} \\right\\rfloor - 10 \\lfloor x \\rfloor + 24 \\geq 24 > 0\n$$\nThis means that if $x$ is a solution to the equation then $\\lfloor x \\rfloor \\geq 1$.\nLet $x$ be a solution to the equation, $m = \\lfloor x \\rfloor \\geq 1$, and $r = x - \\lfloor x \\rfloor \\geq 0$. We have\n$$\n0 = \\left\\lfloor x^{2} \\right\\rfloor - 10 \\lfloor x \\rfloor + 24 = \\lfloor r(r + 2m) \\rfloor + (m - 4)(m - 6).\n$$\nBecause $r(r + 2m) \\geq 0$, we deduce that $(m - 4)(m - 6) \\leq 0$, which is equivalent to $m = 4, 5$, or $6$.\n\n1. If $m = 4$, the equation becomes $\\lfloor r(r + 8) \\rfloor = 0$. This is equivalent to $(r + 4)^{2} - 17 < 0$ and $(r + 4)^{2} - 16 \\geq 0$, and its solutions are $0 \\leq r < \\sqrt{17} - 4$. This means that the solutions to the equation in this case are $4 \\leq x < \\sqrt{17}$.\n\n2. If $m = 5$, the equation becomes $\\lfloor r(r + 10) \\rfloor = 1$. This is equivalent to $(r + 5)^{2} - 27 < 0$ and $(r + 5)^{2} - 26 \\geq 0$, and its solutions are $\\sqrt{26} - 5 \\leq r < \\sqrt{27} - 5$. This means that the solutions to the equation in this case are $\\sqrt{26} \\leq x < \\sqrt{27}$.\n\n3. If $m = 6$, the equation becomes $\\lfloor r(r + 12) \\rfloor = 0$. This is equivalent to $(r + 6)^{2} - 37 < 0$ and $(r + 6)^{2} - 36 \\geq 0$, and its solutions are $0 \\leq r < \\sqrt{37} - 6$. This means that the solutions to the equation in this case are $6 \\leq x < \\sqrt{37}$.\n\nThus, the set of solutions to this equation is\n$$\n[4, \\sqrt{17}) \\cup [\\sqrt{26}, \\sqrt{27}) \\cup [6, \\sqrt{37}) .\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40114,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle, $M$ be the midpoint of $BC$ and $P$ be a point on line segment $AM$. Lines $BP$ and $CP$ meet the circumcircle of $ABC$ again at $X$ and $Y$, respectively, and sides $AC$ at $D$ and $AB$ at $E$, respectively. Prove that the circumcircles of $AXD$ and $AYE$ have a common point $T \\neq A$ on line $AM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By applying Ceva to the concurrent cevians $AM$, $BD$ and $CE$, we obtain\n$$\n\\frac{AE}{EB} = \\frac{AD}{DC} \\cdot \\frac{CM}{MB} = \\frac{AD}{DC}.\n$$\nWe deduce from Thales' theorem that segments $ED$ and $BC$ are parallel.\n\n![](attached_image_1.png)\n\nSince quadrilateral $BCXY$ is cyclic, we have $\\angle BXY = \\angle BCY = \\angle CED$. We deduce that quadrilateral $EDXY$ is cyclic and therefore\n$$\nPD \\cdot PX = PE \\cdot PY\n$$\nHence, point $P$ lies on the radical axis of circumcircles of triangles $AXD$ and $AYE$ which passes through $A$.\n\nIf these two circles are tangent to $AP$ at $A$ then\n$$\n\\angle MAC = \\angle AXB = \\angle ACB, \\text{ and } \\angle BAM = \\angle CYA = \\angle CBA,\n$$\nwhich implies that triangle $ABC$ is a right triangle at $A$ and this is a contradiction.\n\nWe conclude that the two circles intersect in a second point $T \\neq A$ on line $AM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40115,
    "subject": "Mathematics (Multi-modal)",
    "question": "The positive integer $a$ is relatively prime with $10$. Prove that for any positive integer $n$, there exists a power of $a$ whose last $n$ digits are $\\underbrace{0 \\cdots 0}_{n-1} 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "This is equivalent to prove that there exists a positive integer $k$ such that $10^{n}$ divides $a^{k}-1$.\n\nFirst solution. Consider the remainders of the division of the $10^{n}+1$ powers $a^{1}, a^{2}, \\ldots, a^{10^{n}+1}$ of $a$ by $10^{n}$. Since there are at most $10^{n}$ possible remainders, by the pigeonhole principle there exist at least two powers $a^{i}, a^{j}, i<j$, having the same remainder. Therefore $10^{n}$ divides $a^{j}-a^{i}=a^{i}\\left(a^{j-i}-1\\right)$. But $10^{n}$ is relatively prime with $a^{i}$. This proves that $10^{n}$ divides $a^{j-i}-1$.\n\n\nSecond solution. Since $a$ and $10^{n}$ are relatively prime, by Euler's theorem\n$$\na^{\\phi\\left(10^{n}\\right)}=a^{4 \\cdot 10^{n-1}} \\equiv 1 \\quad \\bmod 10^{n} .\n$$\nTherefore, $10^{n}$ divides $a^{4 \\cdot 10^{n-1}}-1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40116,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, we consider all its divisors (including $1$ and itself). Suppose that $p\\%$ of these divisors have their unit digit equal to $3$ (For example $n=117$, has six divisors, namely $1, 3, 9, 13, 39, 117$. Two of these divisors, namely $3$ and $13$, have unit digits equal to $3$. Hence for $n=117$, $p=33.33\\cdots$). Find, when $n$ is any positive integer, the maximum possible value of $p$.",
    "options": [],
    "answer": "50",
    "solution": "Let $n$ be a positive integer. Consider $D_{1}$, the set of all divisors of $n$ with unit digit $3$, and $D_{2}$ the set of all the other divisors of $n$.\n\nIf the unit digit of $n$ is different from $9$, consider the map $\\delta: D_{1} \\longrightarrow D_{2}$ defined by $\\delta(d) = \\frac{n}{d}$ for all $d \\in D_{1}$. If $d \\equiv 3 \\bmod 10$ and $\\delta(d) \\equiv 3 \\bmod 10$ for some $d \\in D_{1}$, then $n = d \\cdot \\delta(d) \\equiv 9 \\bmod 10$ which is a contradiction. Then the map $\\delta$ is well defined. Clearly, the map $\\delta$ is an injective map. Therefore the cardinality of $D_{1}$ is less than or equal to the cardinality of $D_{2}$. Hence $p \\leq 50$.\n\nIf the unit digit of $n$ is equal to $9$, the integer $n$ is neither divisible by $2$ nor by $5$. This means that all prime divisors of $n$ have unit digit equal to $1, 3, 5$, or $9$.\n\nIf all prime divisors of $n$ have unit digits equal to $1$ or $9$, all divisors of $n$ have unit digits equal to $1$ or $9$. Therefore $p = 0$.\n\nIf the integer $n$ has a prime divisor $p$ with unit digit $3$ or $7$, consider the map $\\delta: D_{1} \\longrightarrow D_{2}$ defined for $d \\in D_{1}$ by $\\delta(d) = \\frac{d}{p}$ if $p$ divides $d$ and by $\\delta(d) = p d$ otherwise. We can see easily in both cases, that the unit digit of $\\delta(d)$ is equal to $1$ or $9$. Hence $\\delta$ is well defined.\n\nAssume that there exist $d_{1}, d_{2} \\in D_{1}$ such that $\\delta\\left(d_{1}\\right) = \\delta\\left(d_{2}\\right)$. It is clear that if both $d_{1}, d_{2}$ are divisible by $p$ or both $d_{1}, d_{2}$ are not divisible by $p$, we have $d_{1} = d_{2}$. Assume that $d_{1}$ is divisible by $p$ and $d_{2}$ is not divisible by $p$. We have $\\frac{d_{1}}{p} = d_{2} p$, which is equivalent to $d_{1} = p^{2} d_{2}$. Because $d_{1} \\equiv d_{2} \\equiv 3 \\bmod 10$, we deduce that $p^{2} \\equiv 1 \\bmod 10$, which contradicts the fact that $p^{2} \\equiv -1 \\bmod 10$ since $p \\equiv 3, 7 \\bmod 10$. Hence $\\delta$ is injective and therefore the cardinality of $D_{1}$ is less than or equal to the cardinality of $D_{2}$. Thus $p \\leq 50$.\n\nNotice that for $n = 3$, its divisors are $1, 3$ and $p = 50$. This proves that the maximum possible value of $p$ is $50$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40117,
    "subject": "Mathematics (Multi-modal)",
    "question": "The points of the plane have been colored by $2013$ different colors. We say that a triangle $\\triangle ABC$ has the color $X$ if its three vertices $A$, $B$, $C$ have the color $X$. Prove that there are infinitely many triangles with the same color and the same area.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider $2014$ parallel lines. Each line contains infinitely many points. Since the number of the colors is finite, by the pigeonhole principle, there exist on each line infinitely many points of the same color. Choose for each line one color for which there exist infinitely many points. Since there are $2013$ colors and $2014$ lines, by the pigeonhole principle there exist at least two lines $\\ell_1$, $\\ell_2$ for which the same color $\\mathcal{C}$ has been chosen. Choose two points $B$, $C$ from the first line $\\ell_1$ of this color $\\mathcal{C}$. Choose infinitely many points $A_1$, $A_2$, $\\ldots$ from the second line $\\ell_2$ of this color $\\mathcal{C}$. Triangles $A_1BC$, $A_2BC$, $\\ldots$ are all of the same color $\\mathcal{C}$ and have the same area since $\\ell_1$, $\\ell_2$ are parallel.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40118,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $T$ be a real number satisfying the property: For any nonnegative real numbers $a, b, c, d, e$ with their sum equal to $1$, it is possible to arrange them around a circle such that the products of any two neighboring numbers are no greater than $T$. Determine the minimum value of $T$.",
    "options": [],
    "answer": "1/9",
    "solution": "Assume, without loss of generality, that\n$$\n0 \\leq a \\leq b \\leq c \\leq d \\leq e\n$$\nBecause\n$$\ne a \\leq \\max \\{e b, c d\\} \\leq e c \\leq e d\n$$\nto get the smallest possible maximum, the best arrangement around the circle is\n![](attached_image_1.png)\nIn this case, the maximum is given by\n$$\n\\max \\{e b, c d\\} .\n$$\nWe have\n$$\ne b \\leq e \\frac{b+c+d}{3} \\leq \\frac{e(1-e)}{3} \\leq \\frac{1}{12},\n$$\nand the equality holds when $a=0$, $b=c=d=\\frac{1}{6}$, and $e=\\frac{1}{2}$.\nWe have, on the other hand,\n$$\nc d=\\sqrt[3]{c^{2} d^{2}(c d)} \\leq (\\sqrt[3]{c d e})^{2} \\leq \\left(\\frac{c+d+e}{3}\\right)^{2} \\leq \\frac{1}{9}\n$$\nand the equality holds when $a=b=0$, and $c=d=e=\\frac{1}{3}$.\nTherefore, the minimum possible value of $T$ is $\\frac{1}{9}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40119,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ten students take a test consisting of 4 different papers in Algebra, Geometry, Number Theory and Combinatorics. First, the proctor distributes randomly the Algebra paper to each student. Then the remaining papers are distributed one at a time in the following order: Geometry, Number Theory, Combinatorics in such a way that no student receives a paper before he finishes the previous one. In how many ways can the proctor distribute the test papers given that a student may for example finish the Number Theory paper before another student receives the Geometry paper, and that he receives the Combinatorics paper after that the same other student receives the Combinatorics papers.",
    "options": [],
    "answer": "10! · 30! / (3!)^10",
    "solution": "First, since the proctor distributes randomly the Algebra paper to each student, he has $10!$ ways to do it depending on how he orders the students.\n\nFor the other three papers, we order the students from $1$ to $30$, each student receiving three positions corresponding to the three papers. Since the first position a student receives corresponds to the Geometry paper, the second to the Number Theory paper and the third to the Combinatorics paper, the problem is equivalent to count the number of partitions of the set $\\{1,2, \\ldots, 30\\}$ into $10$ subsets, each of $3$ elements. This is equal to\n$$\n\\binom{30}{3, \\ldots, 3} = \\frac{30!}{3!^{10}}.\n$$\nTherefore, the number of ways the proctor can distribute the test papers is\n$$\n\\frac{10!\\,30!}{3!^{10}}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Other"
  },
  {
    "id": 40120,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest integer $k$ such that $k$ divides $n^{55} - n$ for all integer $n$.",
    "options": [],
    "answer": "798",
    "solution": "Let $p$ be a prime divisor of $n^{55} - n$ for all integer $n$. Whenever $n$ is not divisible with $p$, we have\n$$\nn^{54} \\equiv 1 \\quad \\bmod p\n$$\nIn this case, the order of $n$ modulo $p$ divides $54$. But there exists an integer $n$ of order $p-1$ modulo $p$. We deduce that $p-1$ divides $54$. But the only primes $p$ such that $p-1$ divides $54$ are $p=2,3,7$ and $19$. Conversely, all these four primes $2,3,7$ and $19$ divide $n^{55} - n$ for all integer $n$ by Fermat's little theorem.\nNotice that for $p$, $p^{55} - 1$ is not divisible by $p^2$ for all prime numbers $p$ since $p^{54} - 1$ is relatively prime with $p$. Therefore, the greatest integer $k$ which divides $n^{55} - n$ for all integer $n$ is $2 \\times 3 \\times 7 \\times 19 = 798$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40121,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{Z}_{\\geq 0} \\rightarrow \\mathbb{Z}_{\\geq 0}$ be a function which satisfies for all integer $n \\geq 0$ :\n(a) $f(2n+1)^2 - f(2n)^2 = 6f(n) + 1$,\n(b) $f(2n) \\geq f(n)$;\nwhere $\\mathbb{Z}_{\\geq 0}$ is the set of nonnegative integers. Solve the equation $f(n) = 1000$.",
    "options": [],
    "answer": "105",
    "solution": "Let $n$ be a nonnegative integer. We have\n$$\nf(2n)^2 < f(2n)^2 + 6f(n) + 1 = f(2n+1)^2 < f(2n)^2 + 6f(2n) + 9 = (f(2n) + 3)^2.\n$$\nTherefore,\n$$\nf(2n) < f(2n+1) < f(2n) + 3.\n$$\nAssume that $f(2n+1) = f(2n) + 2$. In this case\n$$\n6f(n) + 1 = f(2n+1)^2 - f(2n)^2 = 4f(2n) + 4.\n$$\nThis is impossible since the left hand side is odd while the right hand side is even. Therefore $f(2n+1) = f(2n) + 1$.\n\nOn the other hand,\n$$\n6f(n) + 1 = f(2n+1)^2 - f(2n)^2 = 2f(2n) + 1.\n$$\nWe deduce that $f(2n) = 3f(n)$, and $f(0) = 0$.\n\nNow, let $n \\geq 0$ and write $n = \\overline{a_1 a_2 \\cdots a_k}_{(2)}$ in basis 2. We prove by induction on $k$ that $f(n) = \\overline{a_1 a_2 \\cdots a_k}_{(3)}$ in basis 3.\n\nFor $k = 1$, we have\n$$\nf\\left(\\overline{0}_{(2)}\\right) = f(0) = 0 = \\overline{0}_{(3)} \\quad \\text{and} \\quad f\\left(\\overline{1}_{(2)}\\right) = f(1) = f(0) + 1 = 1 = \\overline{1}_{(3)}.\n$$\nAssume this true for $k$. We have\n$$\n\\begin{aligned}\nf\\left(\\overline{a_1 a_2 \\cdots a_k a_{k+1}}_{(2)}\\right) & = f\\left(2\\, \\overline{a_1 a_2 \\cdots a_k}_{(2)} + a_{k+1}\\right) \\\\\n& = f\\left(2\\, \\overline{a_1 a_2 \\cdots a_k}_{(2)}\\right) + a_{k+1} \\\\\n& = 3 f\\left(\\overline{a_1 a_2 \\cdots a_k}_{(2)}\\right) + a_{k+1} \\\\\n& = 3 \\overline{a_1 a_2 \\cdots a_k}_{(3)} + a_{k+1} \\\\\n& = \\overline{a_1 a_2 \\cdots a_k a_{k+1}}_{(3)} \\\\\n& .\n\\end{aligned}\n$$\nThis completes the induction.\n\nApplying this to 1000, we have\n$$\n1000 = \\overline{1101001}_{(3)} = f\\left(\\overline{1101001}_{(2)}\\right) = f(105).\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40122,
    "subject": "Mathematics (Multi-modal)",
    "question": "$\\triangle ABC$ is a triangle with $AB < BC$, $\\mathcal{C}$ its circumcircle, $K$ the midpoint of the minor arc $\\overparen{CA}$ of the circle $\\mathcal{C}$ and $T$ a point on $\\mathcal{C}$ such that $KT$ is perpendicular to $BC$. If $A'$, $B'$ are the intouch points of the incircle of $\\triangle ABC$ with the sides $BC$, $AC$, prove that the lines $AT$, $BK$, $A'B'$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $E$ be the intersection point of $BK$ with $AT$. The problem is equivalent to prove that points $A'$, $E$, $B'$ are collinear.\n\nFirst solution. Let $F$ be the intersection point of $BC$ with $KT$. Since $K$ is the midpoint of $\\widetilde{CA}$, by expressing the angles in terms of arc lengths we obtain\n$$\n\\measuredangle KEA = \\frac{1}{2}(\\overparen{KA} + \\overparen{BT}) = \\frac{1}{2}(\\overparen{CK} + \\overparen{BT}) = \\measuredangle CFK = 90^\\circ\n$$\nBut $B'$ is an intouch point. This implies that\n$$\n\\measuredangle AB'I = 90^\\circ = \\measuredangle AEI,\n$$\nand the points $A$, $I$, $E$, $B'$ are concyclic.\n\n![](attached_image_1.png)\n\nWe deduce from this that\n$$\n\\measuredangle IB'E = \\measuredangle IAE = 90^\\circ - \\measuredangle EIA = 90^\\circ - \\frac{1}{2} \\measuredangle BAC - \\frac{1}{2} \\measuredangle CBA = \\frac{1}{2} \\measuredangle ACB .\n$$\nOn the other hand, triangle $IA'B'$ is isosceles ($IA' = IB'$). Therefore,\n$$\n\\measuredangle IB'A' = \\frac{1}{2}\\left(180^\\circ - \\measuredangle A'I B'\\right) = \\frac{1}{2} \\measuredangle ACB = \\measuredangle IB'E,\n$$\nsince $\\measuredangle CA'I = 90^\\circ$. This proves that the points $A'$, $E$, $B'$ are collinear.\n\n\nSecond solution. We prove as in the first solution that $\\measuredangle KEA = 90^\\circ$. Let $D$ be the intersection point of $AT$ with $BC$. The pedal triangle of $I$ with respect to triangle $ADC$ is $A'B'E$. Proving that points $A'$, $B'$, $E$ are collinear is equivalent to proving that $I$ is on the circumcircle of triangle $ADC$ and therefore, the line $A'B'E$ will be the Simson line of point $I$ with respect to triangle $ADC$.\n\n![](attached_image_2.png)\n\nWe have\n$$\n\\measuredangle CDA = \\measuredangle DBA + \\measuredangle BAD = \\measuredangle CBA + 90^\\circ - \\frac{1}{2} \\measuredangle CBA = 90^\\circ + \\frac{1}{2} \\measuredangle CBA .\n$$\nOn the other hand\n$$\n\\measuredangle CIA = 180^\\circ - \\frac{1}{2} \\measuredangle BAC - \\frac{1}{2} \\measuredangle ACB = 90^\\circ + \\frac{1}{2} \\measuredangle CBA = \\measuredangle CDA .\n$$\nThis proves that $I$ is on the circumcircle of triangle $ADC$ and thus $A'$, $E$, $B'$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40123,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine whether it is possible to place the integers $1,2, \\ldots, 2012$ in a circle in such a way that the 2012 products of adjacent pairs of numbers leave pairwise distinct remainders when divided by 2013.",
    "options": [],
    "answer": "Not possible",
    "solution": "Assume that it is possible to place the integers $1,2, \\ldots, 2012$ in a circle in such a way that the 2012 products of adjacent pairs of numbers leave pairwise distinct remainders when divided by 2013. Let $a_{1}, a_{2}, \\ldots, a_{2012}$ be such a reordering of the integers $1,2, \\ldots, 2012$ on the circle.\n\nBecause $2013=3 \\times 11 \\times 61$, a number is a multiple of $3$ or $11$ or $61$ if and only if its remainder when divided by $2013$ is a multiple of $3$ or $11$ or $61$.\n\nBy the pigeonhole principle, there are at least two adjacent numbers in the list $a_{1}, a_{2}, \\ldots, a_{2012}$ which are not multiples of $3$. Make this list starting from these two adjacent numbers and consider the list $b_{1}, b_{2}, \\ldots, b_{2012}$ of their products, where $b_{i}=a_{i} \\cdot a_{i+1}$, for $i=1, \\ldots, 2012$ with $a_{2013}=a_{1}$.\n\nConsider $a_{i_{1}}, a_{i_{2}}, \\ldots, a_{i_{670}}$, all the multiples of $3$ with $2<i_{1}<i_{2}<\\cdots<i_{670}$. It is clear that $b_{i_{1}}, b_{i_{2}}, \\ldots, b_{i_{670}}$ are all multiples of $3$ and that $b_{i_{1}-1}, b_{i_{2}-1}, \\ldots, b_{i_{670}-1}$ are also all multiples of $3$. So their remainders when divided by $2013$ are all multiples of $3$. But there are only $671$ different multiples of $3$ between $0$ and $2012$ included. Therefore $i_{j+1}=i_{j}+1$ for all $j=1, \\ldots, 669$. This means that the multiples of $3$ in the list $a_{1}, a_{2}, \\ldots, a_{2012}$ form a block and are not separated by any non-multiple of $3$.\n\nIn a similar way we prove that multiples of $11$ form a block in this list and that multiples of $61$ form also a block in this list. But since there are common multiples of $3$ and $11$, their blocks must be connected to each other. For the same reason the $3$ blocks must be connected by pairs to each other.\n\nBut, in each block, there are numbers which are in none of the two other blocks, for example there are multiples of $3$ which are neither multiples of $11$ nor multiples of $61$ like $3,6,9$, and the same thing happens for $11$ and $61$. So, these numbers are in the middle of each of the blocks and make the blocks intersecting only in their sides. There are also numbers which are not multiples of any of these $3$ numbers, like $1,2,4$. So, these numbers will prevent two of the three blocks to intersect, and here is the contradiction.\n\nHence, it is not possible to place the integers $1,2, \\ldots, 2012$ in a circle under the given condition.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40124,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n < 589$ for which $589$ divides $n^2 + n + 1$.",
    "options": [],
    "answer": "87, 273, 315, 501",
    "solution": "Because $589 = 19 \\times 31$, we will find all positive integers $n < 589$ such that both $19$ and $31$ divide $n^2 + n + 1$.\nLet $n$ be such an integer. We have\n$$\n0 \\equiv n^2 + n + 1 \\equiv n^2 + 20n + 1 \\equiv (n + 10)^2 - 2^2 \\equiv (n + 8)(n + 12) \\pmod{19}.\n$$\nHence, $19$ divides $n^2 + n + 1$ if and only if $n \\equiv 7$ or $11 \\pmod{19}$.\nOn the other hand, we have\n$$\n0 \\equiv n^2 + n + 1 \\equiv n^2 + 32n + 1 \\equiv (n + 16)^2 - 10^2 \\equiv (n + 26)(n + 6) \\pmod{31}.\n$$\nHence, $31$ divides $n^2 + n + 1$ if and only if $n \\equiv 5$ or $25 \\pmod{31}$.\nUsing the fact that $8 \\times 31 \\equiv 1 \\pmod{19}$ and $18 \\times 19 \\equiv 1 \\pmod{31}$, we consider the following four cases:\n\n1. If $n \\equiv 7 \\pmod{19}$ and $n \\equiv 5 \\pmod{31}$. There exists an integer $k$ such that\n$$\nn = 7 \\times 8 \\times 31 + 5 \\times 18 \\times 19 + k \\times 19 \\times 31 = 501 + 589(k + 5).\n$$\nBut $0 < n < 589$. We deduce that $n = 501$.\n\n2. If $n \\equiv 7 \\pmod{19}$ and $n \\equiv 25 \\pmod{31}$. There exists an integer $k$ such that\n$$\nn = 7 \\times 8 \\times 31 + 25 \\times 18 \\times 19 + k \\times 19 \\times 31 = 273 + 589(k + 17).\n$$\nBut $0 < n < 589$. We deduce that $n = 273$.\n\n3. If $n \\equiv 11 \\pmod{19}$ and $n \\equiv 5 \\pmod{31}$. There exists an integer $k$ such that\n$$\nn = 11 \\times 8 \\times 31 + 5 \\times 18 \\times 19 + k \\times 19 \\times 31 = 315 + 589(k + 7).\n$$\nBut $0 < n < 589$. We deduce that $n = 315$.\n\n4. If $n \\equiv 11 \\pmod{19}$ and $n \\equiv 25 \\pmod{31}$. There exists an integer $k$ such that\n$$\nn = 11 \\times 8 \\times 31 + 25 \\times 18 \\times 19 + k \\times 19 \\times 31 = 87 + 589(k + 19).\n$$\nBut $0 < n < 589$. We deduce that $n = 87$.\n\nTherefore, the possible values for $n$ are $87$, $273$, $315$, and $501$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40125,
    "subject": "Mathematics (Multi-modal)",
    "question": "$\\triangle ABC$ is a triangle, $M$ the midpoint of $BC$, $D$ the projection of $M$ on $AC$ and $E$ the midpoint of $MD$. Prove that the lines $AE$, $BD$ are orthogonal if and only if $AB = AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First solution. Let $N$ be the midpoint of $CD$ and $F$ the intersection point of lines $AE$ and $MN$. Since $M$ is the midpoint of $BC$, the segment $MN$ is parallel to $BD$.\n\n![](attached_image_1.png)\n\nTherefore, $AE$, $BD$ are perpendicular if and only if $AF$, $MN$ are perpendicular.\nEquivalently, triangles $AFN$ and $MDN$ are similar since they already share the same angle at $N$.\nSince $AFN$ and $ADE$ are similar, then triangles $ADE$ and $MDN$ are similar, which is equivalent to the fact that triangles $ADM$ and $MDC$ are similar since $E$ is the midpoint of $DM$ and $N$ is the midpoint of $DC$.\nIn conclusion, $AE$ and $BD$ are perpendicular if and only if the median $AM$ of triangle $ABC$ is also an altitude. That is $AB = AC$.\n\n\nSecond solution. By coordinates or equivalently by complex numbers. Fix the origin at $C$, and the $x$-axis to be $CA$. The affixes of the points are\n$$\nC(0), \\quad A(a) \\quad \\text{ and } \\quad B(b+ic),\n$$\nwhere $a, b, c$ are positive real numbers. Hence, the affixes of the other points are\n$$\nM\\left(\\frac{b}{2}+i \\frac{c}{2}\\right), \\quad D\\left(\\frac{b}{2}\\right) \\quad \\text{ and } \\quad E\\left(\\frac{b}{2}+i \\frac{c}{4}\\right)\n$$\n![](attached_image_2.png)\n\nTherefore, the affixes of the vectors are\n$$\n\\overrightarrow{DB}\\left(\\frac{b}{2}+i c\\right) \\quad \\text{ and } \\quad \\overrightarrow{EA}\\left(a-\\frac{b}{2}-i \\frac{c}{4}\\right) .\n$$\nOn the other hand, the lengths of the sides are equal to\n$$\nAB=\\sqrt{(a-b)^2+c^2} \\quad \\text{ and } \\quad AC=a\n$$\nThe segments $AE$, $BD$ are perpendicular if and only if\n$$\n\\operatorname{Re}\\left(\\left(\\frac{b}{2}+i c\\right)\\left(a-\\frac{b}{2}+i \\frac{c}{4}\\right)\\right)=0\n$$\nwhich is equivalent to $2ab-b^2-c^2=0$, or, after a straight computation, to $AB=AC$.\n\n\nThird solution. By using scalar products of vectors. Because $M$ is the midpoint of $BC$ and $E$ the midpoint of $MD$, we have\n$$\n\\overrightarrow{DM}=\\frac{1}{2}(\\overrightarrow{DB}+\\overrightarrow{DC}), \\text{ or, equivalently } \\overrightarrow{BD}=2 \\overrightarrow{MD}+\\overrightarrow{DC}\n$$\nand\n$$\n\\overrightarrow{AE}=\\overrightarrow{AM}+\\overrightarrow{ME}=\\frac{1}{2}(\\overrightarrow{AM}+\\overrightarrow{AD}) .\n$$\nBecause $MD$ and $AC$ are perpendicular, we have\n$$\n\\overrightarrow{AD} \\cdot \\overrightarrow{MD}=\\overrightarrow{ME} \\cdot \\overrightarrow{DC}=0 .\n$$\nTherefore\n$$\n\\begin{aligned}\n\\overrightarrow{AE} \\cdot \\overrightarrow{BD} & =2 \\overrightarrow{AE} \\cdot \\overrightarrow{MD}+\\overrightarrow{AE} \\cdot \\overrightarrow{DC} \\\\\n& =(\\overrightarrow{AM}+\\overrightarrow{AD}) \\cdot \\overrightarrow{MD}+(\\overrightarrow{AM}+\\overrightarrow{ME}) \\cdot \\overrightarrow{DC} \\\\\n& =\\overrightarrow{AM} \\cdot \\overrightarrow{MD}+\\overrightarrow{AM} \\cdot \\overrightarrow{DC}=\\overrightarrow{AM} \\cdot \\overrightarrow{MC} \\\\\n& =\\frac{1}{2} \\overrightarrow{AM} \\cdot \\overrightarrow{BC}\n\\end{aligned}\n$$\nHence, $AE$ and $BD$ are perpendicular if and only if the median $AM$ of the triangle $ABC$ is an altitude. This is equivalent to $AB=AC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40126,
    "subject": "Mathematics (Multi-modal)",
    "question": "In acute triangle $A B C$, points $D$ and $E$ are the feet of the perpendiculars from $A$ to $B C$ and $B$ to $C A$, respectively. Segment $A D$ is a diameter of circle $\\omega$. Circle $\\omega$ intersects sides $A C$ and $A B$ at $F$ and $G$ (other than $A$), respectively. Segment $B E$ intersects segments $G D$ and $G F$ at $X$ and $Y$ respectively. Ray $D Y$ intersects side $A B$ at $Z$. Prove that lines $X Z$ and $B C$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have $\\angle A F G=\\angle A D G$ since $A G D F$ is cyclic. On the other hand $\\angle D G A=\\angle A F D=90^\\circ$, since $A D$ is a diameter. We deduce that triangles $A G D$ and $A D B$ are similar, and therefore $\\angle A F G=\\angle C B A$.\n\nBecause $\\angle A E B=\\angle A D B=90^\\circ$, quadrilateral $A B D E$ is cyclic. Therefore $\\angle D E B=\\angle C B A=\\angle E F Y$. But $\\angle E F D=\\angle Y E F=90^\\circ$. We deduce that $D F E Y$ is a rectangle and therefore $B Y$ is an altitude in triangle $B D Z$. Line segment $D G$ is also an altitude in triangle $B D Z$ which intersects $B Y$ at $X$. We deduce that $Z X$ and $B C$ are perpendicular.\n\n![](attached_image_1.png)\nBecause $\\angle A D B=\\angle A E B=90^\\circ$, the quadrilateral $A B D E$ is cyclic. Therefore, the projections of the point $D$ on the lines $A B, B E$, and $E A$ are collinear (Simson line). But $G$ and $F$ are the projections of $D$ on lines $A B$ and $E A$, respectively, since $A D$ is a diameter of the circle $\\omega$. Then the projection of $D$ on $B E$ is $Y$, the intersection point of $B E$ with $G F$. Therefore, $B Y$ is perpendicular to $D Z$. Hence, in triangle $B D Z$, line segments $B Y$ and $D G$ are altitudes and intersect at $X$. Thus $Z X$ is also an altitude and therefore $Z X$ and $B C$ are perpendicular.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40127,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $p(x)$ with integer coefficients such that for each positive integer $n$, the number $2^{n}-1$ is divisible by $p(n)$.",
    "options": [],
    "answer": "p(x) = 1 or p(x) = -1",
    "solution": "Suppose there is some value of $n$ such that $p(n) \\neq \\pm 1$. Let $q$ be a prime divisor of $p(n)$. Because $q = (n+q) - n$ divides $p(n+q) - p(n)$, we deduce that $q$ divides $p(n+q)$. Therefore, $q$ divides both $2^{n}-1$ and $2^{n+q}-1$, which implies in particular that $q$ is an odd prime. We have\n$$\n1 \\equiv 2^{n+q} \\equiv 2^{n} \\cdot 2^{q} \\equiv 2^{q} \\equiv 2 \\quad \\bmod q\n$$\nby Fermat's little theorem. This is a contradiction.\n\nHence, for all values of $n$, $p(n)= \\pm 1$. Since $p$ is a polynomial and takes infinitely many times the same value, either $1$ or $-1$, it is constant. This proves that the only polynomials with integer coefficients which have the required property are $p(X)=1$ and $p(X)=-1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40128,
    "subject": "Mathematics (Multi-modal)",
    "question": "$\\triangle ABC$ is a triangle and $I_{b}, I_{c}$ its excenters opposite to $B, C$. Prove that $\\triangle ABC$ is right at $A$ if and only if its area is equal to $\\frac{1}{2} AI_{b} \\cdot AI_{c}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First solution. Let $\\triangle ABC$ be any triangle. Considering triangle $A I_{c} C$, we have\n$$\n\\begin{aligned}\n\\measuredangle C I_{c} A & =180^{\\circ}-\\left(\\measuredangle A C I_{c}+\\measuredangle I_{c} A C\\right) \\\\\n& =180^{\\circ}-\\left(\\frac{1}{2} \\measuredangle A C B+\\measuredangle B A C+\\frac{1}{2}\\left(180^{\\circ}-\\measuredangle B A C\\right)\\right) \\\\\n& =90^{\\circ}-\\frac{1}{2}(\\measuredangle A C B+\\measuredangle B A C) \\\\\n& =\\measuredangle I_{b} B A .\n\\end{aligned}\n$$\n\n![](attached_image_1.png)\n\nOn the other hand, $\\measuredangle I_{c} A C=\\measuredangle B A I_{b}$. We deduce that triangles $A B I_{b}, A I_{c} C$ are similar and therefore\n$$\n\\frac{A I_{c}}{A B}=\\frac{A C}{A I_{b}} \\quad \\Leftrightarrow \\quad A I_{b} \\cdot A I_{c}=A B \\cdot A C .\n$$\nThus, the area of triangle $A B C$ is equal to $\\frac{1}{2} A I_{b} \\cdot A I_{c} \\sin \\measuredangle B A C$. Hence, triangle $\\triangle ABC$ is right at $A$ if and only if its area is equal to $\\frac{1}{2} A I_{b} \\cdot A I_{c}$.\n\n\nSecond solution. Let $\\triangle ABC$ be any triangle. Denote $a, b, c$ the lengths of the opposite sides to the vertices $A, B, C$ respectively, $s$ the semiperimeter, $r$ the inradius, $r_{b}, r_{c}$ the exradius opposite to the vertices $B, C$ respectively. We have\n$$\nA I_{b}^{2}=r_{b}^{2}+(s-c)^{2} \\quad \\text{ and } \\quad A I_{c}^{2}=r_{c}^{2}+(s-b)^{2} .\n$$\n\n![](attached_image_2.png)\n\nBut the area of the triangle $\\triangle ABC$ is equal to\n$$\nK=rs=r_{b}(s-b)=r_{c}(s-c)=\\sqrt{s(s-a)(s-b)(s-c)},\n$$\nand that $r_{b} r_{c}=s(s-a)$ (which follows from the above formulas for the area). Hence\n$A I_{b}^{2} \\cdot A I_{c}^{2}=\\left(r_{b}^{2}+(s-c)^{2}\\right)\\left(r_{c}^{2}+(s-b)^{2}\\right)=s^{2}(s-a)^{2}+2 K^{2}+(s-b)^{2}(s-c)^{2}$.\nTherefore, $K=\\frac{1}{2} A I_{b} \\cdot A I_{c}$ is equivalent to\n$$\n2 K^{2}=s^{2}(s-a)^{2}+(s-b)^{2}(s-c)^{2} .\n$$\nBut $2 K^{2}=2 s(s-a)(s-b)(s-c)$. Hence, this is equivalent to\n$$\n((s-b)(s-c)-s(s-a))^{2}=0\n$$\nwhich simplifies to\n$$\na^{2}=b^{2}+c^{2} .\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40129,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABC$ is a triangle, $G$ its centroid and $A'$, $B'$, $C'$ the midpoints of its sides $BC$, $CA$, $AB$, respectively. Prove that if the quadrilateral $AC'GB'$ is cyclic then\n$$\nAB \\cdot CC' = AC \\cdot BB'.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First solution. Because $AC'GB'$ is cyclic, we have $\\measuredangle GAB' = \\measuredangle GC'B'$. Because $B'C'$ is parallel to $BC$, we have $\\measuredangle CC'B' = \\measuredangle C'CB$. We deduce that\n$$\n\\sin \\measuredangle A'A C = \\sin \\measuredangle C'C B.\n$$\n![](attached_image_1.png)\nBut triangles $AA'C$, $BCC'$ have the same area which is half of the area of triangle $ABC$. We deduce that\n$$\n\\frac{1}{2} AA' \\cdot AC \\sin \\measuredangle A'A C = \\frac{1}{2} CC' \\cdot BC \\sin \\measuredangle C'C B\n$$\nand therefore\n$$\n\\frac{AC}{CC'} = \\frac{BC}{AA'}.\n$$\nWe proceed in a similar way, by considering triangles $ABA'$ and $BCB'$, obtaining\n$$\n\\frac{AB}{BB'} = \\frac{BC}{AA'},\n$$\nand deduce the relation\n$$\nAB \\cdot CC' = AC \\cdot BB'.\n$$\n\n\nSecond solution. Lets us consider the power of the point $B$ with respect to the circumcircle of the cyclic quadrilateral $AC'GB'$. We have\n$$\n\\mathcal{P}(B) = \\overline{BC'} \\cdot \\overline{BA} = \\overline{BG} \\cdot \\overline{BB'}.\n$$\nBut $\\overline{BC'} = \\frac{1}{2} \\overline{BA}$ and $\\overline{BG} = \\frac{2}{3} \\overline{BB'}$. We deduce that\n$$\nBB' = \\frac{\\sqrt{3}}{2} AB.\n$$\nSimilarly, by considering the power of the point $C$ with respect to the circumcircle of the cyclic quadrilateral $AC'GB'$, we obtain\n$$\nCC' = \\frac{\\sqrt{3}}{2} AC\n$$\nFrom these two relations we deduce easily that\n$$\nAB \\cdot CC' = AC \\cdot BB'.\n$$\n\n\nThird solution (Contains also a proof for the converse). We can see from the first solution that the quadrilateral $AC'GB'$ is cyclic if and only if\n$$\n\\measuredangle GAC = \\measuredangle GCB.\n$$\nThis is equivalent to saying that the line $BC$ is tangent to the circumcircle of triangle $AGC$.\n![](attached_image_2.png)\nThis property is equivalent to saying that $A'C^2 = \\overline{A'G} \\cdot \\overline{A'A}$ by using the power of the point $A'$ with respect to this circumcircle. But\n$$\nA'C = \\frac{1}{2} BC, \\quad \\overline{A'G} = \\frac{1}{3} \\overline{A'A} \\quad \\text{and} \\quad AA' = \\sqrt{\\frac{2AB^2 + 2AC^2 - BC^2}{4}}.\n$$\nWe deduce that $AC'GB'$ is cyclic if and only if\n$$\n2BC^2 = AB^2 + AC^2\n$$\nNow, using the formulas for the medians\n$$\nBB' = \\sqrt{\\frac{2BC^2 + 2AB^2 - CA^2}{4}} \\quad \\text{and} \\quad CC' = \\sqrt{\\frac{2CA^2 + 2BC^2 - AB^2}{4}},\n$$\nwe deduce that the relation\n$$\nAB \\cdot CC' = AC \\cdot BB'\n$$\nis equivalent to\n$$\nAB^2 \\cdot \\left(2CA^2 + 2BC^2 - AB^2\\right) = AC^2 \\cdot \\left(2BC^2 + 2AB^2 - CA^2\\right),\n$$\nwhich is equivalent to\n$$\n\\left(AB^2 - AC^2\\right) \\cdot \\left(2BC^2 - AB^2 - AC^2\\right) = 0\n$$\nwhich is equivalent to either $AB = AC$ or $AC'GB'$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40130,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $(a, b)$ such that $a^{2} + b^{2}$ divides both $a^{3} + 1$ and $b^{3} + 1$.",
    "options": [],
    "answer": "(1,1)",
    "solution": "We have\n$$\n0 \\equiv (a^{3} + 1) - (b^{3} + 1) \\equiv (a - b)(a^{2} + a b + b^{2}) \\equiv (a - b) a b \\pmod{a^{2} + b^{2}}.\n$$\nLet $d$ be a common divisor of $a$ and $a^{2} + b^{2}$. Then $d$ divides $a^{3} + 1$ and $a^{3}$, so it divides $1$. Hence $a$ and $a^{2} + b^{2}$ are coprime. In a similar way $b$ and $a^{2} + b^{2}$ are coprime. Thus $a - b \\equiv 0 \\pmod{a^{2} + b^{2}}$.\n\nIf $a \\neq b$ then $a^{2} + b^{2} \\leq |a - b| \\leq (a - b)^{2} < a^{2} + b^{2}$, since $a b \\geq 1$, which is a contradiction. Hence $a = b = 1$, since $a$ and $a^{2} + b^{2}$ are coprime.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40131,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine if there exists an infinite sequence of positive integers\n$$\na_{1}, a_{2}, a_{3}, \\ldots\n$$\nsuch that\n(i) each positive integer occurs exactly once in the sequence, and\n(ii) each positive integer occurs exactly once in the sequence $\\left|a_{1}-a_{2}\\right|, \\left|a_{2}-a_{3}\\right|, \\ldots, \\left|a_{k}-a_{k+1}\\right|, \\ldots$",
    "options": [],
    "answer": "Yes, such an infinite sequence exists.",
    "solution": "We will construct such a sequence by induction:\nDefine $a_{1}=1$ and $a_{2}=2$. In this case we have $b_{1}=|a_{1}-a_{2}|=1$.\nAssume that $a_{1}, a_{2}, \\ldots, a_{2n}$ are defined such that there is no positive integer which occurs at least twice neither in the finite sequence $a_{1}, a_{2}, \\ldots, a_{2n}$ nor in the finite sequence $b_{1}=|a_{1}-a_{2}|, b_{2}=|a_{2}-a_{3}|, \\ldots, b_{2n-1}=|a_{2n-1}-a_{2n}|$.\nLet $M_{n}$ be the maximum element of the set of integers $\\{a_{1}, a_{2}, \\ldots, a_{2n}\\}$, $c_{n}$ the maximum of the integers $k$ such that $\\{1,2, \\ldots, k\\} \\subseteq \\{a_{1}, a_{2}, \\ldots, a_{2n}\\}$, and $d_{n}$ the maximum of the integers $k$ such that $\\{1,2, \\ldots, k\\} \\subseteq \\{b_{1}, b_{2}, \\ldots, b_{2n-1}\\}$.\nNotice that for all $1 \\leq i \\leq 2n-1$, we have\n$$\nb_{i}=|a_{i}-a_{i+1}| \\leq \\max_{1 \\leq j \\leq 2n} a_{j} - \\min_{1 \\leq j \\leq 2n} a_{j} = M_{n}-1.\n$$\nIf $c_{n}<d_{n}$ define $a_{2n+1}=2M_{n}+c_{n}+1$ and $a_{2n+2}=c_{n}+1$. If $c_{n} \\geq d_{n}$ define $a_{2n+1}=2M_{n}+1$ and $a_{2n+2}=2M_{n}+d_{n}+2$.\nIt is clear that none of the two possible values for $a_{2n+1}$ and for $a_{2n+2}$ occur in the finite sequence $a_{1}, a_{2}, \\ldots, a_{2n}$.\nIn the first case, we have\n$$\nb_{2n}=2M_{n}+c_{n}+1-a_{2n} \\geq M_{n}+c_{n}+1 > M_{n}-1 \\geq b_{i}\n$$\nand\n$$\nb_{2n+1}=2M_{n} > M_{n}-1 \\geq b_{i}\n$$\nfor all $1 \\leq i \\leq 2n-1$, and $b_{2n+1} \\neq b_{2n}$ since $a_{2n} \\neq c_{n}+1$.\nIn the second case, we have\n$$\nb_{2n}=2M_{n}+1-a_{2n} \\geq M_{n}+1 > M_{n}-1 \\geq b_{i}\n$$\nand\n$$\nb_{2n+1}=d_{n}+1 \\neq b_{i}\n$$\nfor all $1 \\leq i \\leq 2n-1$, and $b_{2n+1}=d_{n}+1 \\leq M_{n} < M_{n}+1 \\leq b_{2n}$.\nTherefore, there is no positive integer which occurs at least twice in the finite sequence $a_{1}, a_{2}, \\ldots, a_{2n+2}$ or in the finite sequence $b_{1}, b_{2}, \\ldots, b_{2n+1}$.\nThis proves that there is no positive integer which occurs at least twice in the infinite sequence $a_{1}, a_{2}, \\ldots, a_{n}, \\ldots$ or in the infinite sequence $b_{1}, b_{2}, \\ldots, b_{n}, \\ldots$.\nNotice moreover, that in the first case $c_{n+1} \\geq c_{n}+1$ and in the second case $d_{n+1} \\geq d_{n}+1$. Therefore, if $e_{n}=\\min\\{c_{n}, d_{n}\\}$ then $e_{n+2} \\geq e_{n}+1$. But $c_{1}=2$ and $d_{1}=1$. We deduce that $e_{2n} \\geq n$ for all integer $n$. Therefore, any positive integer $n$ occurs in both finite sequences $a_{1}, a_{2}, \\ldots, a_{4n}$ and $b_{1}, b_{2}, \\ldots, b_{4n-1}$.\nThis proves that any positive integer $n$ occurs exactly once in each infinite sequence $a_{1}, a_{2}, \\ldots$ and $b_{1}, b_{2}, \\ldots$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40132,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $F_{0}=0$, $F_{1}=1$ and $F_{n+1}=F_{n}+F_{n-1}$, for all positive integer $n$, be the Fibonacci sequence. Prove that for any positive integer $m$ there exist infinitely many positive integers $n$ such that\n$$\nF_{n}+2 \\equiv F_{n+1}+1 \\equiv F_{n+2} \\quad \\bmod m\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $m$ be a positive integer and consider the infinite set of pairs $(F_{k}, F_{k+1})$, for $k \\in \\mathbb{N}$. By the pigeonhole principle, there exists a pair $(a, b)$ of integers $0 \\leq a, b \\leq m-1$ and an infinite sequence of integers $0<k_{1}<k_{2}<\\cdots$ such that\n$$\n(F_{k_{i}}, F_{k_{i}+1}) \\equiv (a, b) \\quad \\bmod m, \\text{ for all } i \\geq 1.\n$$\nTherefore\n$$\n(F_{k_{i}-1}, F_{k_{i}}) = (F_{k_{i}+1}-F_{k_{i}}, F_{k_{i}}) \\equiv (b-a, a) \\bmod m, \\text{ for all } i \\geq 1.\n$$\nWe keep descending in this way until we get\n$$\n(F_{k_{i}-k_{1}+2}, F_{k_{i}-k_{1}+3}) \\equiv (F_{2}, F_{3}) \\equiv (1,2) \\quad \\bmod m, \\quad \\text{ for all } i \\geq 1.\n$$\nLet $n_{i}=k_{i}-k_{1}+2$, for all $i \\geq 1$. Clearly, the infinite sequence $2=n_{1}< n_{2}<\\cdots$ is increasing and we have\n$$\n\\begin{gathered}\nF_{n_{i}}+2 \\equiv F_{2}+2 \\equiv 3 \\bmod m, \\quad F_{n_{i}+1}+1 \\equiv F_{3}+1 \\equiv 3 \\bmod m \\\\\n\\text{ and } F_{n_{i}+2} \\equiv F_{n_{i}+1}+F_{n_{i}} \\equiv F_{3}+F_{2} \\equiv 3 \\bmod m\n\\end{gathered}\n$$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 40133,
    "subject": "Mathematics (Multi-modal)",
    "question": "$A B C D E F$ is an equiangular hexagon of perimeter $21$. Given that $A B = 3$, $C D = 4$, and $E F = 5$, compute the area of hexagon $A B C D E F$.",
    "options": [],
    "answer": "71√3/4",
    "solution": "We extend sides $F A$ and $B C$ to intersect at $A'$, and sides $B C$ and $D E$ to intersect at $B'$, and sides $D E$ and $F A$ to intersect at $C'$.\n\n![](attached_image_1.png)\n\nTriangles $A' B' C'$, $A' B A$, $B' D C$, and $C' F E$ are equilateral with side lengths $11$, $3$, $4$, and $5$ respectively. Therefore, the area of hexagon $A B C D E F$ is\n\n$$\n\\frac{\\sqrt{3}}{4} \\cdot 11^{2} - \\frac{\\sqrt{3}}{4} \\cdot 3^{2} - \\frac{\\sqrt{3}}{4} \\cdot 4^{2} - \\frac{\\sqrt{3}}{4} \\cdot 5^{2} = \\frac{71 \\sqrt{3}}{4}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40134,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a real number such that the product of real roots of the equation\n$$\nX^{4}+2 X^{3}+(2+2 k) X^{2}+(1+2 k) X+2 k=0\n$$\nis $-2013$. Find the sum of the squares of these real roots.",
    "options": [],
    "answer": "4027",
    "solution": "Notice first that\n$$\nX^{4}+2 X^{3}+(2+2 k) X^{2}+(1+2 k) X+2 k = (X^{2}+X+1)(X^{2}+X+2k).\n$$\nBecause the factor $X^{2}+X+1$ has no real roots, we deduce from Vieta relations that $r_{1}+r_{2}=-1$ and $r_{1} r_{2}=2k=-2013$, where $r_{1}, r_{2}$ are the real roots of the equation $X^{4}+2 X^{3}+(2+2 k) X^{2}+(1+2 k) X+2 k=0$. Therefore,\n$$\nr_{1}^{2}+r_{2}^{2}=(r_{1}+r_{2})^{2}-2 r_{1} r_{2}=1+2 \\times 2013=4027.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40135,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call a positive integer good if it doesn't have a zero digit and the sum of the squares of its digits is a perfect square. For example, $122$ and $34$ are good and $304$ and $12$ are not good. Prove that there exists a $n$-digit good number for every positive integer $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove this by induction on $n$.\n\nFor $n=1,2,3$, the numbers $1$, $34$, and $122$ are good numbers and all their odd digits are less than $5$.\n\nAssume that $a_{n}$ is an $n$-digit good number and all its odd digits are less than $5$.\n\nIf $a_{n}$ has an even digit $2m$, remove this digit and replace it by $4$ digits $m m m m$ to obtain an $n+3$-digit number. The sum of squares of the digits of this new number is equal to the sum of squares of the digits of $a_{n}$ and is therefore a perfect square. Notice that all the odd digits of this new $n+3$-digit good number are less than $5$. For example, replace the $4$ in the $2$-digit good number $34$ by $2222$ to obtain the $5$-digit good number $32222$.\n\nIf $a_{n}$ has no even digit, all its digits are less than $5$. We can multiply all its digits by $2$ to obtain a new $n$-digit good number. Choose one of the new even digits $2m$, remove it and replace it by four digits $m m m m$ to obtain a new $n+3$-digit good number and its odd digits $m$ are less than $5$. For example, for the $4$-digit good number $3333$ multiply all its digits by $2$ to obtain the $4$-digit good number $6666$ and then replace one of the $6$ by $3333$ to obtain the $7$-digit good number $3333666$.\n\nThis proves by induction that for any positive integer $n$, there exists an $n$-digit good number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40136,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with incenter $I$, and let $D, E, F$ be the midpoints of sides $BC, CA, AB$, respectively. Lines $BI$ and $DE$ meet at $P$, and lines $CI$ and $DF$ meet at $Q$. Line $PQ$ meets sides $AB$ and $AC$ at $T$ and $S$, respectively. Prove that $AS = AT$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because sides of triangles $ABC$ and $DEF$ are parallel (homothetic triangles), we have\n$$\n\\begin{aligned}\n\\angle BPD & = 180^\\circ - \\angle EDF - \\angle FDB - \\frac{1}{2} \\angle CBA \\\\\n& = 180^\\circ - \\angle BAC - \\angle ACB - \\frac{1}{2} \\angle CBA \\\\\n& = \\frac{1}{2} \\angle CBA = \\angle DBP.\n\\end{aligned}\n$$\nWe deduce that triangle $DPB$ is isosceles and therefore $DP = DB$.\n\n![](attached_image_1.png)\n\nSimilarly, we prove that $DQ = DC$. But $DB = DC$. We conclude that triangle $DPQ$ is isosceles, and therefore, its bisector at $D$ is perpendicular to $PQ$.\n\nBut the bisector of triangle $ABC$ at $A$ is parallel to the bisector of triangle $DEF$ at $E$ since the two triangles are homothetic. We deduce that the bisector of triangle $ATS$ at $A$ is perpendicular to $ST$ and therefore $AS = AT$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40137,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ayman wants to color the cells of a $50 \\times 50$ chessboard into black and white so that each $2 \\times 3$ or $3 \\times 2$ rectangle contains an even number of white cells. Determine the number of ways Ayman can color the chessboard.",
    "options": [],
    "answer": "32",
    "solution": "Let us associate a $1$ to each cell with a black color and a $0$ to each cell with a white color. The condition is equivalent to the sum of numbers in each $2 \\times 3$ rectangle and in each $3 \\times 2$ rectangle being even. Let $a_{i, j}$ be this number at the cell in the $i^{\\text{th}}$ row and $j^{\\text{th}}$ column, for $1 \\leq i, j \\leq 50$.\n\nConsider, for a fixed pair $i, j$, with $1 \\leq i \\leq 48$ and $1 \\leq j \\leq 47$, the $3 \\times 4$ rectangle:\n| $a_{i, j}$ | $a_{i, j+1}$ | $a_{i, j+2}$ | $a_{i, j+3}$ |\n| :--- | :--- | :--- | :--- |\n| $a_{i+1, j}$ | $a_{i+1, j+1}$ | $a_{i+1, j+2}$ | $a_{i+1, j+3}$ |\n| $a_{i+2, j}$ | $a_{i+2, j+1}$ | $a_{i+2, j+2}$ | $a_{i+2, j+3}$ |\n\nBy applying the condition to the two $2 \\times 3$ rectangles which contain cells from the second and the third rows, we get\n$$\na_{i+1, j}+a_{i+1, j+1}+a_{i+1, j+2}+a_{i+2, j}+a_{i+2, j+1}+a_{i+2, j+2} \\equiv 0 \\pmod{2},\n$$\nand\n$$\na_{i+1, j+1}+a_{i+1, j+2}+a_{i+1, j+3}+a_{i+2, j+1}+a_{i+2, j+2}+a_{i+2, j+3} \\equiv 0 \\pmod{2}\n$$\nBy applying the condition to all the $3 \\times 2$ rectangles, we get\n$$\n\\begin{gathered}\na_{i, j}+a_{i, j+1}+a_{i+1, j}+a_{i+1, j+1}+a_{i+2, j}+a_{i+2, j+1} \\equiv 0 \\pmod{2}, \\\\\na_{i, j+1}+a_{i, j+2}+a_{i+1, j+1}+a_{i+1, j+2}+a_{i+2, j+1}+a_{i+2, j+2} \\equiv 0 \\pmod{2},\n\\end{gathered}\n$$\nand\n$$\na_{i, j+2}+a_{i, j+3}+a_{i+1, j+2}+a_{i+1, j+3}+a_{i+2, j+2}+a_{i+2, j+3} \\equiv 0 \\pmod{2} .\n$$\nBy adding these 5 relations and cancelling all even numbers we get\n$$\na_{i, j}+a_{i, j+3} \\equiv 0 \\quad \\pmod{2} .\n$$\nThis proves that\n$$\na_{i, j+3}=a_{i, j}\n$$\nWe prove in a similar way that\n$$\na_{i+3, j}=a_{i, j}\n$$\nTherefore, it is enough to know the numbers in the $3 \\times 3$ rectangle\n| $a_{1,1}$ | $a_{1,2}$ | $a_{1,3}$ |\n| :---: | :---: | :---: |\n| $a_{2,1}$ | $a_{2,2}$ | $a_{2,3}$ |\n| $a_{3,1}$ | $a_{3,2}$ | $a_{3,3}$ |\nto deduce, by periodicity, the numbers in all the other cells.\n\nApplying the condition to the first $2 \\times 3$ rectangle, we will get\n$$\na_{2,3} \\equiv a_{1,1}+a_{1,2}+a_{1,3}+a_{2,1}+a_{2,2} \\quad \\pmod{2}\n$$\nApplying the condition to the second $2 \\times 3$ rectangle and to the second $3 \\times 2$ rectangle and adding the two relations, we will get\n$$\na_{3,1} \\equiv a_{2,1}+a_{1,2}+a_{1,3} \\quad \\pmod{2}\n$$\nWe obtain in a similar way\n$$\na_{3,2} \\equiv a_{2,2}+a_{1,3}+a_{1,1} \\quad \\pmod{2}\n$$\n$$\na_{3,3} \\equiv a_{2,3}+a_{1,1}+a_{1,2} \\quad \\pmod{2}\n$$\nTherefore, it is enough to know the 5 numbers\n$$\na_{1,1}, a_{1,2}, a_{1,3}, a_{2,1}, a_{2,2}\n$$\nto deduce all the numbers in the cells of the $50 \\times 50$ chessboard.\n\nConversely, choose a value in $\\{0,1\\}$ for each number $a_{1,1}, a_{1,2}, a_{1,3}, a_{2,1}, a_{2,2}$, and deduce the values in all the other cells of the chessboard. It is easy to check that the condition on the two $2 \\times 3$ rectangles and the two $3 \\times 2$ rectangles in the $3 \\times 3$ rectangle above is satisfied. We deduce, by periodicity, that it is satisfied in all the chessboard. Hence there are $2^{5}$ ways to color the chessboard.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40138,
    "subject": "Mathematics (Multi-modal)",
    "question": "The excircle $\\omega_{B}$ of triangle $ABC$ opposite $B$ touches side $AC$, rays $BA$ and $BC$ at $B_{1}$, $C_{1}$ and $A_{1}$, respectively. Point $D$ lies on major arc $\\widetilde{A_{1}C_{1}}$ of $\\omega_{B}$. Rays $DA_{1}$ and $C_{1}B_{1}$ meet at $E$. Lines $AB_{1}$ and $BE$ meet at $F$. Prove that line $FD$ is tangent to $\\omega_{B}$ (at $D$).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $F'$ be the intersection point of the tangents to circle $\\omega_{B}$ at $B_{1}$ and $D$.\n\n![](attached_image_1.png)\n\nLet $G$ be the intersection point of lines $B_{1}A_{1}$ and $DC_{1}$.\n\nConsider the six cyclic points $C_{1}, C_{1}, B_{1}, A_{1}, A_{1}, D$. Because the tangent lines to $\\omega_{B}$ at $C_{1}$ and $A_{1}$ intersect at $B$, lines $C_{1}B_{1}$ and $A_{1}D$ intersect at $E$, and lines $B_{1}A_{1}$ and $DC_{1}$ intersect at $G$, from Pascal theorem, the three points $B$, $E$, and $G$ are collinear.\n\nConsider the six cyclic points $C_{1}, B_{1}, B_{1}, A_{1}, D, D$. Because lines $C_{1}B_{1}$ and $A_{1}D$ intersect at $E$, the tangent lines to $\\omega_{B}$ at $B_{1}$ and $D$ intersect at $F'$, and lines $B_{1}A_{1}$ and $DC_{1}$ intersect at $G$, from Pascal theorem, the three points $E$, $F'$, and $G$ are collinear.\n\nWe deduce that $B$, $E$, and $F'$ are collinear and therefore $F' = F$. This proves that $FD$ is tangent to $\\omega_{B}$ at $D$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40139,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive proper divisor is a positive divisor of a number, excluding itself. For positive integers $n \\geq 2$, let $f(n)$ denote the number that is one more than the largest proper divisor of $n$. Determine all positive integers $n$ such that $f(f(n))=2$.",
    "options": [],
    "answer": "All primes n, and all n of the form n = 2(p − 1) where p is a prime.",
    "solution": "Let $n \\geq 2$ such that $f(f(n))=2$. The largest proper divisor of $f(n)$ is 1 if and only if $f(n)$ is a prime number.\n\nIf $f(n)=2$, this is equivalent to $n$ being a prime number.\n\nIf $f(n)=p \\geq 3$, then $p-1$, the largest proper divisor of $n$ is an even number. This means that $n$ is even and therefore, its largest proper divisor is $\\frac{n}{2}$. This is equivalent to $n=2(p-1)$.\n\nHence, the positive integers $n$ such that $f(f(n))=2$ are prime numbers or numbers $n=2(p-1)$ for any prime number $p$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40140,
    "subject": "Mathematics (Multi-modal)",
    "question": "A perfect number is an integer that equals half the sum of its positive divisors. For example, because $2 \\cdot 28 = 1 + 2 + 4 + 7 + 14 + 28$, $28$ is a perfect number.\n\na. A square-free integer is an integer not divisible by a square of any prime number. Find all square-free integers that are perfect numbers.\n\nb. Prove that no perfect square is a perfect number.",
    "options": [],
    "answer": "6; no perfect square is a perfect number",
    "solution": "Recall that if\n$$\nn = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_k^{\\alpha_k}\n$$\nwhere $p_1 < p_2 < \\cdots < p_k$ are prime numbers and $\\alpha_1, \\alpha_2, \\ldots, \\alpha_k$ are positive integers, then the sum of the positive divisors of $n$ is\n$$\n\\sigma(n) = \\prod_{i=1}^{k} \\left( \\sum_{j=0}^{\\alpha_i} p_i^j \\right )\n$$\n\na. Let $n$ be a square-free perfect number. In this case, $n = p_1 p_2 \\cdots p_k$, for some prime numbers $p_1 < p_2 < \\ldots < p_k$, and\n$$\n2 p_1 p_2 \\cdots p_k = (p_1 + 1)(p_2 + 1) \\cdots (p_k + 1)\n$$\nIf $p_1 \\neq 2$ then $p_1 + 1, p_2 + 1, \\ldots, p_k + 1$ are all even numbers while $p_1 p_2 \\cdots p_k$ is an odd number. We deduce that $k = 1$ and $2 p_1 = p_1 + 1$\nwhich is impossible.\nIf $p_1 = 2$ then $p_1 + 1 = 3 = p_2$ and if $k > 2$, we have\n$$\np_3 p_4 \\cdots p_k = (p_3 + 1)(p_4 + 1) \\cdots (p_k + 1),\n$$\nwhich is impossible since both sides have different parity. Hence $k = 2$ and $n = 6$.\nThis proves that the only square-free perfect number is $6$.\n\nb. Let $n$ be a perfect square. There exist prime numbers $p_1 < p_2 < \\ldots < p_k$ and positive integers $\\alpha_1, \\alpha_2, \\ldots, \\alpha_k$, such that\n$$\nn = p_1^{2 \\alpha_1} p_2^{2 \\alpha_2} \\cdots p_k^{2 \\alpha_k}\n$$\nBecause $\\sum_{j=0}^{2 \\alpha_i} p_i^j$ is an odd number, for $i = 1, 2, \\ldots, k$, the sum $\\sigma(n)$ of the positive divisors of $n$ is odd while $2n$ is even. Hence $n$ cannot be a perfect number.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40141,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $\\angle A < \\angle B \\leq \\angle C$, $M$ and $N$ the midpoints of sides $CA$ and $AB$, respectively, and $P$ and $Q$ the projections of $B$ and $C$ on the medians $CN$ and $BM$, respectively. Prove that the quadrilateral $MNPQ$ is cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because $\\angle BPC = \\angle BQC = 90^\\circ$, quadrilateral $BCQP$ is cyclic and therefore\n$$\n\\angle CPQ = \\angle CBQ.\n$$\nBecause $M$ and $N$ are midpoints of sides $AC$ and $AB$, segment $MN$ is parallel to side $BC$. Hence $\\angle NMQ = \\angle CBQ$.\n\nWe deduce that $\\angle NMQ = \\angle CPQ$. This proves that quadrilateral $MNPQ$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40142,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $2014$ divides $53 n^{55} - 57 n^{53} + 4 n$ for all integer $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice first that $2014 = 2 \\times 19 \\times 53$. Applying Fermat's little theorem for the prime numbers $2, 19, 53$ we obtain:\n$$\n\\begin{gathered}\n53 n^{55} - 57 n^{53} + 4 n \\equiv n - n + 0 \\equiv 0 \\quad \\bmod 2 \\\\\n53 n^{55} - 57 n^{53} + 4 n \\equiv 15 n^{17} (n^{19})^{2} - 0 + 4 n \\equiv 15 n^{19} + 4 n \\equiv 19 n \\equiv 0 \\quad \\bmod 19 \\\\\n\\text{and } 53 n^{55} - 57 n^{53} + 4 n \\equiv 0 - 4 n + 4 n \\equiv 0 \\quad \\bmod 53\n\\end{gathered}\n$$\nfor all integer $n$. Therefore $2014$ divides $53 n^{55} - 57 n^{53} + 4 n$ for all integer $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40143,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{1}, a_{2}, \\ldots, a_{2 n}$ be positive real numbers such that $a_{i}+a_{n+i}=1$, for all $i=1, \\cdots, n$. Prove that there exist two different integers $1 \\leq j, k \\leq 2 n$ for which\n$$\n\\sqrt{a_{j}^{2}-a_{k}^{2}}<\\frac{1}{\\sqrt{n}+\\sqrt{n-1}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume, without loss of generality, that\n$$\n0<a_{1} \\leq a_{2} \\leq \\cdots \\leq a_{n} \\leq \\frac{1}{2} \\leq a_{2 n} \\leq a_{2 n-1} \\leq \\cdots \\leq a_{n+1}<1,\n$$\nand let\n$$\nK=\\min \\left\\{\\left|a_{i}^{2}-a_{j}^{2}\\right|, \\text{ with } i, j=1, \\ldots, 2 n, \\text{ and } i \\neq j\\right\\} .\n$$\nWe have\n$$\nK \\leq a_{2 n}^{2}-a_{n}^{2}=\\left(a_{2 n}+a_{n}\\right)\\left(a_{2 n}-a_{n}\\right)=1-2 a_{n},\n$$\nwhich means that\n$$\n\\frac{1-K}{2} \\geq a_{n} .\n$$\nWe deduce that\n$$\n\\frac{(1-K)^{2}}{4} \\geq a_{n}^{2}>a_{n}^{2}-a_{1}^{2}=\\sum_{i=1}^{n-1}\\left(a_{i+1}^{2}-a_{i}^{2}\\right) \\geq(n-1) K .\n$$\nThis implies that\n$$\nK^{2}-2(2 n-1) K+1>0 .\n$$\nBecause $K<1$, we deduce that\n$$\nK<2 n-1-\\sqrt{(2 n-1)^{2}-1}=(\\sqrt{n}-\\sqrt{n-1})^{2},\n$$\nand therefore\n$$\n\\sqrt{K}<\\sqrt{n}-\\sqrt{n-1}=\\frac{1}{\\sqrt{n}+\\sqrt{n-1}} .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40144,
    "subject": "Mathematics (Multi-modal)",
    "question": "Tarik and Sultan are playing the following game. Tarik thinks of a number that is greater than $100$. Then Sultan is telling a number greater than $1$. If Tarik's number is divisible by Sultan's number, Sultan wins, otherwise Tarik subtracts Sultan's number from his number and Sultan tells his next number. Sultan is forbidden to repeat his numbers. If Tarik's number becomes negative, Sultan loses. Does Sultan have a winning strategy?",
    "options": [],
    "answer": "Yes",
    "solution": "Yes, Sultan has winning strategies. Here are two examples:\n\nFirst winning strategy. Sultan plays the following numbers in order: $2$, $3$, $4$, $6$, $8$, $20$, $24$ and he wins. Indeed, let $n \\geq 100$ be Tarik's number and assume that Sultan will not win before playing the last number $24$.\n- $n$ is not divisible by $2$, so it is congruent to one of the twelve numbers $1,3,5,7,9,11,13,15,17,19,21,23$, in modulo $24$.\n- $n-2$ is not divisible by $3$, so it is congruent to one of the eight numbers $1,5,7,11,13,17,19,23$, in modulo $24$.\n- $n-5=(n-2)-3$ is not divisible by $4$, so it is congruent to one of the four numbers $2,10,14,22$, in modulo $24$.\n- $n-9=(n-5)-4$ is not divisible by $6$, so it is congruent to one of the two numbers $10,22$, in modulo $24$.\n- $n-15=(n-9)-6$ is not divisible by $8$ and is congruent to $4$ in modulo $24$.\n- $n-23=(n-15)-8$ is not divisible by $20$ and is congruent to $20$ in modulo $24$.\n- Finally, Sultan ends up with $n-43 \\equiv 0 \\bmod 24$, which is still a positive number divisible by $24$, and Sultan wins.\n\n\nSecond winning strategy. Sultan plays the following numbers in order: $2$, $3$, $4$, $6$, $16$, $12$ and he wins. Indeed, let $n \\geq 100$ be Tarik's number and assume that Sultan will not win before playing the last number $12$.\n- $n$ is not divisible by $2$, so it is congruent to one of the six numbers $1,3,5,7,9,11$, in modulo $12$.\n- $n-2$ is not divisible by $3$, so it is congruent to one of the four numbers $1,5,7,11$, in modulo $12$.\n- $n-5=(n-2)-3$ is not divisible by $4$, so it is congruent to one of the two numbers $2,10$, in modulo $12$.\n- $n-9=(n-5)-4$ is not divisible by $6$, so it is congruent to $10$ in modulo $12$.\n- $n-15=(n-9)-6$ is not divisible by $16$ and is congruent to $4$ in modulo $12$.\n- Finally, Sultan ends up with $n-31 \\equiv 0 \\bmod 12$, which is still a positive number divisible by $12$, and Sultan wins.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40145,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 14 students who have participated in a 3 hour test consisting of 15 short problems. Each student has solved a different number of problems and each problem has been solved by a different number of students. Prove that there exists a student who has solved exactly 5 problems.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because there are 14 students, the possible numbers of students solving a problem are $0, 1, \\ldots, 14$. Because there are 15 problems and each problem has been solved by a different number of students, there is a problem which has been solved by all the 14 students and another problem which has not been solved by any of the 14 students.\n\nIf we cancel both problems, the one solved by all the students and the one not solved by any of the students, we still have that each of the 14 students has solved a different number of the 13 remaining problems. But from these 13 remaining problems, the possible number of problems that a student has solved are $0, 1, \\ldots, 13$. Again, because there are 14 students and each student has solved a different number of problems, there is a student who has solved exactly 4 problems from these 13 remaining problems. But this student has also solved the canceled problem solved by all the students and has not solved the canceled problem which has not been solved by any of the students. Therefore, this student has solved exactly 5 problems.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40146,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$, $C$ be colinear points in this order, $\\mathcal{C}$ an arbitrary circle passing through $B$ and $C$, and $l$ an arbitrary line different from $BC$, passing through $A$ and intersecting $\\mathcal{C}$ at $M$ and $N$. The bisectors of the angles $\\angle C M B$ and $\\angle C N B$ intersect $BC$ at $P$ and $Q$. Prove that $AP \\cdot AQ = AB \\cdot AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D$ be the midpoint of the arc $\\overparen{BC}$ of the circle $\\mathcal{C}$ opposite to both $M$, $N$. Because $\\overparen{CD} = \\overparen{DB}$, the bisectors of angles $\\angle C M B$ and $\\angle C N B$, both, intersect the arc $\\widehat{CB}$ at $D$.\n\n![](attached_image_1.png)\n\nTherefore, we have\n$$\n\\angle N Q C = \\frac{1}{2}(\\overparen{NC} + \\overparen{DB}) = \\frac{1}{2}(\\overparen{NC} + \\overparen{CD}) = \\frac{1}{2} \\overparen{ND} = \\angle N M P.\n$$\nThis proves that quadrilateral $M N Q P$ is cyclic. But quadrilateral $M N C B$ is also cyclic and lines $M N$, $BC$ intersect at $A$, hence by the power of the point $A$ with respect to these two circles we have\n$$\nAP \\cdot AQ = AM \\cdot AN = AB \\cdot AC.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40147,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{1}, \\ldots, a_{n}$ be a non increasing sequence of positive real numbers. Prove that\n$$\n\\sqrt{a_{1}^{2}+a_{2}^{2}+\\cdots+a_{n}^{2}} \\leq a_{1}+\\frac{a_{2}}{\\sqrt{2}+1}+\\cdots+\\frac{a_{n}}{\\sqrt{n}+\\sqrt{n-1}}\n$$\nWhen does equality hold?",
    "options": [],
    "answer": "Equality holds if and only if all terms are equal.",
    "solution": "We prove the inequality by induction on $n \\geq 2$.\n\nFor $n=2$, let $a_{1} \\geq a_{2}>0$. We have\n$$\n\\begin{aligned}\n\\sqrt{a_{1}^{2}+a_{2}^{2}} & =\\sqrt{2 a_{2}^{2}+\\left(a_{1}-a_{2}\\right)^{2}+2 a_{2}\\left(a_{1}-a_{2}\\right)} \\\\\n& \\leq \\sqrt{2 a_{2}^{2}+\\left(a_{1}-a_{2}\\right)^{2}+2 \\sqrt{2} a_{2}\\left(a_{1}-a_{2}\\right)}=\\left(a_{1}-a_{2}\\right)+\\sqrt{2} a_{2} \\\\\n& \\leq a_{1}+\\frac{a_{2}}{\\sqrt{2}+1}\n\\end{aligned}\n$$\n\nAssume the inequality true for any non increasing sequence of positive $n$ real numbers and let $a_{1} \\geq a_{2} \\geq \\cdots \\geq a_{n+1}>0$. Denote $q_{n}=\\sqrt{a_{1}^{2}+a_{2}^{2}+\\cdots+a_{n}^{2}}$. It is clear that $q_{n} \\geq \\sqrt{n} a_{n+1}$. We have\n$$\n\\begin{aligned}\n\\sqrt{a_{1}^{2}+a_{2}^{2}} & +\\cdots+a_{n+1}^{2}=\\sqrt{q_{n}^{2}+a_{n+1}^{2}}= \\\\\n& =\\sqrt{(n+1) a_{n+1}^{2}+\\left(q_{n}-\\sqrt{n} a_{n+1}\\right)^{2}+2 \\sqrt{n} a_{n+1}\\left(q_{n}-\\sqrt{n} a_{n+1}\\right)} \\\\\n& \\leq \\sqrt{(n+1) a_{n+1}^{2}+\\left(q_{n}-\\sqrt{n} a_{n+1}\\right)^{2}+2 \\sqrt{n+1} a_{n+1}\\left(q_{n}-\\sqrt{n} a_{n+1}\\right)} \\\\\n& \\leq \\sqrt{n+1} a_{n+1}+\\left(q_{n}-\\sqrt{n} a_{n+1}\\right)=q_{n}+\\frac{a_{n+1}}{\\sqrt{n+1}+\\sqrt{n}} \\\\\n& \\leq a_{1}+\\frac{a_{2}}{\\sqrt{2}+1}+\\cdots+\\frac{a_{n+1}}{\\sqrt{n+1}+\\sqrt{n}}\n\\end{aligned}\n$$\n\nThe equality holds when $q_{i}=\\sqrt{i} a_{k+1}$, for $i=1,2, \\ldots, n-1$, that is $a_{1}= a_{2}=\\cdots=a_{n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40148,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{N} \\rightarrow (0, \\infty)$ such that $f(4) = 4$ and\n$$\n\\frac{1}{f(1) f(2)} + \\frac{1}{f(2) f(3)} + \\cdots + \\frac{1}{f(n) f(n+1)} = \\frac{f(n)}{f(n+1)}, \\quad \\forall n \\in \\mathbb{N},\n$$\nwhere $\\mathbb{N} = \\{1, 2, \\ldots\\}$ is the set of positive integers.",
    "options": [],
    "answer": "f(n) = n for all positive integers n",
    "solution": "Taking $n = 1$, we obtain $\\frac{1}{f(1)} = f(1)$. Because $f(1) > 0$, we deduce that $f(1) = 1$.\n\nTaking $n = 2$, we obtain $f(3) + 1 = f(2)^2$, and taking $n = 3$ we obtain $4 f(3) + 4 + f(2) = f(2) f(3)^2$. Because $f(2) \\neq 0$, this is equivalent to\n$$\nf(3) + 1 = f(2)^2, \\quad \\text{ and } \\quad 4 f(2) + 1 = f(3)^2\n$$\nWe deduce that $4 f(2) + 1 = (f(2)^2 - 1)^2$, and therefore\n$$\nf(2)(f(2) - 2)(f(2)^2 + 2 f(2) + 2) = 0\n$$\nBut $f(2) \\neq 0$. Hence $f(2) = 2$ and $f(3) = 3$.\n\nNow assume, for all $1 \\leq k \\leq n$, that $f(k) = k$. We have\n$$\n\\frac{n-1}{n} = \\frac{1}{1 \\cdot 2} + \\frac{1}{2 \\cdot 3} + \\cdots + \\frac{1}{(n-1) \\cdot n} = \\frac{n}{f(n+1)} - \\frac{1}{n \\cdot f(n+1)} = \\frac{n^2 - 1}{n f(n+1)}\n$$\nand therefore $f(n+1) = n + 1$.\n\nThis proves that $f(n) = n$, for all $n \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40149,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the minimum of $\\sum_{k=0}^{40}\\left(x+\\frac{k}{2}\\right)^{2}$ where $x$ is a real number.",
    "options": [],
    "answer": "1435",
    "solution": "We have\n$$\n\\begin{aligned}\n\\sum_{k=0}^{40}\\left(x+\\frac{k}{2}\\right)^{2} & =(x+10)^{2}+\\sum_{k=1}^{20}\\left(((x+10)+\\frac{k}{2})^{2}+((x+10)-\\frac{k}{2})^{2}\\right) \\\\\n& =(x+10)^{2}+2 \\sum_{k=1}^{20}\\left((x+10)^{2}+\\left(\\frac{k}{2}\\right)^{2}\\right) \\\\\n& =41(x+10)^{2}+\\frac{1}{2} \\sum_{k=1}^{20} k^{2}=41(x+10)^{2}+\\frac{20 \\cdot 21 \\cdot 41}{2 \\cdot 6} \\\\\n& =41(x+10)^{2}+1435\n\\end{aligned}\n$$\nTherefore, the minimum value is $1435$, which is obtained when $x=-10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40150,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\Gamma$ be a circle with center $O$ and $A E$ be a diameter. Point $D$ lies on segment $O E$ and point $B$ is the midpoint of one of the $\\operatorname{arcs} \\overparen{A E}$ of $\\Gamma$. Construct point $C$ such that $A B C D$ is a parallelogram. Lines $E B$ and $C D$ meet at $F$. Line $O F$ meets the minor $\\operatorname{arc} \\overparen{E B}$ at $I$. Prove that $E I$ bisects $\\angle B E C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because $D C$ and $A B$ are parallel, $\\angle C D E = \\angle B A O = 45^\\circ$. Because $B C$ and $A E$ are parallel, $\\angle C B E = \\angle O E B = 45^\\circ$. Hence quadrilateral $B C E D$ is cyclic and therefore $\\angle B E C = \\angle B D C$.\n\nOn the other hand, $\\angle D F B = \\angle D E F + \\angle F D E = 45^\\circ + 45^\\circ = 90^\\circ = \\angle A O B$. We deduce that quadrilateral $B F D O$ is cyclic and therefore $\\angle B D C = \\angle B O I$.\n\nBut $\\angle B O I = 2 \\angle B F I$ since both angles intercept the same $\\operatorname{arc} \\overparen{B I}$. We conclude that $\\angle B E C = 2 \\angle B E I$, which means that $E I$ bisects $\\angle B E C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40151,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number and $n \\geq 2$ a positive integer, such that $p \\mid (n^{6}-1)$. Prove that $n > \\sqrt{p} - 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because $p$ is prime and divides\n$$\nn^{6}-1 = (n-1)(n+1)(n^{2}-n+1)(n^{2}+n+1),\n$$\nit divides at least one of these positive factors. The prime number $p$ is therefore less or equal to at least one of these factors. Because\n$$\nn-1 < n+1 \\leq (n-1)n+1 = n^{2}-n+1 < n^{2}+n+1 < (n+1)^{2},\n$$\nwe have $p < (n+1)^{2}$ and therefore $\\sqrt{p} - 1 < n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40152,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all ordered triples $(a, b, c)$ of positive integers which satisfy\n$$\n5^{a} + 3^{b} - 2^{c} = 32\n$$",
    "options": [],
    "answer": "(2, 2, 1)",
    "solution": "By considering this equation modulo $3$, we get\n$$\n5^{a} + 3^{b} - 2^{c} \\equiv 2^{a} - 2^{c} \\equiv 2 \\pmod{3}\n$$\nThis occurs only when $a$ is even and $c$ is odd.\n\nIf $c = 1$, the equation becomes $5^{a} + 3^{b} = 34$, which is equivalent to $a = 2$ and $b = 1$.\n\nIf $c \\neq 1$, then $c \\geq 3$, since $c$ is odd. By considering this equation modulo $8$, and using the fact that $a$ is even, we get\n$$\n1 + 3^{b} \\equiv 0 \\pmod{8}\n$$\nwhich can never occur since $3^{b}$ is congruent to either $1$ or $3$ modulo $8$.\n\nHence, the unique solution to this equation is $(a, b, c) = (2, 1, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40153,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circles $\\omega_{1}$ and $\\omega_{2}$ meet at $P$ and $Q$. Segments $AC$ and $BD$ are chords of $\\omega_{1}$ and $\\omega_{2}$ respectively, such that segment $AB$ and ray $CD$ meet at $P$. Ray $BD$ and segment $AC$ meet at $X$. Point $Y$ lies on $\\omega_{1}$ such that $PY \\parallel BD$. Point $Z$ lies on $\\omega_{2}$ such that $PZ \\parallel AC$. Prove that points $Q, X, Y, Z$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because quadrilateral $BPDQ$ is cyclic, we have $\\angle PQD = \\angle PBD$. Because quadrilateral $ACQP$ is cyclic, we have $\\angle PQC = 180^{\\circ} - \\angle CAP$. We deduce that\n$$\n\\begin{aligned}\n\\angle DQC & = \\angle PQC - \\angle PQD = 180^{\\circ} - \\angle CAP - \\angle PBD \\\\\n& = 180^{\\circ} - \\angle XAB - \\angle ABX = \\angle BXA = \\angle DXA.\n\\end{aligned}\n$$\nThis proves that quadrilateral $CQDX$ is cyclic and therefore\n$$\n\\angle XQC = \\angle XDC.\n$$\nBecause $PA$ is parallel to $DX$, we have $\\angle YPC = \\angle XDC$. Because $CQPY$ is cyclic, we have $\\angle YPC = \\angle YQC$. Therefore, $\\angle YQC = \\angle XQC$, which means that points $Q, X, Y$ are collinear.\n\n![](attached_image_1.png)\n\nBecause quadrilateral $BPQZ$ is cyclic, we have $\\angle ZQB = \\angle ZPB$. Because lines $PZ$ and $AC$ are parallel, we have $\\angle ZPB = \\angle CAP$.\nLet $C'$ be the second intersection point of line $CQ$ with circle $\\omega_{1}$. Because quadrilateral $ACQP$ is cyclic, we have $\\angle C'QP = \\angle CAP$. Therefore, $\\angle ZQB = \\angle C'QP$. We deduce, by cyclicity of quadrilaterals $BPDQ$ and $CQDX$, that\n$$\n\\angle ZQC' = \\angle BQP = \\angle BDP = \\angle XDC = \\angle XQC,\n$$\nwhich means that points $Q, X, Z$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40154,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a set of positive real numbers with five elements such that for any distinct $a, b, c$ in $S$, the number $a b + b c + c a$ is rational. Prove that for any $a$ and $b$ in $S$, $\\frac{a}{b}$ is a rational number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a, b, c$ be three distinct elements in $S$.\nIf we denote by $\\mathcal{S}_i$, the set of subsets of $S$ of $i$ elements, $i=2,3$, we notice that\n$$\n\\sum_{\\{x, y\\} \\in \\mathcal{S}_2} x y = \\frac{1}{3} \\sum_{\\{x, y, z\\} \\in \\mathcal{S}_3} (x y + y z + z x) \\in \\mathbb{Q} .\n$$\nIf we denote by $\\mathcal{S}_{x, i}$, the set of subsets of $S \\setminus \\{x\\}$ of $i$ elements, $i=2,3$, for $x \\in S$, we notice that\n$$\na c = \\sum_{\\{x, y\\} \\in \\mathcal{S}_2} x y - \\frac{1}{2} \\left( \\sum_{\\{x, y, z\\} \\in \\mathcal{S}_{a, 3}} (x y + y z + z x) + \\sum_{\\{x, y, z\\} \\in \\mathcal{S}_{c, 3}} (x y + y z + z x) \\right) .\n$$\nHence $a c \\in \\mathbb{Q}$. Similarly $b c \\in \\mathbb{Q}$. We deduce that $\\frac{a}{b} = \\frac{a c}{b c} \\in \\mathbb{Q}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40155,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p \\geq 2$ be a prime number and $\\frac{a_{p}}{b_{p}} = 1 + \\frac{1}{2} + \\cdots + \\frac{1}{p^{2}-1}$, where $a_{p}$ and $b_{p}$ are two relatively prime positive integers. Compute $\\operatorname{gcd}\\left(p, b_{p}\\right)$.",
    "options": [],
    "answer": "gcd(p, b_p) = 2 if p = 2, and gcd(p, b_p) = 1 if p is odd",
    "solution": "For $p = 2$ we have $\\frac{a_{2}}{b_{2}} = \\frac{11}{6}$, and therefore $\\operatorname{gcd}\\left(2, b_{2}\\right) = 2$.\n\nFor $p$ an odd prime we propose two solutions to this problem:\n\nFirst solution. We have\n$$\n\\frac{a_{p}}{b_{p}} = \\frac{\\sum_{k=1}^{p^{2}-1} \\frac{(p^{2}-1)!}{k}}{(p^{2}-1)!} .\n$$\nBecause the only multiples of $p$ appearing in $1 \\times 2 \\times \\cdots \\times (p^{2}-1)$ are $p, 2p, \\ldots, (p-1)p$, and $p$ is a prime number, $p^{p-1}$ is the largest power of $p$ dividing $(p^{2}-1)!$.\nFor $1 \\leq k \\leq p^{2}-1$, if $k$ is not a multiple of $p$ then $p^{p-1}$ divides $\\frac{(p^{2}-1)!}{k}$. If $k$ is a multiple of $p$ then $p^{p-2}$ is the largest power of $p$ dividing $\\frac{(p^{2}-1)!}{k}$.\nSumming up the terms where $k$ is a multiple of $p$ we obtain\n$$\n\\sum_{j=1}^{p-1} \\frac{(p^{2}-1)!}{jp} = \\frac{(p^{2}-1)!}{p!} \\sum_{j=1}^{p-1} \\frac{(p-1)!}{j}\n$$\nNotice that all these fractions are integers. Using the fact that for each integer $1 \\leq j \\leq p-1$ there is a unique inverse integer $1 \\leq \\psi(j) \\leq p-1$ modulo $p$ (i.e. $j \\psi(j) \\equiv 1 \\bmod p$) and that this inverse map $j \\rightarrow \\psi(j)$ is a permutation of $\\{1,2, \\ldots, p-1\\}$, we get\n$$\n\\begin{aligned}\n\\sum_{j=1}^{p-1} \\frac{(p-1)!}{j} &\\equiv \\sum_{j=1}^{p-1} \\frac{(p-1)!}{j} j \\psi(j) \\equiv (p-1)! \\sum_{j=1}^{p-1} \\psi(j) \\\\\n&\\equiv (p-1)! \\sum_{j=1}^{p-1} j \\equiv (p-1)! \\frac{p(p-1)}{2} \\equiv 0, \\quad (\\bmod p)\n\\end{aligned}\n$$\nThis proves that $p^{p-1}$ divides $\\sum_{j=1}^{p-1} \\frac{(p^{2}-1)!}{jp}$ and so it divides $\\sum_{k=1}^{p^{2}-1} \\frac{(p^{2}-1)!}{k}$.\nWe deduce that $p^{p-1}$ cancels from the denominator and therefore\n$$\n\\operatorname{gcd}\\left(p, b_{p}\\right) = 1 .\n$$\n\n\nSecond solution. Because $p$ is odd, we have\n$$\n\\frac{a_{p}}{b_{p}} = \\sum_{k=1}^{\\frac{p^{2}-1}{2}} \\left( \\frac{1}{k} + \\frac{1}{p^{2}-k} \\right ) = \\sum_{k=1}^{\\frac{p^{2}-1}{2}} \\frac{p^{2}}{k(p^{2}-k)}\n$$\nFor $1 \\leq k \\leq \\frac{p^{2}-1}{2}$, if $k$ is not a multiple of $p$ then the denominator $k(p^{2}-k)$ is not divisible by $p$. If $k$ is a multiple of $p$ then\n$$\n\\frac{p^{2}}{k(p^{2}-k)} = \\frac{1}{j(p-j)},\n$$\nwhere $k = jp$ for some integer $1 \\leq j \\leq \\frac{p-1}{2}$. But the denominator $j(p-j)$ is not divisible by $p$. Therefore $\\frac{a_{p}}{b_{p}}$ is a sum of fractions where none of the denominators is divisible by $p$ and, since $\\frac{a_{p}}{b_{p}}$ is irreducible, then $\\operatorname{gcd}\\left(p, b_{p}\\right) = 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40156,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$, $a_{5}$ be nonzero real numbers. Prove that the polynomial\n$$\nP(X) = \\prod_{k=0}^{4} \\left( a_{k+1} X^{4} + a_{k+2} X^{3} + a_{k+3} X^{2} + a_{k+4} X + a_{k+5} \\right),\n$$\nwhere $a_{5+i} = a_{i}$ for $i = 1, 2, 3, 4$, has a root with negative real part.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume, to the contrary, that all roots of the polynomial $P(X)$ have nonegative real parts. We deduce that the real parts of the sums of the roots of its factors\n$$\na_{k} X^{4} + a_{k+1} X^{3} + a_{k+2} X^{2} + a_{k+3} X + a_{k+4}\n$$\nfor $k = 1, 2, 3, 4$, are nonegative. Therefore, by Vieta's relations, we have\n$$\n\\operatorname{Re} \\left( -\\frac{a_{k+1}}{a_{k}} \\right) \\geq 0\n$$\nfor $k = 1, 2, 3, 4$. Hence\n$$\n-1 = \\operatorname{Re} \\prod_{k=1}^{4} \\left( -\\frac{a_{k+1}}{a_{k}} \\right) = \\prod_{k=1}^{4} \\operatorname{Re} \\left( -\\frac{a_{k+1}}{a_{k}} \\right) \\geq 0,\n$$\nwhich is a contradiction.\n\nThis proves that the polynomial $P(X)$ has a root with negative real part.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40157,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S=\\{(a, b) \\mid a, b=1,2,3,4$, and $a \\neq b\\}$, and consider all nonzero polynomials $p(X, Y)$ with integer coefficients such that $p(a, b)=0$ for every element $(a, b)$ in $S$.\n(a) What is the minimal degree of such polynomial $p(X, Y)$?\n(b) Determine all such polynomials $p(X, Y)$ with minimal degree.",
    "options": [],
    "answer": "(a) 3. (b) Exactly the integer multiples of (X+Y−5)(X^2+Y^2−5X−5Y+10).",
    "solution": "(a) Let $p(X, Y)$ be such a polynomial. There are two possible cases:\n\nCase 1. If $p(1, Y)=p(2, Y)=p(3, Y)=p(4, Y)=0$. In this case\n$$\n(X-1)(X-2)(X-3)(X-4)\n$$\nis a factor of $p(X, Y)$ and therefore $\\deg p(X, Y) \\geq 4$.\n\nCase 2. If there exists $a_{0} \\in \\{1,2,3,4\\}$ such that $p\\left(a_{0}, Y\\right) \\neq 0$. Because $p\\left(a_{0}, b\\right)=0$ for $b \\in \\{1,2,3,4\\} \\setminus \\{a_{0}\\}$, the nonzero polynomial $p\\left(a_{0}, Y\\right)$ has 3 different roots and therefore $\\deg p(X, Y) \\geq \\deg p\\left(a_{0}, Y\\right) \\geq 3$.\n\nHence $\\deg p(X, Y) \\geq 3$.\n\nTo construct such a polynomial of degree 3, consider, as shown in the figure, the line of equation $x+y-5=0$ and the circle of equation $\\left(x-\\frac{5}{2}\\right)^{2}+\\left(y-\\frac{5}{2}\\right)^{2}-\\frac{5}{2}=0$.\n\n![](attached_image_1.png)\n\nThe graph obtained by union of this line and this circle passes by all the points of $S$. Therefore, the polynomial\n$$\np_{0}(X, Y)=(X+Y-5)\\left(X^{2}+Y^{2}-5 X-5 Y+10\\right)\n$$\nsatisfies the condition $p_{0}(a, b)=0$ for all $(a, b) \\in S$ and is of degree 3. Hence the minimal degree of such polynomials $p(X, Y)$ is 3.\n\n(b) Let $p(X, Y)$ be a nonzero polynomial of degree 3 with integer coefficients such that $p(a, b)=0$ for every element $(a, b)$ in $S$. From (a), there exists $a_{0} \\in \\{1,2,3,4\\}$ such that $p\\left(a_{0}, Y\\right) \\neq 0$. Let $p_{0}(X, Y)=(X+Y-5)\\left((2 X-5)^{2}+(2 Y-5)^{2}-10\\right)$ and define the polynomial\n$$\nq(X, Y)=p_{0}\\left(a_{0}, a_{0}\\right) p(X, Y)-p\\left(a_{0}, a_{0}\\right) p_{0}(X, Y).\n$$\nPolynomial $q(X, Y)$ is of degree at most 3. But polynomial $q\\left(a_{0}, Y\\right)$ has at least $1,2,3,4$ as roots. Therefore $q\\left(a_{0}, Y\\right)=0$, which means that $q(X, Y)=\\left(X-a_{0}\\right) q_{1}(X, Y)$ for some polynomial $q_{1}(X, Y)$ of degree at most 2. But for any $a \\in \\{1,2,3,4\\} \\setminus \\{a_{0}\\}, q_{1}(a, Y)$ has at least 3 roots, which implies that $q_{1}(a, Y)=0$. Hence $(X-1)(X-2)(X-3)(X-4)$ divides $q(X, Y)$ of degree 3. Therefore $q(X, Y)=0$. Because $p_{0}(X, Y)$ is a unitary polynomial, we deduce that\n$$\np(X, Y)=k p_{0}(X, Y), \\quad \\text{ for some } k \\in \\mathbb{Z}.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40158,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number. Prove that there exist infinitely many positive integers $n$ such that $p$ divides\n$$\n1^{n}+2^{n}+\\cdots+(p+1)^{n}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $k$ be a positive integer. Using Fermat's little theorem we have\n$$\n1^{k(p-1)}+2^{k(p-1)}+\\cdots+(p+1)^{k(p-1)} \\equiv \\underbrace{1^{k}+1^{k}+\\cdots+1^{k}}_{p-1 \\text{ times}}+0^{k}+1^{k} \\equiv 0 \\pmod{p} .\n$$\nTherefore, for $n=k(p-1)$, and $k=1,2,3, \\ldots$, the prime number $p$ divides\n$$\n1^{n}+2^{n}+\\cdots+(p+1)^{n} .\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40159,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded 1 point, the loser got 0 points, and each of the two players earned $\\frac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned in games against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of his points against the other nine of the ten). What was the total number of players in the tournament?",
    "options": [],
    "answer": "25",
    "solution": "Suppose that there were $n$ players and assume $s_{1} \\geq s_{2} \\geq \\cdots \\geq s_{n}$ be the scores of each player. Because each player played exactly one game against each of the other players and each game worths 1 point, the total sum of scores is\n$$\ns_{1}+s_{2}+\\cdots+s_{n}=\\binom{n}{2}=\\frac{n(n-1)}{2}\n$$\nBecause each of the ten lowest scoring players earned half of his points against the other 9 of the 10, he earned the other half of his points against the $n-10$ highest scoring players. Therefore, the total sum of the scores earned in the games between the first $n-10$ players and the last 10 players is\n$$\n10(n-10)=\\frac{1}{2} s_{1}+\\frac{1}{2} s_{2}+\\cdots+\\frac{1}{2} s_{n}=\\frac{n(n-1)}{4} .\n$$\nThis implies that $n^{2}-41 n+400=0$, and thus $n=16$ or $25$.\n\nBecause each of the ten lowest scoring players earned half of his points against the other 9 of the 10, the total sum of the scores earned in the games between the last 10 players is\n$$\n\\frac{1}{2} s_{n-9}+\\frac{1}{2} s_{n-8}+\\cdots+\\frac{1}{2} s_{n}=\\binom{10}{2}=45 .\n$$\nThis gives an average of 9 points of score for each of the ten lowest scoring players. Therefore, each of the first $n-10$ players has earned at least 9 points.\n\nBecause each of the first $n-10$ has earned half of his score from the games against the $n-11$ of the $n-10$ players, we have\n$$\n\\frac{9}{2}(n-10) \\leq \\frac{1}{2} s_{1}+\\frac{1}{2} s_{2}+\\cdots+\\frac{1}{2} s_{n-10}=\\binom{n-10}{2}=\\frac{(n-10)(n-11)}{2} .\n$$\nWe deduce that $n \\geq 20$ and hence $n=25$.\n\nFinally, we construct such a tournament with 25 players. In this tournament, there are 15 winners, each of them having 14 points of scores and 10 losers each of them having 9 points of score. Any game played between two winners or two losers results in a tie. For games played between winners and losers, each winner wins in 6 games, loses in 2 games, and 2 games result in a tie and conversely each loser wins 3 games, loses 9 games and 3 games result in a tie. This makes such a tournament possible.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40160,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$ be positive real numbers. Find the minimum of\n$$\nx^{2} + x y + \\frac{y^{2}}{2} + \\frac{2^{6}}{x + y} + \\frac{3^{4}}{x^{3}}.\n$$",
    "options": [],
    "answer": "63/2 (achieved at x=3, y=1)",
    "solution": "$$\nx^{2} + x y + \\frac{y^{2}}{2} + \\frac{2^{6}}{x + y} + \\frac{3^{4}}{x^{3}} = \\frac{x^{2}}{2} + \\frac{(x + y)^{2}}{2} + \\frac{2^{6}}{x + y} + \\frac{3^{4}}{x^{3}}.\n$$\nBy applying AM-GM inequality we have\n$$\n\\frac{x^{2}}{2} + \\frac{3^{4}}{x^{3}} = \\frac{x^{2}}{6} + \\frac{x^{2}}{6} + \\frac{x^{2}}{6} + \\frac{3^{4}}{2 x^{3}} + \\frac{3^{4}}{2 x^{3}} \\geq 5 \\sqrt[5]{\\left(\\frac{x^{2}}{6}\\right)^{3} \\left(\\frac{3^{4}}{2 x^{3}}\\right)^{2}} = \\frac{15}{2}\n$$\nand the equality holds when $\\frac{x^{2}}{6} = \\frac{3^{4}}{2 x^{3}}$. This is when $x = 3$.\nBy applying AM-GM inequality we have\n$$\n\\frac{(x + y)^{2}}{2} + \\frac{2^{6}}{x + y} = \\frac{(x + y)^{2}}{2} + \\frac{2^{5}}{x + y} + \\frac{2^{5}}{x + y} \\geq 3 \\sqrt[3]{\\frac{(x + y)^{2}}{2} \\left(\\frac{2^{5}}{x + y}\\right)^{2}} = 24\n$$\nand the equality holds when $\\frac{(x + y)^{2}}{2} = \\frac{2^{5}}{x + y}$. This is when $x + y = 4$.\nTherefore, the minimum of\n$$\nx^{2} + x y + \\frac{y^{2}}{2} + \\frac{2^{6}}{x + y} + \\frac{3^{4}}{x^{3}}\n$$\nis\n$$\n\\frac{15}{2} + 24 = \\frac{63}{2}\n$$\nand it is reached when $x = 3$ and $y = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40161,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{N}$ denote the set of positive integers, and let $S$ be a set. There exists a function $f: \\mathbb{N} \\rightarrow S$ such that if $x$ and $y$ are a pair of positive integers with their difference being a prime number, then $f(x) \\neq f(y)$. Determine the minimum number of elements in $S$.",
    "options": [],
    "answer": "4",
    "solution": "Let $f: \\mathbb{N} \\rightarrow S$ be such a function. Because the difference of any two numbers in $\\{1, 3, 6, 8\\}$ is a prime number, the cardinality of $\\{f(1), f(3), f(6), f(8)\\}$ is $4$. Hence, the minimum number of elements in $S$ is greater than or equal to $4$.\n\nNow, consider the function $f: \\mathbb{N} \\rightarrow \\{0, 1, 2, 3\\}$ that associates to each $n \\in \\mathbb{N}$ its remainder when divided by $4$. This function satisfies the condition of the problem since when $f(n_1) = f(n_2)$, for $n_1, n_2 \\in \\mathbb{N}$, then $n_1 - n_2$ is not a prime number since it is a multiple of $4$. Therefore, the minimum number of elements in $S$ is $4$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40162,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $x \\geq 0$, prove that\n$$\n\\frac{\\left(x^{2}+1\\right)^{6}}{2^{7}}+\\frac{1}{2} \\geq x^{5}-x^{3}+x\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because $x^{2}+1 \\geq 2x$, we have\n$$\n\\frac{\\left(x^{2}+1\\right)^{7}}{2^{7}} \\geq \\frac{(2x)^{7}}{2^{7}} = x^{7}, \\quad \\text{and} \\quad \\frac{x^{2}+1}{2} \\geq x,\n$$\nand deduce\n$\\left(x^{2}+1\\right) \\cdot \\left(\\frac{\\left(x^{2}+1\\right)^{6}}{2^{7}}+\\frac{1}{2}\\right) = \\frac{\\left(x^{2}+1\\right)^{7}}{2^{7}} + \\frac{x^{2}+1}{2} \\geq x^{7} + x = x\\left(x^{2}+1\\right)\\left(x^{4}-x^{2}+1\\right)$.\nTherefore\n$$\n\\frac{\\left(x^{2}+1\\right)^{6}}{2^{7}}+\\frac{1}{2} \\geq x^{5}-x^{3}+x\n$$\nand the equality holds when $x=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40163,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f:[0, \\infty) \\rightarrow \\mathbb{R}$ such that $f(0)=0$ and\n$$\nf(x)=1+5 f\\left(\\left\\lfloor\\frac{x}{2}\\right\\rfloor\\right)-6 f\\left(\\left\\lfloor\\frac{x}{4}\\right\\rfloor\\right)\n$$\nfor all $x>0$.",
    "options": [],
    "answer": "f(0)=0; f(x)=1 for 0<x<2; and for n>=1, f(x) = -2^{n+2} + (3^{n+2}+1)/2 for x in [2^n, 2^{n+1}).",
    "solution": "Let $x \\geq 0$. If $x \\in (0,2)$ then $f(x)=1+5 f(0)-6 f(0)=1$. \nIf $x \\in [2,4)$ then $f(x)=1+5 f(1)-6 f(0)=6=a_{1}$. \nIf $x \\in [4,8)$ then $\\left\\lfloor\\frac{x}{2}\\right\\rfloor \\in [2,4)$ and $\\left\\lfloor\\frac{x}{4}\\right\\rfloor \\in [1,2)$, and therefore $f(x)=1+5 \\cdot 6-6 \\cdot 1=25=a_{2}$.\n\nAssume for $n \\geq 1$, that function $f$ is constant on $[2^{n}, 2^{n+1})$ taking a value $a_{n}$, and constant on $[2^{n+1}, 2^{n+2})$ taking a value $a_{n+1}$, and let $x \\in [2^{n+2}, 2^{n+3})$. Because $\\left\\lfloor\\frac{x}{2}\\right\\rfloor \\in [2^{n+1}, 2^{n+2})$ and $\\left\\lfloor\\frac{x}{4}\\right\\rfloor \\in [2^{n}, 2^{n+1})$, we deduce that $f(x)= 1+5 a_{n+1}-6 a_{n}$. Therefore, function $f$ is also constant on $[2^{n+2}, 2^{n+3})$ taking the value $a_{n+2}=1+5 a_{n+1}-6 a_{n}$, which can be rewritten\n$$\na_{n+2}-\\frac{1}{2}=5\\left(a_{n+1}-\\frac{1}{2}\\right)-6\\left(a_{n}-\\frac{1}{2}\\right) .\n$$\nBecause the roots of the characteristic polynomial $X^{2}-5 X+6$ are $2$ and $3$, there exist two real numbers $u, v$ such that\n$$\na_{n}=\\frac{1}{2}+2^{n} u+3^{n} v, \\quad \\text{ for all } n \\geq 1\n$$\nBecause $a_{1}=6$ and $a_{2}=25$, we have $u=-4$ and $v=\\frac{9}{2}$. We deduce that\n$$\nf(x)= \\begin{cases}0 & \\text{ if } x=0 \\\\ 1 & \\text{ if } x \\in (0,2) \\\\ -2^{n+2}+\\frac{3^{n+2}+1}{2} & \\text{ if } x \\in [2^{n}, 2^{n+1}) \\text{ and } n \\geq 1\\end{cases}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40164,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be the midpoint of side $B C$ of triangle $A B C$ and $E$ the midpoint of median $A D$. Line $B E$ intersects side $C A$ at $F$. Prove that the area of quadrilateral $C D E F$ is $\\frac{5}{12}$ the area of triangle $A B C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K$ be the midpoint of $A B$ and $L$ the intersection point of $D K$ with $B E$.\n\n![](attached_image_1.png)\n\nBecause $D$ and $K$ are midpoints of $B C$ and $B A$, respectively, segment $D K$ is parallel to side $A B$ and we have\n$$\n\\frac{E F}{L E}=\\frac{E A}{D E}=1 .\n$$\nBecause $E$ and $K$ are midpoints of $A D$ and $A B$, respectively, the point $L$ is the centroid of triangle $A B D$. Therefore\n$$\n\\frac{L E}{B E}=\\frac{1}{3} .\n$$\nWe deduce that the ratio of the areas of the triangles\n$$\n\\frac{[A E F]}{[A B E]}=\\frac{E F}{B E}=\\frac{1}{3} .\n$$\nTherefore\n$$\n\\frac{[A E F]}{[A B C]}=\\frac{[A E F]}{[A B E]} \\cdot \\frac{[A B E]}{[A B D]} \\cdot \\frac{[A B D]}{[A B C]}=\\frac{1}{12},\n$$\nsince $D$ and $E$ are midpoints of $B C$ and $A D$, respectively. We deduce that\n$$\n\\frac{[C F E D]}{[A B C]}=\\frac{[A D C]-[A E F]}{[A B C]}=\\frac{1}{2}-\\frac{1}{12}=\\frac{5}{12} .\nLet\n$$\nx=\\frac{[C F D]}{[A B C]} \\quad \\text{ and } \\quad y=\\frac{[D F E]}{[A B C]}\n$$\nbe the ratios of the areas.\nBecause $E$ is the midpoint of $A D$, we have\n$$\n[A B E]=[B D E]=\\frac{1}{2}[A B D] \\quad \\text{ and } \\quad[A E F]=[D F E] .\n$$\nBecause $D$ is the midpoint of $B C$, we have\n$$\n[A B D]=[A D C]=\\frac{1}{2}[A B C] \\quad \\text{ and } \\quad[F B D]=[F D C] .\n$$\n![](attached_image_2.png)\n\nWe deduce that\n$$\nx+2 y=\\frac{1}{2} \\quad \\text{ and } \\quad x=y+\\frac{1}{4},\n$$\nand therefore\n$$\nx=\\frac{1}{3} \\quad \\text{ and } \\quad y=\\frac{1}{12} .\n$$\nThis proves that\n$$\n\\frac{[C F E D]}{[A B C]}=x+y=\\frac{5}{12} .",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40165,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that among any 16 perfect cubes we can always find two cubes whose difference is divisible by 91.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice first that any perfect cube is congruent to either $0$, $1$ or $6$ modulo $7$ and it is congruent to either $0$, $1$, $5$, $8$, or $12$ modulo $13$. Because $7$ and $13$ are relatively prime numbers, by the Chinese Remainder theorem, a perfect cube is congruent to precisely one of $15 = 3 \\times 5$ different residues modulo $91 = 7 \\times 13$. Therefore, by the Pigeonhole principle, we can always find among any $16$ perfect cubes, two cubes which are congruent modulo $91$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40166,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define a domino to be an ordered pair of distinct positive integers. A proper sequence of dominoes is a list of distinct dominoes in which the first coordinate of each pair after the first equals the second coordinate of the immediately preceding pair, and in which $(i, j)$ and $(j, i)$ do not both appear for any $i$ and $j$. Let $D_{n}$ be the set of all dominoes whose coordinates are no larger than $n$. Find the length of the longest proper sequence of dominoes that can be formed using the dominoes of $D_{n}$.",
    "options": [],
    "answer": "If n is odd, the maximum length is n(n−1)/2. If n is even, the maximum length is (n^2 − 2n + 2)/2.",
    "solution": "Consider the complete graph $\\mathcal{G}_n$, over the set of vertices $\\{1,2, \\ldots, n\\}$. To each domino $(i, j)$ in $D_n$, we associate the edge joining the two vertices $i, j$. In this way, to each edge in $\\mathcal{G}_n$ we have associated exactly two dominoes $(i, j)$ and $(j, i)$, for some $1 \\leq i \\neq j \\leq n$. A proper sequence of dominoes corresponds exactly to a path in $\\mathcal{G}_n$ which visits every edge at most once. The problem asks for the length of the longest such path. We will use and prove the following:\n\n> If a finite graph $\\mathcal{G}$ is connected and each of its vertices has an even degree, there exists a closed path which visits all its edges exactly once. This closed path is called Eulerian.\n\nWe will prove this fact by induction on the number of edges of $\\mathcal{G}$.\n\nIf $\\mathcal{G}$ has 3 edges, then it is a triangle and the statement is obvious. Assume there exists a closed Eulerian path in any connected graph with all vertices of even degree having less than or equal to $m$ edges.\n\nLet $\\mathcal{G}$ be a connected graph with $m+1$ edges and all its vertices are of even degree. Let $\\mathcal{P}$ be a longest path visiting each edge of $\\mathcal{G}$ at most once.\n\nAssume that $\\mathcal{P}$ starts and ends at different vertices $v, v'$. Path $\\mathcal{P}$ visits an odd number of edges incident to $v$. But the degree of $v$ is even. Then, there exists an edge $e$ incident to $v$ which is not visited by $\\mathcal{P}$ and so, we can extend $\\mathcal{P}$ with $e$. This contradicts the maximality of $\\mathcal{P}$. Hence path $\\mathcal{P}$ starts and ends at the same vertex.\n\nAssume that there exists an edge $e$ not visited by $\\mathcal{P}$ and let $v$ be a vertex incident to $e$ and $v_1$ a vertex visited by $\\mathcal{P}$. Because $\\mathcal{G}$ is connected, there exists a path from $v$ to $v_1$. Replace $v_1$ by the first vertex in this connecting path visited by $\\mathcal{P}$. Let $\\mathcal{G}'$ be the graph obtained by deleting from the graph $\\mathcal{G}$ all edges visited by $\\mathcal{P}$ and taking the connected component of $v_1$. Graph $\\mathcal{G}'$ contains edge $e$ and has less than $m$ edges. Each vertex $v'$ of $\\mathcal{G}'$ has even degree since path $\\mathcal{P}$ has visited precisely an even number of edges incident to $v'$. Hence, there exists a closed Eulerian path $\\mathcal{P}'$ in $\\mathcal{G}'$. By considering in $\\mathcal{G}$ the path starting at $v_1$ and walking along all the path $\\mathcal{P}$ and then along all the path $\\mathcal{P}'$ we obtain a longer path in $\\mathcal{G}$ visiting each edge at most once. This contradicts the maximality of $\\mathcal{P}$. This proves that edge $e$ does not exist and that $\\mathcal{P}$ is Eulerian.\n\nBack to our problem, we will consider two cases:\n\nIf $n$ is an odd number, the degree of each vertex of the complete graph $\\mathcal{G}_n$ is $n-1$, even, and $\\mathcal{G}_n$ is connected. So there exists a path visiting all the edges of $\\mathcal{G}_n$ exactly once. Hence the longest proper sequence of dominoes that can be formed using the dominoes of $D_n$ uses all dominoes. It has length $\\binom{n}{2} = \\frac{n(n-1)}{2}$.\n\nIf $n$ is an even number, the degree of each vertex of the complete graph $\\mathcal{G}_n$ is odd. Let $\\mathcal{P}$ be a longest path visiting each edge at most once. For each vertex $v$ different from the starting and the ending vertices of $\\mathcal{P}$, path $\\mathcal{P}$ visits an even number of edges incident to $v$. So there exists at least one edge incident to $v$ not visited by $\\mathcal{P}$. Because each edge is incident to two vertices, there exists at least $\\frac{n-2}{2}$ edges not visited by $\\mathcal{P}$. Hence the length of $\\mathcal{P}$ is less than or equal to $\\binom{n}{2} - \\frac{n-2}{2} = \\frac{n^2 - 2n + 2}{2}$.\n\nNow, consider the graph $\\mathcal{G}_n'$ obtained from $\\mathcal{G}_n$ by deleting the edges between $1$ and $3$, $2$ and $3$, and $2i-1$ and $2i$, for $2 \\leq i \\leq \\frac{n}{2}$. Graph $\\mathcal{G}_n'$ is connected and all its vertices have even degree. So there exists an Eulerian closed path $\\mathcal{P}'$ starting and ending at $3$. Extending path $\\mathcal{P}'$ in $\\mathcal{G}_n$ at the beginning with the edge between $1$ and $3$ and at the end with the edge between $2$ and $3$, we obtain a path $\\mathcal{P}$ of length $\\frac{n^2 - 2n + 2}{2}$ visiting each edge at most once. Hence the longest proper sequence of dominoes that can be formed using the dominoes of $D_n$ has length $\\frac{n^2 - 2n + 2}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40167,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, $I$ its incenter, and $\\omega$ a circle of center $I$. Points $A'$, $B'$, $C'$ are on $\\omega$ such that rays $IA'$, $IB'$, $IC'$ starting from $I$ intersect perpendicularly sides $BC$, $CA$, $AB$, respectively. Prove that lines $AA'$, $BB'$, $CC'$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Define $A''$, $B''$, $C''$ to be the intersection points of the lines $AA'$, $BB'$, $CC'$ with the sides $BC$, $CA$, $AB$, respectively, $B_a$, $C_a$ the intersection points of the tangent line to $\\omega$ at $A'$ with the lines $AB$, $AC$, respectively, $C_b$, $A_b$ the intersection points of the tangent line to $\\omega$ at $B'$ with the lines $BC$, $BA$, respectively, $A_c$, $B_c$ the intersection points of the tangent line to $\\omega$ at $C'$ with the lines $CA$, $CB$, respectively, $A_1$, $B_1$, $C_1$ the intersection points of the three tangent lines to $\\omega$ at $A'$, $B'$, $C'$, as shown in the figure, $r$ the radius of $\\omega$ and $r_0$ the inradius of triangle $ABC$.\nThe distance from $A_1$ to each line $AB$, $AC$ is equal to $|r - r_0|$. Therefore, $A_1$ is on the bisector of angle $\\angle BAC$. But $AA_bA_1A_c$ is a parallelogram. We deduce that $AA_bA_1A_c$ is a rhombus and therefore $A_1A_b = A_1A_c$. On the other hand, we have $A_1B' = A_1C'$ as tangents to $\\omega$ from $A_1$. We deduce that $A_bB' = A_cC'$. We deduce in a similar way that $B_cC' = B_aA'$ and $C_aA' = C_bB'$.\n\n![](attached_image_1.png)\n\nBecause lines $B_aC_a$ and $BC$ are parallel, we have from Thales\n$$\n\\frac{BA''}{A''C} = \\frac{B_aA'}{A'C_a}.\n$$\nWe have similarly\n$$\n\\frac{CB''}{B''A} = \\frac{C_bB'}{B'A_b},\n$$\nand\n$$\n\\frac{AC''}{C''B} = \\frac{A_cC'}{C'B_c}.\n$$\nTherefore\n$$\n\\frac{BA''}{A''C} \\cdot \\frac{CB''}{B''A} \\cdot \\frac{AC''}{C''B} = \\frac{B_aA'}{A'C_a} \\cdot \\frac{C_bB'}{B'A_b} \\cdot \\frac{A_cC'}{C'B_c} = 1,\n$$\nand by Ceva's theorem, lines $AA'$, $BB'$, $CC'$ are concurrent.\nBecause $IA'$ and $IC'$ are perpendicular to $BC$ and $AB$, respectively, at the intouch points of the incircle of $ABC$, and $IA' = IC'$, we have $\\angle CBA' = \\angle C'BA$ and $\\angle A'BA = \\angle CBC'$ as oriented angles. We have in a similar way $\\angle BAC' = \\angle B'AC$, $\\angle C'AC = \\angle BAB'$, $\\angle ACB' = \\angle A'CB$ and $\\angle B'CB = \\angle ACA'$ as oriented angles.\nBecause $AA'$, $BA'$, $CA'$ intersect in a point $A'$, we have from the trigonometric form of Ceva's theorem\n$$\n\\frac{\\sin \\angle BAA'}{\\sin \\angle A'AC} = \\frac{\\sin \\angle A'BA}{\\sin \\angle CBA'} \\cdot \\frac{\\sin \\angle A'CB}{\\sin \\angle ACA'}.\n$$\nSimilarly, we have\n$$\n\\frac{\\sin \\angle CBB'}{\\sin \\angle B'BA} = \\frac{\\sin \\angle B'CB}{\\sin \\angle ACB'} \\cdot \\frac{\\sin \\angle B'AC}{\\sin \\angle BAB'},\n$$\nand\n$$\n\\frac{\\sin \\angle ACC'}{\\sin \\angle C'CB} = \\frac{\\sin \\angle C'AC}{\\sin \\angle BAC'} \\cdot \\frac{\\sin \\angle C'BA}{\\sin \\angle CBC'}.\n$$\n![](attached_image_2.png)\n\nAfter multiplication and cancellation, we obtain\n$$\n\\frac{\\sin \\angle BAA'}{\\sin \\angle A'AC} \\cdot \\frac{\\sin \\angle CBB'}{\\sin \\angle B'BA} \\cdot \\frac{\\sin \\angle ACC'}{\\sin \\angle C'CB} = 1.\n$$\nWe deduce from the trigonometric form of Ceva's theorem that $AA'$, $BB'$, $CC'$ are concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40168,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that it is possible to write a $n \\times n$ array of non-negative numbers (not necessarily distinct) such that the sums of entries on each row and each column are pairwise distinct perfect squares.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First solution suggested by the student Salman Saleh. We will construct an $n \\times n$ array by induction on $n \\geq 2$.\nFor $n=2$, consider the following example:\n| 0 | 9 |\n| :---: | :---: |\n| 0 | 16 |\nAssume there exists an $n \\times n$ array\n| $a_{1,1}$ | $a_{1,2}$ | $\\ldots$ | $a_{1, n}$ |\n| :---: | :---: | :---: | :---: |\n| $a_{2,1}$ | $a_{2,2}$ | $\\ldots$ | $a_{2, n}$ |\n| $\\vdots$ | $\\vdots$ | $\\ddots$ | $\\vdots$ |\n| $a_{n, 1}$ | $a_{n, 2}$ | $\\ldots$ | $a_{n, n}$ |\nsuch that the sums of entries on each row and each column are pairwise distinct perfect squares and such that the sum of the last column is greater than all the other sums. Let\n$$\nr_{i}^{2}=\\sum_{j=1}^{n} a_{i, j}, \\text{ for } i=1,2, \\ldots, n\n$$\nand\n$$\nc_{j}^{2}=\\sum_{i=1}^{n} a_{i, j}, \\text{ for } j=1,2, \\ldots, n,\n$$\nbe the distinct perfect squares. Notice that\n$$\n\\sum_{i=1}^{n} r_{i}^{2}=\\sum_{j=1}^{n} c_{j}^{2}\n$$\nConsider the following $(n+1) \\times(n+1)$ array\n| $4 a_{1,1}$ | $4 a_{2,1}$ | … | $4 a_{n, 1}$ | $12 c_{1}^{2}$ |\n| :--- | :--- | :--- | :--- | :--- |\n| $4 a_{1,2}$ | $4 a_{2,2}$ | … | $4 a_{n, 2}$ | $12 c_{2}^{2}$ |\n| $\\vdots$ | $\\vdots$ | $\\ddots$ | $\\vdots$ | $\\vdots$ |\n| $4 a_{1, n-1}$ | $4 a_{2, n-1}$ | … | $4 a_{n, n-1}$ | $12 c_{n-1}^{2}$ |\n| $4 a_{1, n}$ | $4 a_{2, n}$ | $\\cdots$ | $4 a_{n, n}$ | $4\\left(25 n^{2}-1\\right) c_{n}^{2}$ |\n| $12 r_{1}^{2}$ | $12 r_{2}^{2}$ | … | $12 r_{n}^{2}$ | $x_{n}$ |\nwhere\n$$\nx_{n}=\\left(\\left(25 n^{2}-4\\right) c_{n}^{2}-1\\right)^{2}-12 \\sum_{i=1}^{n} r_{i}^{2} .\n$$\nNotice that\n$$\nx_{n}>\\left(\\left(25 n^{2}-4\\right) c_{n}^{2}-1\\right)^{2}-12 n c_{n}^{2}>\\left(21 n^{2}-4\\right) c_{n}^{2}>0 .\n$$\nThe sums of entries of each row and each columns are\n$$\n\\left(4 c_{1}\\right)^{2},\\left(4 c_{2}\\right)^{2}, \\ldots,\\left(4 c_{n-1}\\right)^{2},\\left(4 r_{1}\\right)^{2},\\left(4 r_{2}\\right)^{2}, \\ldots,\\left(4 r_{n}\\right)^{2},\n$$\nwhich are all distinct and less than the three other sums\n$$\n\\left(10 n c_{n}\\right)^{2}<\\left(\\left(25 n^{2}-4\\right) c_{n}^{2}-1\\right)^{2}<\\left(\\left(25 n^{2}-4\\right) c_{n}^{2}+1\\right)^{2},\n$$\nand clearly, the sum of the last column is the greatest perfect square.\n\n\nSecond solution suggested by the student Alhamza Alnufayli. If $n=2 k \\geq 2$, we consider the following $n \\times n$ array:\n| $48 \\cdot 2^{2(k-1)}$ | 0 | $\\cdots$ | 0 | 0 | $\\cdots$ | 0 | $33 \\cdot 2^{2(k-1)}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| 0 | $\\ddots$ | $\\ddots$ | $\\vdots$ | $\\vdots$ | $\\cdot$ | $\\cdot$ | 0 |\n| $\\vdots$ | $\\ddots$ | $48 \\cdot 2^{2}$ | 0 | 0 | $33 \\cdot 2^{2}$ | $\\cdot$ | $\\vdots$ |\n| 0 | $\\cdots$ | 0 | 48 | 33 | 0 | $\\cdots$ | 0 |\n| 0 | $\\cdots$ | 0 | 1 | 3 | 0 | $\\cdots$ | 0 |\n| $\\vdots$ | $\\cdot$ | $2^{2}$ | 0 | 0 | $3 \\cdot 2^{2}$ | $\\ddots$ | $\\vdots$ |\n| 0 | $\\cdot$ | $\\cdot$ | $\\vdots$ | $\\vdots$ | $\\ddots$ | $\\ddots$ | 0 |\n| $2^{2(k-1)}$ | 0 | $\\cdots$ | 0 | 0 | $\\cdots$ | 0 | $3 \\cdot 2^{2(k-1)}$ |\nThe sums of entries of each row and each column are\n$$\n\\begin{aligned}\n& \\left(9 \\cdot 2^{k-1}\\right)^{2}, \\ldots,(9 \\cdot 2)^{2},(9 \\cdot 1)^{2},(2 \\cdot 1)^{2},(2 \\cdot 2)^{2}, \\ldots,\\left(2 \\cdot 2^{k-1}\\right)^{2}, \\\\\n& \\left(7 \\cdot 2^{k-1}\\right)^{2}, \\ldots,(7 \\cdot 2)^{2},(7 \\cdot 1)^{2},(6 \\cdot 1)^{2},(6 \\cdot 2)^{2}, \\ldots,\\left(6 \\cdot 2^{k-1}\\right)^{2},\n\\end{aligned}\n$$\nwhich are all distinct perfect squares.\nIf $n=2 k+1 \\geq 2$, we consider the following $n \\times n$ array:\n| $48 \\cdot 2^{2(k-1)}$ | 0 | $\\cdots$ | 0 | 0 | 0 | $\\cdots$ | 0 | $33 \\cdot 2^{2(k-1)}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| 0 | $\\ddots$ | $\\ddots$ | $\\vdots$ | $\\vdots$ | $\\vdots$ | $\\cdot$ | $\\cdot$ | 0 |\n| $\\vdots$ | $\\ddots$ | $48 \\cdot 2^{2}$ | 0 | 0 | 0 | $33 \\cdot 2^{2}$ | $\\cdot$ | $\\vdots$ |\n| 0 | $\\cdots$ | 0 | 48 | 360 | 33 | 0 | $\\cdots$ | 0 |\n| 0 | $\\cdots$ | $\\cdots$ | 0 | $13^{2}$ | 0 | $\\cdots$ | $\\cdots$ | 0 |\n| 0 | $\\cdots$ | 0 | 1 | 0 | 3 | 0 | $\\cdots$ | 0 |\n| $\\vdots$ | $\\cdot$ | $2^{2}$ | 0 | 0 | 0 | $3 \\cdot 2^{2}$ | $\\ddots$ | $\\vdots$ |\n| 0 | $\\cdot$ | $\\cdot$ | $\\vdots$ | $\\vdots$ | $\\ddots$ | $\\ddots$ | 0 |\n| $2^{2(k-1)}$ | 0 | $\\cdots$ | 0 | 0 | 0 | $\\cdots$ | 0 | $3 \\cdot 2^{2(k-1)}$ |\nThe sums of entries on each row and each column are\n$$\n\\begin{gathered}\n\\left(9 \\cdot 2^{k-1}\\right)^{2}, \\ldots,(9 \\cdot 2)^{2}, 21^{2}, 13^{2},(2 \\cdot 1)^{2},(2 \\cdot 2)^{2}, \\ldots,\\left(2 \\cdot 2^{k-1}\\right)^{2} \\\\\n\\left(7 \\cdot 2^{k-1}\\right)^{2}, \\ldots,(7 \\cdot 2)^{2},(7 \\cdot 1)^{2}, 23^{2},(6 \\cdot 1)^{2},(6 \\cdot 2)^{2}, \\ldots,\\left(6 \\cdot 2^{k-1}\\right)^{2}\n\\end{gathered}\n$$\nwhich are all distinct perfect squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40169,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which $1-5^{n}+5^{2 n+1}$ is a perfect square.",
    "options": [],
    "answer": "1",
    "solution": "Assume that $1-5^{n}+5^{2 n+1}=m^{2}$, for a positive integer $m$. We have\n$$\n5^{n}\\left(5^{n+1}-1\\right)=(m-1)(m+1) .\n$$\nBecause $(m+1)-(m-1)=2$, the number $5$ cannot divide both $m-1$ and $m+1$. Therefore, we have two cases:\n\nFirst case when $5$ divides $m-1$. In this case, there exists a positive integer $k$ such that $m-1=5^{n} k$ and $5^{n+1}-1=(m+1) k=\\left(5^{n} k+2\\right) k$. We, therefore, obtain the equation\n$$\n5^{n}\\left(5-k^{2}\\right)=2 k+1\n$$\nFor $k=1$, this equation has no solution. For $k=2$, $n=1$ is a solution for the equation. For $k \\geq 3$ this equation has no solution since its left hand side is negative.\n\nSecond case when $5$ divides $m+1$. In this case, there exists a positive integer $k$ such that $m+1=5^{n} k$ and $5^{n+1}-1=(m-1) k=\\left(5^{n} k-2\\right) k$. We, therefore, obtain the equation\n$$\n5^{n}\\left(k^{2}-5\\right)=2 k-1\n$$\nThis equation has no solution since for $k=1,2$ or $3$ the left hand side is either negative or even, and for $k \\geq 4$ we have\n$$\n5^{n}\\left(k^{2}-5\\right)>5(k+2)(k-3) \\geq 5 k+10>2 k-1 .\n$$\nTherefore, the unique solution to this problem is $n=1$ since we have\n$$\n1-5+5^{3}=11^{2} .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40170,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $2015$ coins on a table. For $i=1,2, \\ldots, 2015$ in succession, one must turn over exactly $i$ coins. Prove that it is always possible either to make all of the coins face up or to make all of the coins face down, but not both.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We start by proving that it is always possible either to make all of the coins face up or to make all of the coins face down.\nNotice that if we proceed by choosing the numbers $i=1,2, \\ldots, 2015$ in this order or in any other order, the final result will depend only on the $i$ coins chosen at each step. For the sake of simplicity, we will follow the order\n$$\ni=1007,1008,1006,1009,1005,1010, \\ldots, 2,2013,1,2014,2015 .\n$$\nBy symmetry of the face up/face down positions, we can assume without loss of generality that at the beginning, the number of coins face up is odd equal to $2k+1$. Choose $k$ of these face up coins and $1007-k$ of the face down coins and turn them over to get $1008$ coins face up and $1007$ coins face down. Turn these new $1008$ coins face up to get all the coins face down. After this, at each step, whenever we choose $i$ coins, for $i=1006,1005, \\ldots, 1,2015$, in the next step we choose the $2015-i$ remaining coins to make at the end all the coins face up.\n\nNow we prove that for each starting position, it is not possible to choose to make all the coins face up or all the coins face down.\nAssume that starting with an odd number of face up coins we can make at the end all the coins face down. This means that each face up coin will be turned over an odd number of times while each face down coin will be turned over an even number of times. Therefore, the total number of turning over coins is\n$$\n\\text{odd} \\times \\text{odd} + \\text{even} \\times \\text{even} = \\text{odd} .\n$$\nBut the total number of turning over coins is\n$$\n1+2+\\cdots+2015=2015 \\times 1008=\\text{even},\n$$\nwhich is a contradiction. So it is only possible to make all the coins face up and not to make all of them face down. In a similar way, if we start with an odd number of face down coins it is only possible to make them all face down.\nThe statement works for any odd number of coins. We prove it by induction on the number $n$ of coins, $n$ being odd. The case $n=1$ is obvious. Suppose the statement is true for $n=2k-1$, for some positive integer $k$. If we are given $n=2k+1$ coins, we consider the following cases.\n\n- Case 1. There is a coin $C_{1}$ facing up and another coin $C_{2}$ facing down. We consider the other $2k-1$ coins first. By our induction, we can turn the coins $1,2, \\ldots, 2k-1$ times in succession so all the $2k-1$ coins are in one direction. Without loss of generality, we assume that all the $2k-1$ coins are facing up. Then we turn all these coins together with $C_{1}$ and then turn all the $2k+1$ coins so they will be all facing up.\n\n- Case 2: All the coins are in the same direction. We arrange the coins around a circle and number them $1,2, \\ldots, 2k+1$ in clockwise order. We first turn coin $1$, then coins $2$ and $3$, and then $4,5$, and $6$, and so on along the circle. Then we make a total of $1+2+\\cdots+(2k+1)=(k+1)(2k+1)$ turnings and each coin has been turned $k+1$ times. Since they start in the same direction, they end in the same direction.\n\nFrom the above argument, we can find a way to make all of the $2k+1$ coins facing in one direction after $2k+1$ operations regardless of the initial configuration. Hence our induction is complete.\n\nTo prove that it is impossible to achieve both final configurations, we proceed as in the first solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40171,
    "subject": "Mathematics (Multi-modal)",
    "question": "The 2013 numbers\n$$\n\\frac{1}{1 \\times 2}, \\frac{1}{2 \\times 3}, \\frac{1}{3 \\times 4}, \\ldots, \\frac{1}{2013 \\times 2014}\n$$\nare arranged randomly on a circle.\n\na. Prove that there exist ten consecutive numbers on the circle whose sum is less than $\\frac{1}{4000}$.\n\nb. Prove that there exist ten consecutive numbers on the circle whose sum is less than $\\frac{1}{10000}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nConsider 201 disjoint blocks $B_{1}, B_{2}, \\ldots, B_{201}$ consisting each of 10 numbers, consecutive on the circle. By the pigeonhole principle, there exists a block $B_{k}$, for some $1 \\leq k \\leq 201$, not containing any of the 200 numbers\n$$\n\\frac{1}{1 \\times 2}, \\frac{1}{2 \\times 3}, \\ldots, \\frac{1}{200 \\times 201}\n$$\nThe sum of the ten consecutive numbers in this block $B_{k}$ is less than or equal to\n$$\n\\begin{aligned}\n\\frac{1}{201 \\times 202} + & \\frac{1}{202 \\times 203} + \\cdots + \\frac{1}{210 \\times 211} \\\\\n& = \\left(\\frac{1}{201} - \\frac{1}{202}\\right) + \\left(\\frac{1}{202} - \\frac{1}{203}\\right) + \\cdots + \\left(\\frac{1}{210} - \\frac{1}{211}\\right) \\\\\n& = \\frac{1}{201} - \\frac{1}{211} = \\frac{10}{201 \\times 211} < \\frac{1}{4000}\n\\end{aligned}\n$$\n\nb.\nAgain, consider the 201 disjoint blocks $B_{1}, B_{2}, \\ldots, B_{201}$ like in (a). By the pigeonhole principle, there exists at least 101 blocks $B_{k_{1}}, B_{k_{2}}, \\ldots, B_{k_{201}}$, not containing any of the 100 numbers\n$$\n\\frac{1}{1 \\times 2}, \\frac{1}{2 \\times 3}, \\ldots, \\frac{1}{100 \\times 101}.\n$$\nThe sum of the 1010 numbers in these 101 blocks $B_{k_{1}}, B_{k_{2}}, \\ldots, B_{k_{201}}$ is less than or equal to\n$$\n\\begin{aligned}\n\\sum_{i=1}^{201} \\left( \\sum_{b \\in B_{k_{i}}} b \\right) & \\leq \\frac{1}{101 \\times 102} + \\frac{1}{102 \\times 103} + \\cdots + \\frac{1}{210 \\times 211} \\\\\n& \\leq \\left(\\frac{1}{101} - \\frac{1}{102}\\right) + \\left(\\frac{1}{102} - \\frac{1}{103}\\right) + \\cdots + \\left(\\frac{1}{210} - \\frac{1}{211}\\right) \\\\\n& < \\frac{1}{101}\n\\end{aligned}\n$$\nTherefore, there exists at least one block $B_{k_{i_{0}}}$ among these 101 blocks with the sum of its ten consecutive numbers less than\n$$\n\\frac{1}{101 \\times 101} < \\frac{1}{10000}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40172,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDE$ be a cyclic pentagon such that the diagonals $AC$ and $AD$ intersect $BE$ at $P$ and $Q$, respectively, with $BP \\cdot QE = PQ^{2}$. Prove that $BC \\cdot DE = CD \\cdot PQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because triangles $PBC$ and $PAE$ are similar, and triangles $QED$ and $QAB$ are similar, we have\n$$\n\\frac{BC}{AE} = \\frac{PB}{PA}, \\quad \\text{and} \\quad \\frac{DE}{BA} = \\frac{EQ}{AQ}.\n$$\n![](attached_image_1.png)\nTherefore\n$$\n\\begin{equation*}\n\\frac{BC \\cdot DE}{AE \\cdot BA} = \\frac{PB \\cdot EQ}{PA \\cdot AQ} = \\frac{PQ^{2}}{PA \\cdot AQ}. \\tag{*}\n\\end{equation*}\n$$\nOn the other hand, applying sine law on triangles $ABQ$ and $APQ$, we have\n$$\n\\frac{AB}{AQ} = \\frac{\\sin \\angle AQB}{\\sin \\angle EBA}, \\quad \\text{and} \\quad \\frac{PQ}{AP} = \\frac{\\sin \\angle PAQ}{\\sin \\angle AQP}.\n$$\nFinally, by applying sine law on the circumcircle of the pentagon $ABCDE$, we get\n$$\n\\frac{AB}{AQ} \\cdot \\frac{PQ}{AP} = \\frac{\\sin \\angle AQB}{\\sin \\angle EBA} \\cdot \\frac{\\sin \\angle PAQ}{\\sin \\angle AQP} = \\frac{\\sin \\angle CAD}{\\sin \\angle EBA} = \\frac{CD}{EA}. \\quad \\text{(**)}\n$$\nCombining the two relations ($*$) and ($**$), we obtain after cancellation\n$$\nBC \\cdot DE = CD \\cdot PQ.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40173,
    "subject": "Mathematics (Multi-modal)",
    "question": "Turki has divided a square into finitely many white and green rectangles, each with sides parallel to the sides of the square. Within each white rectangle, he writes down its width divided by its height. Within each green rectangle, he writes down its height divided by its width. Finally, he calculates $S$, the sum of these numbers. If the total area of white rectangles equals the total area of green rectangles, determine the minimum possible value of $S$.",
    "options": [],
    "answer": "5/2",
    "solution": "Let $s$ be the sidelength of the original square, $m$ and $n$ the numbers of white and green rectangles $W_{1}, W_{2}, \\ldots, W_{m}$ and $G_{1}, G_{2}, \\ldots, G_{n}$, respectively, $a_{i}$ and $b_{i}$ the width and height of the white rectangle $W_{i}$, respectively, for $i=1,2, \\ldots, m$, and $c_{j}$ and $d_{j}$ the width and height of the green rectangle $G_{j}$, respectively, for $j=1,2, \\ldots, n$.\n\nThe total area of the white rectangles and the total area of the green rectangles satisfy\n$$\na_{1} b_{1}+a_{2} b_{2}+\\cdots+a_{m} b_{m}=c_{1} d_{1}+c_{2} d_{2}+\\cdots+c_{n} d_{n}=\\frac{s^{2}}{2}\n$$\nWe have\n$$\n\\begin{aligned}\nS & =\\frac{a_{1}}{b_{1}}+\\frac{a_{2}}{b_{2}}+\\cdots+\\frac{a_{m}}{b_{m}}+\\frac{d_{1}}{c_{1}}+\\frac{d_{2}}{c_{2}}+\\cdots+\\frac{d_{n}}{c_{n}} \\\\\n& =\\frac{a_{1}^{2}}{a_{1} b_{1}}+\\frac{a_{2}^{2}}{a_{2} b_{2}}+\\cdots+\\frac{a_{m}^{2}}{a_{m} b_{m}}+\\frac{d_{1}^{2}}{c_{1} d_{1}}+\\frac{d_{2}^{2}}{c_{2} d_{2}}+\\cdots+\\frac{d_{n}^{2}}{c_{n} d_{n}} \\\\\n& \\geq \\frac{\\left(a_{1}+a_{2}+\\cdots+a_{m}\\right)^{2}}{\\frac{s^{2}}{2}}+\\frac{\\left(d_{1}+d_{2}+\\cdots+d_{n}\\right)^{2}}{\\frac{s^{2}}{2}}\n\\end{aligned}\n$$\nby Cauchy-Schwarz inequality.\n\nAssume that $a_{1}+a_{2}+\\cdots+a_{m}<\\frac{s}{2}$. Because $b_{i} \\leq s$, for $i=1,2, \\ldots, m$, we have\n$$\n\\frac{s^{2}}{2}=a_{1} b_{1}+a_{2} b_{2}+\\cdots+a_{m} b_{m} \\leq\\left(a_{1}+a_{2}+\\cdots+a_{m}\\right) s<\\frac{s^{2}}{2},\n$$\nwhich is a contradiction. Therefore $a_{1}+a_{2}+\\cdots+a_{m} \\geq \\frac{s}{2}$ and, similarly, $d_{1}+d_{2}+\\cdots+d_{n} \\geq \\frac{s}{2}$.\n\nNow assume that $a_{1}+a_{2}+\\cdots+a_{m}<s$. There exists a vertical line $L$ crossing the original square and not intersecting any of the white rectangles. Therefore, the sum of the heights of the green rectangles intersecting $L$ is greater than or equal to $s$. Hence, either $a_{1}+a_{2}+\\cdots+a_{m} \\geq s$ or $d_{1}+d_{2}+\\cdots+d_{n} \\geq s$.\n\nWe deduce that\n$$\nS \\geq \\frac{2}{s^{2}}\\left(s^{2}+\\left(\\frac{s}{2}\\right)^{2}\\right)=\\frac{5}{2}\n$$\nIf Turki divides the square into two horizontal rectangles of the same area then $S=\\frac{2}{1}+\\frac{1}{2}=\\frac{5}{2}$. Thus the minimum possible value of $S$ is $\\frac{5}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40174,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\n\\frac{1}{n}=0.a_{1}a_{2}\\ldots\n$$\nSuppose that $n=a_{1}+a_{2}+\\cdots$. Determine all possible values of $n$.",
    "options": [],
    "answer": "8",
    "solution": "Let $n \\geq 2$ be such an integer. Because $n=a_{1}+a_{2}+\\cdots$, there exists a positive integer $k>0$ such that $a_{k} \\neq 0$ and $a_{i}=0$ for all integers $i>k$.\n\nBecause $\\frac{10^{k}}{n}=\\overline{a_{1}a_{2}\\cdots a_{k}}$ is an integer and $\\frac{10^{k-1}}{n}=\\overline{a_{1}a_{2}\\cdots a_{k-1},a_{k}}$ is not an integer, there exist two non-negative integers $a, b \\geq 0$ such that $n=2^{a} \\cdot 5^{b}$, with $\\max\\{a, b\\}=k$. We have three cases:\n\na. If $k=a=b$, then $n=10^{k} \\geq 10$ and $a_{1}+a_{2}+\\cdots+a_{k}=1 \\neq n$, contradiction.\n\nb. If $k=b>a$, then\n$$\n5^{k} \\leq 2^{a} \\cdot 5^{k}=n=a_{1}+a_{2}+\\cdots+a_{k} \\leq 9k.\n$$\nFor $k=2$, we have $5^{2}=25>18=9 \\times 2$. If $k \\geq 2$ and $5^{k}>9k$, then $5^{k+1}>5 \\times 9k=9(k+4k)>9(k+1)$. This implies that $k=1$ and therefore $a=0$, that is $n=5$. But $\\frac{1}{n}=0.2$ and $n \\neq 2$, contradiction.\n\nc. If $k=a>b$, then\n$$\n2^{k} \\leq 2^{k} \\cdot 5^{b}=n=a_{1}+a_{2}+\\cdots+a_{k} \\leq 9k.\n$$\nFor $k=6$, we have $2^{6}=64>54=9 \\times 6$. If $k \\geq 6$ and $2^{k}>9k$, then $2^{k+1}=2^{k}+2^{k}>9k+2^{6}>9(k+1)$. This implies that $k \\leq 5$.\n\nNow, assume that $b \\geq 1$. We have\n$$\n10 \\times 2^{k-1} \\leq n=a_{1}+a_{2}+\\cdots+a_{k} \\leq 9k.\n$$\nFor $k=1$, this is not satisfied. Assume that $10 \\times 2^{k-1}>9k$. We have $10 \\times 2^{k}>9k+10 \\times 2^{k-1} \\geq 9k+10>9(k+1)$, contradiction.\n\nHence, $b=0$ and $n=2^{k}$.\n\nBy checking for $k=1,2,3,4,5$ we find that only $n=2^{3}$ has this property.\n\nHence $n=8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40175,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fatima and Asma are playing the following game. First, Fatima chooses $2013$ pairwise different numbers, called $a_{1}, a_{2}, \\ldots, a_{2013}$. Then, Asma tries to know the value of each number $a_{1}, a_{2}, \\ldots, a_{2013}$. At each time, Asma chooses $1 \\leq i < j \\leq 2013$ and asks Fatima \"What is the set $\\{a_{i}, a_{j}\\}$?\" (For example, if Asma asks what is the set $\\{a_{1}, a_{2}\\}$, and $a_{1} = 17$ and $a_{2} = 13$, Fatima will answer $\\{13, 17\\}$). Find the least number of questions Asma needs to ask, to know the value of all the numbers $a_{1}, a_{2}, \\ldots, a_{2013}$.",
    "options": [],
    "answer": "1342",
    "solution": "Let $n$ be the number of different questions Asma asks to know the values of each number $a_{1}, a_{2}, \\ldots, a_{2013}$. Let us make the following remarks:\n\na. Because Asma needs to know the value of each number $a_{i}$, each number $a_{i}$ must appear at least in one of the sets that Asma chooses to ask about.\n\nb. If Asma chooses a set $\\{a_{i_{1}}, a_{i_{2}}\\}$, she needs to choose another set that contains only one of the numbers $a_{i_{1}}, a_{i_{2}}$, so that she can distinguish between the values of $a_{i_{1}}$ and $a_{i_{2}}$.\n\nc. If Asma asks about the sets $\\{a_{i_{1}}, a_{i_{2}}\\}$ and $\\{a_{i_{2}}, a_{i_{3}}\\}$, for different indices $i_{1}, i_{2}, i_{3}$, she will deduce the value of each number $a_{i_{1}}, a_{i_{2}}, a_{i_{3}}$, by taking the intersection of the two sets and their differences.\n\nd. If for some $k \\geq 3$, Asma asks about the sets $\\{a_{i_{1}}, a_{i_{2}}\\}, \\{a_{i_{2}}, a_{i_{3}}\\}, \\ldots, \\{a_{i_{k-1}}, a_{i_{k}}\\}$, for different indices $i_{1}, i_{2}, \\ldots, i_{k}$, she will deduce the value of each number $a_{i_{1}}, a_{i_{2}}, \\ldots, a_{i_{k}}$ and does not need to ask after this about the set $\\{a_{i_{k}}, a_{i_{1}}\\}$ since she already knows the answer.\n\nFrom now on, we assume, without loss of generality, that Asma never asks about a set that she already knows.\n\nAccording to these remarks, we can make a partition of the set $\\{a_{1}, a_{2}, \\ldots, a_{2013}\\}$ into non-empty disjoint subsets $A_{1}, A_{2}, \\ldots, A_{m}$ such that two numbers $a_{i}, a_{j}$ are in the same subset $A_{k}$ if and only if there exist sets $\\{a_{i_{1}}, a_{i_{2}}\\}, \\{a_{i_{2}}, a_{i_{3}}\\}, \\ldots, \\{a_{i_{l-1}}, a_{i_{l}}\\}$ that Asma has asked about each one of them, and such that $a_{i_{1}} = a_{i}, a_{i_{l}} = a_{j}$. Notice that if $a_{i_{1}}, a_{i_{2}}, \\ldots, a_{i_{l}}$ are different, because of remark (d) they are unique.\n\nFor each $A_{i}$, there are exactly $\\#A_{i} - 1$ sets in $A_{i}$ that Asma has asked about. To see this, fix $a_{i_{0}} \\in A_{i}$, and for each $a_{j} \\in A_{i}$ with $a_{j} \\neq a_{i_{0}}$, associate the unique set $\\{a_{i_{l-1}}, a_{i_{l}}\\}$ in the previous paragraph when $a_{i_{1}} = a_{i_{0}}, a_{i_{l}} = a_{j}$ and $a_{i_{1}}, a_{i_{2}}, \\ldots, a_{i_{l}}$ are different. It is not hard to see from remark (d) that this map is a bijection between $A_{i} \\setminus \\{a_{i_{0}}\\}$ and the set of subsets of $A_{i}$ in the questions of Asma.\n\nBecause each set in the questions of Asma is contained exactly in one of the subsets $A_{1}, A_{2}, \\ldots, A_{m}$, we have\n$$\nn = \\sum_{i=1}^{m} (\\#A_{i} - 1) = 2013 - m.\n$$\nBecause each set $A_{i}$ must contain at least $3$ elements (remarks (a) and (b)),\n$$\nm \\leq \\left\\lfloor \\frac{2013}{3} \\right\\rfloor = 671\n$$\nThis proves that $n \\geq 1342$.\n\nConversely, if Asma asks about the $1342$ sets\n$$\n\\{a_{3i+1}, a_{3i+2}\\}, \\{a_{3i+2}, a_{3i+3}\\}, \\quad \\text{for} \\quad i = 0, 1, \\ldots, 670\n$$\nshe will know the values of all the numbers $a_{1}, a_{2}, \\ldots, a_{2013}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40176,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and let $P$ be a point on $BC$. Points $M$ and $N$ lie on $AB$ and $AC$, respectively, such that $MN$ is not parallel to $BC$ and $AMPN$ is a parallelogram. Line $MN$ meets the circumcircle of $ABC$ at $R$ and $S$. Prove that the circumcircle of triangle $RPS$ is tangent to $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because $BM$ and $NP$ are parallel, and $AN$ and $MP$ are parallel, we have\n$$\n\\frac{QP}{QB} = \\frac{NP}{MB} = \\frac{AM}{MB} = \\frac{CP}{PB},\n$$\nand\n$$\n\\frac{QC}{QP} = \\frac{CN}{PM} = \\frac{CN}{NA} = \\frac{CP}{PB}.\n$$\n\n![](attached_image_1.png)\n\nWe deduce that\n$$\nQP^2 = QB \\cdot QC.\n$$\nBut from the power of point $Q$ with respect to the circumcircle of $ABC$, we have $QB \\cdot QC = QR \\cdot QS$. We deduce that\n$$\nQP^2 = QR \\cdot QS,\n$$\nthat is, line $BC$ is tangent to the circumcircle of triangle $PRS$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40177,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b$ be two nonnegative real numbers and $n$ a positive integer. Prove that\n$$\n\\left(1-2^{-n}\\right)\\left|a^{2^{n}}-b^{2^{n}}\\right| \\geq \\sqrt{a b}\\left|a^{2^{n}-1}-b^{2^{n}-1}\\right|\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First Solution. If $a = b$, the inequality is satisfied. If $a \\neq b$, we can assume without loss of generality that $a > b$. In this case, the inequality can be written\n$$\n\\left(2^{n}-1\\right)\\left(a^{2^{n}}-b^{2^{n}}\\right) \\geq 2^{n} \\sqrt{a b}\\left(a^{2^{n}-1}-b^{2^{n}-1}\\right),\n$$\nor equivalently\n$$\n2^{n}(\\sqrt{a}-\\sqrt{b})\\left(a^{2^{n}-\\frac{1}{2}}+b^{2^{n}-\\frac{1}{2}}\\right) \\geq a^{2^{n}}-b^{2^{n}} .\n$$\nBy dividing in both sides by $\\sqrt{a}-\\sqrt{b}$, the inequality becomes\n$$\n2^{n}\\left(A^{2^{n+1}-1}+B^{2^{n+1}-1}\\right) \\geq (A+B)\\left(A^{2}+B^{2}\\right)\\left(A^{4}+B^{4}\\right) \\cdots\\left(A^{2^{n}}+B^{2^{n}}\\right),\n$$\nwhere $A = \\sqrt{a}$ and $B = \\sqrt{b}$. We prove this last inequality by induction on $n \\geq 1$.\n\nFor $n = 1$, we have\n$$\n\\begin{aligned}\n(A+B)\\left(A^{2}+B^{2}\\right) & = A^{3}+B^{3}+A^{2} B+A B^{2} \\\\\n& \\leq A^{3}+B^{3}+\\frac{2 A^{3}+B^{3}}{3}+\\frac{A^{3}+2 B^{3}}{3} \\\\\n& \\leq 2\\left(A^{3}+B^{3}\\right),\n\\end{aligned}\n$$\nby AM-GM inequality.\n\nAssume the inequality true for $n$. We have\n$$\n\\begin{aligned}\n\\left(A^{2^{n+1}-1}+B^{2^{n+1}-1}\\right)\\left(A^{2^{n+1}}+B^{2^{n+1}}\\right) & \\\\\n= & A^{2^{n+2}-1}+B^{2^{n+2}-1}+A^{2^{n+1}-1} B^{2^{n+1}}+A^{2^{n+1}} B^{2^{n+1}-1} \\\\\n\\leq & A^{2^{n+2}-1}+B^{2^{n+2}-1}+\\frac{\\left(2^{n+1}-1\\right) A^{2^{n+2}-1}+2^{n+1} B^{2^{n+2}-1}}{2^{n+2}-1} \\\\\n& +\\frac{2^{n+1} A^{2^{n+2}-1}+\\left(2^{n+1}-1\\right) B^{2^{n+2}-1}}{2^{n+2}-1} \\\\\n\\leq 2\\left(A^{2^{n+2}-1}+B^{2^{n+2}-1}\\right) &\n\\end{aligned}\n$$\nby AM-GM inequality. We deduce the inequality for $n+1$.\n\n\nSecond Solution. Consider the polynomial\n$$\nP(X) = \\left(1-2^{-n}\\right)\\left(X^{2^{n+1}}-1\\right) - X\\left(X^{2^{n+1}-2}-1\\right) .\n$$\nIts derivative polynomial is\n$$\nP^{\\prime}(X) = \\left(\\left(2^{n+1}-2\\right) X - \\left(2^{n+1}-1\\right)\\right) X^{2^{n+1}-2} + 1 .\n$$\nTherefore, for all $x \\geq 1$, $P^{\\prime}(x) \\geq P^{\\prime}(1) = 0$. Hence, the polynomial function $P$ is increasing on $[1, \\infty[$.\n\nLet $a, b$ be two positive real numbers with $a \\geq b$. Because $\\sqrt{\\frac{a}{b}} \\geq 1$, we have\n$$\n\\begin{aligned}\nb^{2^{n}} P\\left(\\sqrt{\\frac{a}{b}}\\right) & = \\left(1-2^{-n}\\right) b^{2^{n}}\\left(\\left(\\frac{a}{b}\\right)^{2^{n}}-1\\right) - b^{2^{n}} \\sqrt{\\frac{a}{b}}\\left(\\left(\\frac{a}{b}\\right)^{2^{n}-1}-1\\right) \\\\\n& = \\left(1-2^{-n}\\right)\\left(a^{2^{n}}-b^{2^{n}}\\right) - \\sqrt{a b}\\left(a^{2^{n}-1}-b^{2^{n}-1}\\right) \\geq b^{2^{n}} P(1) = 0\n\\end{aligned}\n$$\nwhich proves the inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40178,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and $I$ its incenter. The line $AI$ intersects the side $BC$ at $D$ and the perpendicular bisector of $BC$ at $E$. Let $J$ be the incenter of triangle $CDE$. Prove that triangle $CIJ$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is well known that the bisector of the angle at vertex $A$ and the perpendicular bisector of side $BC$ intersect on the circumcircle of triangle $ABC$. We give here two solutions.\n\nSolution 1. We have\n$$\n\\begin{aligned}\n\\measuredangle JDI & = 180^\\circ - \\measuredangle EDJ = 180^\\circ - \\frac{1}{2} \\measuredangle ADB = 180^\\circ - \\frac{1}{2}(\\measuredangle DAC + \\measuredangle ACB) \\\\\n& = 180^\\circ - \\left(\\frac{1}{4} \\measuredangle BAC + \\frac{1}{2} \\measuredangle ACB\\right)\n\\end{aligned}\n$$\nOn the other hand\n$\\measuredangle ICJ = \\frac{1}{2} \\measuredangle ACB + \\frac{1}{2} \\measuredangle BCE = \\frac{1}{2} \\measuredangle ACB + \\frac{1}{2} \\measuredangle BAE = \\frac{1}{2} \\measuredangle ACB + \\frac{1}{4} \\measuredangle BAC$.\nTherefore, quadrilateral $C I D J$ is cyclic.\n\n![](attached_image_1.png)\n\nWe deduce that\n$$\n\\measuredangle JIC = \\measuredangle JDC = \\measuredangle EDJ = \\measuredangle ICJ,\n$$\nand triangle $CIJ$ is isosceles.\n\n\nSolution 2. It is well known that $EC = EI$. This is because\n$$\n\\measuredangle ICE = \\measuredangle ICD + \\measuredangle DCE = \\measuredangle ACI + \\measuredangle BAE = \\measuredangle ACI + \\measuredangle IAC = \\measuredangle EIC.\n$$\n![](attached_image_2.png)\n\nTherefore, the bisector of angle $\\angle CEI$ is the perpendicular bisector of segment $CI$. But $J$ is on this bisector. Hence, $JC = JI$ and triangle $CIJ$ is isosceles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40179,
    "subject": "Mathematics (Multi-modal)",
    "question": "We are given a lattice and two pebbles $A$ and $B$ that are placed at two lattice points. At each step we are allowed to relocate one of the pebbles to another lattice point with the condition that the distance between pebbles is preserved. Is it possible after finite number of steps to switch positions of the pebbles?",
    "options": [],
    "answer": "No, it is impossible to switch the positions of the pebbles under the given moves.",
    "solution": "Solution suggested by the student Mahdi Al-Shaikh Saleh. Consider the lattice $\\mathbb{Z}^{2}$ and $(x_{a}, y_{a}),(x_{b}, y_{b})$ the coordinates of positions of pebbles $A, B$ respectively. If we relocate one of the pebbles, say $B$, at a new position, $(x_{b}^{\\prime}, y_{b}^{\\prime})$, because $(x_{b}-x_{a})^{2}+(y_{b}-y_{a})^{2}=(x_{b}^{\\prime}-x_{a})^{2}+(y_{b}^{\\prime}-y_{a})^{2}$, by considering this equality in modulo $4$, we deduce that the number of even terms among $x_{b}-x_{a}, y_{b}-y_{a}$ is preserved after each relocation. Whenever both terms are even, one can cancel the common $2$ by considering a sublattice to end up with two terms with at most one even term. In this way, we can consider two cases:\n\n- If only one term among $x_{b}-x_{a}, y_{b}-y_{a}$ is odd, we color the vertices of the lattice in blue and red by alternating the colors between each two consecutive vertices in a horizontal line and two consecutive vertices in a vertical line.\n\n![](attached_image_1.png)\n\n- If both terms $x_{b}-x_{a}, y_{b}-y_{a}$ are odd, we color the vertices of the lattice in blue and red by making the colors of all vertices in a same horizontal line the same and alternating the colors between each two consecutive horizontal lines.\n\n![](attached_image_2.png)\n\nIn both cases, the colors of the vertex positions of pebbles $A$ and $B$ are different, and at each relocation, the colors of their vertex new positions remains unchanged. So, in both cases, it is impossible after finite number of steps to switch positions of the pebbles $A$ and $B$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40180,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exists a positive integer $n$ such that the last digits of $n^{3}$ are $\\square 201320132013$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that $a = 201320132013$ and $10^{12}$ are coprime. By Euler's theorem\n$$\na^{\\phi\\left(10^{12}\\right)} \\equiv 1 \\quad \\bmod \\left(10^{12}\\right)\n$$\nLet $b = a^{\\phi\\left(10^{12}\\right) - 1}$. We have $a b \\equiv 1 \\bmod \\left(10^{12}\\right)$. Because $b$ and $10^{12}$ are coprime, again by Euler's theorem\n$$\na \\equiv a b^{\\phi\\left(10^{12}\\right)} \\equiv b^{\\phi\\left(10^{12}\\right) - 1} \\quad \\bmod \\left(10^{12}\\right) .\n$$\nBut\n$$\n\\phi\\left(10^{12}\\right) - 1 = 2^{13} \\times 5^{11} - 1 \\equiv (-1)^{13} \\times (-1)^{11} - 1 \\equiv 0 \\bmod (3) .\n$$\nTherefore, there exists a positive integer $m$ such that $\\phi\\left(10^{12}\\right) - 1 = 3 m$.\nLet $n = b^{m}$. We have\n$$\nn^{3} = b^{\\phi\\left(10^{12}\\right) - 1} \\equiv 201320132013 \\bmod \\left(10^{12}\\right) .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40181,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 3$ be an odd positive integer not divisible by $3$. Determine if it is possible to form an $n \\times n$ array of numbers such that\n\na. The set of the numbers in each row is a permutation of $0, 1, \\ldots, n-1$; the set of the numbers in each column is a permutation of $0, 1, \\ldots, n-1$;\n\nb. The board is totally non-symmetric: for $1 \\leq i < j \\leq n$ and $1 \\leq i' < j' \\leq n$, if $(i, j) \\neq (i', j')$ then $(a_{i, j}, a_{j, i}) \\neq (a_{i', j'}, a_{j', i'})$, where $a_{i, j}$ denotes the entry in the $i^{\\text{th}}$ row and $j^{\\text{th}}$ column.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $1 \\leq i, j \\leq n$, choose $a_{i, j}$ equal to the remainder of $3i + j$ when divided by $n$. Clearly, $0 \\leq a_{i, j} \\leq n-1$, for $1 \\leq i, j \\leq n$.\n\nIf we fix a row $1 \\leq i \\leq n$, $a_{i, j_1} = a_{i, j_2}$, for $1 \\leq j_1, j_2 \\leq n$, implies $3i + j_1 \\equiv 3i + j_2 \\pmod{n}$, that is $j_1 \\equiv j_2 \\pmod{n}$ and therefore $j_1 = j_2$.\n\nHence, $a_{i, 1}, \\ldots, a_{i, n}$ is a permutation of $0, 1, \\ldots, n-1$.\n\nIf we fix a column $1 \\leq j \\leq n$, $a_{i_1, j} = a_{i_2, j}$, for $1 \\leq i_1, i_2 \\leq n$, implies $3i_1 + j \\equiv 3i_2 + j \\pmod{n}$, that is $3i_1 \\equiv 3i_2 \\pmod{n}$. But $n$ is not divisible by $3$. Therefore $i_1 = i_2$. This proves that $a_{1, j}, \\ldots, a_{n, j}$ is a permutation of $0, 1, \\ldots, n-1$.\n\nNow, let $1 \\leq i < j \\leq n$ and $1 \\leq i' < j' \\leq n$ such that $(a_{i, j}, a_{j, i}) = (a_{i', j'}, a_{j', i'})$ this implies that $3i + j \\equiv 3i' + j' \\pmod{n}$ and $3j + i \\equiv 3j' + i' \\pmod{n}$. This is equivalent to $8i \\equiv 8i' \\pmod{n}$ and $8j \\equiv 8j' \\pmod{n}$. But $n$ is an odd number. Therefore, $(i, j) = (i', j')$. Hence, the board is totally non-symmetric.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40182,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, $D$ the midpoint of side $BC$ and $E$ the intersection point of the bisector of angle $\\angle BAC$ with side $BC$. The perpendicular bisector of $AE$ intersects the bisectors of angles $\\angle CBA$ and $\\angle CDA$ at $M$ and $N$, respectively. The bisectors of angles $\\angle CBA$ and $\\angle CDA$ intersect at $P$. Prove that points $A, M, N, P$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The bisector of angle $\\angle EBA$ and the perpendicular bisector of side $EA$, both bisect the $\\operatorname{arc}\\overparen{EA}$ of the circumcircle of triangle $ABE$ opposite to the vertex $B$. Therefore quadrilateral $ABEM$ is cyclic and hence\n$$\n\\angle EAM = \\angle EBM\n$$\nThe segment defined by the two intersection points of the perpendicular bisector of $AE$ with the circumcircle of triangle $AED$ is a diameter and intersects the bisector of the angle $\\angle ADE$ on this circumcircle. But the bisectors of angles $\\angle CDA$ and $\\angle ADE$ are perpendicular. Therefore quadrilateral $AEDN$ is cyclic and hence\n$$\n\\angle CDN = \\angle EAN.\n$$\n![](attached_image_1.png)\nWe deduce that\n$$\n\\angle MAN = \\angle EAN - \\angle EAM = \\angle CDN - \\angle EBM = \\angle MPN,\n$$\nfrom triangle $BDP$. This proves that points $A, M, N, P$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40183,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $\\angle B \\leq \\angle C$, $I$ its incenter and $D$ the intersection point of line $AI$ with side $BC$. Let $M$ and $N$ be points on sides $BA$ and $CA$, respectively, such that $BM = BD$ and $CN = CD$. The circumcircle of triangle $CMN$ intersects again line $BC$ at $P$. Prove that quadrilateral $DIM P$ is cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the bisector theorem we have\n$$\n\\frac{MB}{AB} = \\frac{DB}{AB} = \\frac{DC}{AC} = \\frac{NC}{AC}\n$$\nWe deduce that segments $MN$ and $BC$ are parallel.\n\n![](attached_image_1.png)\n\nBecause $CNMP$ is a cyclic trapezoid, it is isosceles. Hence\n$$\n\\angle DPM = \\angle ACB\n$$\nBecause $BD = BM$ and $I$ is on the bisector of $\\angle DBM$, the quadrilateral $BDIM$ is a kite. We deduce that\n$$\n\\begin{aligned}\n\\angle MID & = 360^\\circ - \\angle DBM - 2 \\angle ADB \\\\\n& = 360^\\circ - \\angle CBA - 2\\left(\\angle ACB + \\frac{1}{2} \\angle BAC\\right) \\\\\n& = 180^\\circ - \\angle ACB = 180^\\circ - \\angle DPM.\n\\end{aligned}\n$$\nThis proves that quadrilateral $DIM P$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40184,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles $\\omega_{1}$ and $\\omega_{2}$, with center $O_{1}$ and $O_{2}$ respectively, meet at points $A$ and $B$. Let $X$ and $Y$ be points on $\\omega_{1}$. Lines $X A$ and $Y A$ meet $\\omega_{2}$ at $Z$ and $W$, respectively, such that $A$ lies between $X$ and $Z$ and between $Y$ and $W$. Let $M$ be the midpoint of $O_{1} O_{2}$, $S$ be the midpoint of $X A$ and $T$ be the midpoint of $W A$. Prove that $M S = M T$ if and only if $X, Y, Z$ and $W$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution suggested by the Student Alzubair Habibullah. Because $M$ is the midpoint of $O_{1} O_{2}$, we have from Apollonius' theorem\n$$\n4 M S^{2} = 2 S O_{1}^{2} + 2 S O_{2}^{2} - O_{1} O_{2}^{2}\n$$\nBecause $S$ is the midpoint of segment $A X$, segment $O_{1} S$ is perpendicular to segment $A X$ and therefore $S O_{1}^{2} = O_{1} A^{2} - A S^{2}$.\nOn the other hand, we have from the power of point $S$ with respect to circle $\\omega_{2}$, $S O_{2}^{2} = S A \\cdot S Z + O_{2} A^{2} = S A^{2} + S A \\cdot A Z + O_{2} A^{2}$.\nWe deduce that\n$$\n4 M S^{2} = 2 O_{1} A^{2} + 2 O_{2} A^{2} + 2 S A \\cdot A Z - O_{1} O_{2}^{2} = 2 O_{1} A^{2} + 2 O_{2} A^{2} + X A \\cdot A Z - O_{1} O_{2}^{2}\n$$\n![](attached_image_1.png)\nWe deduce in a similar way that\n$$\n4 M T^{2} = 2 O_{2} A^{2} + 2 O_{1} A^{2} + W A \\cdot A Y - O_{2} O_{1}^{2}\n$$\nTherefore\n$$\n4\\left(M S^{2} - M T^{2}\\right) = X A \\cdot A Z - W A \\cdot A Y\n$$\nThis proves that $M S = M T$ if and only if $X A \\cdot A Z = W A \\cdot A Y$, which is equivalent to saying that $X, Y, Z$ and $W$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40185,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $X$ be a set of rational numbers satisfying the following two conditions:\n(a) The set $X$ contains at least two elements,\n(b) For any $x, y$ in $X$, if $x \\neq y$ then there exists $z$ in $X$ such that either $\\left|\\frac{x-z}{y-z}\\right|=2$ or $\\left|\\frac{y-z}{x-z}\\right|=2$.\nProve that $X$ contains infinitely many elements.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The condition (b) tells that if $x \\neq y$ are in $X$, then at least one of the four numbers\n$z_1 = x - (y - x), \\quad z_2 = x + \\frac{1}{3}(y - x), \\quad z_3 = x + \\frac{2}{3}(y - x), \\quad z_4 = x + 2(y - x)$,\n![](attached_image_1.png)\nis in $X$.\n\nAssume that $X$ is finite and let $x_0$ be the smallest element of $X$. By condition (a), the set $X \\backslash \\{x_0\\}$ is nonempty. Consider a map\n$$\nf: X \\backslash \\{x_0\\} \\rightarrow X \\backslash \\{x_0\\}\n$$\nwhich satisfies, for $y \\in X \\backslash \\{x_0\\}$, the condition\n$$\nf(y) \\in \\{x_0 + \\frac{1}{3}(y - x_0), x_0 + \\frac{2}{3}(y - x_0), x_0 + 2(y - x_0)\\} \\cap X .\n$$\nNotice that $f(y)$ is defined for all $y \\in X \\backslash \\{x_0\\}$ by condition (b) since $x_0 - (y - x_0) < x_0$ is not in $X$.\n\nNow, fix an element $y_0 \\in X \\backslash \\{x_0\\}$ and define by induction $a_0 = y_0 - x_0$ and $a_{n+1} = f(x_0 + a_n) - x_0$ for all $n \\in \\mathbb{N}_0$. Clearly, the sequence ($a_n$) is well defined and positive and we have, for each $n \\in \\mathbb{N}_0$, either\n$$\na_{n+1} = \\frac{1}{3} a_n, \\quad \\text{or} \\quad a_{n+1} = \\frac{2}{3} a_n, \\quad \\text{or} \\quad a_{n+1} = 2 a_n .\n$$\nTherefore, for each positive integers $m, n$, there exist non-negative integers $p, q$ with $p + q > 0$ such that\n$$\na_{m+n} = \\frac{2^p}{3^q} a_m .\n$$\nBecause $X \\backslash \\{x_0\\}$ is finite, there exist two positive integers $m_0, n_0$ such that $a_{m_0 + n_0} = a_{m_0}$. Hence the corresponding two non-negative integers $p_0, q_0$ with $p_0 + q_0 > 0$ satisfy\n$$\n1 = \\frac{2^{p_0}}{3^{q_0}},\n$$\nwhich is impossible.\n\nThis proves that $X$ contains infinitely many elements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40186,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ be an injective function such that $f(1)=2$, $f(2)=4$ and\n$$\nf(f(m)+f(n))=f(f(m))+f(n)\n$$\nfor all $m, n \\in \\mathbb{N}$. Prove that $f(n)=n+2$ for all $n \\geq 2$.",
    "options": [],
    "answer": "f(n) = n + 2 for all n ≥ 2",
    "solution": "Taking $n=1$, we obtain $f(f(m)+2)=f(f(m))+2$, for all $m \\in \\mathbb{N}$. Taking $m=1$, we obtain $f(2+f(n))=4+f(n)$, for all $n \\in \\mathbb{N}$. Therefore\n$$\nf(f(n))=f(n)+2, \\quad \\text{for all } n \\in \\mathbb{N} .\n$$\nBecause $f(2)=f(f(1))=f(1)+2=4$, we prove by induction that\n$$\nf(n)=n+2\n$$\nfor all positive even integer $n \\geq 2$.\nBecause the set $\\{f(1)\\} \\cup \\{f(2), f(4), f(6), \\ldots\\}$ contains all even positive integers and because $f$ is an injective function, the image by $f$ of any odd number $n \\geq 3$ is an odd number.\nLet\n$$\nk=\\min \\{f(n) \\mid f(n) \\text{ is an odd integer }\\}\n$$\nand let $f\\left(n_{0}\\right)=k$ for some odd positive number $n_{0} \\geq 3$. We prove by induction that\n$$\nf(n)=n+2, \\quad \\text{for any odd number } n \\geq k .\n$$\nIndeed, $f(k)=f\\left(f\\left(n_{0}\\right)\\right)=f\\left(n_{0}\\right)+2=k+2$. Assume that $f(n)=n+2$, for some odd number $n \\geq k$. We deduce that $f(n+2)=f(f(n))=f(n)+2= (n+2)+2$.\nClearly, $n_{0}<k$, otherwise $k=f\\left(n_{0}\\right)=n_{0}+2 \\geq k+2$, which is impossible. Therefore,\n$$\n\\left\\{f\\left(n_{0}\\right), f(k), f(k+2), f(k+4), \\ldots\\right\\}=\\{k, k+2, k+4, k+6, \\ldots\\}\n$$\ncovers all possible odd values taken by $f$. Because, $f$ is injective, and $f(3), f(5) \\geq k$ are odd integers, we deduce that $n_{0}=3, k=5$ and\n$$\nf(n)=n+2\n$$\nfor all positive odd integer $n \\geq 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40187,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute triangle, with $\\widehat{A} > \\widehat{B} \\geq \\widehat{C}$. Let $D$, $E$ and $F$ be the tangency points between the incircle of the triangle and sides $BC$, $CA$, $AB$, respectively. Let $J$ be a point on $(BD)$, $K$ a point on $(DC)$, $L$ a point on $(EC)$ and $M$ a point on $(FB)$, such that\n$$\nAF = FM = JD = DK = LE = EA\n$$\nLet $P$ be the intersection point between $AJ$ and $KM$ and let $Q$ be the intersection point between $AK$ and $JL$. Prove that $PJKQ$ is cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $I$ be the incenter of triangle $ABC$. Because $ID = IE = IF$, $AF = FM = JD = DK = LE = EA$, and the angles at $D$, $E$, and $F$ are right, we have by Pythagoras $IA = IM = IJ = IK = IL$. Therefore, the pentagon $AMJKL$ is cyclic.\n\n![](attached_image_1.png)\n\nOn the other hand, $AM = 2AF = 2EA = LA$. We deduce for the circumcircle of the pentagon $AMJKL$ that $\\overparen{AM} = \\overparen{LA}$. This implies that $\\measuredangle QKP = \\measuredangle QJP$, and therefore the quadrilateral $PJKQ$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40188,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{1} \\geq a_{2} \\geq \\cdots \\geq a_{n}>0$ be real numbers. Prove that\n$$\na_{1} a_{2}\\left(a_{1}-a_{2}\\right)+a_{2} a_{3}\\left(a_{2}-a_{3}\\right)+\\cdots+a_{n-1} a_{n}\\left(a_{n-1}-a_{n}\\right) \\geq a_{1} a_{n}\\left(a_{1}-a_{n}\\right) .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First solution. Assume that there exists $1 \\leq i_{0}<n$ such that $a_{i_{0}}= a_{i_{0}+1}$. The sum of the two terms in the left hand side $a_{i_{0}-1} a_{i_{0}}\\left(a_{i_{0}-1}-a_{i_{0}}\\right)+ a_{i_{0}} a_{i_{0}+1}\\left(a_{i_{0}}-a_{i_{0}+1}\\right)$ is equal to $a_{i_{0}-1} a_{i_{0}+1}\\left(a_{i_{0}-1}-a_{i_{0}-1}\\right)$. This means that we can skip $i_{0}$ in the left hand side. This means that it is enough to prove this inequality when $a_{1}>a_{2}>\\cdots>a_{n}>0$. In this case we have\n$$\n\\begin{aligned}\n& a_{1} a_{2}\\left(a_{1}-a_{2}\\right)+a_{2} a_{3}\\left(a_{2}-a_{3}\\right)+\\cdots+a_{n-1} a_{n}\\left(a_{n-1}-a_{n}\\right) \\\\\n& \\quad=\\frac{\\left(a_{1}-a_{2}\\right)^{2}}{\\frac{1}{a_{1}}-\\frac{1}{a_{2}}}+\\frac{\\left(a_{2}-a_{3}\\right)^{2}}{\\frac{1}{a_{2}}-\\frac{1}{a_{3}}}+\\cdots+\\frac{\\left(a_{n-1}-a_{n}\\right)^{2}}{\\frac{1}{a_{n-1}}-\\frac{1}{a_{n}}} \\\\\n& \\quad \\geq \\frac{\\left(\\left(a_{1}-a_{2}\\right)+\\left(a_{2}-a_{3}\\right)+\\cdots+\\left(a_{n-1}-a_{n}\\right)\\right)^{2}}{\\left(\\frac{1}{a_{1}}-\\frac{1}{a_{2}}\\right)+\\left(\\frac{1}{a_{2}}-\\frac{1}{a_{3}}\\right)+\\cdots+\\left(\\frac{1}{a_{n-1}}-\\frac{1}{a_{n}}\\right)}=a_{1} a_{n}\\left(a_{1}-a_{n}\\right),\n\\end{aligned}\n$$\nby Cauchy-Schwarz inequality.\nThe equality holds here when\n$$\n\\frac{a_{1}-a_{2}}{\\frac{1}{a_{1}}-\\frac{1}{a_{2}}}=\\frac{a_{2}-a_{3}}{\\frac{1}{a_{2}}-\\frac{1}{a_{3}}}=\\cdots=\\frac{a_{n-1}-a_{n}}{\\frac{1}{a_{n-1}}-\\frac{1}{a_{n}}},\n$$\nwhich is equivalent to\n$$\na_{1} a_{2}=a_{2} a_{3}=\\cdots=a_{n-1} a_{n} .\n$$\nThis implies that $n=2$, otherwise $a_{1}=a_{3}$ which is rejected. Therefore, in the general case, the equality holds when at least $n-2$ of the $n-1$ inequalities $a_{1} \\geq a_{2} \\geq \\cdots \\geq a_{n}$ are in fact equalities.\n\n\nSecond solution. We prove this inequality by induction on $n \\geq 2$. For $n=2$, clearly both sides are the same. Assume this inequality true for $n$. Let $a_{1} \\geq a_{2} \\geq \\ldots \\geq a_{n} \\geq a_{n+1}>0$. We have\n$$\n\\begin{aligned}\n& a_{1} a_{2}\\left(a_{1}-a_{2}\\right)+\\cdots+a_{n-1} a_{n}\\left(a_{n-1}-a_{n}\\right)+a_{n} a_{n+1}\\left(a_{n}-a_{n+1}\\right) \\\\\n& \\quad \\geq a_{1} a_{n}\\left(a_{1}-a_{n}\\right)+a_{n} a_{n+1}\\left(a_{n}-a_{n+1}\\right) \\geq a_{1} a_{n+1}\\left(a_{1}-a_{n+1}\\right),\n\\end{aligned}\n$$\nsince\n$$\n\\begin{aligned}\na_{1} a_{n}\\left(a_{1}-a_{n}\\right) & +a_{n} a_{n+1}\\left(a_{n}-a_{n+1}\\right)-a_{1} a_{n+1}\\left(a_{1}-a_{n+1}\\right) \\\\\n= & \\left(a_{1}-a_{n}\\right)\\left(a_{n}-a_{n+1}\\right)\\left(a_{1}-a_{n+1}\\right) \\geq 0\n\\end{aligned}\n$$\nThe equality holds here when $a_{1}=a_{n}$ or when $a_{n}=a_{n+1}$ and there is equality in the hypothesis of induction. Therefore, in the general case, the equality holds when at least $n-2$ of the $n-1$ inequalities $a_{1} \\geq a_{2} \\geq \\cdots \\geq a_{n}$ are in fact equalities.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40189,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ such that\n$$\nf(n+1) > \\frac{f(n) + f(f(n))}{2}\n$$\nfor all $n \\in \\mathbb{N}$, where $\\mathbb{N}$ is the set of strictly positive integers.",
    "options": [],
    "answer": "All functions f: N -> N of the form: (i) f(n) = n for all n; (ii) f(n) = n + 1 for all n; (iii) for some integer m > 1, f(n) = n for n < m and f(n) = n + 1 for n >= m.",
    "solution": "It is clear that $f(n) \\geq 1$ for all $n \\geq 1$. Assume that $f(n) \\geq m$ for all $n \\geq m$, for some $m \\geq 1$. Let $n \\geq m+1$. Because $n-1 \\geq m$, we have $f(n-1) \\geq m$ and therefore $f(f(n-1)) \\geq m$. We deduce that $f(n) > \\frac{f(n-1) + f(f(n-1))}{2} \\geq m$. Hence, $f(n) \\geq m+1$. In particular, $f(n) \\geq n$ for all $n \\in \\mathbb{N}$.\n\nAssume there exists $m$ for which $f(m) \\geq m+2$. Then $f(f(m)) \\geq f(m+2) \\geq m+2$, and $f(m+1) > \\frac{f(m) + f(f(m))}{2} \\geq m+2$. This gives $f(m+1) \\geq (m+1)+2$. It follows by induction that $f(n) \\geq n+2$ for all $n \\geq m$. Therefore, for all $n \\geq m$, $f(n+1) - f(n) > f(f(n)) - f(n+1) \\geq f(n+2) - f(n+1)$, which means that the sequence $f(n+1) - f(n)$ becomes decreasing from $m$ and this contradicts the fact $f(n) \\geq n$ for all $n \\in \\mathbb{N}$. We deduce that\n$$\nn \\leq f(n) \\leq n+1 \\text{ for all } n \\in \\mathbb{N}\n$$\nIt is clear that the identity function is a solution. Assume that there exists $m \\in \\mathbb{N}$ such that $f(m) = m+1$. We have $m+2 \\geq f(m+1) > f(m) + f(f(m)) - f(m+1) = m+1$. This proves that $f(m+1) = (m+1)+1$ and by induction $f(n) = n+1$, for all $n \\geq m$. After checking, we conclude that the solutions are the identity function, $f(n) = n+1$ for all $n \\in \\mathbb{N}$, and all the functions which can be written\n$$\nf(n) = \\begin{cases} n & \\text{ if } n < m \\\\ n+1 & \\text{ if } n \\geq m \\end{cases}\n$$\nfor some $m > 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40190,
    "subject": "Mathematics (Multi-modal)",
    "question": "Majid wants to color the cells of an $n \\times n$ chessboard into white and black so that each $2 \\times 2$ subsquare contains two white cells and two black cells. In how many ways can Majid color this $n \\times n$ chessboard?",
    "options": [],
    "answer": "2^{n+1} - 2",
    "solution": "We have three possible cases.\n\nThe first case is when there are no two adjacent cells with the same color. In this case there are exactly two possible ways to color this $n \\times n$ chessboard depending on the color of the first cell on the top left of the chessboard:\n\n![](attached_image_1.png)\n![](attached_image_2.png)\n\nThe second case is when there are two adjacent cells with the same color on the same row.\n\n![](attached_image_3.png)\n![](attached_image_4.png)\n\nIn this case the two adjacent cells in the previous row and the two adjacent cells in the next row must be of the other color.\n\n![](attached_image_5.png)\n![](attached_image_6.png)\n\nIf the previous cell or the next cell in this row is of the same color, the adjacent cells in the previous or the next row must be of the other color. If the previous cell or the next cell in this row is of the other color, the adjacent cells in the previous or the next row must be of this color.\n\n![](attached_image_7.png)\n![](attached_image_8.png)\n\nTherefore, once fixing the colors of the cells of this row, the colors of the cells of the other rows are fixed by just alternating vertically the colors of adjacent cells.\n\n![](attached_image_9.png)\n\nHence, in this case any coloring of the first row with at least two adjacent cells with the same color will fix the coloring of the cells of the whole chessboard. There are precisely $2^{n}-2$ possible such coloring corresponding to the number of subsets of $\\{1,2, \\ldots, n\\}$ (the positions of the black cells in the row) minus the two alternating coloring of the first case.\n\nThe third case is when there are two adjacent cells with the same color on the same column. This case is similar to the previous one and in the same way we obtain $2^{n}-2$ possible such coloring.\n\nNotice that the second and the third cases cannot occur at the same time. Therefore, there are precisely $2+2\\left(2^{n}-2\\right)=2^{n+1}-2$ possible coloring of the $n \\times n$ chessboard.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40191,
    "subject": "Mathematics (Multi-modal)",
    "question": "Aws plays a solitaire game on a fifty-two card deck: whenever two cards of the same color are adjacent, he can remove them. Aws wins the game if he removes all the cards. If Aws starts with the cards in a random order, what is the probability for him to win?",
    "options": [],
    "answer": "binom(26,13)^2 / binom(52,26)",
    "solution": "Let us consider the positions $1,2, \\ldots, 52$, from left to right, of the fifty-two cards at the beginning of the game. Whenever Aws removes two adjacent cards, we subtract $2$ from the positions of all the cards on the right of these adjacent removed cards.\n\nWe notice that, at each operation, if a card is not removed, even if its position changes, it will keep the same parity. We notice also that whenever two adjacent cards of the same color are removed, the positions of these two cards have different parity.\n\nDefine $R$ to be the number of red cards with even positions at the beginning subtracted from the number of red cards with odd positions at the beginning. Clearly, $R$ is an invariant under Aws' operations. Hence, if at the beginning, $R \\neq 0$, Aws can never remove all his cards.\n\nNow assume that $R=0$ and that Aws has started playing. If there are no more red cards, then Aws is left with only black cards, so he can keep removing adjacent cards until he ends up with no card. If there is at least one red card, because $R=0$, there is at least one red card with even position $e_{1}$ and one red card with odd position $o_{1}$. Assume, without loss of generality, that $e_{1}<o_{1}$.\n\nConsider $o=\\min \\mathcal{O}$, where $\\mathcal{O}$ is the set of odd positions $o'$ of red cards with $o'>e_{1}$. Number $o$ exists since $o_{1} \\in \\mathcal{O}$. Consider $e=\\max \\mathcal{E}$, where $\\mathcal{E}$ is the set of even positions $e'$ of red cards with $o>e'$. Number $e$ exists since $e_{1} \\in \\mathcal{E}$. If $o=e+1$, then there are two adjacent red cards that Aws can remove. If $o>e+1$ then $o \\geq e+3$ and the positions between $e$ and $o$ are all occupied by adjacent black cards. In this case, Aws can remove adjacent black cards. This proves that if $R=0$, Aws can keep removing cards until he removes all cards.\n\nHence, Aws can win if and only if $R=0$. In this situation, there must be $13$ red cards with even positions and $13$ red cards with odd positions. There are precisely $\\binom{26}{13}^{2} \\cdot 26!^{2}$ possible such starting positions and the probability for Aws to win is\n\n$$\n\\frac{\\binom{26}{13}^{2}}{\\binom{52}{26}}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40192,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with $\\angle A < \\angle B \\leq \\angle C$, and $O$ its circumcenter. The perpendicular bisector of side $AB$ intersects side $AC$ at $D$. The perpendicular bisector of side $AC$ intersects side $AB$ at $E$. Express the angles of triangle $DEO$ in terms of the angles of triangle $ABC$.",
    "options": [],
    "answer": "∠ODE = 90° − ∠ACB, ∠DEO = 90° − ∠CBA, ∠EOD = ∠CBA + ∠ACB",
    "solution": "Let $M$ be the midpoint of $AC$. Because $\\measuredangle COM = \\frac{1}{2} \\measuredangle COA = \\measuredangle CBE$, the quadrilateral $BCOE$ is cyclic.\n\n![](attached_image_1.png)\n\nLet $N$ be the midpoint of $AB$. Because $\\measuredangle NOB = \\frac{1}{2} \\measuredangle AOB = \\measuredangle DCB$, the quadrilateral $BCDO$ is cyclic. Therefore, the pentagon $BCDOE$ is cyclic and we have:\n$$\n\\measuredangle ODE = \\measuredangle OBE = 90^{\\circ} - \\measuredangle ACB, \\quad \\measuredangle DEO = \\measuredangle DCO = 90^{\\circ} - \\measuredangle CBA,\n$$\nand\n$$\n\\measuredangle EOD = 180^{\\circ} - \\measuredangle ODE - \\measuredangle DEO = \\measuredangle CBA + \\measuredangle ACB.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40193,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $A_{1}, B_{1}, C_{1}$ lie on the sides $BC$, $AC$ and $AB$ of a triangle $ABC$, respectively, such that $AB_{1} - AC_{1} = CA_{1} - CB_{1} = BC_{1} - BA_{1}$. Let $I_{A}, I_{B}, I_{C}$ be the incenters of triangles $AB_{1}C_{1}$, $A_{1}BC_{1}$ and $A_{1}B_{1}C$ respectively. Prove that the circumcenter of triangle $I_{A}I_{B}I_{C}$ is the incenter of triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If we orient positively the line $AB$ from $A$ to $B$, the line $BC$ from $B$ to $C$ and the line $CA$ from $C$ to $A$, we get\n$$\n\\overline{B_{1}A} - \\overline{AC_{1}} = \\overline{A_{1}C} - \\overline{CB_{1}} = \\overline{C_{1}B} - \\overline{BA_{1}}.\n$$\nLet $A_{0}, B_{0}, C_{0}$ be the intouch points of the incircle of triangle $ABC$ and the sides $BC$, $CA$, $AB$, respectively. We have\n$$\n\\overline{B_{1}A} - \\overline{AC_{1}} = \\overline{B_{1}B_{0}} + \\overline{B_{0}A} - \\overline{AC_{0}} - \\overline{C_{0}C_{1}} = \\overline{B_{1}B_{0}} + \\overline{C_{1}C_{0}},\n$$\nand similarly $\\overline{A_{1}C} - \\overline{CB_{1}} = \\overline{A_{1}A_{0}} + \\overline{B_{1}B_{0}}$, and $\\overline{C_{1}B} - \\overline{BA_{1}} = \\overline{C_{1}C_{0}} + \\overline{A_{1}A_{0}}$. We deduce that\n$$\n\\overline{B_{1}B_{0}} + \\overline{C_{1}C_{0}} = \\overline{A_{1}A_{0}} + \\overline{B_{1}B_{0}} = \\overline{C_{1}C_{0}} + \\overline{A_{1}A_{0}},\n$$\nwhich is equivalent to\n$$\n\\overline{A_{0}A_{1}} = \\overline{B_{0}B_{1}} = \\overline{C_{0}C_{1}}.\n$$\nBecause $IA_{0} = IB_{0} = IC_{0}$ and $\\angle A_{1}A_{0}I = \\angle B_{1}B_{0}I = \\angle C_{1}C_{0}I$, the triangles $IA_{0}A_{1}$, $IB_{0}B_{1}$ and $IC_{0}C_{1}$ are congruent and $I$ is the circumcenter of triangle $A_{1}B_{1}C_{1}$.\n\n![](attached_image_1.png)\n\nBecause $\\angle AB_{0}I = \\angle BC_{0}I = 90^{\\circ}$, quadrilateral $AC_{0}IB_{0}$ is cyclic. Because $\\angle B_{0}IB_{1} = \\angle C_{0}IC_{1}$, we have $\\angle B_{1}IC_{1} = \\angle B_{0}IC_{0} = 180^{\\circ} - \\angle B_{1}AC_{1}$. We deduce that quadrilateral $AC_{1}IB_{1}$ is cyclic.\nBecause $I$ is the midpoint of the minor $\\operatorname{arc} \\widehat{C_{1}B_{1}}$ of the circumcircle of triangle $AC_{1}B_{1}$ not containing $A$, we deduce that $IB_{1} = IC_{1} = II_{1}$. Therefore $I_{A}$ is on the circumcircle of triangle $A_{1}B_{1}C_{1}$ of center $I$.\nSimilarly, one can prove that $I_{B}, I_{C}$ are on the circumcircle of triangle $A_{1}B_{1}C_{1}$. Hence the circumcenter of triangle $I_{A}I_{B}I_{C}$ is the incenter of triangle $ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40194,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that\n$$\n3^{n} + 4^{n} + \\cdots + (n+2)^{n} = (n+3)^{n}.\n$$",
    "options": [],
    "answer": "n = 2, 3",
    "solution": "Notice first that\n$$\n3^{2} + 4^{2} = 5^{2}, \\quad \\text{and} \\quad 3^{3} + 4^{3} + 5^{3} = 6^{3}.\n$$\nHowever\n$$\n3^{4} + 4^{4} + 5^{4} + 6^{4} = 2258 < 2401 = 7^{4}\n$$\nAssume that\n$$\n3^{n} + 4^{n} + \\cdots + (n+2)^{n} < (n+3)^{n}\n$$\nfor some $n \\geq 4$. We have\n$$\n\\begin{aligned}\n3^{n+1} + 4^{n+1} + & \\cdots + (n+2)^{n+1} + (n+3)^{n+1} \\\\\n& < (n+2)\\left(3^{n} + 4^{n} + \\cdots + (n+2)^{n}\\right) + (n+3)^{n+1} \\\\\n& < (n+2)(n+3)^{n} + (n+3)^{n+1} < 2(n+3)^{n+1}\n\\end{aligned}\n$$\nFor $n=4$, we have $(n+2)(n+3)^{n} + (n+3)^{n+1} = 31213 < 32768 = (n+4)^{n+1}$. For $n \\geq 5$, we have\n$$\n\\begin{aligned}\n\\left(\\frac{n+4}{n+3}\\right)^{n+1} &= \\left(1 + \\frac{1}{n+3}\\right)^{n+1} > 1 + \\frac{n+1}{n+3} + \\frac{n(n+1)}{2(n+3)^{2}} \\\\\n& > 2 + \\frac{(n-5)(n+2) + 2}{2(n+3)^{2}} > 2\n\\end{aligned}\n$$\nTherefore, $3^{n+1} + 4^{n+1} + \\cdots + (n+2)^{n+1} + (n+3)^{n+1} < (n+4)^{n+1}$.\nHence, we have proved by induction that $3^{n} + 4^{n} + \\cdots + (n+2)^{n} < (n+3)^{n}$ for all $n \\geq 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40195,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. Circle $\\Omega$ passes through points $B$ and $C$. Circle $\\omega$ is tangent internally to $\\Omega$ and also to sides $AB$ and $AC$ at $T$, $P$, and $Q$, respectively. Let $M$ be midpoint of $\\operatorname{arc} \\overparen{BC}$ (containing $T$) of $\\Omega$. Prove that lines $PQ$, $BC$, and $MT$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $R$ be the intersection point of lines $PQ$ and $BC$, and $S$ the intersection point of lines $MT$ and $BC$.\n\n![](attached_image_1.png)\n\nBecause circle $\\omega$ is tangent to both sides $AB$ and $AC$ at $P$ and $Q$, we have $AP = AQ$. Applying Menelaus' theorem to triangle $ABC$ and line $PQ$ we obtain\n$$\n\\frac{\\overline{BR}}{\\overline{RC}} = -\\frac{BP}{PA} \\cdot \\frac{AQ}{QC} = -\\frac{BP}{QC}.\n$$\nBecause $M$ is the midpoint of arc $\\widehat{BC}$ of circle $\\Omega$ containing $T$, line $MT$ is the external bisector of angle $\\angle BTC$. We deduce that\n$$\n\\frac{\\overline{BS}}{\\overline{SC}} = -\\frac{TB}{TC}\n$$\nTo prove that $S = R$, we have to prove that $\\frac{\\overline{BS}}{\\overline{SC}} = \\frac{\\overline{BR}}{\\overline{RC}}$, which is equivalent to proving that\n$$\n\\frac{BP}{QC} = \\frac{TB}{TC}.\n$$\nLet $B'$ and $C'$ be the second intersection points of lines $TB$ and $TC$ with circle $\\omega$, respectively. Because circles $\\Omega$ and $\\omega$ are tangent at $T$, segments $BC$ and $B'C'$ are parallel, and therefore\n$$\n\\frac{TB}{TC} = \\frac{BB'}{CC'}.\n$$\nBecause lines $AB$ and $AC$ are tangent to $\\omega$ at $P$ and $Q$, respectively, we have from the power of points $B$ and $C$ with respect to circle $\\omega$\n$$\n\\left(\\frac{BP}{CQ}\\right)^2 = \\frac{BT \\cdot BB'}{CT \\cdot CC'} = \\left(\\frac{BT}{CT}\\right)^2.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40196,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and $D$ a point on the side $BC$. Point $E$ is the symmetric of $D$ with respect to $AB$. Point $F$ is the symmetric of $E$ with respect to $AC$. Point $P$ is the intersection of line $DF$ with line $AC$. Prove that the quadrilateral $AEDP$ is cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\alpha = \\angle BAC$ and $\\theta = \\angle BAD$. Because $E$ is the symmetric of $D$ with respect to $AB$, we have $AD = AE$ and $DE$ is perpendicular to $AB$. We deduce that $\\angle EAD = 2\\theta$ and $\\angle DEA = 90^\\circ - \\theta$.\n\nBecause $F$ is the symmetric of $E$ with respect to $AC$, we have $AE = AF$ and $EF$ is perpendicular to $AC$. We deduce that\n$$\n\\angle CAF = \\angle EAC = \\angle EAB + BAC = \\theta + \\alpha\n$$\nand\n$$\n\\begin{aligned}\n\\angle PDA & = 90^\\circ - \\frac{1}{2} \\angle DAF = 90^\\circ - \\frac{1}{2}(\\angle DAC + CAF) \\\\\n& = 90^\\circ - \\frac{1}{2}((\\alpha - \\theta) + (\\alpha + \\theta)) = 90^\\circ - \\alpha.\n\\end{aligned}\n$$\nWe deduce that\n$$\n\\angle DPC = \\angle PDA + \\angle DAC = (90^\\circ - \\alpha) + (\\alpha - \\theta) = 90^\\circ - \\theta = \\angle DEA.\n$$\nThis proves that quadrilateral $AEDP$ is cyclic.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40197,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, $I$ its incenter, and $D$ a point on the $\\operatorname{arc}\\overparen{BC}$ of the circumcircle of $ABC$ not containing $A$. The bisector of the angle $\\angle ADB$ intersects the segment $AB$ at $E$. The bisector of the angle $\\angle CDA$ intersects the segment $AC$ at $F$. Prove that the points $E$, $F$, $I$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First solution. Let $P$ be the intersection point of the bisector of the angle $\\angle BAC$ with the segment $EF$. Because both $I$ and $P$ are on the bisector of $\\angle BAC$, the problem is equivalent to showing that $AP = AI$.\nBecause $DE$ is the bisector of $\\angle ADB$ we have from the bisector theorem\n$$\n\\frac{AE}{AD} = \\frac{BE}{BD} = \\frac{AB}{AD + BD}\n$$\nBecause $DF$ is the bisector of $\\angle CDA$ we have from the bisector theorem\n$$\n\\frac{AF}{AD} = \\frac{CF}{CD} = \\frac{AC}{AD + CD}.\n$$\nWe deduce that\n$$\n\\begin{aligned}\n\\frac{1}{AE} + \\frac{1}{AF} & = \\frac{AD \\cdot AC + BD \\cdot AC + AD \\cdot AB + CD \\cdot AB}{AB \\cdot AC \\cdot AD} \\\\\n& = \\frac{AD \\cdot AC + AD \\cdot AB + AD \\cdot BC}{AB \\cdot AC \\cdot AD} = \\frac{AB + BC + CA}{AB \\cdot AC}\n\\end{aligned}\n$$\nfrom Ptolemy's theorem applied to the cyclic quadrilateral $ABDC$.\nNow from the areas of triangles $AEP$, $APF$ and $AEF$, we have\n$$\n\\frac{1}{2} AE \\cdot AP \\sin \\frac{\\alpha}{2} + \\frac{1}{2} AF \\cdot AP \\sin \\frac{\\alpha}{2} = \\frac{1}{2} AE \\cdot AF \\sin \\alpha,\n$$\nfrom which we deduce\n$$\n\\begin{gathered}\nAP = \\frac{2 AE \\cdot AF \\cos \\frac{\\alpha}{2}}{AE + AF} = \\frac{2 AB \\cdot AC \\cos \\frac{\\alpha}{2}}{AB + BC + CA} = \\frac{AB \\cdot AC \\cos \\frac{\\alpha}{2}}{s} \\\\\n= \\frac{4R \\Delta \\cos \\frac{\\alpha}{2}}{s \\cdot BC} = \\frac{2r \\cos \\frac{\\alpha}{2}}{\\sin \\alpha} = \\frac{r}{\\sin \\frac{\\alpha}{2}} = AI,\n\\end{gathered}\n$$\n![](attached_image_1.png)\n\nSaudi Arabia Mathematical Competitions 2015\nwhere $\\alpha = \\angle BAC$, $s = \\frac{AB + BC + CA}{2}$ is the semi-perimeter of triangle $ABC$, $R$ its circumradius, $r$ its inradius and $\\Delta = \\frac{AB \\cdot BC \\cdot CA}{4R} = rs$ its area.\n\n\nSecond Solution. Let $B'$ and $B''$ be the intersection points of the bisector of the angle $\\angle CBA$ with the side $BC$ and the $\\operatorname{arc}\\overparen{BC}$. Denote $\\beta = \\angle CBA$, $\\gamma = \\angle ACB$ and $\\theta = \\angle BAD$.\nWe have $\\angle B'B''F = \\angle BB''D = \\angle BAD = \\theta$ and $\\angle FB'B'' = \\angle CAB'' + \\angle AB''B = \\angle CBB'' + \\angle ACB = \\frac{\\beta}{2} + \\gamma$. We deduce from sine law in triangle $B'FB''$ that\n$$\n\\frac{B'F}{FB''} = \\frac{\\sin \\theta}{\\sin \\left(\\frac{\\beta}{2} + \\gamma\\right)}\n$$\nWe have $\\angle AB''F = \\angle AB''B + \\angle BB''D = \\angle ACB + \\angle BAD = \\gamma + \\theta$ and $\\angle FAB'' = \\angle CAB'' = \\angle CBB'' = \\frac{\\beta}{2}$. We deduce from sine law in triangle $AFB''$ that\n$$\n\\frac{FA}{FB''} = \\frac{\\sin (\\gamma + \\theta)}{\\sin \\frac{\\beta}{2}}\n$$\n![](attached_image_2.png)\n\nWe deduce that\n$$\n\\frac{\\overline{B'F}}{\\overline{FA}} = -\\frac{\\sin \\theta \\sin \\frac{\\beta}{2}}{\\sin \\left(\\frac{\\beta}{2} + \\gamma\\right) \\sin (\\gamma + \\theta)}\n$$\nBecause $DE$ is the bisector of angle $\\angle ADB$, we have from the theorem of bisectors\n$$\n\\frac{\\overline{AE}}{\\overline{EB}} = \\frac{AD}{DB} = \\frac{\\sin \\angle ABB''D}{\\sin \\angle DAB} = \\frac{\\sin (\\gamma + \\theta)}{\\sin \\theta}.\n$$\nBecause $AI$ is the bisector of angle $\\angle BAB'$, we have from the theorem of bisectors and sine law in triangle $ABB'$\n$$\n\\frac{\\overline{BI}}{\\overline{IB'}} = \\frac{AB}{AB'} = \\frac{\\sin \\angle ABB'B}{\\sin \\angle B'BA} = \\frac{\\sin \\left(\\frac{\\beta}{2} + \\gamma\\right)}{\\sin \\frac{\\beta}{2}}.\n$$\n\nTherefore\n$$\n\\frac{\\overline{AE}}{\\overline{EB}} \\cdot \\frac{\\overline{BI}}{\\overline{IB'}} \\cdot \\frac{\\overline{B'F}}{\\overline{FA}} = -1\n$$\nand from the theorem of Menelaus applied to the triangle $ABB'$, the points $E$, $I$, $F$ are collinear.\n\n\nThird Solution. Let $B'$ and $C'$ be the midpoints of the $\\operatorname{arcs} \\widehat{CA}$ and $\\widehat{AB}$, respectively. Because $DB'$ and $DC'$ are the bisectors of the angles $\\angle ADB$ and $\\angle CDA$, respectively, point $E$ is the intersection of $AB$ and $DC'$, point $I$ is the intersection of $BB'$ and $C'C$, and point $F$ is the intersection of $B'D$ and $CA$. We deduce from Pascal's theorem applied to the cyclic hexagon $ABB'DC'C$ that the three points $E$, $I$ and $F$ are collinear.\n![](attached_image_3.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40198,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ satisfying the following conditions\n(a) $f(1)=1$\n(b) $\\forall (x, y) \\in \\mathbb{R}^2$, $f(x+y)=f(x)+f(y)$\n(c) $\\forall x \\in \\mathbb{R} \\setminus \\{0\\}$, $f\\left(\\frac{1}{x}\\right)=\\frac{f(x)}{x^2}$.",
    "options": [],
    "answer": "f(x) = x for all real x",
    "solution": "Put $x=y=0$ into equation (b) to get $f(0)=0$. Put $y=-x$ into equation (b) to get $f(-x)=-f(x)$ and deduce that $f(x-y)=f(x)-f(y)$.\n\nLet $x \\in \\mathbb{R} \\setminus \\{0,1\\}$. We have\n$$\nf\\left(\\frac{1}{x-1}\\right)=\\frac{f(x-1)}{(x-1)^2}=\\frac{f(x)-1}{(x-1)^2}\n$$\nOn the other hand,\n$$\n\\begin{aligned}\nf\\left(\\frac{1}{x-1}\\right) & = f\\left(\\frac{1}{x-1}+1\\right)-1 = f\\left(\\frac{x}{x-1}\\right)-1 = \\frac{f\\left(\\frac{x-1}{x}\\right)}{\\left(\\frac{x-1}{x}\\right)^2}-1 \\\\\n& = \\frac{1-f\\left(\\frac{1}{x}\\right)}{\\left(\\frac{x-1}{x}\\right)^2}-1 = \\frac{x^2-f(x)}{(x-1)^2}-1 = \\frac{2x-1-f(x)}{(x-1)^2}\n\\end{aligned}\n$$\nWe deduce from these two relations that for all $x \\in \\mathbb{R} \\setminus \\{0,1\\}$, we have $f(x)-1=2x-1-f(x)$ which is equivalent to $f(x)=x$. Therefore, $f(x)=x$ for all $x \\in \\mathbb{R}$.\n\nConversely, this function satisfies all the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40199,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all strictly increasing functions $f: \\mathbb{Z} \\rightarrow \\mathbb{R}$ such that for any $m, n \\in \\mathbb{Z}$ there exists a $k \\in \\mathbb{Z}$ such that $f(k)=f(m)-f(n)$.",
    "options": [],
    "answer": "All functions of the form f(n) = a(n − n0) with a > 0 real and n0 an integer.",
    "solution": "Let $f: \\mathbb{Z} \\rightarrow \\mathbb{R}$ be such a function. We prove that $f(n+1)-f(n)$ is a positive constant $a$ independent of the integer $n$.\n\nIndeed, assume that there exist two integers $n_{0}, n_{1}$ such that\n$$\nf\\left(n_{0}+1\\right)-f\\left(n_{0}\\right)<f\\left(n_{1}+1\\right)-f\\left(n_{1}\\right) .\n$$\nLet $k_{0}, k_{1}$ be integers such that $f\\left(k_{0}\\right)=f\\left(n_{0}+1\\right)-f\\left(n_{0}\\right)$ and $f\\left(k_{1}\\right)= f\\left(n_{1}+1\\right)-f\\left(k_{0}\\right)$. Because $f$ is strictly increasing, $f\\left(k_{0}\\right)>0$. We deduce that $f\\left(n_{1}\\right)<f\\left(k_{1}\\right)=f\\left(n_{1}+1\\right)-f\\left(k_{0}\\right)<f\\left(n_{1}+1\\right)$, which is impossible since $f$ is strictly increasing and $n_{1}, n_{1}+1$ are consecutive integers.\n\nMoreover, there exists an integer $n_{0}$ such that $f\\left(n_{0}\\right)=f(0)-f(0)=0$.\nLet $n>n_{0}$. We have\n$$\nf(n)=f\\left(n_{0}\\right)+\\sum_{m=0}^{n-n_{0}-1}\\left(f\\left(n_{0}+m+1\\right)-f\\left(n_{0}+m\\right)\\right)=a\\left(n-n_{0}\\right) .\n$$\nWe obtain a similar formula for $n<n_{0}$ and therefore\n$$\nf(n)=a\\left(n-n_{0}\\right), \\quad \\text{ for all } n \\in \\mathbb{Z} .\n$$\nConversely, let $a>0$ be a positive real number and $n_{0}$ an integer and define $f(n)=a\\left(n-n_{0}\\right)$ for all $n \\in \\mathbb{Z}$. The function $f$ is strictly increasing and for any integers $m, n \\in \\mathbb{Z}$ we have $f(m)-f(n)=f\\left(m-n+n_{0}\\right)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40200,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and $D$ a point on the side $BC$. The tangent line to the circumcircle of the triangle $ABD$ at the point $D$ intersects the side $AC$ at $E$. The tangent line to the circumcircle of the triangle $ACD$ at the point $D$ intersects the side $AB$ at $F$. Prove that the point $A$ and the circumcenters of the triangles $ABC$ and $DEF$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\alpha = \\angle BAC$, $\\beta = \\angle CBA$, $\\gamma = \\angle ACB$ and $\\theta = BAD$. Because line $DE$ is tangent to the circumcircle of triangle $ABD$, we have $\\angle EDA = \\angle DBA = \\beta$. Because line $DF$ is tangent to the circumcircle of triangle $ADC$, we have $\\angle ADF = \\angle ACD = \\gamma$.\n\n![](attached_image_1.png)\n\nTherefore\n$$\n\\angle EDF + \\angle FAE = \\angle EDA + \\angle ADF + \\angle BAC = \\gamma + \\beta + \\alpha = 180^\\circ.\n$$\nThis proves that quadrilateral $AFDE$ is cyclic and hence $\\angle EFA = \\angle EDA = \\beta$ and $\\angle AEF = \\angle ADF = \\gamma$. This proves that sides $EF$ and $BC$ are parallel. Let $O$ and $O'$ be circumcenters of triangle $ABC$ and $AFE$, respectively. We have\n$$\n\\angle FAO' = 90^\\circ - \\angle AEF = 90^\\circ - \\gamma = \\angle BAO.\n$$\nThis proves that $A$, $O$ and $O'$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40201,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any integer $n \\geq 2$, there exists a unique finite sequence $x_{0}, x_{1}, \\ldots, x_{n}$ of real numbers which satisfies $x_{0}=x_{n}=0$ and $x_{i+1}-8 x_{i}^{3}- 4 x_{i}+3 x_{i-1}+1=0$ for all $i=1,2, \\ldots, n-1$. Prove moreover that $\\left|x_{i}\\right| \\leq \\frac{1}{2}$ for all $i=1,2, \\ldots, n-1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P_{1}(X)=X$, $P_{2}(X)=8 X^{3}+4 X-1$ and define by induction $P_{k+1}(X)$ by $P_{k+1}(X)=8 P_{k}(X)^{3}+4 P_{k}(X)-3 P_{k-1}(X)-1$ for all integer $k \\geq 2$. Clearly, $P_{k}(X)$ is a polynomial of odd degree for all $k \\geq 1$.\n\nLet $n \\geq 2$ be an integer and $a$ a real zero of the polynomial $P_{n}(X)$. The real $a$ exists since the degree of $P_{n}(X)$ is odd.\n\nConsider the finite sequence $x_{0}, x_{1}, \\ldots, x_{n}$ of real numbers defined by $x_{0}=0$, $x_{1}=a$ and $x_{i+1}=8 x_{i}^{3}+4 x_{i}-3 x_{i-1}-1$, for all $1 \\leq i \\leq n-1$. Clearly, $x_{1}=P_{1}(a)$ and $x_{2}=P_{2}(a)$. Assume that for $2 \\leq k \\leq n-1$, $x_{k-1}=P_{k-1}(a)$ and $x_{k}=P_{k}(a)$. We have\n\n$x_{k+1}=8 x_{k}^{3}+4 x_{k}-3 x_{k-1}-1=8 P_{k}(a)^{3}+4 P_{k}(a)-3 P_{k-1}(a)-1=P_{k+1}(a)$.\n\nThis proves that $x_{k}=P_{k}(a)$ for all $1 \\leq k \\leq n$ and in particular $x_{n}= P_{n}(a)=0$. This proves the existence of the sequence $x_{0}, x_{1}, \\ldots, x_{n}$ satisfying $x_{0}=x_{n}=0$ and $x_{i+1}-8 x_{i}^{3}-4 x_{i}+3 x_{i-1}+1=0$ for all $i=1,2, \\ldots, n-1$.\n\nConversely, if such a sequence $x_{0}, x_{1}, \\ldots, x_{n}$ exists then $x_{1}$ is a real zero of the polynomial $P_{n}(X)$. Therefore, proving the uniqueness of the sequence is equivalent to proving that $P_{n}(X)$ has a unique real zero.\n\nLet $x, y$ be two real numbers. We have $\\left|P_{2}(x)-P_{2}(y)\\right|=4|x-y|\\left|2\\left(x^{2}+x y+y^{2}\\right)+1\\right| \\geq |x-y|=\\left|P_{1}(x)-P_{1}(y)\\right|$, since $x^{2}+x y+y^{2} \\geq 0$. Assume that $\\left|P_{k}(x)-P_{k}(y)\\right| \\geq \\left|P_{k-1}(x)-P_{k-1}(y)\\right|$, for some integer $k \\geq 2$. We have\n\n$$\n\\begin{gathered}\n\\left|P_{k+1}(x)-P_{k+1}(y)\\right| \\geq \\\\\n\\geq 4\\left|P_{k}(x)-P_{k}(y)\\right| \\cdot \\left|2\\left(P_{k}^{2}(x)+P_{k}(x) P_{k}(y)+P_{k}^{2}(y)\\right)+1\\right| \\\\\n-3\\left|P_{k-1}(x)-P_{k-1}(y)\\right| \\\\\n\\geq 4\\left|P_{k}(x)-P_{k}(y)\\right|-3\\left|P_{k-1}(x)-P_{k-1}(y)\\right| \\geq \\left|P_{k}(x)-P_{k}(y)\\right|\n\\end{gathered}\n$$\n\nHence, the sequence $\\left(\\left|P_{k}(x)-P_{k}(y)\\right|\\right)_{k \\geq 1}$ is non-decreasing and if $x \\neq y$ then $P_{k}(x) \\neq P_{k}(y)$ for all $k \\geq 1$. We deduce that $P_{n}(X)$ has a unique real zero, and therefore the sequence $x_{0}, x_{1}, \\ldots, x_{n}$ is unique.\n\nLet $c=\\max \\left\\{\\left|x_{i}\\right|: 1 \\leq i \\leq n-1\\right\\}$. There exists $k \\in \\{1,2, \\ldots, n-1\\}$ such that $c=\\left|x_{k}\\right|$. Because $x_{k}$ and $x_{k}^{3}$ have the same sign, it follows that\n\n$$\n\\begin{aligned}\nc+2 c^{3} & =\\left|x_{k}\\right|+2\\left|x_{k}\\right|^{3}=\\left|x_{k}+2 x_{k}^{3}\\right|=\\frac{1}{4}\\left|1+x_{k+1}+3 x_{k-1}\\right| \\\\\n& \\leq \\frac{1}{4}+\\frac{1}{4}\\left(\\left|x_{k+1}\\right|+3\\left|x_{k-1}\\right|\\right) \\leq \\frac{1}{4}+c,\n\\end{aligned}\n$$\n\nwhich implies that $c \\leq \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40202,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = 1$. Prove that\n$$\n2\\left(\\frac{a b}{a + b} + \\frac{b c}{b + c} + \\frac{c a}{c + a}\\right) + 1 \\geq 6(a b + b c + c a).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using the condition $a + b + c = 1$, we have\n$$\n\\begin{aligned}\n\\left(\\frac{a b}{a + b} + \\frac{b c}{b + c} + \\frac{c a}{c + a}\\right) &= (a + b + c)\\left(\\frac{a b}{a + b} + \\frac{b c}{b + c} + \\frac{c a}{c + a}\\right) \\\\\n&= (a b + b c + c a) + a b c\\left(\\frac{1}{a + b} + \\frac{1}{b + c} + \\frac{1}{c + a}\\right) \\\\\n&\\geq (a b + b c + c a) + \\frac{9 a b c}{2(a + b + c)} \\\\\n&\\geq (a b + b c + c a) + \\frac{9}{2} a b c,\n\\end{aligned}\n$$\nby Cauchy-Schwarz inequality. Therefore, it remains to prove that\n$$\n9 a b c + 1 \\geq 4(a b + b c + c a).\n$$\nUsing again the condition $a + b + c = 1$, the last inequality becomes\n$$\n9 a b c + (a + b + c)^2 \\geq 4(a + b + c)(a b + b c + c a),\n$$\nwhich is equivalent, by simple algebraic manipulations, to\n$$\n9 a b c + a^2 + b^2 + c^2 \\geq 2(a b + b c + c a)\n$$\nUsing a third time the condition $a + b + c = 1$, the obtained inequality becomes\n$$\n9 a b c + (a + b + c)(a^2 + b^2 + c^2) \\geq 2(a + b + c)(a b + b c + c a)\n$$\nwhich is equivalent, by simple algebraic manipulations, to\n$$\n3 a b c + a^3 + b^3 + c^3 \\geq a^2(b + c) + b^2(c + a) + c^2(a + b).\n$$\nThis last inequality is nothing but Schur's inequality. This solves the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40203,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R}_{>0} \\rightarrow \\mathbb{R}$ such that\n$$\nf\\left(\\frac{x}{y}\\right)=f(x)+f(y)-f(x) f(y)\n$$\nfor all $x, y \\in \\mathbb{R}_{>0}$. Here, $\\mathbb{R}_{>0}$ denotes the set of all positive real numbers.",
    "options": [],
    "answer": "Either f(x) = 0 for all x > 0, or f(x) = 1 for all x > 0.",
    "solution": "Define $g: \\mathbb{R}_{>0} \\rightarrow \\mathbb{R}$ by $g(x)=1-f(x)$ for all $x>0$. The functional equation can be rewritten $g(x / y)=g(x) g(y)$ for all $x, y>0$.\n\nPutting $y=x=1$, we get $g(1)=g(1)^2$. This means that $g(1)=0$ or $g(1)=1$.\n\n1. Assume $g(1)=0$. For all $x>0$, we have $g(x)^2=g(x / x)=g(1)=0$. Hence $g(x)=0$ for all $x>0$. Conversely, $g=0$ is a solution of the given functional equation.\n\n2. Assume $g(1)=1$. In this case, for all $x>0$, we have $g(x)=g(1) g(x)= g(1 / x)$.\nLet $x>0$. We have\n$$\ng(x)=g\\left(\\frac{\\sqrt{x}}{1 / \\sqrt{x}}\\right)=g(\\sqrt{x}) g(1 / \\sqrt{x})=g(\\sqrt{x})^2=g(\\sqrt{x} / \\sqrt{x})=g(1)=1 .\n$$\nConversely, $g=1$ is a solution of the given functional equation.\n\nIn conclusion, $f(x)$ is identically 0 or identically 1 .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40204,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let be given the sequence $\\left(x_{n}\\right)$ defined by $x_{1}=1$ and $x_{n+1}=3 x_{n}+\\left\\lfloor x_{n} \\sqrt{5}\\right\\rfloor$ for all $n=1,2,3, \\ldots$, where $\\lfloor x\\rfloor$ denotes the greatest integer that does not exceed $x$. Prove that for any positive integer $n$ we have\n$$\nx_{n} x_{n+2}-x_{n+1}^{2}=4^{n-1}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is easy to prove by induction that $x_{n}$ is an integer for all $n \\geq 1$. Therefore, we have $x_{n+1}=\\left\\lfloor(3+\\sqrt{5}) x_{n}\\right\\rfloor$ for all $n \\geq 1$, and we deduce that\n$$\n\\begin{aligned}\nx_{n+2} & =\\left\\lfloor(3+\\sqrt{5}) x_{n+1}\\right\\rfloor=6 x_{n+1}+\\left\\lfloor-(3-\\sqrt{5})\\left\\lfloor(3+\\sqrt{5}) x_{n}\\right\\rfloor\\right\\rfloor \\\\\n& =6 x_{n+1}+\\left\\lfloor-4 \\frac{\\left\\lfloor(3+\\sqrt{5}) x_{n}\\right\\rfloor}{(3+\\sqrt{5})}\\right\\rfloor=6 x_{n+1}-4 x_{n}\n\\end{aligned}\n$$\nIndeed, for the last equality we have\n$$\n(3+\\sqrt{5}) x_{n}-1<\\left\\lfloor(3+\\sqrt{5}) x_{n}\\right\\rfloor \\leq(3+\\sqrt{5}) x_{n},\n$$\nfrom which we deduce\n$$\n-4 x_{n} \\leq-4 \\frac{\\left\\lfloor(3+\\sqrt{5}) x_{n}\\right\\rfloor}{(3+\\sqrt{5})}<-4 x_{n}+\\frac{4}{3+\\sqrt{5}}<-4 x_{n}+1 .\n$$\nLet $y_{n}=x_{n} x_{n+2}-x_{n+1}^{2}$, for all $n \\geq 1$. We will prove by induction that $y_{n}=4^{n-1}$, for all $n \\geq 1$.\nFor $n=1$, we have $x_{1}=1, x_{2}=\\lfloor(3+\\sqrt{5})\\rfloor=5$ and $x_{3}=\\lfloor 5(3+\\sqrt{5})\\rfloor=26$, and therefore $y_{1}=4^{0}$. Assume $y_{n}=4^{n-1}$ for some positive integer $n$. We have\n$$\n\\begin{aligned}\ny_{n+1} & =x_{n+1} x_{n+3}-x_{n+2}^{2}=x_{n+1}\\left(6 x_{n+2}-4 x_{n+1}\\right)-x_{n+2}\\left(6 x_{n+1}-4 x_{n}\\right) \\\\\n& =4 y_{n}=4^{n}\n\\end{aligned}\n$$\nThis ends the induction and solves the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40205,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be positive integers such that $a c + b d$ is divisible by $a^{2} + b^{2}$. Prove that $\\operatorname{gcd}\\left(c^{2} + d^{2}, a^{2} + b^{2}\\right) > 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose for the sake of contradiction that $\\operatorname{gcd}\\left(c^{2} + d^{2}, a^{2} + b^{2}\\right) = 1$. We first notice that\n$$\n(a c + b d)^{2} + (a d - b c)^{2} = \\left(a^{2} + b^{2}\\right)\\left(c^{2} + d^{2}\\right).\n$$\nSince $a c + b d$ is divisible by $a^{2} + b^{2}$, it follows that $(a d - b c)^{2}$ is also divisible by $a^{2} + b^{2}$.\nLet $p$ be a prime divisor of $a^{2} + b^{2}$ and $n = v_{p}\\left(a^{2} + b^{2}\\right)$. We have\n$$\nv_{p}\\left((a c + b d)^{2} + (a d - b c)^{2}\\right) = n\n$$\nsince $c^{2} + d^{2}$ is not divisible by $p$. Note that,\n$$\nv_{p}\\left((a c + b d)^{2}\\right) = 2 v_{p}(a c + b d) \\geq 2 v_{p}\\left(a^{2} + b^{2}\\right) = 2 n > n.\n$$\nThis implies that $v_{p}\\left((a d - b c)^{2}\\right) = n$, which means that $n$ is even. This proves that $a^{2} + b^{2} = u^{2}$ for some positive integer $u$, and therefore both $a c + b d$ and $a d - b c$ are divisible by $u$.\nBut\n$$\na\\left(c^{2} + d^{2}\\right) = c(a c + b d) + d(a d - b c)\n$$\nand\n$$\nb\\left(c^{2} + d^{2}\\right) = d(a c + b d) - c(a d - b c)\n$$\nare also divisible by $u$. Because $\\operatorname{gcd}\\left(u, c^{2} + d^{2}\\right) = 1$, we deduce that $u$ divides both $a$ and $b$. But $u^{2} = a^{2} + b^{2}$. We have a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40206,
    "subject": "Mathematics (Multi-modal)",
    "question": "The total number of languages used in KAUST is $n$. For each positive integer $k \\leq n$, let $A_{k}$ be the set of all those people in KAUST who can speak at least $k$ languages; and let $B_{k}$ be the set of all people $P$ in KAUST with the property that, for any $k$ pairwise different languages (used in KAUST), $P$ can speak at least one of these $k$ languages. Prove that\n\na. If $2k \\geq n+1$ then $A_{k} \\subseteq B_{k}$;\n\nb. If $2k \\leq n+1$ then $A_{k} \\supseteq B_{k}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\mathcal{L}$ be the set of all the languages used in KAUST.\n\na.\nAssume $2k \\geq n+1$. Consider $P \\in A_{k}$, and let $\\mathcal{L}_P \\subseteq \\mathcal{L}$ be the set of all languages spoken by $P$. The set $\\mathcal{L}_P$ contains at least $k$ languages.\n\nLet $\\mathcal{L}' \\subseteq \\mathcal{L}$ be a set of $k$ languages. We have\n$$\n|\\mathcal{L}_P \\cap \\mathcal{L}'| = |\\mathcal{L}_P| + |\\mathcal{L}'| - |\\mathcal{L}_P \\cup \\mathcal{L}'| \\geq 2k - |\\mathcal{L}| = 2k - n \\geq 1.\n$$\nThis means that $\\mathcal{L}_P \\cap \\mathcal{L}' \\neq \\emptyset$ and therefore $P$ can speak at least one language in $\\mathcal{L}'$. Hence $P \\in B_{k}$.\n\nb.\nAssume $2k \\leq n+1$. Consider $P \\in B_{k}$, and let $\\mathcal{L}_P \\subseteq \\mathcal{L}$ be the set of all languages spoken by $P$ and $\\mathcal{L}' = \\mathcal{L} \\setminus \\mathcal{L}_P$. Because there is no language in $\\mathcal{L}'$ spoken by $P$ we have $|\\mathcal{L}'| \\leq k-1$. We deduce that\n$$\n|\\mathcal{L}_P| = |\\mathcal{L}| - |\\mathcal{L}'| \\geq n - (k-1) \\geq k.\n$$\nThis means that $P$ speaks at least $k$ languages and therefore $P \\in A_{k}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40207,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the maximum number of bishops that can be placed on an $8 \\times 8$ chessboard such that at most three bishops lie on any diagonal?",
    "options": [],
    "answer": "38",
    "solution": "If the chessboard is colored black and white as usual, then any diagonal is a solid color. So we may consider bishops on black and white squares separately.\n\nIn one direction, the lengths of the black diagonals are $2, 4, 6, 8, 6, 4$, and $2$. Each of these can have at most three bishops, except the first and last diagonals which can have at most two, giving a total of at most $2 + 3 + 3 + 3 + 3 + 3 + 2 = 19$ bishops on black squares. Likewise there can be at most $19$ bishops on white squares for a total of at most $38$ bishops.\n\n![](attached_image_1.png)\n\nConversely, if we place $38$ bishops on the four boundaries of the table and on the second and seventh rows except the second and seventh square of the second row, as shown in the picture, one can check that this arrangement satisfies the condition of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40208,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many polynomials $P$ of integer coefficients and degree at most $4$ satisfy $0 \\leq P(x) < 72$ for all $x \\in \\{0,1,2,3,4\\}$?",
    "options": [],
    "answer": "6718464",
    "solution": "Remember that if $P_{0}(X), P_{1}(X), P_{2}(X), \\ldots$ are monic polynomials with integer coefficients such that $\\deg P_{k}(X) = k$ for all $k = 0, 1, 2, \\ldots$, then any polynomial $P(X)$ with integer coefficients can be written in a unique way as\n$$\nP(X) = a_{0} P_{0}(X) + a_{1} P_{1}(X) + \\cdots + a_{n} P_{n}(X),\n$$\nwhere $a_{0}, a_{1}, \\ldots, a_{n}$ are integers and $a_{n} \\neq 0$. This can be proved by induction on the degree of $P(X)$ using the Euclidean algorithm.\n\nTherefore, let us write our polynomial as:\n$$\nP(x) = a_{0} x(x-1)(x-2)(x-3) + a_{1} x(x-1)(x-2) + a_{2} x(x-1) + a_{3} x + a_{4}.\n$$\nSince $0 \\leq a_{4} = P(0) < 72$, there are $72$ possible choices for $a_{4}$.\n\nFor any choice of $a_{4}$, we have $-a_{4} \\leq a_{3} = P(1) - a_{4} < 72 - a_{4}$. Hence, we also have $72$ possible choices for $a_{3}$.\n\nFor each choice of $a_{3}$ and $a_{4}$ we have\n$$\n-\\frac{2 a_{3} + a_{4}}{2} \\leq a_{2} = \\frac{P(2) - 2 a_{3} - a_{4}}{2} < \\frac{72}{2} - \\frac{2 a_{3} + a_{4}}{2}.\n$$\nHence, there are $36$ possible choices for $a_{2}$.\n\nKeep doing that way, the total number of polynomials is\n$$\n72 \\times 72 \\times (72 / 2) \\times (72 / 6) \\times (72 / 24) = 72 \\times 72 \\times 36 \\times 12 \\times 3 = 2592^{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40209,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the polynomial $P(X)=\\left(X^{2}-12 X+11\\right)^{4}+23$ can not be written as the product of three non-constant polynomials with integer coefficients.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose for the sake of contradiction that\n$$\nP(X)=Q(X) H(X) R(X),\n$$\nwhere $Q(X)$, $H(X)$, $R(X)$ are non-constant polynomials with integer coefficients. Since $P(x)>0$ for every $x \\in \\mathbb{R}$, the degrees of $Q(X)$, $H(X)$, $R(X)$ are all even. It implies that two of these three polynomials are quadratic. Suppose that $\\operatorname{deg} Q(X)=\\operatorname{deg} H(X)=2$.\nNow, $P(1)=P(11)=23$, implies that $Q(1)$, $Q(11)$ are divisors of $23$. This means that $Q(1)$, $Q(11) \\in\\{ \\pm 1, \\pm 23\\}$. But because $10$ divides $Q(11)-Q(1)$ we have $Q(1)=Q(11)$. Similarly, we have $H(1)=H(11)$.\nBesides, $Q(1) H(1)$ is a divisor of $23$ so at least one of $Q(1)$ or $H(1)$ is $\\pm 1$. Suppose without loss of generality that $Q(1)= \\pm 1$ then $Q(11)=Q(1)= \\pm 1$. This implies that $Q(X)=(X-1)(X-11) \\pm 1$. But this implies that $Q(X)$ has a real root while $P(x)$ is positive for all $x \\in \\mathbb{R}$, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40210,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ that satisfy\n$$\nf\\left(x+y^{2}-f(y)\\right)=f(x)\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "All constant functions f(x) = c for any real constant c, and the function f(x) = x^2.",
    "solution": "Notice that the function $f$, defined by $f(x)=x^{2}$ for all $x \\in \\mathbb{R}$, is a solution of the equation.\n\nAssume that there exists $a \\in \\mathbb{R}$ such that $f(a) \\neq a^{2}$ and let $b=a^{2}-f(a) \\neq 0$. Then for all $x \\in \\mathbb{R}$ we have $f(x+b)=f(x)$. Therefore, for all $x, y \\in \\mathbb{R}$, let $z=\\frac{y-x-b^{2}}{2b}$ we have\n$$\n\\begin{aligned}\nf(x) & =f\\left(x+(z+b)^{2}-f(z+b)\\right)=f\\left(x+2bz+b^{2}+z^{2}-f(z)\\right) \\\\\n& =f\\left(x+2bz+b^{2}\\right)=f(y)\n\\end{aligned}\n$$\nWe deduce that $f$ is constant. Conversely, any constant function satisfies the given functional equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40211,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given 2015 subsets $A_{1}, A_{2}, \\ldots, A_{2015}$ of the set $\\{1,2, \\ldots, 1000\\}$ such that $|A_{i}| \\geq 2$ for every $i \\geq 1$ and $|A_{i} \\cap A_{j}| \\geq 1$ for every $1 \\leq i < j \\leq 2015$. Prove that $k=3$ is the smallest number of colors such that we can always color the elements of the set $\\{1,2, \\ldots, 1000\\}$ by $k$ colors with the property that the subset $A_{i}$ has at least two elements of different colors for every $i \\geq 1$.",
    "options": [],
    "answer": "3",
    "solution": "Consider the collection: $A_{1} = \\{1,2\\}$, $A_{2} = \\{2,3\\}$, $A_{3} = \\{1,3\\}$, $A_{4}, \\ldots, A_{2015}$ be any 2011 subsets of $\\{1,2, \\ldots, 1000\\}$ that contain all $1,2,3$. One can check that this collection satisfies the given condition. For any $2$-coloring of the set $\\{1,2, \\ldots, 1000\\}$, at least one of $A_{1}, A_{2}, A_{3}$ will contain two elements of the same color. Hence, we need more than 2 colors.\n\nNow we show that we always can color $\\{1,2, \\ldots, 1000\\}$ by 3 colors such that the set $A_{i}$ contains at least two elements of different color for all $i$. We choose a set $A_{i_{0}}$ with least number of elements. We color one element of $A_{i_{0}}$ by red and all the rest by blue. We color by green all the elements $\\{1,2, \\ldots, 1000\\}$ which do not belong to $A_{i_{0}}$. It is clear that $A_{i_{0}}$ contains two elements of different color. For any subset $A_{i}$ with $i \\neq i_{0}$, $|A_{i} \\cap A_{i_{0}}| \\geq 1$ so $A_{i}$ contains at least one red or blue color element. Moreover, because $|A_{i}| \\geq |A_{i_{0}}|$, either $A_{i}$ contains another element of $A_{i_{0}}$ with the other color or contains a green element. Hence, the subset $A_{i}$ contains two elements of different color.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40212,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the number of 6-tuples $(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6})$ of distinct positive integers satisfying the following two conditions:\n\na. $a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} = 30$;\n\nb. We can write $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ on sides of a hexagon such that after a finite number of times choosing a vertex of the hexagon and adding 1 to the two numbers written on two sides adjacent to the vertex, we obtain a hexagon with equal numbers on its sides.",
    "options": [],
    "answer": "13680",
    "solution": "We label the vertex of the hexagon by $1,2,3,4,5,6$ and suppose that six numbers are written in the order $a, b, c, d, e, f$ on the edges $(1,2), (2,3), \\ldots, (6,1)$, respectively.\n\nWe first notice that by choosing a vertex of the hexagon and adding 1 to the two numbers written on two adjacent sides to the vertex, the difference $(a + c + e) - (b + d + f)$ is an invariant. Thus, if we want to obtain a hexagon with equal numbers on its sides, we must have $a + c + e = b + d + f$ at the beginning.\n\nWe will show that it is also sufficient. Let $N$ be a big integer, for example, we can take $N > 30$. We choose the vertices $1,2,4,5$ and add 1 to the two numbers written on two sides of those vertices by $N - f, N - b, N - c$ and $N - e$ times, respectively. Then, we obtain a hexagon with six numbers\n\n$a + 2N - b - f, N, N, d + 2N - e - c, N, N$.\n\nSince $a + c + e = b + d + f$, we have $a + 2N - b - f = d + 2N - e - c = M > N$ (since $N > 30 = a + b + c + d + e + f$). We now can choose the vertices 3, 6 for $M - N$ times each, then we obtain a hexagon with numbers $M$ on all of its sides.\n\nNow, we count how many 6-tuples $(a, b, c, d, e, f)$ of distinct positive integers with $a + c + e = b + d + f = 15$.\n\nWe list all triples of distinct positive integers with sum 15: $(1,2,12), (1,3,11), (1,4,10), (1,5,9), (1,6,8), (2,3,10), (2,4,9), (2,5,8), (2,6,7), (3,4,8), (3,5,7), (4,5,6)$. We then check that there are exactly 19 times we can pair these triples to obtain 6 distinct numbers.\n\nSince we can permute the numbers on each tuple, the total number of 6-tuples satisfying the given conditions is $19 \\times 6!$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 40213,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A B C$ be a triangle with orthocenter $H$. Let $P$ be any point of the plane of the triangle. Let $\\Omega$ be the circle with the diameter $A P$. The circle $\\Omega$ cuts $C A$ and $A B$ again at $E$ and $F$, respectively. The line $P H$ cuts $\\Omega$ again at $G$. The tangent lines to $\\Omega$ at $E, F$ intersect at $T$. Let $M$ be the midpoint of $B C$ and $L$ be the point on $M G$ such that $A L$ and $M T$ are parallel. Prove that $L A$ and $L H$ are orthogonal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $B Y, C Z$ be altitudes of $A B C$. Points $Y, Z$ lie on the circle of diameter $A H$. The line $H P$ cuts the circle $\\Omega$ again at $G$. Since $A P$ is a diameter in $\\Omega$, the lines $A G$ and $P H$ are perpendicular and therefore point $G$ lies on the circle of diameter $A H$.\n\n![](attached_image_1.png)\n\nBecause $A, E, G, F$ are concyclic and $A, G, Z, H, Y$ are concyclic, we have $\\angle G E F=\\angle G A F=\\angle G A Z=\\angle G Y Z$ and $\\angle F G E=\\angle Z A Y=\\angle Z G Y$. We deduce that triangles $G E F$ and $G Y Z$ are similar.\nWe know that $\\angle M Z C=\\angle Z C B=\\angle Z A H$. We deduce that $Z M$ is tangent to the circumcircle of triangle $G Z Y$. Similarly, $M Y$ is tangent to the circumcircle of triangle $G Z Y$. But $F T$ and $E T$ are tangent to the circumcircle of triangle $G F E$. We deduce that quadrilaterals $G F T E$ and $G Z M Y$ are similar and therefore $\\angle G Z A=\\angle G Z F=\\angle G M T=\\angle G L A$, since $L A$ is parallel to $M T$. This means that point $L$ lies on the circle of diameter $A H$ and therefore $L A$ and $L H$ are orthogonal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40214,
    "subject": "Mathematics (Multi-modal)",
    "question": "In his bag, Salman has a number of stones. The weight of each stone is not greater than $0.5$ kg and the total weight of the stones is not greater than $2.5$ kg. Prove that Salman can divide his stones into $4$ groups, each group has a total weight not greater than $1$ kg.\n\nSuggested by Trân Nam Dũng",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $k$ be the number of stones. Note that the sum of weights of any two stones is not greater than $1$ kg.\n\nThus, if $k \\leq 8$ we can divide these stones into $4$ (or less) groups, each group has $1$ or $2$ stones. This division obviously fulfills the condition.\n\nIf $k=9$, we take $3$ lightest stones. Their total weight must not be greater than $\\frac{2.5}{3} < 1$ kg. We put these stones together into one group and distribute the other $6$ into $3$ groups, $2$ stones in each. This division again fulfills the condition.\n\nNow, consider the general case. Whenever two or more stones have the total weight less or equal to $0.5$ kg then we merge them into one new stone. Since the number of stones is finite, this process must terminate and our new stones have the property that the sum of the weights of any two stones is greater than $0.5$ kg. We deduce that the number of our new stones is less or equal to $9$, otherwise, we will have at least $5$ pairs of stones with total weight greater than $0.5$ kg each, and the total weight of all the stones will be greater than $2.5$ kg. Applying what we have done in the first two cases, we can divide these new stones as required. Obviously, the same division applies for the original unmerged stones.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40215,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the number of strictly increasing sequences of nonnegative integers with the first term $0$ and the last term $15$, and among any two consecutive terms, exactly one of them is even.",
    "options": [],
    "answer": "610",
    "solution": "Let $A_{n}$ be the set of such sequences with the last term is $n$ instead of $15$, and $a_{n}=|A_{n}|$. We will show that $(a_{n})$ is in fact the Fibonacci sequence and deduce that $a_{15}=610$.\n\nWe can check easily that $a_{1}=a_{2}=1$. For $n \\geq 1$, we consider the second-last term of each sequence in $A_{n+2}$. We have two cases.\n\nCase 1. The second-last term is $n+1$. Then we can leave out the last term to get an element of $A_{n+1}$. Conversely, for each element of $A_{n+1}$, we can add $n+2$ to the end to have an element of $A_{n+2}$. Hence, in this case, we have $a_{n+1}$ sequences.\n\nCase 2. The second-last term is at most $n$. But $n$ and $n+2$ have the same parity, so the second-last term is at most $n-1$. When we substitute the last term $n+2$ with $n$, we have an element of $A_{n}$. Conversely, if we replace $n$ in an element of $A_{n}$ by $n+2$, we have an element of $A_{n+2}$. Hence, in this case, we have $a_{n}$ sequences.\n\nThis proves that $a_{n+2}=a_{n+1}+a_{n}$, and therefore $(a_{n})$ is the Fibonacci sequence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40216,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integer solutions of the equation $14^{x} - 3^{y} = 2015$.",
    "options": [],
    "answer": "(x, y) = (3, 6)",
    "solution": "Notice that $14^{x} \\geq 2015$ which implies that $x \\geq 3$. On the other hand $3^{y} = 14^{x} - 2015 \\geq 14^{3} - 2015 = 729 = 3^{6}$. We deduce that $y \\geq 6$.\n\nWe have $3^{y} = 14^{x} - 2015 \\equiv 1 \\pmod{7}$. But $\\operatorname{Ord}_{7} 3 = 6$. We deduce that $y = 6k$ for some positive integer $k$.\n\nWe have $0 \\equiv 2015 \\equiv 14^{x} - 3^{6k} \\equiv (-1)^{x} - (-1)^{k} \\pmod{5}$. We deduce that $x$ and $k$ have the same parity.\n\nBecause $y \\geq 6$, we have $5^{x} \\equiv 14^{x} - 3^{y} \\equiv 2015 \\equiv -1 \\pmod{9}$. This implies that $x \\equiv 3 \\pmod{6}$ and therefore both $x$ and $k$ are odd numbers.\n\nNow, assume that $x > 3$. This means that $14^{x} \\equiv 0 \\pmod{16}$ which implies that $9^{k} \\equiv 3^{6k} \\equiv 1 \\pmod{16}$. But $\\operatorname{Ord}_{16} 9 = 2$, which means that $k$ is even, which is a contradiction.\n\nWe deduce that $x = 3$ and hence $y = 6$ which gives the unique solution of the equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40217,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\left(a_{n}\\right)_{n \\geq 0}$ be a sequence of positive integers such that $a_{n}^{2}$ divides $a_{n-1} a_{n+1}$, for all $n \\geq 1$. Prove that if there exists an integer $k \\geq 2$ such that $a_{k}$ and $a_{1}$ are relatively prime, then $a_{1}$ divides $a_{0}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume, for the sake of contradiction, that there exists an integer $k \\geq 2$ such that $a_{k}$ and $a_{1}$ are relatively prime and that $a_{1}$ does not divide $a_{0}$. We deduce that there exists a prime number $p$ such that $0 \\leq v_{p}\\left(a_{0}\\right)<v_{p}\\left(a_{1}\\right)$, where $v_{p}(a)$ is the $p$-adic valuation of the integer $a$, that is the greatest integer $m$ such that $p^{m}$ divides $a$.\n\nAssume that there exists an integer $m \\geq 0$, for which we have $v_{p}\\left(a_{m}\\right)< v_{p}\\left(a_{m+1}\\right)$. Because $a_{m+1}^{2}$ divides $a_{m} a_{m+2}$, we have $2 v_{p}\\left(a_{m+1}\\right) \\leq v_{p}\\left(a_{m}\\right)+ v_{p}\\left(a_{m+2}\\right)$, and therefore $v_{p}\\left(a_{m+1}\\right)<v_{p}\\left(a_{m+2}\\right)$. This proves by induction that the sequence $v_{p}\\left(a_{n}\\right)$ is increasing, and therefore $0<v_{p}\\left(a_{1}\\right)<v_{p}\\left(a_{k}\\right)$. This implies that $p$ divides $\\operatorname{gcd}\\left(a_{1}, a_{k}\\right)=1$, which is a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40218,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ and $k$ be two positive integers. Prove that if $n$ is relatively prime with $30$, then there exist two integers $a$ and $b$, each relatively prime with $n$, such that $\\frac{a^{2}-b^{2}+k}{n}$ is an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will proceed by steps.\n\n- First Step: We prove it for $n = p$ a prime number, with $p \\geq 7$:\n\nIf $p$ divides $k$ then $a = b = 1$ satisfy the condition. If $p$ does not divide $k$ then let us consider two cases:\n\na. If $k$ is a quadratic residue modulo $p$, and $c$ and $c+1$ are two consecutive quadratic residues modulo $p$ such that $1 \\leq c < c+1 \\leq p-1$, then both $k c$ and $k c + k$ are quadratic residues relatively prime with $p$. This means that there exist $a$ and $b$, each relatively prime with $p$, such that $a^{2} \\equiv c k$ and $b^{2} \\equiv c k + k$ modulo $p$ and therefore $a^{2} - b^{2} + k \\equiv 0 \\pmod{p}$.\n\nb. If $k$ is a quadratic nonresidue modulo $p$, and $c$ and $c+1$ are two consecutive quadratic nonresidues modulo $p$ such that $1 \\leq c < c+1 \\leq p-1$, then both $k c$ and $k c + k$ are quadratic residues relatively prime with $p$. This means that there exist $a$ and $b$, each relatively prime with $p$, such that $a^{2} \\equiv c k$ and $b^{2} \\equiv c k + k$ modulo $p$ and therefore $a^{2} - b^{2} + k \\equiv 0 \\pmod{p}$.\n\nThus, this first step reduces to proving that among $1, 2, \\ldots, p-1$ there are two consecutive quadratic residues and two consecutive quadratic nonresidues modulo $p$. We will prove this fact in two different ways:\n\nFirst way. Notice that both $1$ and $4$ are quadratic residues.\nIf $2$ or $3$ is a quadratic residue then there exists at least two consecutive quadratic residues among $1, 2, 3, 4$. But exactly half of the numbers $1, 2, \\ldots, p-1$ are quadratic nonresidues. So there exist at least $\\frac{p-3}{2}$ quadratic nonresidues among the $p-5$ consecutive numbers $5, 6, \\ldots, p-1$. So, by a pigeonhole principle, at least two of them must be consecutive.\nIf $2$ and $3$ are consecutive quadratic nonresidues, we have two possibilities: Either $-1$ is a quadratic nonresidue and in this case $p-3$ and $p-2$ are consecutive quadratic residues, or $-1$ is a quadratic residue, which means that $p \\equiv 1 \\pmod{4}$, $p \\geq 13$, and for each integer $m$ the integers $m$ and $p-m$ are either both quadratic residues or both quadratic nonresidues. This means that there are $\\frac{p-1}{4}$ quadratic residues between $1, 2, \\ldots, \\frac{p-1}{2}$.\nIf $\\frac{p-1}{2}$ is a quadratic residue then $\\frac{p-1}{2}, \\frac{p+1}{2}$ are two consecutive quadratic residues.\nIf $\\frac{p-1}{2}$ is a quadratic nonresidue then there are $\\frac{p-5}{4}$ quadratic residues among the $\\frac{p-9}{2}$ consecutive numbers $4, 5, \\ldots, \\frac{p-3}{2}$. But $\\frac{p-9}{2}$ is an even number. So, by a pigeonhole principle, at least two of the quadratic residues must be consecutive.\n\nSecond way. In this second way we will compute the exact number of consecutive pairs of quadratic residues and the exact number of consecutive pairs of quadratic nonresidues.\nConsider the partition\n$$\n\\{1, 2, \\ldots, p-1\\} = Q_{1} \\cup N_{1} \\cup Q_{2} \\cup N_{2} \\cup \\cdots \\cup Q_{k} \\cup N_{k}\n$$\nwhere $Q_{1}$ is the set of all the first consecutive quadratic residues, possibly reduced to the singleton $\\{1\\}$ when $2$ is a quadratic nonresidue, $N_{1}$ is the set of all the first consecutive quadratic nonresidues, possibly reduced to a singleton, $Q_{2}$ is the set of all the first consecutive quadratic residues after the elements of $N_{1}$, and so on... Notice that $N_{k}$ is the only possibly empty set when $p-1$ is a quadratic residue.\nLet $a \\in \\{1, 2, \\ldots, p-2\\}$. Notice that $\\left(\\frac{a}{p}\\right)\\left(\\frac{a+1}{p}\\right) = 1$ if and only if $a, a+1$ are both quadratic residues or both quadratic nonresidues, where $\\left(\\frac{a}{p}\\right)$ is the Legendre symbol. This means that $\\left(\\frac{a}{p}\\right)\\left(\\frac{a+1}{p}\\right) = -1$ if and only if $a, a+1$ do not belong to the same subset of the partition. This means that there are either $2k-2$ or $2k-1$ pairs $(a, a+1)$, with $1 \\leq a \\leq p-2$ such that $\\left(\\frac{a}{p}\\right)\\left(\\frac{a+1}{p}\\right) = -1$, depending on whether $N_{k}$ is empty or not. But $\\left(\\frac{a}{p}\\right)\\left(\\frac{a+1}{p}\\right) = \\left(\\frac{a}{p}\\right)^{2}\\left(\\frac{1+a^{-1}}{p}\\right) = \\left(\\frac{1+a^{-1}}{p}\\right)$. But when $a$ covers all the residues $1, 2, \\ldots, p-2$, its inverse $a^{-1}$ covers all the residues $1, 2, \\ldots, p-2$. Because there are exactly $\\frac{p-1}{2}$ quadratic nonresidues among the residues $2, 3, \\ldots, p-1$, we deduce that $k = \\frac{p+3}{4}$ if $p \\equiv 1 \\pmod{4}$ and $k = \\frac{p+1}{4}$ if $p \\equiv 3 \\pmod{4}$.\nNow, for $1 \\leq i \\leq k$, there are $\\operatorname{Card} Q_{i} - 1$ consecutive pairs of quadratic residues in $Q_{i}$ and $\\operatorname{Card} N_{i} - 1$ consecutive pairs of quadratic nonresidues in $N_{i}$. We deduce that the total number of consecutive pairs of quadratic residues in $\\{1, 2, \\ldots, p-1\\}$ is\n$$\n\\sum_{i=1}^{k} (\\operatorname{Card} Q_{i} - 1) = \\operatorname{Card} \\bigcup_{i=1}^{k} Q_{i} - k = \\frac{p-1}{2} - k = \\left\\lfloor \\frac{p-3}{4} \\right\\rfloor \\geq 1,\n$$\nand the total number of consecutive pairs of quadratic nonresidues in $\\{1, 2, \\ldots, p-1\\}$ is\n$\\sum_{i=1}^{k-1} (\\operatorname{Card} N_{i} - 1) = \\operatorname{Card} \\bigcup_{i=1}^{k-1} N_{i} - k + 1 = \\frac{p-1}{2} - k + 1 = \\left\\lfloor \\frac{p-1}{4} \\right\\rfloor \\geq 1$, if $p \\equiv 1 \\pmod{4}$ and\n$$\n\\sum_{i=1}^{k} (\\operatorname{Card} N_{i} - 1) = \\operatorname{Card} \\bigcup_{i=1}^{k} N_{i} - k = \\frac{p-1}{2} - k = \\left\\lfloor \\frac{p-1}{4} \\right\\rfloor \\geq 1,\n$$\nif $p \\equiv 3 \\pmod{4}$.\n\n- Second Step: We prove it for $n = p^{\\alpha}$, with $p \\geq 7$ is a prime number and $\\alpha \\geq 1$. We proceed by induction on $\\alpha$:\n\nFor $\\alpha = 1$, we have already proved in the first step that there exist two integers $a_{1}, b_{1}$, each relatively prime with $p$, such that $p$ divides $a_{1}^{2} - b_{1}^{2} + k$.\nFor $\\alpha \\geq 1$, assume that there exists an integer $a_{\\alpha} \\equiv a_{1} \\pmod{p}$ such that $p^{\\alpha}$ divides $a_{\\alpha}^{2} - b_{1}^{2} + k$. Consider the polynomial $P(X) = X^{2} - b_{1}^{2} + k$. Because $P(a_{\\alpha}) \\equiv 0 \\pmod{p^{\\alpha}}$ and $P'(a_{\\alpha}) = 2 a_{\\alpha} \\equiv 2 a_{1} \\not\\equiv 0 \\pmod{p}$, there exists, from Hensel's lemma, an integer $a_{\\alpha+1} \\equiv a_{\\alpha} \\pmod{p^{\\alpha}}$, and therefore $a_{\\alpha+1} \\equiv a_{1} \\pmod{p}$, such that $P(a_{\\alpha+1}) = a_{\\alpha+1}^{2} - b_{1}^{2} + k \\equiv 0 \\pmod{p^{\\alpha+1}}$. This proves the induction.\n\n- Last Step: We prove the general case when $n = p_{1}^{\\alpha_{1}} p_{2}^{\\alpha_{2}} \\cdots p_{l}^{\\alpha_{l}}$, where $7 \\leq p_{1} < p_{2} < \\cdots < p_{l}$ are prime numbers. We know from the second step, that for each $i = 1, 2, \\ldots, l$, there exist two integers $a_{i}, b_{i}$, relatively prime with $p_{i}$, such that $p_{i}^{\\alpha_{i}}$ divides $a_{i}^{2} - b_{i}^{2} + k$. From the Chinese remainder theorem, there exist two integers $a, b$ such that $a \\equiv a_{i} \\pmod{p_{i}^{\\alpha_{i}}}$ and $b \\equiv b_{i} \\pmod{p_{i}^{\\alpha_{i}}}$, for $i = 1, 2, \\ldots, l$. Therefore,\n$$\na^{2} - b^{2} + k \\equiv a_{i}^{2} - b_{i}^{2} + k \\equiv 0 \\quad \\pmod{p_{i}^{\\alpha_{i}}}\n$$\nfor all $i = 1, 2, \\ldots, l$, and hence\n$$\na^{2} - b^{2} + k \\equiv 0 \\quad (\\bmod n).\n$$\nOn the other hand,\n$$\n\\gcd(n, a) = \\prod_{i=0}^{l} \\gcd(p_{i}^{\\alpha_{i}}, a_{i}) = 1 \\text{ and } \\gcd(n, b) = \\prod_{i=0}^{l} \\gcd(p_{i}^{\\alpha_{i}}, b_{i}) = 1.\n$$\nThis completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40219,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $BD$ and $CE$ be altitudes of an arbitrary scalene triangle $ABC$ with orthocenter $H$ and circumcenter $O$. Let $M$ and $N$ be the midpoints of sides $AB$, respectively $AC$, and $P$ the intersection point of lines $MN$ and $DE$. Prove that lines $AP$ and $OH$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because $\\angle ADH = \\angle HEA = 90^{\\circ}$, quadrilateral $AEHD$ is cyclic with $AH$ a diameter of its circumcircle $\\omega_{H}$.\n\nBecause $\\angle ANO = \\angle OMA = 90^{\\circ}$, quadrilateral $AMON$ is cyclic with $AO$ a diameter of its circumcircle $\\omega_{O}$.\n\nLet $\\omega_{9}$ be the nine point circle of the triangle $ABC$. It is the circumcircle of quadrilateral $DEMN$.\n\n![](attached_image_1.png)\n\nBecause the circles $\\omega_{H}$ and $\\omega_{9}$ intersect at $D$ and $E$, the line $DE$ is radical axis of $\\omega_{H}$ and $\\omega_{9}$. Because the circles $\\omega_{O}$ and $\\omega_{9}$ intersect at $M$ and $N$, the line $MN$ is radical axis of $\\omega_{O}$ and $\\omega_{9}$. We deduce that $P$ is the radical center of the circles $\\omega_{H}, \\omega_{O}$ and $\\omega_{9}$, and therefore, the line $AP$ is the radical axis of the circles $\\omega_{H}$ and $\\omega_{O}$, which is perpendicular to the line $H' O'$, where $H'$ is the center of $\\omega_{H}$ and $O'$ the center of $\\omega_{O}$. But $H'$ is the midpoint of $AH$ and $O'$ is the midpoint of $AO$. We deduce that $AP$ is perpendicular to $HO$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40220,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and $G$ its centroid. Let $G_{a}$, $G_{b}$ and $G_{c}$ be the orthogonal projections of $G$ on sides $BC$, $CA$, respectively $AB$. If $S_{a}$, $S_{b}$ and $S_{c}$ are the symmetrical points of $G_{a}$, $G_{b}$, respectively $G_{c}$ with respect to $G$, prove that $AS_{a}$, $BS_{b}$ and $CS_{c}$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H_{a}$ be the foot of altitude from $A$, $M_{a}$ the midpoint of side $BC$, $A'$ the intersection point of line $AS_{a}$ and side $BC$ and $A''$ the intersection point of the parallel line to $BC$ passing through $S_{a}$ with $AM_{a}$.\n\n![](attached_image_1.png)\n\nBecause $GG_{a}$ and $AH_{a}$ are parallel, we have\n$$\n\\frac{H_{a}M_{a}}{G_{a}M_{a}} = \\frac{AM_{a}}{GM_{a}} = 3\n$$\nBecause $S_{a}A''$ and $BC$ are parallel, we have\n$$\n\\frac{G_{a}M_{a}}{S_{a}A''} = \\frac{G_{a}G}{S_{a}G} = 1\n$$\nWe also have\n$$\n\\frac{S_{a}A''}{A'M_{a}} = \\frac{AA''}{AM_{a}} = 1 - \\frac{M_{a}A''}{AM_{a}} = 1 - 2\\frac{M_{a}G}{AM_{a}} = \\frac{1}{3}\n$$\nMultiplying these three relations we deduce that $A'M_{a} = M_{a}H_{a}$, which means that $AS_{a}$ and $AH_{a}$ are isotomic conjugate. Similarly, $BS_{b}$ and $BH_{b}$ are isotomic conjugate and $CS_{c}$ and $CH_{c}$ are isotomic conjugate. Since the altitudes of a triangle are concurrent, so are $AS_{a}$, $BS_{b}$ and $CS_{c}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40221,
    "subject": "Mathematics (Multi-modal)",
    "question": "أثبت أنّه لا يمكن كتابة كثيرة الحدود $P(x)=\\left(x^{2}-12 x+11\\right)^{4}+23$ كاصل ضرب ثلاث كثيرات حدود غير ثابتة، معاملاتُها صحيحة.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40222,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integer solutions of the equation $x^{2} y^{5} - 2^{x} 5^{y} = 2015 + 4 x y$.",
    "options": [],
    "answer": "(5, 3)",
    "solution": "Notice that if $x < 0$ or $y < 0$ then the term $2^{x} 5^{y}$ is not an integer while $x^{2} y^{5} - 2015 - 4 x y$ is an integer which is impossible. If $x = 0$ or $y = 0$ then the equation becomes $-2^{x} 5^{y} = 2015$ which has no solution. We deduce that both $x, y$ are positive integers. Moreover, because $2^{x} 5^{y} + 2015 + 4 x y$ is an odd integer, both $x, y$ are odd integers.\n\nWorking modulo $5$, we get $x^{2} y \\equiv 4 x y$ and therefore either $x \\equiv 0, 4$ or $y \\equiv 0 \\pmod{5}$.\n\nWorking modulo $8$, and because $x, y$ are odd, we get $y - 2^{x} \\cdot 5 \\equiv -1 + 4$ which is equivalent to $y \\equiv 5(2^{x} - 1) \\pmod{8}$. Therefore, either $x = 1$ and $y \\geq 5$ or $x \\geq 3$ and $y \\equiv 3 \\pmod{8}$.\n\nIf $x = 1$ and $y \\geq 5$ then the equation becomes $y^{5} - 2 \\cdot 5^{y} = 2015 + 4 y$ and has no solution since the left hand side is negative (We prove it by induction on $y \\geq 5$).\n\nIf $x = 3$ then $y \\equiv 3 \\pmod{8}$. But $y \\equiv 0 \\pmod{5}$. Therefore $y \\geq 35$ and $9 \\cdot y^{5} - 8 \\cdot 5^{y}$ is negative so the equation has no solution.\n\nIf $x \\geq 5$ and $y \\neq 3$ then $y \\geq 5$. In this case $x^{2} < 2^{x}$ and $y^{5} \\leq 5^{y}$ (We prove it by induction) and therefore $x^{2} y^{5} - 2^{x} 5^{y}$ is negative and the equation has no solution.\n\nIf $x \\geq 7$ and $y = 3$ then because $x \\equiv 0, 4 \\pmod{5}$, $x \\geq 9$. Therefore $x^{2} \\cdot 3^{5} - 2^{x} 5^{3} < 3^{5}(x^{2} - 2^{x-2})$ is negative and therefore the equation has no solution.\n\nIf $x = 5$ and $y = 3$, the equation is satisfied.\n\nHence the unique solution is $(5, 3)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40223,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, $H_{a}, H_{b}$ and $H_{c}$ the feet of its altitudes from $A, B$ and $C$, respectively, $T_{a}, T_{b}, T_{c}$ its intouch points on the sides $BC, CA$ and $AB$, respectively. The circumcircles of triangles $A H_{b} H_{c}$ and $A T_{b} T_{c}$ intersect again at $A'$. The circumcircles of triangles $B H_{c} H_{a}$ and $B T_{c} T_{a}$ intersect again at $B'$. The circumcircles of triangles $C H_{a} H_{b}$ and $C T_{a} T_{b}$ intersect again at $C'$. Prove that the points $A', B', C'$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet $H$ and $I$ be the orthocenter and the incenter of triangle $ABC$, respectively. Because $\\angle A H_{b} H = \\angle H H_{c} A = 90^{\\circ}$, $AH$ is a diameter of the circumcircle of $A H_{c} H H_{b}$, and therefore $\\angle A A' H = 90^{\\circ}$. Because $\\angle A T_{b} I = \\angle I T_{c} A = 90^{\\circ}$, $AI$ is a diameter of the circumcircle of $A T_{c} I T_{b}$, and therefore $\\angle A A' I = 90^{\\circ}$. We deduce that point $A'$ is on the line $HI$.\n\nSimilarly, we prove that the points $B'$ and $C'$ are on the line $HI$, which prove that the points $A', B', C'$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40224,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, with $AB < AC$, $D$ the foot of the altitude from $A$, $M$ the midpoint of $BC$, and $B'$ the symmetric of $B$ with respect to $D$. The perpendicular line to $BC$ at $B'$ intersects $AC$ at point $P$. Prove that if $BP$ and $AM$ are perpendicular then triangle $ABC$ is right-angled.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $E$ be the intersection point of $AD$ and $BP$. Because $AD$ is perpendicular to $BM$ and $BP$ is perpendicular to $AM$, the point $E$ is the orthocenter of triangle $ABM$ and therefore $ME$ is perpendicular to $AB$.\n\n![](attached_image_1.png)\n\nBecause point $D$ is the midpoint of the segment $BB'$ and $DE$ and $B'P$ are parallel, we deduce that point $E$ is the midpoint of segment $BP$. But $M$ is the midpoint of segment $BC$. We deduce that lines $ME$ and $BC$ are parallel.\n\nIt follows that $AB$ and $AC$ are perpendicular and triangle $ABC$ is right-angled.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40225,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, $\\Gamma$ its circumcircle, $I$ its incenter, and $\\omega$ a tangent circle to the line $AI$ at $I$ and to the side $BC$. Prove that the circles $\\Gamma$ and $\\omega$ are tangent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ be the midpoint of arc $BC$ not containing $A$, $E$ the tangent point of $BC$ to $\\omega$, $F$ the second intersection point of $EM$ with $\\omega$. Remember that $M$ is the circumcenter of triangle $BCI$ and therefore $MI = MB$.\n\n![](attached_image_1.png)\n\nBecause $MI$ is tangent to $\\omega$, we have from the power of the point $M$ with respect to $\\omega$\n$$\nME \\cdot MF = MI^2 = MB^2.\n$$\nWe deduce that the line $BM$ is tangent to the circumcircle of triangle $BEF$. Therefore\n$$\n\\angle BFM = \\angle MBE = \\angle MAC = \\angle BAM.\n$$\nThis means that point $F$ is on the circle $\\Gamma$.\n\nBecause the tangent line to $\\Gamma$ at $M$ is parallel to the tangent line to $\\omega$ at $E$ and $F, E, M$ are collinear and $F$ is an intersection point of $\\Gamma$ and $\\omega$, $F$ is the center of the homothety of the two circles $\\omega$ and $\\Gamma$ and therefore, they are tangent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40226,
    "subject": "Mathematics (Multi-modal)",
    "question": "We color each unit square of a $8 \\times 8$ table into green or blue such that there are $a$ green unit squares in each $3 \\times 3$ square and $b$ green unit squares in each $2 \\times 4$ rectangle. Find all possible values of ($a, b$).",
    "options": [],
    "answer": "a = 0, b = 0 or a = 9, b = 8",
    "solution": "By tiling our $8 \\times 8$ table by eight $2 \\times 4$ rectangles like in Tiling (1) we find that the total number of green unit squares in the table is $8b$.\n\nBy tiling our $8 \\times 8$ table by four $3 \\times 3$ squares, three $2 \\times 4$ rectangles and one $2 \\times 2$ square, like in Tiling (2), we find that the total number of green unit squares in the table is $4a + 3b + x$, where $x$ is the number of green unit squares in the $2 \\times 2$ square in the left upper corner. Notice that by just rotating our tiling (2) so that the $2 \\times 2$ square occupies at each time the four corners of the table, we deduce that the four $2 \\times 2$ unit squares in the four corners, all have the same number of green unit squares $x$.\n\nFinally, by tiling our $8 \\times 8$ table by six $2 \\times 4$ rectangles and four $2 \\times 2$ squares, like in Tiling (3), we find that the total number of green unit squares in the table is $6b + 4x$\n\n![](attached_image_1.png)\nTiling (1)\n![](attached_image_2.png)\nTiling (2)\n![](attached_image_3.png)\nTiling (3)\n\nPutting together all these totals, we obtain the relations\n$$\n8b = 4a + 3b + x = 6b + 4x\n$$\nfrom which we deduce that $b = 2x$ and $9x = 4a$. Since $0 \\leq a \\leq 9$, either $a = 0$ and $b = 0$ or $a = 9$ and $b = 8$. In other words, either we color the whole table or we don't color at all.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40227,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$, $q$ be two different odd prime numbers and $n$ an integer such that $p q$ divides $n^{p q} + 1$. Prove that if $p^{3} q^{3}$ divides $n^{p q} + 1$ then either $p^{2}$ divides $n + 1$ or $q^{2}$ divides $n + 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because $p q$ divides $n^{p q} + 1$, neither $p$ nor $q$ divides $n$.\nAssume $p < q$. We have from Fermat's little theorem\n$$\n0 \\equiv n^{p q} + 1 \\equiv n^{q} + 1 \\quad (\\bmod p).\n$$\nTherefore $n^{2 q} \\equiv 1 (\\bmod p)$. But $n^{p-1} \\equiv 1 (\\bmod p)$ and $\\gcd(p-1, q) = 1$, since $p-1 < q$ and $q$ is prime. We deduce that $n^{2} \\equiv 1 (\\bmod p)$ and hence $p$ divides $n + 1$.\nLet $n + 1 = a p$ for an integer $a$. We have\n$$\n\\begin{aligned}\nn^{p q - 1} - n^{p q - 2} + \\cdots - n + 1 &\\equiv \\sum_{i=0}^{p q - 1} (1 - a p)^{i} \\equiv \\sum_{i=0}^{p q - 1} (1 - i a p) \\\\\n&\\equiv p q - \\frac{p q - 1}{2} a p^{2} q \\equiv p q \\not\\equiv 0 \\quad \\left(\\bmod p^{2}\\right).\n\\end{aligned}\n$$\nTherefore $p^{2}$ does not divide $n^{p q - 1} - n^{p q - 2} + \\cdots - n + 1$ and since $p^{3}$ divides\n$$\nn^{p q} + 1 = (n + 1) \\left(n^{p q - 1} - n^{p q - 2} + \\cdots - n + 1\\right),\n$$\nthen $p^{2}$ divides $n + 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40228,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exist infinitely many non prime positive integers $n$ such that $7^{n-1} - 3^{n-1}$ is divisible by $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will look for integers of the form $n = 7^{a} - 3^{a}$ with $a$ dividing $n-1$. Clearly, if $a$ exists then $n = 7^{a} - 3^{a}$ divides $7^{n-1} - 3^{n-1}$.\n\nLet $a = 3^{r}$ for $r \\geq 1$. We have $n = 7^{a} - 3^{a} \\equiv (-1)^{3^{r}} - 3^{3^{r}} \\equiv 4 \\pmod{8}$. We deduce that $n$ is a composite number.\n\nOn the other hand, because $3$ divides $7-1$, we have from the Lifting The Exponent theorem that $v_{3}\\left(7^{a} - 1\\right) = v_{3}(7-1) + v_{3}(a) = r+1$. We deduce that $a = 3^{r}$ divides $7^{3^{r}} - 1$ and therefore it divides $n-1 = 7^{3^{r}} - 1 + 3^{3^{r}}$. But there are infinitely many such $a$. This solves the problem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40229,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geq 2$ be an integer and $p_{1} < p_{2} < \\cdots < p_{n}$ prime numbers. Prove that there exists an integer $k$ relatively prime with $p_{1} p_{2} \\cdots p_{n}$ and such that $\\operatorname{gcd}\\left(k + p_{1} p_{2} \\cdots p_{i},\\ p_{1} p_{2} \\cdots p_{n}\\right) = 1$ for all $i = 1, 2, \\ldots, n-1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The problem is equivalent to proving that $k$ is relatively prime with $p_{1} p_{2} \\cdots p_{n}$ and such that for all $i = 2, \\ldots, n$, none of the integers $k,\\ k + p_{1}, \\ldots, k + p_{1} \\cdots p_{i-1}$ is divisible by $p_{i}$.\n\nLet $i \\in \\{2, \\ldots, n\\}$. Notice that $p_{i} \\geq i + 1$. So the numbers $-p_{1}, -p_{1} p_{2}, \\ldots, -p_{1} p_{2} \\cdots p_{i-1}$ will not cover all the non-zero residues modulo $p_{i}$. Let $k_{i}$ be one of these non-covered non-zero residues. Clearly, none of the numbers $k_{i}, k_{i} + p_{1}, \\ldots, k_{i} + p_{1} \\cdots p_{i-1}$ is congruent to $0$ modulo $p_{i}$.\n\nThis defines $k_{2}, k_{3}, \\ldots, k_{n}$ and let $k_{1} = 1$. By the Chinese remainder theorem there exists an integer $k$ such that $k \\equiv k_{i} \\not\\equiv 0 \\pmod{p_{i}}$ for all $i = 1, 2, \\ldots, n$. The integer $k$ is relatively prime with $p_{1} p_{2} \\cdots p_{n}$. On the other hand, for all $1 \\leq j < i \\leq n$, we have $k + p_{1} p_{2} \\cdots p_{j} \\equiv k_{i} + p_{1} p_{2} \\cdots p_{j} \\not\\equiv 0 \\pmod{p_{i}}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40230,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each positive integer $n$, define $s(n) = \\sum_{k=0}^{n} r_{k}$, where $r_{k}$ is the remainder when $\\binom{n}{k}$ is divided by $3$. Find all positive integers $n$ such that $s(n) \\geq n$.",
    "options": [],
    "answer": "All positive integers whose base-three expansion has no zero digits and either (i) has at most one digit equal to one, or (ii) has exactly two digits equal to one with the most significant digit equal to one. In particular, equality s(n) = n holds precisely when the base-three expansion is one, followed by some twos, and ending with one.",
    "solution": "Let $0 \\leq k \\leq n$ be two integers written in base $3$ as\n$$\nn = ^{(3)} \\overline{n_{m} n_{m-1} \\cdots n_{1} n_{0}}, \\quad k = ^{(3)} \\overline{k_{m} k_{m-1} \\cdots k_{1} k_{0}}\n$$\nsuch that $n_{m} \\neq 0$. We know from Lucas's theorem that\n$$\nr_{k} \\equiv \\binom{n}{k} \\equiv \\prod_{i=0}^{m} \\binom{n_{i}}{k_{i}} \\quad (\\bmod 3)\n$$\nIf there exists $0 \\leq i \\leq m$ such that $k_{i} > n_{i}$ then $\\binom{n_{i}}{k_{i}} = 0$ and therefore $r_{k} = 0$. Therefore, from now on we will restrict to $0 \\leq k \\leq n$ with $k_{i} \\leq n_{i}$ for all $0 \\leq i \\leq m$.\nLet $0 \\leq i \\leq m$. If $n_{i} = 0$ then $k_{i} = 0$. If $n_{i} = 1$ then $k_{i} \\in \\{0, 1\\}$. In all these cases $\\binom{n_{i}}{k_{i}} = 1$. If $n_{i} = 2$ then $k_{i} \\in \\{0, 1, 2\\}$ and $\\binom{n_{i}}{k_{i}} = 1$ if $k_{i} \\neq 1$ and $\\binom{n_{i}}{k_{i}} = 2$ if $k_{i} = 1$. Therefore, if the number $a$ of pairs $(n_{i}, k_{i}) = (2, 1)$, when $0 \\leq i \\leq m$, is odd then $\\binom{n}{k} \\equiv 2^{a} \\equiv 2 \\pmod{3}$, otherwise $\\binom{n}{k} \\equiv 1 \\pmod{3}$. We deduce that\n$$\ns(n) = 2A(n) + B(n),\n$$\nwhere $A(n)$ is the number of pairs of integers $(n, k)$ such that $0 \\leq k_{i} \\leq n_{i}$ for all $0 \\leq i \\leq m$ and the number of pairs $(n_{i}, k_{i}) = (2, 1)$, when $0 \\leq i \\leq m$, is odd, while $B(n)$ is the number of pairs of integers $(n, k)$ such that $0 \\leq k_{i} \\leq n_{i}$ for all $0 \\leq i \\leq m$ and the number of pairs $(n_{i}, k_{i}) = (2, 1)$, when $0 \\leq i \\leq m$, is even.\nDenote by $r$ the number of $0$'s in $n_{0}, n_{1}, n_{2}, \\ldots, n_{m}$, by $s$ the number of $1$'s and by $t$ the number of $2$'s. We have\n$$\nA(n) = 2^{s} \\left( \\binom{t}{1} 2^{t-1} + \\binom{t}{3} 2^{t-3} + \\binom{t}{5} 2^{t-5} + \\cdots \\right)\n$$\nand\n$$\nB(n) = 2^{s} \\left( \\binom{t}{0} 2^{t} + \\binom{t}{2} 2^{t-2} + \\binom{t}{4} 2^{t-4} + \\cdots \\right).\n$$\nWe deduce that\n$$\n\\begin{aligned}\ns(n) & = 2^{s} \\left( (\\binom{t}{0} + \\binom{t}{1}) 2^{t} + (\\binom{t}{2} + \\binom{t}{3}) 2^{t-2} + \\cdots \\right) \\\\\n& = 2^{s} \\left( \\binom{t+1}{1} 2^{t} + \\binom{t+1}{3} 2^{t-2} + \\cdots \\right) \\\\\n& = 2^{s} \\cdot \\frac{(2+1)^{t+1} - (2-1)^{t+1}}{2} = 2^{s-1} (3^{t+1} - 1).\n\\end{aligned}\n$$\n\na. If the first digit on the left of $n$ is $2$ then\n$$\n\\begin{aligned}\nn \\geq \\overline{2 \\underbrace{00 \\cdots 0}_{r} \\underbrace{11 \\cdots 1}_{s} \\underbrace{22 \\cdots 2}_{t-1}} & = 2 \\cdot 3^{r+s+t-1} + \\frac{3^{s} - 1}{2} \\cdot 3^{t-1} + 2 \\cdot \\frac{3^{t-1} - 1}{2} \\\\\n& = \\frac{4 \\cdot 3^{r+s} + 3^{s} + 1}{2} \\cdot 3^{t-1} - 1\n\\end{aligned}\n$$\nTherefore\n$$\nn - s(n) \\geq \\frac{(4 \\cdot 3^{r} + 1) \\cdot 3^{s} - 9 \\cdot 2^{s} + 1}{2} \\cdot 3^{t-1} - 1 + 2^{s-1}.\n$$\n- If $r \\geq 1$ then $n - s(n) \\geq \\frac{13 - 9 + 1}{2} \\cdot \\frac{1}{3} - 1 + \\frac{1}{2} > 0$\n- If $r = 0$ and $s \\geq 2$ then\n$$\nn - s(n) \\geq \\frac{45 \\cdot 3^{s-2} - 36 \\cdot 2^{s-2} + 1}{2} \\cdot 3^{t-1} - 1 + 2 > 0.\n$$\n- If $r = 0$ and $s = 1$ then $n \\leq \\overline{22 \\cdots 2 1}_{t} = 3^{t+1} - 2 < 3^{t+1} - 1 = s(n)$.\n- If $r = s = 0$ then $n = \\overline{22 \\cdots 2}_{t} = 3^{t} - 1 < \\frac{3^{t+1} - 1}{2} = s(n)$.\n\nb. If the first digit on the left of $n$ is $1$ then $s \\geq 1$ and\n$$\n\\begin{aligned}\nn \\geq \\overline{1 \\underbrace{00 \\cdots 0}_{r} \\underbrace{11 \\cdots 1}_{s-1} \\underbrace{22 \\cdots 2}_{t}} & = 3^{r+s+t-1} + \\frac{3^{s-1} - 1}{2} \\cdot 3^{t} + 2 \\cdot \\frac{3^{t} - 1}{2} \\\\\n& = \\frac{2 \\cdot 3^{r+s-1} + 3^{s-1} + 1}{2} \\cdot 3^{t} - 1\n\\end{aligned}\n$$\nTherefore\n$$\nn - s(n) \\geq \\frac{(2 \\cdot 3^{r} + 1) \\cdot 3^{s-1} - 6 \\cdot 2^{s-1} + 1}{2} \\cdot 3^{t} + 2^{s-1} - 1.\n$$\n- If $r \\geq 1$ and because $s \\geq 1$ we have\n$$\nn - s(n) \\geq \\frac{7 \\cdot 3^{s-1} - 6 \\cdot 2^{s-1} + 1}{2} \\cdot 3^{t} + 2^{s-1} - 1 \\geq 1 + 1 - 1 > 0.\n$$\n- If $r = 0$ and $s \\geq 3$ then\n$$\nn - s(n) \\geq \\frac{3 \\cdot (3^{s-1} - 2^{s}) + 1}{2} \\cdot 3^{t} + 2^{s-1} - 1 \\geq 2 + 4 - 1 > 0.\n$$\n- If $r = 0$ and $s = 2$ then $n \\leq \\overline{1 \\underbrace{22 \\cdots 2}_{t} 1} = 2(3^{t+1} - 1) = s(n)$, and the equality is satisfied precisely when $n = \\overline{1 \\underbrace{22 \\cdots 2}_{t} 1}$.\n- If $r = 0$ and $s = 1$ then\n$$\nn = \\overline{1 \\underbrace{22 \\cdots 2}_{t}} = 3^{t+1} - 3^{t} - 1 < 3^{t+1} - 1 = s(n).\n$$\n\nFinally, we deduce that if in the digits of $n$ written in base $3$,\n(a) there is at least one $0$, or three $1$'s, or no $0$'s and two $1$'s with the digit on the left is $2$, then $s(n) < n$;\n(b) $n = \\overline{122 \\cdots 21}$, then $s(n) = n$;\n(c) there are no $0$'s and two $1$'s with the digit on the left is $1$ and the digit on the right is $2$, or there are no $0$'s and at most one $1$, then $s(n) > n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40231,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle inscribed in the circle $(O)$, $H$ the foot of the altitude of $ABC$ at $A$ and $P$ a point inside $ABC$ lying on the bisector of $\\angle BAC$. The circle of diameter $AP$ cuts $(O)$ again at $G$. Let $L$ be the projection of $P$ on $AH$. Prove that if $GL$ bisects $HP$ then $P$ is the incenter of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AP$ intersect $BC$ at $F$ and $(O)$ again at $D$. Because $LP$ and $BC$ are parallel, we have\n$$\n\\begin{aligned}\n\\angle LGA &= \\angle LPD = \\angle BFD = \\angle BAD + \\angle CBA \\\\\n&= \\angle DAC + \\angle CBA = \\angle DBA = \\angle DGA\n\\end{aligned}\n$$\nThis implies that $G$, $L$, $D$ are collinear.\n\nLet $GL$ cut $BC$ at $E$. Because $LE$ bisects $PH$ and $\\angle EHL = \\angle HLP = 90^\\circ$, quadrilateral $PLHE$ is a rectangle. We deduce, using Thales's theorem, that\n$$\n\\frac{DP}{DA} = \\frac{PE}{AL} = \\frac{LH}{AL} = \\frac{PF}{AP}\n$$\nTherefore, $\\frac{DP}{DA} = 1 - \\frac{AP}{DA} = 1 - \\frac{PF}{DP} = \\frac{DF}{DP}$. We deduce that $DP^2 = DF \\cdot DA$.\n\n![](attached_image_1.png)\n\nOn the other hand, $\\angle DBF = \\angle DAC = \\angle BAF$. This means that line $BD$ is tangent to the circumcircle of triangle $ABF$ and therefore $DB^2 = DF \\cdot DA = DP^2$. But $D$ is the circumcenter of triangle $BCI$, where $I$ is the incenter of $ABC$. We deduce that $P$ is the incenter of triangle $ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40232,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 22 chairs in a round table. Find the minimum $n$ such that for any group of $n$ people sitting in the table, we always can find two people with exactly 2 or 8 chairs between them.",
    "options": [],
    "answer": "12",
    "solution": "We consider the graph with 22 vertices $0, 1, \\ldots, 21$. Two vertices $x$ and $y$ are connected by an edge if $x - y \\equiv \\pm 3, \\pm 9 \\pmod{22}$. We consider only $\\pm 3$ modulo 22, then we have a cycle through all vertices. If we choose any $n \\geq 12$ vertices, there are two vertices (corresponding to 2 people) with exactly 2 chairs between them. If $n = 11$, we can choose 11 vertices: $0, 6, 12, \\ldots, 60 \\pmod{22}$. It is clear that $10x \\not\\equiv \\pm 3, \\pm 9 \\pmod{22}$ for every $x$. It means there is an arrangement for 11 people such that no two people with exactly 2 or 8 chairs between them.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40233,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many sequences of integers\n$$\n1 \\leq a_{1} \\leq a_{2} \\leq \\ldots \\leq a_{11} \\leq 2015\n$$\nthat satisfy $a_{i} \\equiv i^{2} (\\bmod\\ 12)$ for all $1 \\leq i \\leq 11$ are there?",
    "options": [],
    "answer": "C(173, 11)",
    "solution": "Let $r_{i}$ be the remainder when $i^{2}$ is divided by $12$, and $a_{i} = 12 k_{i} + r_{i}$ for some nonnegative integer $k_{i}$, when $1 \\leq i \\leq 11$. From the following table showing the values of $r_{i}$\n\n| $i$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $r_{i}$ | 1 | 4 | 9 | 4 | 1 | 0 | 1 | 4 | 9 | 4 | 1 |\n\nwe deduce that the inequality\n$$\n1 \\leq a_{1} \\leq a_{2} \\leq \\ldots \\leq a_{11} \\leq 2015\n$$\nis equivalent to the inequality\n$$\n0 \\leq k_{1} \\leq k_{2} \\leq k_{3} < k_{4} < k_{5} < k_{6} \\leq k_{7} \\leq k_{8} \\leq k_{9} < k_{10} < k_{11} \\leq 167\n$$\nwhich is equivalent to\n$$\n\\begin{aligned}\n0 \\leq & k_{1} < k_{2} + 1 < k_{3} + 2 < k_{4} + 2 < k_{5} + 2 < k_{6} + 2 < \\\\\n& < k_{7} + 3 < k_{8} + 4 < k_{9} + 5 < k_{10} + 5 < k_{11} + 5 \\leq 172 .\n\\end{aligned}\n$$\nTherefore, there are $\\binom{173}{11}$ such sequences.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40234,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be an odd prime number. Prove that there exists a unique integer $k$ such that $0 \\leq k \\leq p^{2}$ and $p^{2}$ divides $k(k+1)(k+2) \\cdots(k+p-3)-1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that there exists an integer $k$ such that $0 \\leq k \\leq p^{2}$ and $p^{2}$ divides $k(k+1)(k+2) \\cdots(k+p-3)-1$. Because $p$ does not divide any of the $p-2$ consecutive integers $k,(k+1),(k+2), \\ldots,(k+p-3)$, either $k \\equiv 1$ or $2 \\bmod p$. If $k \\equiv 2 \\bmod p$ then, by Wilson's theorem,\n$$\nk(k+1)(k+2) \\cdots(k+p-3)-1 \\equiv (p-1)!-1 \\equiv -2 \\not \\equiv 0 \\bmod p.\n$$\nWe deduce that $k \\equiv 1 \\bmod p$.\n\nOn the other hand, we know from Wilson's theorem that $(p-2)!-1 \\equiv 0 \\bmod p$. This means that there exists a unique integer $i_{0} \\in \\{1, \\ldots, p\\}$ such that\n$$\n(p-2)!-1 \\equiv i_{0} p \\bmod p^{2}.\n$$\nLet $k = i p + 1$ for some integer $i \\in \\{0,1, \\ldots, p-1\\}$. We have\n$$\n\\begin{aligned}\nk(k+1)(k+2) \\cdots(k+p-3)-1 &\\equiv (i p+1)(i p+2) \\cdots(i p+p-2)-1 \\\\\n&\\equiv (p-2)!-1 + i p \\sum_{j=1}^{p-2} \\frac{(p-2)!}{j} \\bmod p^{2}\n\\end{aligned}\n$$\nand\n$$\n\\sum_{j=1}^{p-2} \\frac{(p-2)!}{j} \\equiv (p-2)! \\sum_{j=1}^{p-2} j^{-1} \\equiv \\sum_{j=1}^{p-2} j \\equiv \\frac{(p-2)(p-1)}{2} \\equiv 1 \\bmod p\n$$\nWe deduce that\n$$\nk(k+1)(k+2) \\cdots(k+p-3)-1 \\equiv (i_{0}+i) p \\bmod p^{2}.\n$$\nThis proves that $k(k+1)(k+2) \\cdots(k+p-3)-1 \\equiv 0 \\bmod p^{2}$ only for $k = (p-i_{0}) p + 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40235,
    "subject": "Mathematics (Multi-modal)",
    "question": "Hamza and Majid play a game on a horizontal $3 \\times 2015$ white board. They alternate turns, with Hamza going first. A legal move for Hamza consists of painting three unit squares forming a horizontal $1 \\times 3$ rectangle. A legal move for Majid consists of painting three unit squares forming a vertical $3 \\times 1$ rectangle. No one of the two players is allowed to repaint already painted squares. The last player to make a legal move wins. Which of the two players, Hamza or Majid, can guarantee a win no matter what strategy his opponent chooses and what is his strategy to guarantee a win?",
    "options": [],
    "answer": "Hamza. Strategy: Always paint a horizontal three-square segment entirely within an untouched three-by-three block until none remain; this claims each block and prevents any vertical move there, guaranteeing at least 1008 moves for Hamza and ensuring he makes the last move.",
    "solution": "Hamza has a winning strategy.\nWe divide the rectangle into $671$ squares of size $3 \\times 3$ and a small rectangle of size $3 \\times 2$.\nHamza will play as follows: He will paint at each time a horizontal $1 \\times 3$ rectangle in a white $3 \\times 3$ square (all unit squares in this $3 \\times 3$ square are not colored yet) until he cannot continue to do so. In each step, Majid can only paint a $3 \\times 1$ rectangle in a square of size $3 \\times 3$ or in the small rectangle $3 \\times 2$. So the number of $3 \\times 3$ squares that will be used by a rectangle of Hamza is at least $336$.\n\nNote that, whenever Hamza paints a horizontal $1 \\times 3$ in a $3 \\times 3$ square then Majid cannot paint a vertical $3 \\times 1$ in such a square. Hence, Hamza can play at least $336 \\times 3 = 1008$ steps. Since the game ends after at most $2015$ steps so Majid can play at most $2015 - 1008 = 1007 < 1008$ steps. In other words, Hamza has a winning strategy.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40236,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{1}, a_{2}, \\ldots, a_{n}$ be positive real numbers such that\n$$\na_{1}+a_{2}+\\cdots+a_{n}=a_{1}^{2}+a_{2}^{2}+\\cdots+a_{n}^{2} .\n$$\nProve that\n$$\n\\sum_{1 \\leq i<j \\leq n} a_{i} a_{j}\\left(1-a_{i} a_{j}\\right) \\geq 0 .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The original inequality is equivalent to\n$$\n2 \\sum_{1 \\leq i<j \\leq n} a_{i} a_{j} \\geq 2 \\sum_{1 \\leq i<j \\leq n} a_{i}^{2} a_{j}^{2},\n$$\nor\n$$\n\\left(\\sum_{i=1}^{n} a_{i}\\right)^{2}-\\sum_{i=1}^{n} a_{i}^{2} \\geq\\left(\\sum_{i=1}^{n} a_{i}^{2}\\right)^{2}-\\sum_{i=1}^{n} a_{i}^{4} .\n$$\nUsing the given hypothesis, it can be rewritten as\n$$\na_{1}^{4}+a_{2}^{4}+\\cdots+a_{n}^{4} \\geq a_{1}^{2}+a_{2}^{2}+\\cdots+a_{n}^{2} .\n$$\nTo prove this last inequality, we have from Holder's inequality\n$$\n\\left(a_{1}^{4}+a_{2}^{4}+\\cdots+a_{n}^{4}\\right)\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)^{2} \\geq\\left(a_{1}^{2}+a_{2}^{2}+\\cdots+a_{n}^{2}\\right)^{3} .\n$$\nUsing the given hypothesis again, we get the result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40237,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a positive integer divisible by all the integers $1, 2, \\ldots, 2015$ and $a_{1}, a_{2}, \\ldots, a_{k}$ numbers in $\\{1, 2, \\ldots, 2015\\}$ such that\n$$\n2S \\leq a_{1} + a_{2} + \\ldots + a_{k},\n$$\nProve that we can select from $a_{1}, \\ldots, a_{k}$ some numbers so that the sum of these selected numbers is equal to $S$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that because $S$ is divisible by $2013$, $2014$ and $2015$, we have $S \\geq 2013 \\times 2014 \\times 2015$.\n\nAssume that each integer $1, 2, \\ldots, 2015$ appears at most $2013$ times in the list of numbers $a_{1}, a_{2}, \\ldots, a_{k}$. We have\n$$\n2S \\leq a_{1} + \\cdots + a_{k} \\leq 2013(1 + 2 + \\cdots + 2015) < 2013 \\times 2014 \\times 2015,\n$$\nwhich is a contradiction.\n\nTherefore, there exists an integer $s$ which appears at least $2014$ times in the list of numbers $a_{1}, a_{2}, \\ldots, a_{k}$. Put $2014$ of this number $s$ aside from the list. The sum of the remaining numbers in the list is equal to $2S - 2014s \\geq 2S - 2014 \\times 2015 > S$.\n\nStart from the empty set and choose randomly $s$ integers $b_{1}, b_{2}, \\ldots, b_{s}$ from the remaining numbers in the list. If there exists $1 \\leq i \\leq s$ such that $b_{1} + b_{2} + \\cdots + b_{i}$ is divisible by $s$, put $b_{1}, b_{2}, \\ldots, b_{i}$ in the set and put back the other numbers in the remaining list of numbers. If not, by the pigeonhole principle, there exist $1 \\leq i < j \\leq s$ such that $b_{1} + b_{2} + \\cdots + b_{i} \\equiv b_{1} + b_{2} + \\cdots + b_{j} \\pmod{s}$ and therefore $b_{i+1} + b_{i+2} + \\cdots + b_{j}$ is divisible by $s$. In this case, put $b_{i+1}, b_{i+2}, \\ldots, b_{j}$ in the set and put back the other numbers in the remaining list of numbers. Keep repeating this process whenever the sum of the numbers in the set is less or equal to $S - 2015s$.\n\nNotice that each time the sum of the numbers in the set is divisible by $s$ and each time it increases by a number which is less or equal to $2015s$. Moreover, whenever the sum of the numbers in the set is less or equal to $S - 2015s$ it is always possible to choose randomly $s$ integers from the remaining numbers. Indeed, the sum of the remaining numbers is less or equal to $2S - 2014s - (S - 2015s) = S + s > 2015 \\times 2015 \\geq 2015s$.\n\nOnce the sum $T$ of the numbers in the set is strictly greater than $S - 2015s$, in this case we have $S \\geq T > 2015s$. Because both $S$ and $T$ are divisible by $s$, we have $S - T = ms$ where $m \\in \\{0, 1, 2, \\ldots, 2014\\}$. Take $m$ from the $2014$ of the number $s$ put aside and put them in the set to obtain a sum equal to $S$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40238,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the number of binary sequences $S$ of length $2015$ such that for any two segments $I_{1}, I_{2}$ of $S$ of the same length, we have\n- The sum of digits of $I_{1}$ differs from the sum of digits of $I_{2}$ by at most $1$;\n- If $I_{1}$ begins on the left end of $S$ then the sum of digits of $I_{1}$ is not greater than the sum of digits of $I_{2}$;\n- If $I_{2}$ ends on the right end of $S$ then the sum of digits of $I_{2}$ is not less than the sum of digits of $I_{1}$.",
    "options": [],
    "answer": "2016",
    "solution": "Note that if the sequence $S$ ends with zero then all digits of $S$ will be zero.\n\nNow suppose that the sequence $S$ ends with one. Let $k$ be the number of $1$'s in $S$ then $1 \\leq k \\leq 2015$. Let $\\alpha = \\frac{2015}{k} \\geq 1$. We consider the sequence $S_{k} = a_{1} a_{2} \\ldots a_{2015}$ where $a_{i} = 1$ if and only if $i \\in \\{ \\lceil \\alpha \\rceil, \\lceil 2 \\alpha \\rceil, \\ldots, \\lceil k \\alpha \\rceil \\}$.\n\nOne can check that this sequence has exactly $k$ ones and satisfies all the given conditions.\n\nNow we need to prove that those are all possible sequences.\n\nSuppose that $S$ is a sequence with exactly $k$ ones and also satisfying all the given conditions. We will show that we have only one possible arrangement for $k$ ones and $(2015 - k)$ zeros.\n\nLet $\\beta = \\frac{2015 - k}{2015}$. Starting from the right, we will show that for any $1 \\leq s \\leq 2015$, among the first $s$ rightmost digits, there are exactly $\\lfloor \\beta s \\rfloor$ zeros.\n\nIt is clear that the statement holds for $s = 1$ since the rightmost digit is a unit.\n\nSuppose that the statement holds for $s$. We consider two cases.\n\nCase 1. If $\\lfloor \\beta(s+1) \\rfloor = \\lfloor \\beta s \\rfloor + 1$. We need to show that the $(s+1)^{\\text{th}}$ rightmost digit of $S$ is zero. Otherwise, it follows that we have exactly $\\lfloor \\beta(s+1) \\rfloor - 1$ in the $(s+1)$ rightmost digits and any $s+1$ consecutive digits, there are at most $\\lfloor \\beta(s+1) \\rfloor$ zeros. We divide the sequence from the right into $t$ segments of length $s+1$ and possibly one leftmost segment of length $s'$ with $0 \\leq s' \\leq s$.\n\nThe number of zeros is at most\n$$\n\\begin{aligned}\n\\lfloor \\beta(s+1) \\rfloor - 1 & + (t-1) \\lfloor \\beta(s+1) \\rfloor + \\left\\lfloor \\beta s' \\right\\rfloor \\\\\n& \\leq \\beta \\left( t(s+1) + s' \\right) - 1 = 2014 - k\n\\end{aligned}\n$$\nwhich is a contradiction.\n\nCase 2. If $\\lfloor \\beta(s+1) \\rfloor = \\lfloor \\beta s \\rfloor$. We need to show that the $(s+1)^{\\text{th}}$ rightmost digit of $S$ is a unit. Otherwise, it follows that for any $s+1$ consecutive digits, there are at least $\\lfloor \\beta(s+1) \\rfloor + 1$ zeros. We divide the sequence from the right into $t$ segments of length $s+1$ and possibly one leftmost segment of length $s'$ with $0 \\leq s' \\leq s$. Note that in the first $s'$ rightmost digits, we have exactly $\\left\\lfloor \\beta s' \\right\\rfloor$ zeros. If in the leftmost segment of length $s'$ we have $\\left\\lfloor \\beta s' \\right\\rfloor + 1$ then the number of zeros is at least\n$$\nt (\\lfloor \\beta(s+1) \\rfloor + 1) + \\left\\lfloor \\beta s' \\right\\rfloor > \\beta ( t(s+1) + s' ) = 2015 - k\n$$\nwhich is a contradiction.\n\nWe deduce that, in each segment of length $s'$ we have exactly $\\left\\lfloor \\beta s' \\right\\rfloor$ zeros. We divide the sequence into segments of length $s'$ and possible one shorter segment from the left. Using the same argument as in the previous case, we do not have enough zeros, which is again a contradiction.\n\nTherefore, the statement holds for all $1 \\leq s \\leq 2015$, which implies that for any $0 \\leq k \\leq 2015$, there exists a unique binary sequence with $k$ units satisfying the given conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40239,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define the sequence $a_{1}, a_{2}, \\ldots$ as follows: $a_{1}=1$, and for every $n \\geq 2$, $a_{n}=n-2$ if $a_{n-1}=0$ and $a_{n}=a_{n-1}-1$, otherwise. Find the number of $1 \\leq k \\leq 2016$ such that there are non-negative integers $r, s$ and a positive integer $n$ satisfying $k=r+s$ and $a_{n+r}=a_{n}+s$.",
    "options": [],
    "answer": "51",
    "solution": "Let $N=n+r$ and $M=n$, then\n$$\nr=N-M,\\ s=a_{N}-a_{M}\\ \\text{ and }\\ k=r+s=\\left(a_{N}+N\\right)-\\left(a_{M}+M\\right) .\n$$\nWe need to find the number of possible values of $\\left(a_{N}+N\\right)-\\left(a_{M}+M\\right)$, where $N \\geq M$ and $a_{N} \\geq a_{M}$.\nIt is easy to see by induction that $a_{2^{k}}=0$ and thus $a_{2^{k}+1}=2^{k}-1$ for all $k \\geq 1$. The sequence is as the following\n$$\n1,0,1,0,3,2,1,0,7,6,5,4,3,2,1,0, \\ldots\n$$\nWe divide the sequence into blocks with $k^{\\text{th}}$ block contains $a_{i}$ for $2^{k-1}< i \\leq 2^{k}$. Within each block, the value $a_{n}+n$ is constant, and for the $k^{th}$ block $(k \\geq 1)$ it equals $2^{k}$.\nTherefore, $d=\\left(a_{N}+N\\right)-\\left(a_{M}+M\\right)$ is the difference of two powers of $2$. Now it's not difficult to show that there are $51$ such possible values. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40240,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P_{i}(x) = x^{2} + b_{i} x + c_{i}$ for $i = 1, 2, \\cdots, n$ be pairwise distinct polynomials of degree $2$ with real coefficients so that for any $0 \\leq i < j \\leq n$; $i, j \\in \\mathbb{N}$, the polynomial $Q_{i, j}(x) = P_{i}(x) + P_{j}(x)$ has only one real root. Find the greatest possible value of $n$.",
    "options": [],
    "answer": "3",
    "solution": "The answer $n = 3$. In fact, the polynomials $P_{1}(x) = x^{2} - 4$, $P_{2}(x) = x^{2} - 4x + 6$ and $P_{3}(x) = x^{2} - 8x + 12$ satisfy the conditions:\n\n$P_{1} + P_{2} = 2(x - 1)^{2}$,\n\n$P_{1} + P_{3} = 2(x - 2)^{2}$,\n\n$P_{2} + P_{3} = 2(x - 3)^{2}$.\n\nSuppose that there are four polynomials $P_{1}, P_{2}, P_{3}, P_{4}$ satisfying the conditions. Then\n\n$P_{1} + P_{2} = 2(x - t_{12})^{2}$,\n\n$P_{3} + P_{4} = 2(x - t_{34})^{2}$,\n\n$P_{1} + P_{3} = 2(x - t_{13})^{2}$,\n\n$P_{2} + P_{4} = 2(x - t_{24})^{2}$,\n\nwhere $t_{ij}$ is the multiple root of the polynomial $P_{i} + P_{j}$.\n\nLet $Q = P_{1} + P_{2} + P_{3} + P_{4}$. Then $Q$ has two representations\n\n$$\nQ = 2(x - t_{12})^{2} + 2(x - t_{34})^{2} \\text{ and } Q = 2(x - t_{13})^{2} + 2(x - t_{24})^{2}.\n$$\n\nBy considering linear and constant terms of both expressions of $Q$, we get\n\n$$\nt_{12} + t_{34} = t_{13} + t_{24} \\text{ and } t_{12}^{2} + t_{34}^{2} = t_{13}^{2} + t_{24}^{2}.\n$$\n\nThis shows that $(t_{12}, t_{34}) = (t_{24}, t_{13})$. But, if $t_{12} = t_{13} \\Rightarrow P_{2} = P_{3}$, and if $t_{12} = t_{24} \\Rightarrow P_{1} = P_{4}$, in both cases, we get a contradiction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40241,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$, $q$ be given primes and the sequence $(p_{n})_{n \\geq 1}$ defined recursively as follows: $p_{1} = p$; $p_{2} = q$ and $p_{n+2}$ is the largest prime divisor of the number $(p_{n} + p_{n+1} + 2016)$ for all $n \\geq 1$. Prove that this sequence is bounded, that is there exists a positive real number $M$ such that $a_{n} < M$ for all positive integers $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Put $k = 2016$ and $a_{n} = \\max \\{p_{n}, p_{n+1}\\}$ for each $n \\geqslant 1$.\nWe shall show that $a_{n+1} \\leqslant a_{n} + k + 2$. In fact, since $p_{n+1} \\leqslant a_{n}$, it remains to show that $p_{n+2} \\leqslant a_{n} + k + 2$.\nIf $p_{n+2} = 2$ then the inequality is trivially true. It suffices to consider the case $p_{n+2} > 2$.\nIf either $p_{n} = 2$ or $p_{n+1} = 2$ then $p_{n+2} \\leqslant p_{n} + p_{n+1} + k = a_{n} + k + 2$. Otherwise, both primes $p_{n}$ and $p_{n+1}$ are odd, in particular $p_{n} + p_{n+1} + k$ is even. But $p_{n+2} > 2$, we get\n$$\np_{n+2} \\leqslant \\frac{p_{n} + p_{n+1} + k}{2} < a_{n} + k + 2 .\n$$\nIt is clear that there is a positive integer $s$ such that $a_{1} \\leqslant s(k+3)! + 1$. We show, by induction that $a_{n} \\leqslant s(k+3)! + 1$. Indeed, suppose this is true for $n$, then\n$$\na_{n+1} \\leqslant a_{n} + k + 2 \\leqslant s(k+3)! + k + 3 .\n$$\nIf $a_{n+1} > s(k+3)! + 1$, then $1 < m \\leqslant k + 3$ with $m = a_{n+1} - q(k+3)!$. Hence, $m$ must be a divisor of $(k+3)!$, that is $m$ is a proper divisor of $s(k+3)! + m = a_{n+1}$. However, this cannot happen since $a_{n+1}$ is a prime.\nThus, $a_{n} \\leqslant s(k+3)! + 1$, and this completes the solution. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40242,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a > b > c > d$ be positive integers such that $a^{2} + a c - c^{2} = b^{2} + b d - d^{2}$. Prove that $a b + c d$ is a composite number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By contrary, assume that $p = a b + c d$ is a prime. In particular $(b, c) = (b, d) = 1$, and $a b \\equiv -c d \\pmod{p}$. We get\n$$\n\\begin{aligned}\n0 & = b^{2} (b^{2} + b d - d^{2}) - b^{2} (a^{2} + a c - c^{2}) \\\\\n& = b^{2} (b^{2} + b d - d^{2}) - (a b)^{2} - (a b)(b c) + b^{2} c^{2} \\\\\n& \\equiv (b^{2} + c^{2})(b^{2} + b d - d^{2}) \\pmod{p}\n\\end{aligned}\n$$\nTherefore, one of the numbers $b^{2} + c^{2}$ and $b^{2} + b d - d^{2}$ is divisible by $p$. If $p \\mid b^{2} + c^{2}$, then $b^{2} + c^{2} = p$ since $0 < b^{2} + c^{2} < 2(a b + c d) = 2p$. This shows $a b + c d = b^{2} + c^{2}$ and then $b \\mid c(c - d)$. But $b, c$ are coprime, we get $b \\mid c - d$ which is impossible.\nIf $p \\mid b^{2} + b d - d^{2}$, then $b^{2} + b d - d^{2} = p$ since $0 < b^{2} + b d - d^{2} < 2 b^{2} < 2 a b < 2p$. This shows that $a b + c d = b^{2} + b d - d^{2}$ and then\n$$\nb(a - b) = b d - c d - d^{2} = d(b - c - d)\n$$\nIn particular, $b - c - d > 0$ and $b \\mid b - c - d$ (since $b$, $d$ are coprime) which is impossible, again.\nHence, $a b + c d$ must be a composite number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40243,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two non-constant polynomials $P(x), Q(x)$ with real coefficients. For a real number $a$, we define\n$$\nP_{a} = \\{ z \\in \\mathbb{C} : P(z) = a \\} ; \\quad Q_{a} = \\{ z \\in \\mathbb{C} : Q(z) = a \\} .\n$$\nDenote by $K$ the set of real numbers $a$ such that $P_{a} = Q_{a}$. Suppose that the set $K$ contains at least two elements, prove that $P(x) = Q(x)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First we see that if the polynomial $P(x)-a$ has a root $\\alpha$ with multiplicity $k$, then $P'(x)$ also has the root $\\alpha$ with multiplicity $k-1$.\n\nAssume that $a, b$ are two distinct elements of $K$ and $r_{1}, r_{2}, \\ldots, r_{i}$ are the roots of $P(x)-a$ with multiplicity $k_{1}, k_{2}, \\ldots, k_{i}$, respectively.\nThen $r_{1}, r_{2}, \\ldots, r_{i}$ are also the roots of $Q(x)-a$.\nLet $t_{1}, t_{2}, \\ldots, t_{j}$ be the roots of $P(x)-b$ with multiplicity $s_{1}, s_{2}, \\ldots, s_{j}$, respectively.\nThen $t_{1}, t_{2}, \\ldots, t_{j}$ are also the roots of $Q(x)-b$.\n\nAssume that $\\deg P(x) \\geq \\deg Q(x)$, and $R(x) = P(x) - Q(x)$ is not identically zero. Thus, $\\deg P(x) = k_{1} + \\cdots + k_{i} = s_{1} + \\cdots + s_{j} \\geq \\deg R(x) \\geq i + j$, then we can get\n$$\n\\begin{aligned}\n\\deg P'(x) & \\geq (k_{1} - 1) + \\cdots + (k_{i} - 1) + (s_{1} - 1) + \\cdots + (s_{j} - 1) \\\\\n& = (k_{1} + \\cdots + k_{i}) + (s_{1} + \\cdots + s_{j}) - (i + j) \\geq \\deg P(x),\n\\end{aligned}\n$$\na contradiction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40244,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given that the polynomial $P(x) = x^{5} - x^{2} + 1$ has $5$ roots $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$. Find the value of the product\n$$\nQ(r_{1}) Q(r_{2}) Q(r_{3}) Q(r_{4}) Q(r_{5}),\n$$\nwhere $Q(x) = x^{2} + 1$.",
    "options": [],
    "answer": "5",
    "solution": "Since $r_{1}, \\ldots, r_{5}$ are the roots of $P(x) = x^{5} - x^{2} + 1$, by factorization theorem we have\n$$\nP(x) = \\prod_{i=1}^{5} (x - r_{i}) .\n$$\nIt follows that\n$$\n\\prod_{j=1}^{5} Q(r_{j}) = \\prod_{j=1}^{5} (r_{j}^{2} + 1) = \\prod_{j=1}^{5} (r_{j} + i) \\prod_{j=1}^{5} (r_{j} - i) = P(i) P(-i),\n$$\nwhere $i^{2} = -1$. This gives $P(i) = i^{5} - i^{2} + 1 = -i + 1 + 1 = 2 - i$ and $P(-i) = (-i)^{5} - (-i)^{2} + 1 = 2 + i$. Hence, $P(i) P(-i) = (2 + i)(2 - i) = 4 - i^{2} = 5$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40245,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle which is inscribed in a circle $(O)$. A point $I$ belongs to the segment $BC$. Denote by $H$ and $K$ the projections of $I$ on $AB$ and $AC$, respectively. Suppose that the line $HK$ intersects $(O)$ at $M, N$ ($H$ is between $M, K$ and $K$ is between $H, N$). Let $X, Y$ be the centers of the circles $(ABK), (ACH)$ respectively. Prove the following assertions:\n\n1. If $I$ is the projection of $A$ on $BC$, then $A$ is the center of circle $(IMN)$.\n\n2. If $XY \\parallel BC$, then the orthocenter of $XOY$ is the midpoint of $IO$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1) We will use the inversion to solve this problem.\nNote that $\\triangle AHI \\sim \\triangle AIB$ and that $\\triangle AKI \\sim \\triangle AIC$, hence\n$$\nAI^{2} = AH \\cdot AB = AK \\cdot AC =: k.\n$$\nLet $f$ be the inversion with center $A$ and power $k$. Then\n$$\nf(I) = I,\\ f(H) = B,\\ f(B) = H,\\ f(K) = C,\\ f(C) = K.\n$$\nSo this inversion sends the line $HK$ to $(O)$, which implies that the intersection of $HK$ and $(O)$ are fixed points under $f$.\nTherefore, $AM^{2} = AN^{2} = k = AI^{2}$; therefore, $A$ is the center of the circle $(IMN)$.\n\n![](attached_image_1.png)\n\n2) First, we will prove the following lemma: Let $D$ be a point on the segment $BC$ and $H$ be the projection of $D$ on $AC$. Suppose that $I$ is the circumcenter of triangle $ABH$. Denote by $M, N$ the midpoints of $AB, AC$ respectively. Let $MN$ meet $AD$ at $K$. Then $IK \\perp AC$.\n\nDenote by $L$ the projection of $B$ on $AC$. It is easy to see that $I, M, O$ are collinear. Further,\n$$\n\\angle AIM = \\angle LHB,\\ \\angle AOM = \\angle LCB,\n$$\nhence $\\triangle AIM \\sim \\triangle BHL$ and $\\triangle AOM \\sim \\triangle CBL$.\n\n![](attached_image_2.png)\n\nIt follows that\n$$\n\\frac{IM}{IO} = \\frac{LH}{LC} = \\frac{BD}{BC} = \\frac{MK}{MN};\n$$\nwhich implies that $IK \\parallel ON$, i.e. $IK \\perp AC$.\n\nGoing back to the original problem, we denote by $B', C'$ the midpoints of $AB, AC$ and by $Z$ the intersection of $B'C'$ and $AI$.\n\n![](attached_image_3.png)\n\nApplying the lemma, we have $XZ \\parallel OC'$ and $YZ \\parallel OB'$; therefore, $OXZY$ is a parallelogram.\nHence, $OZ$ passes through the midpoint of $XY$. But $XY \\parallel B'C'$, so $Z$ is the midpoint of $B'C'$ and $XY$ is a medial line of triangle $OB'C'$.\n\nOn the other hand, $B'C' \\parallel BC$, hence $I$ is the midpoint of $BC$.\nLet $T$ be the midpoint of the segment $OI$. Then $XT \\parallel B'I \\parallel AC$. But $OC' \\perp AC$, thus $OY \\perp XT$.\nSimilarly, $OX \\perp YT$. Therefore, $T$ is the orthocenter of triangle $XOY$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40246,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geq 3$ and $x_{1}, x_{2}, \\ldots, x_{n}$ be $n$ distinct integers. Prove that\n$$\n\\left(x_{1}-x_{2}\\right)^{2}+\\left(x_{2}-x_{3}\\right)^{2}+\\ldots+\\left(x_{n}-x_{1}\\right)^{2} \\geq 4 n-6 .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove by induction on $n$.\n\nWhen $n=3$, we can assume that $x_{1}>x_{2}>x_{3}$ (only when $n=3$) and the LHS $\\geq 1^{2}+1^{2}+2^{2}=6$.\n\nMoving from $n$ to $n+1$, the change in LHS is\n$$\nC_{n+1}=\\left(x_{n}-x_{n+1}\\right)^{2}+\\left(x_{n+1}-x_{1}\\right)^{2}-\\left(x_{n}-x_{1}\\right)^{2} .\n$$\nWe need to show that $C_{n+1} \\geq 4$. We can see that it is not true in general. However, due to the cyclicity, we can assume that $x_{n+1}$ is the least number.\n\nMoreover, since the LHS is unchanged under translation, we can assume also that $x_{n+1}=0$. Thus, we have\n$$\nC_{n+1}=x_{n}^{2}+x_{1}^{2}-\\left(x_{n}-x_{1}\\right)^{2}=2 x_{1} x_{n} \\geq 4 .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40247,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call a positive integer $N \\geq 2$ \"special\" if for every $k$ such that $2 \\leq k \\leq N$, $N$ can be expressed as a sum of $k$ positive integers that are relatively prime to $N$ (although not necessarily relatively prime to each other). Find all special positive integers.",
    "options": [],
    "answer": "All odd positive integers and 2",
    "solution": "We claim that all odd numbers are special, and the only special even number is $2$. For any even $N > 2$, the number relatively to $N$ must be odd, and $N$ cannot be expressed as a sum of $3$ positive odd numbers.\n\nNow, suppose that $N$ is odd. We consider the binary decomposition of $N$\n$$\nN = 2^{a_{1}} + 2^{a_{2}} + \\ldots + 2^{a_{i}}.\n$$\nNote that $(2^{t}, N) = 1$ and $i < \\log_{2} N + 1$.\n\nFor any $i \\leq k < N-1$, suppose that $N$ can be expressed as a sum of $k$ powers of $2$, i.e. $N = 2^{x_{1}} + \\ldots + 2^{x_{k}}$. Then at least one of $x_{j} > 0$ (otherwise, $k = N$). Suppose that $x_{1} > 0$ then\n$$\nN = 2^{x_{1}-1} + 2^{x_{1}-1} + 2^{x_{2}} + \\ldots + 2^{x_{k}}.\n$$\nSo $N$ can be expressed as a sum of $k+1$ powers of $2$. This implies that, for any $k < \\log_{2} N + 1$, $N$ can be expressed as a sum of $k$ positive integers that are relatively prime to $N$.\n\nNow, start from $k = 2$. Let $2^{a}$ be the largest power of $2$ such that $2^{a} < N$. Then $N = 2^{a} + (N - 2^{a})$. Using the above argument, we can write $2^{a}$ as the sum of $k$ powers of $2$ for any $1 < k < 2^{a}$. Since $2^{a} > N / 2$, we have $N$ can be expressed as a sum of $k$ positive integers that are relatively prime to $N$ for any $2 \\leq k \\leq N / 2$.\n\nSince $N / 2 + 1 > \\log_{2} N$ for all $N \\geq 3$, we conclude that all $N > 3$ odd are special.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40248,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle inscribed in the circle $(O)$. The bisector of $\\angle BAC$ cuts the circle $(O)$ again at $D$. Let $DE$ be the diameter of $(O)$. Let $G$ be a point on $\\operatorname{arc} AB$ which does not contain $C$. The lines $GD$ and $BC$ intersect at $F$. Let $H$ be a point on the line $AG$ such that $FG \\parallel AE$. Prove that the circumcircle of triangle $HAB$ passes through the orthocenter of triangle $HAC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $HF$ cut $DE$ at $P$. We have $\\angle HGD = \\angle E = \\angle HPD$ so $HGPD$ is cyclic, we deduce $FH \\cdot FP = FG \\cdot FD = FB \\cdot FC$ so $HBPC$ is cyclic.\n\n![](attached_image_1.png)\n\nEasily seen $DE$ is perpendicular bisector of $BC$ so $PB = PC$. Hence $HP$ is bisector of $\\angle BHC$. From this,\n$$\n\\begin{aligned}\n\\angle HBA + \\angle HCA & = 180^\\circ - \\angle BHA - \\angle BAH + 180^\\circ - \\angle CHA - \\angle ACH \\\\\n& = 360^\\circ - (\\angle AHB + \\angle AHC) - (\\angle HAB + \\angle HAC) \\\\\n& = 360^\\circ - 2\\angle AHF - 2\\angle HAD \\\\\n& = 360^\\circ - 2\\angle GDE - 2(\\angle GAE - 90^\\circ) = 180^\\circ .\n\\end{aligned}\n$$\nNow let $K$ be orthocenter of triangle $HAC$ then\n$$\n\\angle AKH = 180^\\circ - \\angle ACH = \\angle ABH\n$$\nso $K$ lies on $(HAB)$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40249,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle, $AX$, $BY$, $CZ$ are the altitudes with $X$, $Y$, $Z$ belonging to $BC$, $CA$, $AB$ respectively. Respectively denote $(O_{1})$, $(O_{2})$, $(O_{3})$ as the circumcircles of triangles $AYZ$, $BZX$, $CXY$. Suppose that $(K)$ is a circle that is internally tangent to $\\left(O_{1}\right)$, $\\left(O_{2}\right)$, $\\left(O_{3}\right)$. Prove that $(K)$ is tangent to the circumcircle of triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H$ be the orthocenter of triangle $ABC$.\n\n![](attached_image_1.png)\n\nWe can see that $HA \\cdot HX = HB \\cdot HY = HC \\cdot HZ = k$. We consider the inversion with center $H$ and ratio $-k$ as the function $f$.\n\nIt is easy to see that\n$$\nf(A) = X, \\quad f(B) = Y, \\quad f(C) = Z\n$$\nso $f((O)) = (O')$ with $(O')$ being the 9-point circle (which also passes through $X$, $Y$, $Z$).\n\nOn the other hand, because $\\angle HEA = \\angle HFA = 90^{\\circ}$, hence $H \\in (O_{1})$. After the inversion, the circles $(O_{1})$ become a line passing through the images of $X$, $Y$; indeed, this line is $BC$ or $f((O_{1})) = BC$.\n\nSimilarly, we also have $f((O_{2})) = CA$ and $f((O_{3})) = AB$. So the circle $K$ that is tangent to $(O_{1})$, $(O_{2})$, $(O_{3})$ will become the incircle $(I)$ of triangle $ABC$. But based on Feuerbach's theorem, the two circles $(O')$, $(I)$ are tangent to each other.\n\nTherefore, the circles $(K)$ and $(O)$ are also tangent to each other. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40250,
    "subject": "Mathematics (Multi-modal)",
    "question": "1) Prove that there are infinitely many positive integers $n$ such that there exists a permutation of $1,2,3, \\ldots, n$ with the property that the difference between any two adjacent numbers is equal to either $2015$ or $2016$.\n\n2) Let $k$ be a positive integer. Is the statement in 1) still true if we replace the numbers $2015$ and $2016$ by $k$ and $k+2016$, respectively?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1) First, replace $2015$ and $2016$ by $3$ and $4$ respectively for convenience. We consider the permutation of $\\{1,2,3, \\ldots, 7\\}$ as follows:\n$$\n1,5,2,6,3,7,4.\n$$\nThe numbers increase by $4$ and decrease by $3$, then the sequence covers all numbers from $1$ to $7$. Add $1$ to each term of the above sequence, we get\n$$\n8,12,9,13,10,14,11.\n$$\nCombine two sequences, we have\n$$\n1,5,2,6,3,7,4,8,12,9,13,10,14,11.\n$$\nThis is another permutation of $\\{1,2,3, \\ldots, 14\\}$. By this method, we can construct the desired permutation for each number of the form $7k$, $k \\in \\mathbb{Z}$, satisfying the condition.\n\nIn the same way with the two given numbers $2015$ and $2016$:\n$$\n1,1+2016,1+2016-2015, \\ldots\n$$\nIncrease by $2016$ then decrease by $2015$ and so on, we can get the desired permutation for each number of the form $(2015+2016)k$, $k \\in \\mathbb{Z}$.\n\n2) Now it is easy to see that all numbers $k$ such that $\\gcd(k, 2016)>1$ do not satisfy the given condition. Indeed,\n\nSuppose that $\\gcd(k, 2016)=d>1$ and there exists a number $n$ such that the permutation of $\\{1,2,3, \\ldots, n\\}$ satisfies the given condition. Then all numbers in this sequence have the same remainder when divided by $d$, which is a contradiction.\n\nNow we will show that every number coprime with $2016$ will satisfy the given condition by proving the general statement:\n\nIf $m, k$ are two positive numbers and $\\gcd(m, k)=1$, then there are infinitely many numbers $n$ such that there exists a permutation with the property that the difference between any two adjacent numbers is equal to either $m$ or $k$.\n\nSuppose that $m>k$ and $m=ak+b$, $0 \\leq b \\leq k-1$. We will prove that each number $n$ which is a multiple of $t(m+k)=(a+1)k+b$ satisfies the given condition. Indeed,\n\nPartition the set $\\{1,2,3, \\ldots, n\\}$ into $k$ sets and each set contains the numbers sharing the same remainder when divided by $k$. And each of them will be arranged decreasing from the largest to the smallest (except the set containing number $1$).\n$$\n\\begin{aligned}\n& 1 \\rightarrow 1+(ak+b) \\rightarrow 1+(a-1)k+b \\rightarrow \\ldots \\rightarrow 1+b \\rightarrow \\ldots \\\\\n& 1+ak+2b \\rightarrow 1+(a-1)k+2b \\rightarrow \\ldots \\rightarrow 1+2b \\rightarrow \\ldots \\\\\n& \\ldots \\\\\n& 1+ak+(k-1)b \\rightarrow 1+(a-1)k+(k-1)b \\rightarrow \\ldots \\rightarrow 1+(k-1)b \\rightarrow \\ldots\n\\end{aligned}\n$$\nIt is easy to see that the difference between two adjacent numbers in each sequence is $k$ and two sequences can be connected by adding the smallest number in one set to the biggest number in another set. Notice that when we take a number $i \\in \\{2,3,4, \\ldots, k\\}$ and add $m$, we get $i+m$ which has the remainder differing from $i$ when divided by $k$ and since $\\gcd(m, k)=1$ then the set\n$$\n\\{i, i+m, i+2m, \\ldots, i+(k-1)m\\}\n$$\nforms a complete system of residues modulo $k$, then when we connect all sets we get the permutation of $\\{1,2,3, \\ldots, n\\}$. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 40251,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a polynomial $P(x) = a_{n} x^{n} + a_{n-1} x^{n-1} + \\cdots + a_{1} x + a_{0}$ of real coefficients. Suppose that $P(x)$ has $n$ real roots (not necessarily distinct), and there exists a positive integer $k$ such that $a_{k} = a_{k-1} = 0$. Prove that $P(x)$ has a real root of multiplicity $k+1$.\n\n(Note: we call a real number $x_{0}$ a root of multiplicity $s$ of a polynomial $R(x)$ of real coefficients if there exists a polynomial $Q(x)$ such that $R(x) = (x - x_{0})^{s} Q(x)$ and $Q(x_{0}) \\neq 0$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will show that $a_{k} = a_{k-1} = a_{k-2} = \\cdots = a_{0} = 0$ by induction on $n$, the degree of $P(x)$.\n\nIn fact, we may assume that the leading coefficient of $P(x)$ is $1$. For $n = 1, 2$, the result follows immediately.\n\nAssume that the induction hypothesis is true for every $n < m$, we shall prove it is also true for $n = m$. Denote by $P_{m}(x) = x^{m} + \\cdots + a_{1} x + a_{0}$, and for some $k < m$, $a_{k} = a_{k-1} = 0$.\n\nBy taking derivative of $P_{m}(x)$, we obtain $P_{m}'(x) = b_{m-1} x^{m-1} + \\cdots + b_{1} x + b_{0}$, for some real numbers $b_{m-1}, \\ldots, b_{0}$.\n\nSince $a_{k} = a_{k-1} = 0$, we conclude that $b_{k-1} = b_{k-2} = 0$.\n\nThis together with $P_{m}(x)$ has only real roots, implies that $P_{m}'(x)$ also has only real roots.\n\nHence, by induction hypothesis, we get $b_{k-3} = \\cdots = b_{0} = 0$. In other words, $a_{k-2} = \\cdots = a_{1} = 0$. It remains to show that $a_{0} = 0$. Assume that $a_{0} \\neq 0$, then if $r_{1}, \\ldots, r_{m}$ are the roots of $P_{m}(x)$, then by Vieta's theorem,\n$$\nr_{1} \\cdots r_{m} = (-1)^{m} a_{0}, \\quad r_{1} + r_{2} + \\cdots + r_{m} = 0\n$$\nand\n$$\n\\sum_{i, j} \\frac{1}{r_{i} r_{j}} = 0 \\Rightarrow \\sum_{i} \\frac{1}{r_{i}^{2}} = 0\n$$\na contradiction. Therefore, $a_{0} = 0$, the induction process is completed.\n\nObviously from that, we get $0$ is the root of $P(x)$ with multiplicity at least $k+1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40252,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = x^{2} + a x + b$ be a quadratic function with real coefficients $a, b$. It is given that the equation $f(f(x)) = 0$ has 4 distinct real roots and the sum of 2 roots among these roots is equal to $-1$. Prove that $b \\leq \\frac{-1}{4}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFirst, we will prove that $f(x) = 0$ has some solutions (maybe not distinct).\nIndeed, if $f(x) = 0$ has no root, then it can be written as $f(x) = (x - c)^{2} + d$ with $d > 0$ and\n$$\nf(f(x)) = \\left((x - c)^{2} + d - c\\right)^{2} + d > 0.\n$$\nIt means $f(f(x)) = 0$ has no solution, which is a contradiction.\n\nNow, denote $c_{1} \\geq c_{2}$ as the solution of $f(x) = 0$ and $x_{1}, x_{2}$ as the solutions of $f(f(x)) = 0$ in such a way that $x_{1} + x_{2} = -1$. By Vieta's theorem, note that $c_{1} + c_{2} = -a$ and $c_{1} c_{2} = b$.\n\nIt is easy to see that $f(f(x)) = 0$ is equivalent to $f(x) = c_{1}$, $f(x) = c_{2}$. We need to consider 2 cases:\n\n1. If $x_{1}, x_{2}$ are the solutions of one equation, by Vieta's theorem, then $a = 1$. Thus $c_{2} \\leq -\\frac{1}{2}$.\n\nConsider equation $f(x) - c_{2} = 0$, we have $\\Delta = 1 - 4(b - c_{2}) > 0$, which implies that $b < -\\frac{1}{4}$.\n\n2. If $x_{1}, x_{2}$ are solutions of two equations, then $x_{i}^{2} + a x_{i} + b_{i} = c_{i}$ with $i = 1, 2$. Sum these two identities, we get\n$$\nx_{1}^{2} + x_{2}^{2} - a + 2b = -a \\Leftrightarrow x_{1}^{2} + x_{2}^{2} + 2b = 0.\n$$\nHence, $b = -\\frac{x_{1}^{2} + x_{2}^{2}}{2} \\leq -\\frac{(x_{1} + x_{2})^{2}}{4} = -\\frac{1}{4}$.\n\nTherefore, in all case, we always have $b \\leq -\\frac{1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40253,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the Cartesian coordinate system $Oxy$, consider a sequence of points $A_{n}(x_{n}, y_{n})$ in which $(x_{n})_{n=1}^{\\infty}, (y_{n})_{n=1}^{\\infty}$ are two sequences of positive numbers satisfying the following conditions:\n$$\nx_{n+1} = \\sqrt{\\frac{x_{n}^{2} + x_{n+2}^{2}}{2}}, \\quad y_{n+1} = \\left(\\frac{\\sqrt{y_{n}} + \\sqrt{y_{n+2}}}{2}\\right)^{2} \\quad \\forall n \\geq 1\n$$\nSuppose that $O, A_{1}, A_{2016}$ belong to a line $d$ and $A_{1}, A_{2016}$ are distinct. Prove that all the points $A_{2}, A_{3}, \\ldots, A_{2015}$ lie on one side of $d$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From $O, A_{1}, A_{2016}$ are collinear, there exists some positive number $k$ such that\n$$\n\\frac{x_{1}}{y_{1}} = \\frac{x_{2016}}{y_{2016}} = k > 0.\n$$\nWe shall prove that for all $i = \\overline{2,2015}$, $\\frac{x_{i}}{y_{i}} > k$. Indeed,\nNotice that\n$$\nx_{n+1} = \\sqrt{\\frac{x_{n}^{2} + x_{n+2}^{2}}{2}} \\Leftrightarrow x_{n+1}^{2} = \\frac{x_{n}^{2} + x_{n+2}^{2}}{2},\n$$\nwhich implies that $(x_{n}^{2})$ forms an arithmetic sequence.\nHence, for all $i = \\overline{2,2015}$, let\n$$\n\\alpha_{i} = \\frac{2016 - i}{2015}, \\quad \\beta_{i} = \\frac{i - 1}{2015}\n$$\nthen $\\alpha_{i} + \\beta_{i} = 1$, $2 \\leq i \\leq 2015$ and\n$$\nx_{i}^{2} = \\alpha_{i} x_{1}^{2} + \\beta_{i} x_{2016}^{2} \\Leftrightarrow x_{i} = \\sqrt{\\alpha_{i} x_{1}^{2} + \\beta_{i} x_{2016}^{2}}.\n$$\nSimilarly, we also have\n$$\ny_{i} = \\left(\\alpha_{i} \\sqrt{y_{1}} + \\beta_{i} \\sqrt{y_{2016}}\\right)^{2} \\text{ and } k y_{i} = \\left(\\alpha_{i} \\sqrt{x_{1}} + \\beta_{i} \\sqrt{x_{2016}}\\right)^{2}.\n$$\nWe need to prove\n$$\n\\sqrt{\\alpha_{i} x_{1}^{2} + \\beta_{i} x_{2016}^{2}} > \\left(\\alpha_{i} \\sqrt{x_{1}} + \\beta_{i} \\sqrt{x_{2016}}\\right)^{2}\n$$\nfor all $2 \\leq i \\leq 2015$ $(*)$.\nBy applying the Cauchy-Schwarz inequality, we have\n$$\n\\begin{aligned}\n(\\alpha_{i} + \\beta_{i}) (\\alpha_{i} x_{1}^{2} + \\beta_{i} x_{2016}^{2}) & \\geq (\\alpha_{i} x_{1} + \\beta_{i} x_{2016})^{2} \\\\\n& = \\left((\\alpha_{i} + \\beta_{i})(\\alpha_{i} x_{1} + \\beta_{i} x_{2016})\\right)^{2} \\\\\n& \\geq \\left(\\alpha_{i} \\sqrt{x_{1}} + \\beta_{i} \\sqrt{x_{2016}}\\right)^{4}\n\\end{aligned}\n$$\nSo the inequality $(*)$ is true for each $2 \\leq i \\leq 2015$. Since all points are distinct, then $A_{1} \\neq A_{2016} \\Leftrightarrow x_{1} \\neq x_{2016}$, so equality does not occur in $(*)$.\nSo all the points $A_{2}, A_{3}, \\ldots, A_{2015}$ belong to the same side with respect to $d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40254,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle whose incircle $(I)$ touches $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. The line passing through $A$ and parallel to $BC$ cuts $DE$, $DF$ at $M$, $N$, respectively. The circumcircle of triangle $DMN$ cuts $(I)$ again at $L$.\n\n1. Let $K$ be the intersection of $NE$ and $MF$. Prove that $K$ is the orthocenter of the triangle $DMN$.\n\n2. Prove that $A$, $K$, $L$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\n1) Because $MN \\parallel BC$ so $\\angle ANF = \\angle FDB = \\angle DFB = \\angle AFN$, this deduces $AN = AF$. Similarly, $AM = AE = AF = AN$.\n\nThus, $M$, $N$, $E$, $F$ lie on the circle center $A$. Let $MF$ cut $NE$ at $K$, because $E$, $F$ lie on circle diameter $MN$ so\n$$\n\\angle KFD = \\angle KED = 90^\\circ,\n$$\nthis means $DK$ is diameter of $(AEF)$. So $K$ lies on $(I)$. Easily seen, $K$ is orthocenter of triangle $DMN$.\n\n2) Let $P$ be the reflection of $K$ through $A$ then $KMPN$ is parallelogram, so $PM \\parallel NK \\perp MD$ and $PN \\parallel KM \\perp DN$. Thus, $DP$ is diameter of $(DMN)$. But $DK$ is diameter of $(I)$ so $\\angle DLK = 90^\\circ$, we deduce $L$ lies on $AK$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40255,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ be positive real numbers satisfy the condition $x^{2} + y^{2} + z^{2} = 2(xy + yz + zx)$. Prove that\n$$\nx + y + z + \\frac{1}{2xyz} \\geq 4\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, assume that $z = \\min \\{x, y, z\\}$. From the condition $x^{2} + y^{2} + z^{2} = 2(xy + yz + zx)$, we get\n$$\n(x + y)^{2} - 2z(x + y) + z^{2} = 4xy\n$$\nor\n$$\n(x + y - z)^{2} = 4xy\n$$\nUsing the AM-GM inequality, we have\n$$\n\\frac{x + y - z}{2} + \\frac{x + y - z}{2} + 2z + \\frac{1}{2xyz} \\geq 4 \\sqrt[4]{\\frac{(x + y - z)^{2} z}{4xyz}} = 4\n$$\nHence\n$$\nx + y + z + \\frac{1}{2xyz} \\geq 4\n$$\nThe equality holds if $x = 2$, $y = z = \\frac{1}{2}$, or any its cyclic permutation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40256,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle which is inscribed in a circle $(O)$. A point $I$ belongs to the segment $BC$. Denote by $H$ and $K$ the projections of $I$ on $AB$ and $AC$, respectively. Suppose that the line $HK$ intersects $(O)$ at $M, N$ ($H$ is between $M, K$ and $K$ is between $H, N$). Prove the following assertions:\n\n1. If $A$ is the center of the circle $(IMN)$, then $BC$ is tangent to $(IMN)$.\n2. If $I$ is the midpoint of $BC$, then $BC$ is equal to 4 times the distance between the centers of two circles $(ABK)$ and $(ACH)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "See the solution to Problem 1 in the test of level 4+. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40257,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given three numbers $x$, $y$, $z$, and set $x_{1} = |x - y|$, $y_{1} = |y - z|$, $z_{1} = |z - x|$. From $x_{1}$, $y_{1}$, $z_{1}$, form in the same fashion the numbers $x_{2}$, $y_{2}$, $z_{2}$, and so on. It is known that $x_{n} = x$, $y_{n} = y$, $z_{n} = z$ for some $n$. Find all possible values of $(x, y, z)$.",
    "options": [],
    "answer": "All triples that are permutations of (a, a, 0) with a ≥ 0 (including (0, 0, 0)).",
    "solution": "First, consider 3 sequences $\\left(x_{n}\\right)$, $\\left(y_{n}\\right)$, $\\left(z_{n}\\right)$, $n \\geq 1$ with\n$$\nx_{1} = |x - y|, \\quad y_{1} = |y - z|, \\quad z_{1} = |z - x| \\text{ and } \n\\left\\{\n\\begin{array}{l}\nx_{n+1} = \\left|x_{n} - y_{n}\\right| \\\\\ny_{n+1} = \\left|y_{n} - z_{n}\\right| \\\\\nz_{n+1} = \\left|z_{n} - x_{n}\\right|\n\\end{array}\n\\right.\n$$\nFrom this, it is easy to see that $x_{n}$, $y_{n}$, $z_{n} \\geq 0$ for all $n$.\nLet $w_{n} = \\max \\left\\{x_{n}, y_{n}, z_{n}\\right\\}$, $n \\geq 1$. We have\n$$\n\\begin{aligned}\nw_{n+1} = \\max \\left\\{x_{n+1}, y_{n+1}, z_{n+1}\\right\\} & = \\max \\left\\{\\left|x_{n} - y_{n}\\right|, \\left|y_{n} - z_{n}\\right|, \\left|z_{n} - x_{n}\\right|\\right\\} \\\\\n& \\leq \\max \\left\\{x_{n}, y_{n}, z_{n}\\right\\} = w_{n}.\n\\end{aligned}\n$$\nThe equality occurs when there is at least one number among $x_{n}$, $y_{n}$, $z_{n}$ equal to $0$. Hence, the sequence $(w_{n})$ is non-increasing. Suppose that $k$ is a positive integer such that $(x_{k}, y_{k}, z_{k}) = (x, y, z)$, then\n$$\nw_{1} \\geq w_{2} \\geq \\ldots \\geq w_{k} = w_{1} \\text{ which implies that } w_{1} = w_{2} = \\ldots = w_{k}.\n$$\nOn the other hand, for all $i = \\overline{1, k}$, there is at least one number among $(x_{i}, y_{i}, z_{i})$ equal to $0$. Without loss of generality, we may assume that $x \\geq y \\geq z = 0$ and we can see that $x_{1} = |x - y|$, $y_{1} = |y - z| = y$, $z_{1} = |z - x| = x$ so we have some cases:\n\n1. If $|x - y| = 0 \\Leftrightarrow x = y$, we have the tuple $(x, y, z) = (a, a, 0)$ with $a \\geq 0$. It is easy to check this tuple satisfies the given condition.\n\n2. If $y = 0$, to separate from the previous case, we consider $x > 0$, we have the tuple $(x, y, z) = (a, 0, 0)$ with $a > 0$. But $x_{1} = a$, $y_{1} = 0$, $z_{1} = a$ and from this, we cannot obtain $(a, 0, 0)$ anymore. Hence, this case does not satisfy the given condition.\n\n3. If $x = 0$, we have $x = y = z = 0$ which was mentioned above.\n\nTherefore, the tuples $(x, y, z) = (a, a, 0)$ and its permutations with $a \\geq 0$ satisfy the given condition. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40258,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a positive integer. Prove that there exist integers $x, y$, neither of which divisible by $7$ such that $x^{2} + 6 y^{2} = 7^{k}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Take $x_{1} = y_{1} = -1$, and the following relations\n$$\nx_{k+1} = x_{k} - 6 y_{k}, \\quad y_{k+1} = x_{k} + y_{k}.\n$$\nfor $k = 1, 2, 3, \\ldots$\n\nOne can check that\n$$\nx_{k+1}^{2} + 6 y_{k+1}^{2} = 7 \\left( x_{k}^{2} + 6 y_{k}^{2} \\right)\n$$\nand\n$$\nx_{k} \\equiv y_{k} \\equiv (-1)^{k} \\pmod{7}.\n$$\nHence, there exist the integer $x = x_{k}, y = y_{k}$ satisfy the given condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40259,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S = x + y + z$ where $x, y, z$ are three nonzero real numbers satisfying the following system of inequalities:\n$$\n\\left\\{\\begin{array}{rl}\nx y z & > 1 \\\\\nx + y + z & > \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\n\\end{array} .\\right.\n$$\n\nProve that $S$ can take on any real values when $x, y, z$ vary.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, if $z = \\frac{1}{y}$, then the given system becomes\n$$\n\\left\\{\\begin{array}{c}\nx \\cdot y \\cdot \\frac{1}{y} > 1 \\\\\nx + y + \\frac{1}{y} > \\frac{1}{x} + \\frac{1}{y} + y\n\\end{array} \\Leftrightarrow x > 1 .\\right.\n$$\nSo, all triples $\\left(x, y, \\frac{1}{y}\\right)$, with $x > 1$ and $y \\neq 0$, satisfy the given system of inequalities.\n\nNext, note that the range of $f(y) := y + \\frac{1}{y}$ is $(-\\infty, -2] \\cup [2, +\\infty)$ and that, for such triples, $S = x + f(y)$. Then for each value $a \\in \\mathbb{R}$, we need only consider the following cases:\n\n1. If $a > -1$, we can choose $x = a + 2 (> 1), y = -1$, and get $S = a + 2 + f(-1) = a$.\n\n2. If $a < -1$, we let $x = -a (> 1)$ and choose $y \\neq 0$ such that $f(y) = 2a (\\in (-\\infty, -2])$. Then $S = -a + f(y) = a$.\n\n3. If $a = -1$, we choose $x = 1.5 (> 1), y = -2$ and get $S = 1.5 + f(-2) = -1 = a$.\n\nTherefore, $S$ can take on any real values when $x, y, z$ vary.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40260,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(O_{1})$, $(O_{2})$ be given two circles intersecting at $A$ and $B$. The tangent lines of $(O_{1})$ at $A$, $B$ intersect at $O$. Let $I$ be a point on the circle $(O_{1})$ but outside the circle $(O_{2})$. The lines $IA$, $IB$ intersect circle $(O_{2})$ at $C$, $D$. Denote by $M$ the midpoint of $CD$. Prove that $I$, $M$, $O$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $N$ as the midpoint of $AB$. Because $A$, $B$, $C$, $D$ belong to the same circle, then we have\n$$\n\\triangle IAB \\sim \\triangle IDC.\n$$\nSince $M$ is the midpoint of $CD$ and $N$ is the midpoint of $AB$ then $IM$, $IN$ are isogonal conjugate with respect to the angle $\\angle CID$.\n\nIn the other hand, $O$ is the intersection of two tangent lines of $(O_{2})$ at $A$, $B$, then $IO$ is the symmedian of triangle $IAB$. It means $IO$, $IN$ are isogonal conjugate with respect to $\\angle AIB$.\n\nHence, $I$, $M$, $O$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40261,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geq 4$ be a positive integer and there exist $n$ positive integers that are arranged on a circle such that:\n- The product of each pair of two non-adjacent numbers is divisible by $2015 \\cdot 2016$.\n- The product of each pair of two adjacent numbers is not divisible by $2015 \\cdot 2016$.\nFind the maximum value of $n$.",
    "options": [],
    "answer": "6",
    "solution": "See the solution in the test of level 4. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40262,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a non-isosceles triangle inscribed in a circle $(O)$ and $BE$, $CF$ are two angle bisectors intersecting at $I$ with $E$ belonging to segment $AC$ and $F$ belonging to segment $AB$. Suppose that $BE$, $CF$ intersect $(O)$ at $M$, $N$ respectively. The line $d_{1}$ passes through $M$ and is perpendicular to $BM$, intersecting $(O)$ at the second point $P$; the line $d_{2}$ passes through $N$ and is perpendicular to $CN$, intersecting $(O)$ at the second point $Q$. Denote $H$, $K$ as the midpoints of $MP$ and $NQ$ respectively.\n\n1. Prove that triangles $IEF$ and $OKH$ are similar.\n\n2. Suppose that $S$ is the intersection of the two lines $d_{1}$ and $d_{2}$. Prove that $SO$ is perpendicular to $EF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1) Denote the projections of $I$ on $AC$, $AB$ are $X$, $Y$ respectively.\nNote that $B$, $O$, $P$ are collinear because $\\angle BMP = 90^\\circ$. Because $O$, $H$ are the midpoints of $PB$, $PM$ respectively, we have $OH \\parallel BM$ and $OH = \\frac{1}{2} BM$.\nSimilarly, $OK \\parallel CN$ and $OK = \\frac{1}{2} CN$. Hence, $\\angle HOK = \\angle EIF$. On the other hand,\n$$\n\\begin{aligned}\n& \\frac{OH}{OK} = \\frac{BM}{CN} = \\frac{\\sin \\left(A + \\frac{B}{2}\\right)}{\\sin \\left(A + \\frac{C}{2}\\right)} \\\\\n& \\frac{IF}{IE} = \\frac{IY}{\\sin \\angle IFY} : \\frac{IX}{\\sin \\angle IEX} = \\frac{\\sin \\angle IEX}{\\sin \\angle IFY} = \\frac{\\sin \\left(A + \\frac{B}{2}\\right)}{\\sin \\left(A + \\frac{C}{2}\\right)}\n\\end{aligned}\n$$\nSo $\\frac{OH}{OK} = \\frac{IF}{IE}$ implies that $\\triangle OHK \\sim \\triangle IFE$.\n\n![](attached_image_1.png)\n\n2) It is easy to see that\n$$\nOK \\cdot OT = R^2, \\quad OH \\cdot OU = R^2\n$$\nwith $R$ the radius of $(O)$. Then $OK \\cdot OT = OH \\cdot OU$ or $K$, $T$, $H$, $O$ are cyclic.\nFrom this, we can see $\\angle IEF = \\angle IKH = \\angle IUT$ so $EF \\parallel UT$.\nThe antipole line of $T$, $U$ passes through $S$ so the antipole line of $S$ is $TU$, which leads to $TU \\perp SO$. From these results, we get $SO \\perp EF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40263,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a point outside the circle $\\omega$. Two points $B, C$ lie on $\\omega$ such that $A B, A C$ are tangent to $\\omega$. Let $D$ be any point on $\\omega$ ($D$ is neither $B$ nor $C$) and $M$ the foot of perpendicular from $B$ to $C D$. The line through $D$ and the midpoint of $B M$ meets $\\omega$ again at $P$. Prove that $A P \\perp C P$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the center of $\\omega$ and $Q$ the intersection point of $C O$ and $\\omega$.\nSince $Q D$ and $B M$ are perpendicular to $C D$ then $B M \\parallel Q D$. Because $N$ is the midpoint of $B M$, which implies that $D(M, B, N, Q) = -1$.\nConsider this harmonic quartet with circle $\\omega$, one has $C P B Q$ is a harmonic quadrilateral. This implies that $A, P, Q$ are collinear and $A P \\perp C P$.\n\n![](attached_image_1.png)\n\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40264,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ten vertices of a regular 20-gon $A_{1} A_{2} \\ldots A_{20}$ are painted black and the other ten vertices are painted blue. Consider the set consisting of diagonal $A_{1}A_{4}$ and all other diagonals of the same length.\n\n1. Prove that in this set, the number of diagonals with two black endpoints is equal to the number of diagonals with two blue endpoints.\n\n2. Find all possible numbers of the diagonals with two black endpoints.",
    "options": [],
    "answer": "{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}",
    "solution": "![](attached_image_1.png)\n\n1) Define the diagonal with two black endpoints as the black diagonal and the diagonal with two blue endpoints as the blue diagonal. Firstly, consider the sequence of vertices:\n\n$$\n\\begin{aligned}\n& A_{1} \\rightarrow A_{4} \\rightarrow A_{7} \\rightarrow A_{10} \\rightarrow A_{13} \\rightarrow A_{16} \\rightarrow A_{19} \\rightarrow A_{2} \\rightarrow A_{5} \\rightarrow A_{8} \\rightarrow A_{11} \\\\\n& \\rightarrow A_{14} \\rightarrow A_{17} \\rightarrow A_{20} \\rightarrow A_{3} \\rightarrow A_{6} \\rightarrow A_{9} \\rightarrow A_{12} \\rightarrow A_{15} \\rightarrow A_{18}\n\\end{aligned}\n$$\n\nThe diagonal connecting two adjacent vertices has the same length as $A_{1}A_{4}$.\n\nBecause the vertices are painted by black and blue, then they can be divided into some blocks of black vertices and some blocks of blue vertices. It is easy to see that one black block lies between two blue blocks and one blue block lies between two black blocks, which implies that the number of black blocks and blue blocks are equal, denote this number as $k$.\n\nNotice that in each black block of size $a$, the number of black diagonals is $a-1$. By summing all the black diagonals among $k$ black block(s), we can see that the number of black diagonals equals the total black points minus the number of blocks, i.e. $10-k$.\n\nSimilarly, the number of blue diagonals is $10-k$. Therefore, the number of black diagonals is equal to the number of blue diagonals.\n\n\n2) From part 1), the number of black diagonals is equal to $10-k$ with $1 \\leq k \\leq 10$. It is easy to see that with each $k \\in \\{1,2,3, \\ldots, 10\\}$, we can find a way to paint the vertices such that there are exactly $k$ black block(s).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40265,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle inscribed in the circle $(O)$ and $P$ is a point inside the triangle $ABC$. Let $D$ be a point on $(O)$ such that $AD \\perp AP$. The line $CD$ cuts the perpendicular bisector of $BC$ at $M$. The line $AD$ cuts the line passing through $B$ and is perpendicular to $BP$ at $Q$. Let $N$ be the reflection of $Q$ through $M$. Prove that $CN \\perp CP$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nLet $K, L$ be the midpoints of $PQ, PN$.\nSince $PBQA$ is cyclic\n$$\n\\angle KPB = \\angle BAQ = \\angle DCB.\n$$\nHence the isosceles triangles $KPB$ and $MCB$ are similar. The line $BM$ cuts $PQ$ at $R$. We have\n$$\n\\angle LMC = \\angle LMB - \\angle BMC = \\angle MRQ - \\angle BKP = \\angle KBM.\n$$\nEasily seen $KB = KP = ML$ and $BM = MC$.\nHence, $\\triangle KBM = \\triangle LMC$. We deduce $LC = KM = LP = LN$.\nThis, triangle $PCN$ is right at $C$. We are done. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40266,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I_{a}$ be the excenter of triangle $A B C$ with respect to $A$. The line $A I_{a}$ intersects the circumcircle of triangle $A B C$ at $T$. Let $X$ be a point on segment $T I_{a}$ such that $X I_{a}^{2}=X A \\cdot X T$. The perpendicular line from $X$ to $B C$ intersects $B C$ at $A^{\\prime}$. Define $B^{\\prime}$ and $C^{\\prime}$ in the same way. Prove that $A A^{\\prime}$, $B B^{\\prime}$ and $C C^{\\prime}$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the foot of perpendicular from $I_{a}$ to $B C$, $D$ the midpoint of $B C$ and $S_{a}$ the reflection of $A$ with respect to $S$. The line through $I$ parallel to $B C$ cuts $A P$, $A A^{\\prime}$, $A V$ at $Q$, $R$, $L$, respectively.\nRedefine $A^{\\prime}$ as follow: Let $I$ be the incenter and $S$ the Spieker point of triangle $A B C$. Assume that $A^{\\prime} \\equiv A S \\cap B C$ and $V$ is the reflection of $P$ with respect to $A^{\\prime}$. Then we need to prove that $A^{\\prime}$ is also satisfy the given construction.\n![](attached_image_1.png)\nSince $D$ is clearly the midpoint of $I S_{a}$, then $A^{\\prime}$ is the midpoint of $R S_{a}$, so $P S_{a} \\parallel R V$ and $P S_{a} \\perp B C$ and since $D I \\parallel A P$, $P Q D S_{a}$ is a parallelogram. This implies that $D Q \\parallel P S_{a}$ and $D Q \\perp B C$.\nAs a result $Q \\in D T$ and $V R$ is perpendicular bisector of $L Q$ then\n$$\nV(L, Q, R, P) = -1\n$$\nIf $R V$ cuts $A I_{a}$, $A P$ at $X_{a}$, $P_{a}$, we also get\n$$\nV(L, Q, R, P) = (A, Q, P_{a}, P) = (A, T, X_{a}, I_{a}) = -1\n$$\nThen $X$ is midpoint of $I_{a} X_{a}$ which implies that\n$$\nX A^{\\prime} \\parallel I_{a} P \\parallel X_{a} V \\text{ and } X A^{\\prime} \\perp B C \\text{.}\n$$\nIt's similar to $B^{\\prime}$ and $C^{\\prime}$. So $A A^{\\prime}$, $B C^{\\prime}$, $C C^{\\prime}$ are concurrent at the Spieker point $S$ of $\\triangle A B C$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40267,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle with center $O$ passes through points $A$ and $C$ and intersects the sides $AB$ and $BC$ of triangle $ABC$ at points $K$ and $N$, respectively. The circumcircles of triangles $ABC$ and $KBN$ meet at distinct points $B$ and $M$. Prove that $\\angle OMB = 90^{\\circ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nConsider an inversion with center $B$ and radius $r > 0$. The points $A'$, $B'$, $C'$ are collinear and the points $K'$, $N'$, $M'$ are, too.\nThe circumcircle of quadrilateral $ACNK$ becomes the circumcircle $\\omega$ of quadrilateral $A' C' N' K'$. In case $B$ is outside $\\omega$, let $X$, $Y$ be the tangent points of 2 tangents from $B$ to $\\omega$, then $O'$ is the midpoint of $XY$.\nSince $M'$, $X$, $Y$ lies on the polar of $B$ relative to $\\omega$ then $\\angle M' O' B = 90^{\\circ}$, this implies that $\\angle OMB = 90^{\\circ}$.\nIn the other cases, we also have $O'$ lies on the polar of $B$ relative to $\\omega$, and the argument is similar.\n\nRemark. This problem can be solve by using the spiral similarity of center $M$ or by using the Brocard's theorem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40268,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geq 1$ be a fixed positive integer. We consider all the sets $S$ which consist of sub-sequences of the sequence $0,1,2, \\ldots, n$ satisfying the following conditions:\ni) If $\\left(a_{i}\\right)_{i=0}^{k}$ belongs to $S$, then $a_{0}=0, a_{k}=n$ and $a_{i+1}-a_{i} \\leq 2$ for all $0 \\leq i \\leq k-1$.\nii) If $\\left(a_{i}\\right)_{i=0}^{k}$ and $\\left(b_{j}\\right)_{j=0}^{h}$ both belong to $S$, then there exist $0 \\leq i_{0} \\leq k-1$ and $0 \\leq j_{0} \\leq h-1$ such that $a_{i_{0}}=b_{j_{0}}$ and $a_{i_{0}+1}=b_{j_{0}+1}$.\n\nFind the maximum value of $|S|$ (among all the above-mentioned sets $S$ ).",
    "options": [],
    "answer": "F_n (the nth Fibonacci number with F_1 = 1, F_2 = 1)",
    "solution": "Firstly, denote by $A_{n}$ the set of all sub-sequences $\\left(a_{i}\\right)_{i=0}^{k}$ of $(0,1,2, \\ldots, n)$ such that:\n- $a_{0}=0, a_{1}=1, a_{k}=n$,\n- $a_{i+1}-a_{i} \\leq 2$ for all $0 \\leq i \\leq k-1$.\nObviously, $A_{n}$ is one of the sets $S$ satisfying Conditions i)-ii).\nIt is also clear that $|A_{1}|=|A_{2}|=1$.\nIf $n>2$, then for each $\\left(a_{i}\\right)_{i=0}^{k} \\in A_{n}$, let us define $F\\left(\\left(a_{i}\\right)_{i=0}^{k}\\right)=\\left(a_{i}\\right)_{i=0}^{k-1}$. In so doing, we have clearly obtained a bijection $F: A_{n} \\rightarrow A_{n-1} \\cup A_{n-2}$, hence,\n$$\n|A_{n}|=|A_{n-1}|+|A_{n-2}|\n$$\nwhen $n>2$. So, $|A_{n}|$ is the $n^{\\text{th}}$ Fibonacci number, which is denoted by $F_{n}$. We will show that $F_{n}$ is the maximum value of $|S|$, among all the sets $S$ satisfying Conditions i)-ii). Indeed, for such a set $S$, we need only prove that\n$$\n|S| \\leq F_{n} .\n$$\nLet $S_{1}=S \\cap A_{n}$. Then $S=S_{1} \\cup\\left(S \\backslash S_{1}\\right)$. We define a mapping $f: S \\backslash S_{1} \\rightarrow A_{n}$ by partitioning every $w \\in S \\backslash S_{1}$ into parts of the form ( $m, m+2, m+3, \\ldots, m+2k)$ or ( $m, m+2, m+3, \\ldots, m+2k+1$ ) in which $k \\geq 1$ and then $f$ changes each part by the following rule:\n$(m, m+2, m+3, \\ldots, m+2k) \\mapsto (m, m+1, m+2, m+4, m+6, \\ldots, m+2k)$\nand\n$(m, m+2, m+3, \\ldots, m+2k+1) \\mapsto (m, m+1, m+3, m+5, \\ldots, m+2k+1)$.\nWe can see that:\n- $f$ is one-to-one.\n- The two sequences $w$ and $f(w)$ do not satisfy Condition ii); so,\n$$\nf(w) \\notin S \\quad \\forall w \\in S \\backslash S_{1} .\n$$\nHence, $S_{1} \\cap f\\left(S \\backslash S_{1}\\right)=\\emptyset$. Since $S_{1} \\cup f\\left(S \\backslash S_{1}\\right) \\subset A_{n}$, it follows that\n$$\n|S|=|S_{1} \\cup f\\left(S \\backslash S_{1}\\right)| \\leq |A_{n}|=F_{n}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40269,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b$ be given two real numbers with $a \\neq 0$. Find all polynomials $P$ with real coefficients such that\n$$\nx P(x-a) = (x-b) P(x) \\text{ for all } x \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "All real-coefficient polynomials P are:\n- If b = 0: P(x) = c for any real constant c.\n- If k = b/a is a natural number: P(x) = c · x(x − a)(x − 2a) ··· (x − (k − 1)a) for any real constant c.\n- If k = b/a is not a natural number: P(x) ≡ 0.",
    "solution": "Let $P \\in \\mathbb{R}[x]$ satisfy the given condition.\n\nFirst, we notice that if $b=0$ then $x P(x-a) = x P(x) \\Rightarrow P(x-a) = P(x)$ (for all $x \\neq 0$), hence $P(x) \\equiv$ a constant (since $a \\neq 0$), and we can recheck that any constant polynomial satisfies the given condition.\n\nNow, let $b \\neq 0$, and put $\\frac{b}{a} = k$. We need only consider 2 cases.\n\n1. $k \\in \\mathbb{N}$ : Substituting $x=0$ into the given condition, we have\n$$\n-b P(0) = 0 \\Rightarrow P(0) = 0.\n$$\nSubstituting $x=a, 2a, 3a, \\ldots, k a$ (in succession) into the given condition, we also have\n$$\n\\begin{aligned}\n& a P(0) = (a-b) P(a) \\Rightarrow P(a) = 0 \\\\\n& 2a P(a) = (2a-b) P(2a) \\Rightarrow P(2a) = 0 \\\\\n& 3a P(2a) = (3a-b) P(3a) \\Rightarrow P(3a) = 0 \\\\\n& \\ldots \\\\\n& (k-1)a P((k-2)a) = ((k-1)a-b) P((k-1)a) \\Rightarrow P((k-1)a) = 0.\n\\end{aligned}\n$$\nThus, $0, a, 2a, \\ldots, (k-1)a$ are roots of $P(x) = 0$; therefore, by Bezout's theorem, $P(x)$ can be written in the form\n$$\nP(x) = x(x-a)(x-2a) \\cdots (x-(k-1)a) Q(x)\n$$\nfor some $Q \\in \\mathbb{R}[x]$.\n\nThe given condition is then equivalent to:\n$$\n\\begin{aligned}\n& x(x-a) \\cdots (x-ka) Q(x-a) = x(x-a) \\cdots (x-ka) Q(x) \\\\\n& \\Leftrightarrow Q(x-a) = Q(x) \\quad \\forall x \\notin \\{0, a, 2a, \\ldots, ka\\} \\\\\n& \\Leftrightarrow Q(x) \\equiv c \\\\\n& \\Leftrightarrow P(x) = c x(x-a)(x-2a) \\cdots (x-(k-1)a) \\quad \\forall x \\in \\mathbb{R}\n\\end{aligned}\n$$\nfor any real constant $c$.\n\n2. $k \\notin \\mathbb{N}$ : In the same way, it is easy to check that all numbers $ma$, with $m \\in \\mathbb{N}$, are roots of $P(x) = 0$; which implies that $P(x) \\equiv 0$; and we can re-check that the zero polynomial satisfies the given condition.\n\nIn summary:\n- If $b=0$, then $P(x) \\equiv c$.\n- If $\\frac{b}{a} = k \\in \\mathbb{N}$, then $P(x) \\equiv c x(x-a)(x-2a) \\cdots (x-(k-1)a)$.\n- If $\\frac{b}{a} = k \\notin \\mathbb{N}$, then $P(x) \\equiv 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40270,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $c$ be a given real number. Find all polynomials $P$ with real coefficients such that\n$$\n(x+1) P(x-1)-(x-1) P(x)=c \\text{ for all } x \\in \\mathbb{R} .\n$$",
    "options": [],
    "answer": "P(x) = k x(x+1) + c/2 for any real constant k",
    "solution": "In terms of $G(x) := P(x) - \\frac{c}{2}$, the given condition can be rewritten as\n$$\n(x+1) G(x-1) = (x-1) G(x) \\text{ for all } x \\in \\mathbb{R} .\n$$\nIt follows immediately (with $x = \\pm 1$) that $G(0) = 0$ and $G(-1) = 0$. Therefore, $G(x) = x(x+1) Q(x)$ for some $Q \\in \\mathbb{R}[x]$. Then\n$$\nQ(x-1) = Q(x) \\text{ for all } x \\in \\mathbb{R} \\setminus \\{-1, 0, 1\\} .\n$$\nThis is true iff $Q$ is a constant; i.e.,\n$$\nP(x) = k x(x+1) + \\frac{c}{2} \\quad \\forall x \\in \\mathbb{R}\n$$\nwhere $k$ is a real constant.\n\nRemark. We have $(x+2) G(x) = x G(x+1)$ for all $x \\in \\mathbb{R}$. This is a special case of Problem 2 in the test for Level 4+ (where $a = -1, b = -2$ ).",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40271,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are totally $16$ teams participating in a football tournament; each team playing with every other exactly $1$ time. In each match, the winner gains $3$ points, the loser gains $0$ point and each team gains $1$ point for the tie match. Suppose that at the end of the tournament, each team gains the same number of points. Prove that there are at least $4$ teams that have the same number of winning matches, the same number of losing matches and the same number of tie matches.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "See the solution to Problem 1 in the test of level 4. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40272,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a positive integer. Find all prime numbers $p$ with the following property: there exist exactly $p$ ordered pairs of integers $(x, y)$, with $0 \\leq x, y \\leq p-1$, such that $p$ divides $y^{2}-x^{3}-a^{2} x$.",
    "options": [],
    "answer": "p = 2, or p is an odd prime divisor of a, or p ≡ 3 (mod 4).",
    "solution": "The case $p=2$ works with solutions $(0,0)$ and $(1,1)$ if $a$ is even; $(0,0)$ and $(0,1)$ if $a$ is odd.\n\nWe claim that any odd prime divisor $p$ of $a$ also works. Of course for $0 \\leq x, y \\leq p-1$, the congruence $y^{2} \\equiv x^{3} \\pmod{p}$ implies that $x=0$ if and only if $y=0$. For any $0 < x \\leq p-1$, it's clear that $x^{3} = x^{2} \\cdot x$ is a quadratic residue modulo $p$ if and only if $x$ is.\n\nTherefore, if $x$ is not a quadratic residue then we cannot find any $y$ such that $y^{2} \\equiv x^{3} \\pmod{p}$ and if $x$ is a quadratic residue the congruence $y^{2} \\equiv x^{3} \\pmod{p}$ has exactly 2 solutions in the set $\\{1,2, \\ldots, p-1\\}$ (the condition $p$ is odd ensures that 2 solutions are distinct).\n\nSince there are $(p-1)/2$ quadratic residues, the number of $(x, y)$ with the required property is\n$$\n1 + 2 \\cdot (p-1)/2 = p.\n$$\nIf $p \\equiv 3 \\pmod{4}$ then $-1$ is not a quadratic residue, so if $x^{3} + a^{2} x \\neq 0$, then exactly one of $x^{3} + a^{2} x$ and $-x^{3} - a^{2} x$ is a square and it gives two solutions. Together with $(0,0)$, it gives us exactly $p$ solutions.\n\nIf $p \\equiv 1 \\pmod{4}$, let $i$ be a square root of $-1 \\pmod{p}$, i.e. $i^{2} = -1$. Then we have\n$$\ny^{2} \\equiv x(x + a i)(x - a i) \\pmod{p}.\n$$\nFor $x = 0, a i, -a i$, we have one choice of $y = 0$. For other choices of $x$, we have either 0 or 2 choices of $y$. Replacing $x$ by $-x$, since $-1$ is a quadratic residue, we also have two choices for $y$. Hence, the total number of pairs is $3 \\pmod{4}$, which cannot be exactly $p$.\n\nThus, the answer is $p = 2$, all the odd prime divisors of $a$ and $p \\equiv 3 \\pmod{4}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40273,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ satisfying the conditions:\n1. $f(x+1) \\geq f(x)+1$ for all $x \\in \\mathbb{R}$;\n2. $f(x y) \\geq f(x) f(y)$ for all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = x for all real x",
    "solution": "Let $x = y$ in (2), we have $f\\left(x^{2}\\right) \\geq f^{2}(x) \\geq 0$. This implies that $f(x) \\geq 0$ for all $x \\geq 0$.\n\nLet $x = 0$ in (1), we have $f(1) \\geq f(0) + 1 \\geq 1$.\n\nLet $x = y = 1$ in (1), we have $f(1) \\geq f^{2}(1)$, i.e., $0 \\leq f(1) \\leq 1$. These inequalities imply that $f(1) = 1$, and hence $f(0) = 0$.\n\nFrom (1) and (2), by induction, we have\n$$\nf(x+n) \\geq f(x) + n \\tag{*}\n$$\nfor all $x \\in \\mathbb{R}$, $n \\in \\mathbb{N}$, and\n$$\nf\\left(x^{n}\\right) \\geq f^{n}(x) \\tag{**}\n$$\nfor all $x \\geq 0$ and $n \\in \\mathbb{N}^{*}$.\n\nUsing $(*)$, for each $x > 0$, we have\n$$\nf(x) = f(\\{x\\} + \\lfloor x \\rfloor) \\geq f(\\{x\\}) + \\lfloor x \\rfloor \\geq \\lfloor x \\rfloor > x - 1.\n$$\nFrom $(**)$ and (2), for each $x > 1$ and $n \\in \\mathbb{N}^{*}$, we have\n$$\n\\left(x^{n} - 1\\right) f^{n}\\left(\\frac{1}{x}\\right) \\leq f\\left(x^{n}\\right) f\\left(\\frac{1}{x^{n}}\\right) \\leq f(1) = 1.\n$$\nHence,\n$$\nf\\left(\\frac{1}{x}\\right) \\leq \\frac{1}{\\sqrt[n]{x^{n} - 1}}\n$$\nfor all $x > 1$, $n \\in \\mathbb{N}^{*}$. Fix $x > 1$ and let $n \\rightarrow +\\infty$, we obtain $f\\left(\\frac{1}{x}\\right) \\leq \\frac{1}{x}$, or (note that $f(0) = 0$ and $f(1) = 1$),\n$$\nf(x) \\leq x\n$$\nfor $x \\in [0, 1]$.\n\nOn the other hand, we can prove by induction from (1) that\n$$\nf(x-n) \\leq f(x) - n\n$$\nfor $n \\in \\mathbb{N}$. For each $x < 0$, we have $\\lfloor x \\rfloor < 0$ so\n$$\nf(x) = f(\\{x\\} + \\lfloor x \\rfloor) \\leq f(\\{x\\}) + \\lfloor x \\rfloor \\leq \\{x\\} + \\lfloor x \\rfloor = x.\n$$\nIn (2), set $x = y = t$ with $t \\in [-1, 0]$ then\n$$\nf^{2}(t) \\leq f\\left(t^{2}\\right) \\leq t^{2}\n$$\nThis implies that $t \\leq f(t) \\leq -t$. Hence, $f(t) = t$ for all $t \\in [-1, 0]$.\n\nNow, for each $x < -1$, we have\n$$\n\\frac{1}{x^{2}} f^{2}(x) = f^{2}(x) f^{2}\\left(\\frac{1}{x}\\right) \\leq f\\left(x^{2}\\right) f\\left(\\frac{1}{x^{2}}\\right) \\leq f(1) = 1\n$$\nso $f^{2}(x) \\leq x^{2}$, implying that $x \\leq f(x) \\leq -x$. So $f(x) = x$ for all $x \\leq 0$.\n\nLet $y = -1$ in (2) and consider $x > 0$, we have\n$$\n-x = f(-x) \\geq f(x) f(-1) = -f(x)\n$$\nimplying $f(x) \\geq x$ for all $x > 0$. Hence, we have $f(x) = x$ for all $x \\in (0, 1)$.\n\nFinally, suppose that $x > 1$, we have\n$$\n\\frac{1}{x} f(x) = f(x) f\\left(\\frac{1}{x}\\right) \\leq f(1) = 1\n$$\nso $f(x) \\leq x$. Since $f(x) \\geq x$, we have $f(x) = x$ for all $x > 1$.\n\nInclusion, $f(x) = x$ for every $x$ is the only function satisfying given conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40274,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a set of $2^{2016}$ cards with the numbers $1, 2, \\ldots, 2^{2016}$ written on them. We divide the set of cards into pairs arbitrarily; from each pair, we keep the card with larger number and discard the other. We now again divide the $2^{2015}$ remaining cards into pairs arbitrarily; from each pair, we keep the card with smaller number and discard the other. We now have $2^{2014}$ cards, and again divide these cards into pairs and keep the larger one in each pair. We keep doing this way, alternating between keeping the larger number and keeping the smaller number in each pair, until we have just one card left. Find all possible values of this final card.",
    "options": [],
    "answer": "All integers x with 2^{1008} ≤ x ≤ 2^{2016} − 2^{1008} + 1",
    "solution": "Note that the remaining number is kept $1008$ times as the larger one of the pair. So it is bigger than at least $2^{1008}-1$ numbers.\n\nSimilarly, the remaining number is kept $1008$ times as the smaller one of the pair so it is smaller than at least $2^{1008}-1$ numbers.\n\nTherefore, the remaining number $x$ satisfies\n$$\n2^{1008} \\leq x \\leq 2^{2016} - 2^{1008} + 1.\n$$\nTo prove that any $x$ satisfies the above inequalities is true, we can carry out the pairing inductively so that after $2i$ steps, the following condition is satisfied: if the numbers remaining are\n$$\na_{1} < a_{2} < \\ldots < a_{2^{2(n-i)}}\n$$\nthen $x$ is one of these, and there are at least $2^{n-i}-1$ numbers smaller than $x$ and at least $2^{n-i}-1$ numbers larger than $x$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40275,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ such that\n$$\nf(2 m+f(m)+f(m) f(n))=n f(m)+m\n$$\nfor any integers $m, n$.",
    "options": [],
    "answer": "f(n) = n - 2 for all integers n",
    "solution": "Let $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ satisfy the given functional equation. Putting $n=0$ in this equation, we get\n$$\nf(2 m+f(m)+f(m) f(0))=m \\quad \\forall m \\in \\mathbb{Z} ;\n$$\ntherefore, $f$ is surjective, so there exists $u \\in \\mathbb{Z}$ such that $f(u)=-1$.\nWith $m=u$, the given equation gives us\n$$\nf(2 u-1-f(n))=-n+u .\n$$\nNow, if $a, b$ are some integers such that $f(a)=f(b)$, then\n$$\nu-a=f(2 u-1-f(a))=f(2 u-1-f(b))=u-b,\n$$\nwhich implies that $a=b$. Hence, $f$ is also injective.\nNext, putting $n=u$, we have\n$$\nf(2 m+f(m)-f(m))=u f(m)+m \\Leftrightarrow f(2 m)=u f(m)+m(*)\n$$\nfor all $m \\in \\mathbb{Z}$. In ( $*$ ), letting $m=0$, we see that $f(0)=u f(0)$, so $u=1$ or $f(0)=0$.\nIf $f(0)=0$, then with $m=u$, (*) would imply $f(2 u)=-u+u=0= f(0)$, i.e. $2 u=0 \\Leftrightarrow u=0$ (since $f$ is injective), and thus $f(0)=-1$, a contradiction!\nHence, $u=1 \\Leftrightarrow f(1)=-1$ and ( $*$ ) becomes\n$$\nf(2 m)=f(m)+m\n$$\nfor all $m \\in \\mathbb{Z}$. Here, letting $m=1$, we obtain $f(2)=0$.\nIn the given functional equation, putting $m=n=0$, we have\n$$\nf\\left(f(0)+f(0)^2\\right)=0 \\Leftrightarrow f(0)+f(0)^2=2 \\Leftrightarrow f(0)=1 \\text { or } f(0)=-2 .\n$$\nIf $f(0)=1$, then it would follow from the given equation with $m=0, n=2$ that $f(1)=2$, a contradiction (because $f(1)=-1$ )!\nHence, $f(0)=-2$. In the functional equation, putting $n=0$, we get\n$$\nf(2 m-f(m))=m \\quad \\forall m \\in \\mathbb{Z}\n$$\nUsing this, we see that if $f(m)=m-2$ then $f(m+2)=m$; but $f(0)= -2, f(1)=-1$, we can easily prove by induction that $f(n)=n-2$ for all $n \\geq 0$.\nIn the given equation, letting $m=1$, we obtain\n$$\nf(2-1-f(n))=-n+1 \\Leftrightarrow f(1-f(n))=1-n \\quad \\forall n \\in \\mathbb{Z} .\n$$\nNow replacing $n$ by $n+3$ and letting $n \\geq-3$, we have\n$f(1-f(n+3))=-n-2 \\Leftrightarrow f(1-(n+1))=-n-2 \\Leftrightarrow f(-n)=-n-2$.\nSo $f(n)=n-2$ for all $n \\in \\mathbb{Z}$.\nIt is easy to check that this function satisfies the given condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40276,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and $I$ its incenter. The point $D$ is on segment $BC$ and the circle $\\omega$ is tangent to the circumcircle of triangle $ABC$ but is also tangent to $DC$, $DA$ at $E$, $F$, respectively. Prove that $E$, $F$ and $I$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\omega$ the circumcircle of $\\triangle ABC$ and $\\gamma$ the circle tangent to $\\omega$, $DA$, $DC$. Let $\\omega$ touch $\\gamma$ at $K$ and $M$ be the midpoint of $\\operatorname{arc} BC$ on $\\omega$ not containing $K$. One has the dilation with center $K$ sending $\\gamma$ to $\\omega$ and $BC$ to the tangent of $\\omega$ at $M$. Hence $K$, $E$, $M$ are collinear. We also have $A$, $I$, $M$ are collinear and $MI = MC$.\n\nLet $EI$ meet $\\gamma$ again at $F'$. Since $\\omega$, $\\gamma$ have the same tangent at $K$ then $\\angle KF'I = \\angle EF'K = \\angle MAK = \\angle KAI \\Rightarrow AKIF'$ is cyclic. One has $\\angle MCB = \\angle MBC = \\angle MKC \\Rightarrow \\triangle MEC \\sim \\triangle MCK$. Hence\n$$\nMI^2 = MC^2 = ME \\cdot MK \\Rightarrow \\triangle MEI \\sim \\triangle MIK .\n$$\nTherefore, $\\angle KEI = \\angle AIK = \\angle AF'K$. This implies that $AF'$ is tangent to $\\gamma$, and $F = F'$.\nHence $E$, $I$, $F$ are collinear.\n![](attached_image_1.png) $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40277,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB \\neq AC$. The incircle of triangle $ABC$ is tangent to $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. The perpendicular line from $D$ to $EF$ intersects $AB$ at $X$. The second intersection point of circumcircles of triangles $AEF$ and $ABC$ is $T$. Prove that $TX \\perp TF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $G$ be the intersection point of $DX$ and $EF$.\n\n![](attached_image_1.png)\n\nConsider the inversion $I_{I}^{ID^{2}}$, then the circumcircle of $\\triangle ABC$ is sent to the nine-point circle of $\\triangle DEF$ and the line $EF$ is sent to the circumcircle of $\\triangle AEF$. Hence $I$, $T$, and $G$ are collinear.\n\nThen $AI \\parallel GX$ since they are both perpendicular to $EF$. We have\n$$\n\\angle GTF = \\angle ITF = \\angle IFG = \\angle IAE = \\angle FAI = \\angle FXG\n$$\nso, quadrilateral $TXGF$ is cyclic, this implies that\n$$\n\\angle XTF = \\angle XGF = 90^{\\circ}.\n$$\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40278,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEF$ be a convex hexagon with $AB = CD = EF$, $BC = DE = FA$ and $\\angle A + \\angle B = \\angle C + \\angle D = \\angle E + \\angle F$. Prove that all angles of this hexagon are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $BC$, $DE$, $AF$ intersect and bound triangle $MNP$. Because the sum of angles in a hexagon is $720^\\circ$, we have $\\angle A + \\angle B = \\angle C + \\angle D = \\angle E + \\angle F = 240^\\circ$.\n\n![](attached_image_1.png)\n\nTherefore, we easily see triangle $MNP$ is equilateral. To build the equilateral triangle $DEO$ with $O$ lying inside the hexagon, we see $EOAF$ and $DOBC$ are parallelograms so $AB = EF = OA = CD = OB$, thus $OAB$ is an equilateral triangle. Hence,\n$$\n\\angle AFE + \\angle BCD = \\angle AOE + \\angle BOD = 360^\\circ - 60^\\circ - 60^\\circ = 240^\\circ.\n$$\nBut $\\angle EDC + \\angle BCD = 240^\\circ$. These imply $\\angle AFE = \\angle EDC$.\n\nSimilarly, $\\angle EDC = \\angle CBA$. Similarly, $\\angle BAF = \\angle FED = \\angle EDC$.\n\nNow from the condition $\\angle A + \\angle B = \\angle C + \\angle D = \\angle E + \\angle F$ we deduce that all angles of this hexagon are equal to $120^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40279,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of polynomials $P(x), Q(x)$ with integer coefficients such that\n$$\nP(Q(x)) = (x-1)(x-2) \\ldots (x-9)\n$$\nfor all real numbers $x$.",
    "options": [],
    "answer": "All solutions are when one polynomial is linear. Specifically:\n- If Q(x) = x + a with a an integer, then P(x) = ∏_{k=1}^{9} (x − (a + k)). If Q(x) = −x + a, then P(x) = ∏_{k=1}^{9} (a − k − x).\n- If P(x) = x + a with a an integer, then Q(x) = (x − 1)(x − 2)⋯(x − 9) − a. If P(x) = −x + a, then Q(x) = −(x − 1)(x − 2)⋯(x − 9) − a.\nThere are no solutions with both degrees greater than one.",
    "solution": "We consider 3 cases.\n\n1) Suppose that $\\deg Q = 1$. By looking at the leading coefficient, we have $Q(x) = \\pm x + a$. If $Q(x) = x + a$, we have\n$$\nP(x) = (x - (a+1))(x - (a+2)) \\ldots (x - (a+9)).\n$$\nAnd if $Q(x) = -x + a$ we have\n$$\nP(x) = (a-1-x)(a-2-x) \\ldots (a-9-x).\n$$\n\n2) Suppose that $\\deg P = 1$. By looking at the leading coefficient, we have $P(x) = \\pm x + a$. In this case\n$$\nQ(x) = \\pm(x-1)(x-2) \\ldots (x-9) - a.\n$$\n\n3) Suppose $\\deg P, \\deg Q > 1$. Then, the only option is $\\deg P = \\deg Q = 3$.\nSince $1,2, \\ldots, 9$ are 9 roots of $P(Q(x))$, we have $P(x)$ have 3 roots $a, b, c$. For each root, for example $a$, the equation $Q(x) = a$ has 3 roots $a_1, a_2, a_3$ then $a_1, a_2, a_3, b_1, b_2, b_3, c_1, c_2, c_3$ form a permutation of $1,2, \\ldots, 9$.\nBy Vieta's formula, we have\n$$\na_1 + a_2 + a_3 = b_1 + b_2 + b_3 \\text{ and } a_1 a_2 + a_2 a_3 + a_3 a_1 = b_1 b_2 + b_2 b_3 + b_3 b_1\n$$\nso $a_1^2 + a_2^2 + a_3^2 = b_1^2 + b_2^2 + b_3^2$. Since $x^2 \\equiv 0$ or $1 \\pmod{4}$ so the number of odd numbers in $a_1, a_2, a_3$ is the same as the number of odd numbers in $b_1, b_2, b_3$.\nThis implies that the number of odd numbers in $1,2, \\ldots, 9$ is divisible by 3, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40280,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given four numbers $x, y, z, t$, let $(a, b, c, d)$ be a permutation of $(x, y, z, t)$ and set $x_{1}=|a-b|$, $y_{1}=|b-c|$, $z_{1}=|c-d|$, and $t_{1}=|d-a|$. From $x_{1}, y_{1}, z_{1}, t_{1}$, form in the same fashion the numbers $x_{2}, y_{2}, z_{2}, t_{2}$, and so on. It is known that $x_{n}=x$, $y_{n}=y$, $z_{n}=z$, $t_{n}=t$ for some $n$.\nFind all possible values of $(x, y, z, t)$.",
    "options": [],
    "answer": "All permutations of (0, 0, 0, 0), (a, a, 0, 0) with a > 0, and (2a, a, a, 0) with a > 0.",
    "solution": "First, consider 4 sequences $\\left(x_{n}\\right),\\left(y_{n}\\right),\\left(z_{n}\\right),\\left(t_{n}\\right)$ with $x_{1}=|a-b|$, $y_{1}=|b-c|$, $z_{1}=|c-d|$, $t_{1}=|d-a|$, $(a, b, c, d)$ is permutation of $(x, y, z, t)$. And\n$$\n\\left\\{\n\\begin{array}{l}\nx_{n+1}=\\left|a_{n}-b_{n}\\right| \\\\\ny_{n+1}=\\left|b_{n}-c_{n}\\right| \\\\\nz_{n+1}=\\left|c_{n}-d_{n}\\right| \\\\\nt_{n+1}=\\left|d_{n}-a_{n}\\right|\n\\end{array}\n\\right.\n$$\nwith $(a_{n}, b_{n}, c_{n}, d_{n})$ is a permutation of $(x_{n}, y_{n}, z_{n}, t_{n})$.\nFrom this, it is easy to see that $x_{n}, y_{n}, z_{n}, t_{n} \\geq 0$ for all $n$.\nLet $w_{n}=\\max \\left\\{x_{n}, y_{n}, z_{n}, t_{n}\\right\\}, n \\geq 1$. We have\n$$\n\\begin{aligned}\nw_{n} & =\\max \\left\\{x_{n+1}, y_{n+1}, z_{n+1}, t_{n+1}\\right\\} \\\\\n& =\\max \\left\\{\\left|a_{n}-b_{n}\\right|,\\left|b_{n}-c_{n}\\right|,\\left|c_{n}-d_{n}\\right|,\\left|d_{n}-a_{n}\\right|\\right\\} \\\\\n& \\leq \\max \\left\\{a_{n}, b_{n}, c_{n}, d_{n}\\right\\}=\\max \\left\\{x_{n}, y_{n}, z_{n}, t_{n}\\right\\}=w_{n}\n\\end{aligned}\n$$\nThe equality occurs when there is at least one number among $a_{n}, b_{n}, c_{n}, d_{n}$ is equal to $0$.\nHence the sequence $(w_{n})$ is non-increasing. Suppose that $k$ is a positive integer such that $(x_{k}, y_{k}, z_{k}, t_{k})=(x, y, z, t)$, then $w_{1} \\geq w_{2} \\geq \\cdots \\geq w_{k}=w_{1}$ which implies that $w_{1}=w_{2}=\\cdots=w_{k}$. So there is at least one number among $(x, y, z, t)$ equal to $0$. Based on the definition of the sequences, we can see that for each $i=1,2,3, \\ldots, k$, the tuple $(x_{i}, y_{i}, z_{i}, t_{i})$ has some common properties as follows:\n1. There is at least one number that is $0$.\n2. There are two numbers equal.\n3. In the permutation to obtain the next term, two equal numbers must be adjacent.\nIt is easy to check that 3 tuples as follows satisfy the given conditions:\n\ni. If $(x, y, z, t)=(0,0,0,0)$.\nii. If $(x, y, z, t)=(a, a, 0,0)$ with $a>0$ and its permutation.\niii. If $(x, y, z, t)=(2a, a, a, 0)$ with $a>0$ and its permutation.\n\nIndeed, when $(x, y, z, t)$ is one of the 3 above tuples, then the form of tuple $\\left(x_{n}, y_{n}, z_{n}, t_{n}\\right)$ will not change. Continue, we will prove that if $(x, y, z, t)$ is a tuple that satisfies the condition, then it must be one of the 3 above forms. (*)\n\nWithout loss of generality, we may assume that $x \\geq y \\geq z \\geq t=0, x>0$. Since there are two numbers equal among 4 numbers, we consider some cases:\n\n1. If $x=y$ then we have $(x, y, z, t)=(a, a, b, 0)$ with $a \\geq b \\geq 0, a>0$. Based on the 3rd property, $\\left(x_{1}, y_{1}, z_{1}, t_{1}\\right)$ can only be $(0, a-b, b, a)$. Apply the 2nd property, we have:\n- If $a=a-b, b=0$, we have $(x, y, z, t)=(a, a, 0,0)$, which satisfies the given condition.\n- If $a=b, a-b=0$, then we have $(x, y, z, t)=(a, a, a, 0)$ with $a>0$. So $\\left(x_{1}, y_{1}, z_{1}, t_{1}\\right)$ will be $(0,0, a, a)$. But from this, the form will not change, so we cannot obtain the original form anymore. This case doesn't satisfy.\n- If $a-b=b$ then $a=2b$, we have $(x, y, z, t)=(2b, 2b, b, 0)$ and $\\left(x_{1}, y_{1}, z_{1}, t_{1}\\right)=(0, b, b, 2b)$. But the form of this tuple will not change and similarly, this case doesn't satisfy.\n\n2. If $y=z$, then we have $(x, y, z, t)=(a, b, b, 0)$ with $a \\geq b>0$. So $(x_{1}, y_{1}, z_{1}, t_{1})$ can be $(a-b, 0, b, a)$. Apply the 2nd property, we have:\n- If $a-b=0, a=b$, we have $(x, y, z, t)=(a, a, a, 0)$, not satisfy.\n- If $a-b=a, b=0$, we have the tuple $(x, y, z, t)=(a, 0,0,0)$ and $\\left(x_{1}, y_{1}, z_{1}, t_{1}\\right)=(a, 0,0, a)$. It is easy to see that this case doesn't satisfy.\n- If $a-b=b$, we have $(x, y, z, t)=(2b, b, b, 0)$ which satisfies.\n\n3. If $z=0$ then we have $(x, y, z, t)=(a, b, 0,0)$ with $a \\geq b \\geq 0$. So $(x_{1}, y_{1}, z_{1}, t_{1})$ can be $(a-b, 0, b, a)$. Apply the 2nd property, we have:\n- If $a-b=0, a=b$, we have $(x, y, z, t)=(a, a, 0,0)$, satisfies.\n- If $a-b=a, b=0$, we have $(x, y, z, t)=(a, 0,0,0)$, not satisfy.\n- If $a-b=b$, we have the tuple $(x, y, z, t)=(2b, b, 0,0)$ and $\\left(x_{1}, y_{1}, z_{1}, t_{1}\\right)=(b, b, 0,2b)$. It is easy to see that this case doesn't satisfy.\n\nTherefore, there are 3 tuples that satisfy the given condition as above.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 40281,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ be positive numbers such that $a^{2}+b^{2}+c^{2}+a b c=4$. Prove that\n$$\n\\frac{a+b}{c}+\\frac{b+c}{a}+\\frac{c+a}{b} \\geq a+b+c+\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, assume that $a \\geq b \\geq c$. From the condition, we get $a \\geq 1 \\geq c > 0$. Write the inequality as\n$$\n\\frac{a+b+c}{c}+\\frac{a+b+c}{b}+\\frac{a+b+c}{c} \\geq a+b+c+\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+3\n$$\nwhich is equivalent to\n$$\n(a+b+c-1)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}-1\\right) \\geq 4\n$$\nSince $(a-1)(1-c) \\geq 0$, we obtain $a+c \\geq 1+a c$ and thus $\\frac{1}{a}+\\frac{1}{c} \\geq 1+\\frac{1}{a c}$. Therefore,\n$$\n\\begin{aligned}\n& (a+b+c-1)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}-1\\right) \\\\\n& \\geq (b+a c)\\left(\\frac{1}{b}+\\frac{1}{a c}\\right)=2+\\frac{b}{a c}+\\frac{a c}{b} \\geq 2+2=4\n\\end{aligned}\n$$\nThis completes the proof. Equality holds for $a=b=c=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40282,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geq 4$ be a positive integer and there exist $n$ positive integers that are arranged on a circle such that:\n- The product of each pair of two non-adjacent numbers is divisible by $2015 \\cdot 2016$.\n- The product of each pair of two adjacent numbers is not divisible by $2015 \\cdot 2016$.\nFind the maximum value of $n$.",
    "options": [],
    "answer": "8",
    "solution": "Denote $v_{p}(k)$ as the exponent of prime $p$ in the prime factorization of positive integer $k$. It is easy to see that $v_{p}(ab) = v_{p}(a) + v_{p}(b)$ for all positive integers $a, b$.\n\nWe have $2015 = 5 \\cdot 13 \\cdot 31$, $2016 = 2^{5} \\cdot 3^{2} \\cdot 7$.\nDenote $M = 2015 \\cdot 2016$ and\n$$\nN = \\{2, 3, 5, 7, 13, 31\\}\n$$\nSuppose there exist $n$ positive integers $x_{1}, x_{2}, x_{3}, \\ldots, x_{n}$ arranged on the circle satisfying the given condition and denote $x_{n+1} \\equiv x_{1}$ and $x_{n+2} \\equiv x_{2}$.\n\nSince $x_{i} x_{i+1}$ is not divisible by $M$ for $i = 1, \\ldots, n$, then there exists a prime $p_{i} \\in N$ such that\n$$\nv_{p_{i}}(x_{i} x_{i+1}) < v_{p_{i}}(M) \\Leftrightarrow v_{p_{i}}(x_{i}) + v_{p_{i}}(x_{i+1}) < v_{p_{i}}(M) \\tag{*}\n$$\nThese primes form a sequence $p_{1}, p_{2}, \\ldots, p_{n}$ and the elements can be repeated. We will prove that $p_{i} \\neq p_{j}$ with $i, j$ two non-adjacent indices.\n\nIndeed, if there are some non-adjacent indices $i, j$ such that $p_{i} = p_{j} = p$. Denote $a = v_{p}(M)$ and from (*), we have $v_{p}(x_{i}) + v_{p}(x_{i+1}) < a$ and $v_{p}(x_{j}) + v_{p}(x_{j+1}) < a$.\nHence, we have\n$$\nv_{p}(x_{i}) + v_{p}(x_{i+1}) + v_{p}(x_{j}) + v_{p}(x_{j+1}) < 2a.\n$$\nOn the other hand, since $(i, j), (i+1, j+1)$ are pairs of non-adjacent indices then\n$$\nv_{p}(x_{i}) + v_{p}(x_{j}) \\geq a, \\quad v_{p}(x_{i+1}) + v_{p}(x_{j+1}) \\geq a.\n$$\nThis implies that\n$$\nv_{p}(x_{i}) + v_{p}(x_{i+1}) + v_{p}(x_{j}) + v_{p}(x_{j+1}) \\geq 2a,\n$$\nwhich is a contradiction.\n\nContinue, suppose that there exists an index $i$ such that $p_{i} = p_{i+1} = p$. If $v_{p}(M) = 1$, because $x_{i} x_{i+1}$ and $x_{i+1} x_{i+2}$ are not divisible by $M$, we can conclude that $x_{i}, x_{i+1}, x_{i+2}$ are not divisible by $p$. Hence $x_{i} x_{i+2}$ is not divisible by $M$, contradiction because $x_{i}, x_{i+2}$ are not adjacent.\nSo if $p_{i} = p_{i+1} = p$, then we must have $v_{p}(M) \\geq 2$.\n\nNow we can conclude that the sequence $p_{1}, p_{2}, \\ldots, p_{n}$ has the following properties:\n- Each prime $p_{i} \\in N$ appears at most 2 times.\n- If some prime $p_{i} \\in N$ appears 2 times, then $p_{i}^{2} \\mid M$.\n\nHence, we have\n$$\np_{1} p_{2} \\ldots p_{n} \\mid M.\n$$\nSince $M = 2^{5} \\cdot 3^{2} \\cdot 5 \\cdot 7 \\cdot 13 \\cdot 31$, we can see that $n \\leq 8$ (in the sequence $(p_{i})$, prime $2$ appears $2$ times, prime $3$ appears $2$ times and the rest appear $1$ time). It is easy to check that if we choose 8 numbers as follows:\n$$\n\\frac{M}{2^{5}}, \\frac{M}{2 \\cdot 3}, \\frac{M}{3^{2}}, \\frac{M}{3 \\cdot 5}, \\frac{M}{5 \\cdot 7}, \\frac{M}{7 \\cdot 13}, \\frac{M}{13 \\cdot 31}, \\frac{M}{31 \\cdot 2}\n$$\nand the corresponding sequence is\n$$\np_{1} = 2,\\ p_{2} = 3,\\ p_{3} = 3,\\ p_{4} = 5,\\ p_{5} = 7,\\ p_{6} = 13,\\ p_{7} = 31,\\ p_{8} = 2.\n$$\nTherefore, the maximum value of $n$ is $8$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40283,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integer $n$ such that there exists a permutation $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ of $\\left(1,2,3, \\ldots, n\\right)$ satisfying the condition:\n$$\na_{1}+a_{2}+\\cdots+a_{k} \\text{ is divisible by } k \\text{ for each } k=1,2,3, \\ldots, n \\text{. }\n$$",
    "options": [],
    "answer": "1 and 3",
    "solution": "It is easy to see that $n=1$, $n=3$ satisfy the given condition, while $n=2$ does not. We will show that all numbers $n>3$ do not satisfy the condition.\n\nFirst, we can see that\n$$\na_{1}+a_{2}+\\cdots+a_{n}=1+2+\\cdots+n=\\frac{n(n+1)}{2}\n$$\nthen $\\frac{n(n+1)}{2}$ is divisible by $n$, which means $\\frac{n+1}{2} \\in \\mathbb{Z}$ or $n$ is an odd number. We also have\n$$\nn-1 \\left\\lvert\\, a_{1}+a_{2}+\\cdots+a_{n-1}=\\frac{n(n+1)}{2}-a_{n}=\\frac{(n+1)(n-1)}{2}+\\frac{n+1}{2}-a_{n}\\right.\n$$\nThus, $\\frac{n+1}{2}-a_{n}$ is divisible by $n-1$.\n\nOn the other hand,\n$$\n-(n-1)<\\frac{n+1}{2}-a_{n}<n-1\n$$\nthen we must have\n$$\n\\frac{n+1}{2}-a_{n}=0 \\Leftrightarrow a_{n}=\\frac{n+1}{2}\n$$\nIt implies that\n$$\na_{1}+a_{2}+\\cdots+a_{n-1}=\\frac{n^{2}-1}{2} .\n$$\nContinue, we have\n$$\nn-2 \\left\\lvert\\, a_{1}+a_{2}+\\cdots+a_{n-2}=\\frac{n^{2}-1}{2}-a_{n-1}=\\frac{(n-2)(n+1)}{2}+\\frac{n+1}{2}-a_{n-1}\\right.\n$$\nSimilarly, we get $a_{n-1}=\\frac{n+1}{2}$, then $a_{n-1}=a_{n}$, which is a contradiction.\n\nHence, the only two satisfied values are $n=1$ and $n=3$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40284,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $u$ and $v$ be positive rational numbers with $u \\neq v$. Assume that there are infinitely many positive integers $n$ with the property that $u^{n}-v^{n}$ are integers. Prove that $u$ and $v$ are integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $u = x / z$ and $v = y / z$ with $x$, $y$, $z$ positive integers, and $(x, y, z) = 1$. The statement is equivalent to saying that $x^{n} \\equiv y^{n} \\pmod{z^{n}}$.\n\nIf $z > 1$, assume that there is an odd prime divisor $p \\mid z$. Let $r$ be the least positive integer such that $x^{r} \\equiv y^{r} \\pmod{p}$, so $r \\mid n \\Rightarrow n = r k$. Let $a = v_{p}(n)$ and $b = v_{p}(x^{r} - y^{r})$. By LTE, we have\n$$\nv_{p}(x^{n} - y^{n}) = v_{p}((x^{r})^{k} - (y^{r})^{k}) = v_{p}(x^{r} - y^{r}) + v_{p}(k) \\leq b + v_{p}(n) = a + b.\n$$\nBut $v_{p}(x^{n} - y^{n}) \\geq v_{p}(z^{n}) \\leq n \\Rightarrow n \\leq a + b \\Rightarrow p^{n} \\leq p^{a} \\cdots p^{b} = n p^{b}$. This cannot be true for infinitely many positive integers $n$.\n\nIf $z$ has no odd prime divisor, then $z$ is a power of $2$. From this we have $x$, $y$ are both odd.\n\nIf $n$ is odd then\n$$\n2^{n} \\mid x^{n} - y^{n} = (x - y)(x^{n-1} + x^{n-2} y + \\cdots + x y^{n-2} + y^{n-1}).\n$$\nBut the second factor is odd since $n$ is odd, we get $2^{n} \\mid x - y$ for $x \\neq y$. It is clear that there are only finitely many such $n$.\n\nIf $n$ is even, let $s = v_{2}(x^{2} - y^{2})$ and $c = v_{2}(n)$. By LTE we have\n$$\nv_{2}(x^{n} - y^{n}) = c + s - 1 \\Rightarrow n \\leq c + s - 1 \\Rightarrow n \\leq \\log_{2} n + s - 1.\n$$\nWhich does not hold for sufficiently large even values of $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40285,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n$ such that there exists a polynomial $P(x)$ with integer coefficients satisfying\n$$\nP\\left(\\sqrt[3]{n^{2}}+\\sqrt[3]{n}\\right)=2016 n+20 \\sqrt[3]{n^{2}}+16 \\sqrt[3]{n} .\n$$",
    "options": [],
    "answer": "n ∈ {5, 3, 2, 0, -1, -3}",
    "solution": "First, we prove 2 lemmas as the problem in Level 4's Test. Then turn to the problem, since $P\\left(\\sqrt[3]{n^{2}}+\\sqrt[3]{n}\\right)=2016 n+20 \\sqrt[3]{n^{2}}+16 \\sqrt[3]{n}$, then by the second lemma, we obtain $n-1 \\mid 20-16=4$.\nThat is, $n \\in\\{5,3,2,0,-1,-3\\}$, and we can easily check that these numbers satisfy the condition of the problem. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40286,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle with the circumcircle ($O$). Denote $D, E$ as the midpoints of $AB, AC$ respectively. Two circles $(ABE)$ and $(ACD)$ intersect at $K$ differs from $A$. Suppose that the ray $AK$ intersects ($O$) at $L$. The line $LB$ meets $(ABE)$ at the second point $M$ and the line $LC$ meets $(ACD)$ at the second point $N$.\n1. Prove that $M, K, N$ collinear and $MN$ perpendicular to $OL$.\n2. Prove that $K$ is the midpoint of $MN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1) Denote $G$ as the intersection of $BE, CD$, then $G$ is the centroid of triangle $ABC$. We assume that $AB < AC$ and the solution is similar to all other cases.\nSince $ABLC$ is cyclic, we have $180^{\\circ} - \\angle AKN = \\angle ACN = \\angle ABM = \\angle AKM$, hence\n$$\n\\angle AKM + \\angle AKN = 180^{\\circ},\n$$\nwhich means $M, K, N$ are collinear.\nSince $AKNC$ and $AKBM$ are both cyclic, then by the power of a point to circle, we have\n$$\nLN \\cdot LC = LK \\cdot LA = LB \\cdot LM,\n$$\nthus $MBNC$ is concyclic and we can see $\\angle CBL = \\angle LNM$.\nDenote $P$ as the intersection of $OL$ and $MN$. Since $O$ is the circumcenter of $(ABLC)$ then\n$$\n\\angle PLN = 90^{\\circ} - \\frac{\\angle LOC}{2} = 90^{\\circ} - \\angle CBL = 90^{\\circ} - \\angle MNL\n$$\nTherefore, $OL$ is perpendicular to $MN$.\n![](attached_image_1.png)\n\n2) Notice that $ADGEB C$ is a complete quadrilateral and $K$ is the Miquel point so $K$ belongs to two circles ($BDG$), ($CEG$). Thus we have\n$$\n\\angle DKB = \\angle DGB = \\angle EGC = \\angle EKC, \\quad \\angle KDB = \\angle KGB = \\angle KCE.\n$$\nSo $\\triangle BKD \\sim \\triangle EKC$, then by sin law, we have\n$$\n\\frac{AB}{AC} = \\frac{BD}{CE} = \\frac{BK}{KE} = \\frac{\\sin BAK}{\\sin EAK} = \\frac{\\sin BAL}{\\sin CAL}\n$$\nwhich means $ABLC$ is the harmonic quadrilateral and $LA$ is the symmedian of triangle $LBC$.\nFinally, because $MN$ is the antiparallel to $BC$ with respect to $\\angle BAC$ then the symmedian of triangle $LBC$ is the median of triangle $LMN$, which means $LA$ passes through the midpoint of $MN$ or $K$ is the midpoint of the segment $MN$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40287,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b$ be two positive integers such that\n$$\nb+1 \\mid a^{2}+1, \\quad a+1 \\mid b^{2}+1\n$$\nProve that $a, b$ are odd numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose by contrary that one of $a, b$ is even, say $a$. So $a^{2}+1$ is odd. This means that $b+1$ must be odd, or equivalently, $b$ is even. We will prove that this cannot happen.\n\nIndeed, put $d=\\gcd(a+1, b+1)$ then $d \\mid a+1 \\mid b^{2}+1$, hence $d \\mid b^{2}-b = b(b+1) + 2b$. From this, we have $d \\mid 2b$ and $d \\mid 2(b+1) = 2b+2$ so $d \\mid 2$. But $a+1, b+1$ are odd numbers so $d$ odd, which gives $d=1$.\n\nOn the other hand, by assumption $a+1 \\mid a^{2}+b^{2}$, $b+1 \\mid a^{2}+b^{2}$, we get $a^{2}+b^{2} = (a^{2}+1) + (b^{2}-1) = (a^{2}-1) + (b^{2}+1)$ is divisible by both $a+1$ and $b+1$. Therefore $(a+1)(b+1) \\mid a^{2}+b^{2}$ (since $a+1$ and $b+1$ are coprime).\n\nIn other words, there exists a positive integer $k$ such that\n$$\na^{2}+b^{2}=k(a+1)(b+1) .\n$$\nWe now prove that this equation has no integer solution. Fix a value of $n$, assume that this equation has a solution of positive integers. We can choose positive integers $x_{0}, y_{0}$ satisfying the equation with the property that the sum $x_{0}+y_{0}$ is smallest and $x_{0} \\geq y_{0}$.\n\nConsider the quadratic equation\n$$\nX^{2}-X \\cdot n(y_{0}+1)+y_{0}^{2}-n y_{0}-n=0 .\\quad (*)\n$$\nThis equation has a solution $x_{0}$ so it also has another solution, say $x_{1}$. From Vieta's theorem, we have\n$$\nx_{0}+x_{1}=n(y_{0}+1), \\quad x_{0} x_{1}=y_{0}^{2}-n y_{0}-n\n$$\nWe consider the following cases:\n\n1. If $x_{1}<0$, i.e. $x_{1} \\leq -1$, then from $(*)$ we see that $x_{1}^{2}-x_{1} \\cdot n(y_{0}+1)+ y_{0}^{2}-n y_{0}-n=0$ or\n$$\n0 \\geq x_{1}^{2}+n(y_{0}+1)+y_{0}^{2}-n y_{0}-n = x_{1}^{2}+y_{0}^{2}>0,\n$$\nwhich is clearly a contradiction.\n\n2. If $x_{1}>0$ then $(x_{1}, y_{0})$ is also another solution of $(*)$, from the choice of $x_{0}$, we have $x_{1}>x_{0}$. Moreover, since $x_{0} x_{1}=y_{0}^{2}-n y_{0}-n$ it follows that $x_{0}^{2}<x_{0} x_{1}=y_{0}^{2}-n y_{0}-n<y_{0}^{2}$, this is also a contradiction because of the inequality $x_{0} \\geq y_{0}$.\n\nHence, $x_{1}=0$, and in this case $y_{0}^{2}=n(y_{0}+1)$, which implies that $y_{0}+1 \\mid y_{0}^{2}$. This cannot hold for $y_{0} \\geq 2$.\n\nTherefore, we conclude that $a, b$ are odd numbers, this ends the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40288,
    "subject": "Mathematics (Multi-modal)",
    "question": "A lock has 16 keys arranged in a $4 \\times 4$ array, each key oriented either horizontally or vertically. In order to open it, all the keys must be vertically oriented. When a key is switched to another position, all the other keys in the same row and column automatically switch their positions too. Show that no matter what the starting positions are, it is always possible to open this lock. (Only one key at a time can be switched.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The problem is solved if there is a way to change the orientation of any specified single key, without changing any of the others. This is equivalent to finding a way to switch the chosen key an odd number of times, while switching all other keys an even number of times.\nThis can be done by switching all keys on the same row and column of the chosen key (including the chosen key). To see why, choose a key $K$. If we switch it and all of its sisters that share the same row and column, $K$ will be switched 7 times. Now, examine the other 15 keys in the lock. There are two cases: either the key is a sister of $K$, or not.\nSuppose that $L$ is a sister of $K$, say, sharing a row with $K$. Then $L$ will be switched 4 times. For the other case, suppose $M$ is not a sister of $K$.\nThen $M$ will be switched twice, because among the 6 sisters of $K$ which are turned, exactly two of them share a row or column with $M$. Consequently, of all the keys in the lock, only $K$ is switched an odd number of times.\nAll other keys are switched either 2 or 4 times, leaving their orientation unchanged. Thus we will be able to open the lock by selecting each horizontal key one-by-one, and turning it and all of its sister keys. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40289,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a school, there are totally $n$ students, with $n \\geq 2$. The students take part in $m$ clubs and in each club, there are at least $2$ members (a student may take part in more than $1$ club). Eventually, the Principal notices that: If $2$ clubs share at least $2$ common members then they have different numbers of members. Prove that\n$$\nm \\leq (n-1)^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_i$ be the number of clubs that have $i$ members. Here, $2 \\leq i \\leq n$ and\n$$\nm = a_2 + a_3 + \\cdots + a_n.\n$$\nWe will count the tuples $(A, B, C)$ in which the students $A, B$ take part in the same club $C$ that has $i$ members.\n\n1. The first way of counting:\n- Choosing $1$ club among the clubs that have $i$ members, we have $a_i$ ways.\n- Choosing $2$ students that take part in that club, we have $\\binom{i}{2}$ ways.\n\n2. The second way of counting:\n- Choosing $2$ students among all students of the school, we have $\\binom{n}{2}$ ways.\n- Choose $1$ club in which these two students take part. By assumption, there is at most $1$ such club.\n\nSo we get the following relation:\n$$\na_i \\binom{i}{2} \\leq \\binom{n}{2} \\Leftrightarrow a_i \\leq \\frac{\\binom{n}{2}}{\\binom{i}{2}}.\n$$\nHence, we have\n$$\nm = a_2 + a_3 + \\cdots + a_n \\leq \\binom{n}{2} \\left( \\frac{1}{\\binom{2}{2}} + \\frac{1}{\\binom{3}{2}} + \\cdots + \\frac{1}{\\binom{n}{2}} \\right).\n$$\nNote that\n$$\n\\sum_{i=2}^{n} \\frac{1}{\\binom{i}{2}} = \\sum_{i=2}^{n} \\frac{2}{i(i-1)} = 2 \\sum_{i=2}^{n} \\left( \\frac{1}{i-1} - \\frac{1}{i} \\right ) = 2 \\left( 1 - \\frac{1}{n} \\right ) = \\frac{2(n-1)}{n}.\n$$\nTherefore,\n$$\nm \\leq \\frac{n(n-1)}{2} \\cdot \\frac{2(n-1)}{n} = (n-1)^2.\n$$\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40290,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a polynomial $P(x)$ with integral coefficients such that\n\n1. $P(\\sqrt[3]{25}+\\sqrt[3]{5})=220 \\sqrt[3]{25}+284 \\sqrt[3]{5}$ ?\n\n2. $P(\\sqrt[3]{25}+\\sqrt[3]{5})=1184 \\sqrt[3]{25}+1210 \\sqrt[3]{5}$ ?",
    "options": [],
    "answer": "1) Yes; for example P(x) = 16x^2 + 204x − 160. 2) No such polynomial exists.",
    "solution": "First, we shall prove two following lemmas:\n\nLemma 1. If $a$ is an integer number that is not a perfect cube, and $m, n, p$ are integer numbers such that $m+n \\sqrt[3]{a}+p \\sqrt[3]{a^{2}}=0$, then\n$$\nm=n=p=0 .\n$$\nProof of Lemma 1. Since $m+n \\sqrt[3]{a}+p \\sqrt[3]{a^{2}}=0$, then $\\sqrt[3]{a}$ is the root of $P(x)=p x^{2}+n x+m$ with integer coefficients.\nConsider the polynomial $Q(x)=x^{3}-a$, then $\\sqrt[3]{a}$ is also a root of $Q(x)$. Assume that $Q(x)$ is reducible, then one of the factors must have degree 1, that is $Q(x)$ has a rational root. But we know that $\\sqrt[3]{a}$ is irrational since $a$ is not a perfect cube (by assumption). Hence, $Q(x)$ is irreducible.\nIn other words, $Q(x)$ is the minimal polynomial of $\\sqrt[3]{a}$, and this leads to $P(x)=0$, or $m=n=p=0$.\n\nLemma 2. Given that $P(x)$ is a polynomial with integer coefficients, and $a$ is an integer such that $\\sqrt[3]{a}$ is not an integer. If\n$$\nP\\left(\\sqrt[3]{a^{2}}+\\sqrt[3]{a}\\right)=m+n \\sqrt[3]{a}+p \\sqrt[3]{a^{2}}\n$$\nthen $a-1 \\mid n-p$.\n\nProof of Lemma 2. We will prove by induction on the degree of $P(x)$ which is denoted by $k$.\nFor $k=1$, it is trivial.\nAssume that the statement is true for $k=t$, consider $P(x)$ of degree $t+1$, we write $P(x)=x Q(x)+s$, with $Q(x)$ of degree $t$ and $s=P(0)$ is an integer.\nAssume that $Q\\left(\\sqrt[3]{a^{2}}+\\sqrt[3]{a}\\right)=d+e \\sqrt[3]{a}+f \\sqrt[3]{a^{2}}$, then by induction hypothesis $a-1 \\mid e-f$. This gives\n$$\n\\begin{aligned}\nP\\left(\\sqrt[3]{a^{2}}+\\sqrt[3]{a}\\right) & =\\left(\\sqrt[3]{a^{2}}+\\sqrt[3]{a}\\right)\\left(d+e \\sqrt[3]{a}+f \\sqrt[3]{a^{2}}\\right)+s \\\\\n& =s+e a+f a+(d+f a) \\sqrt[3]{a}+(d+e) \\sqrt[3]{a^{2}}\n\\end{aligned}\n$$\nHence, $(d+f a)-(d+e)=f a-e=(f a-f)+(f-e)$, which is a multiple of $a-1$.\n\nNow we turn to the problem.\n\nFor part a), we show that $P(x)$ does exist, for example we choose $P(x)= u x^{2}+v x+w$, where $u, v, w$ must satisfy\n$$\nu+v=220,\\;5 u+v=284,\\;10 u+w=0 .$$\nThat is, $P(x)=16 x^{2}+204 x-160$.\n\nFor part b), since $1210-1184=26$, which is not a multiple of $4$, in this case there is no $P(x)$ satisfying the condition of the problem.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40291,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive integer $n$, show that there exists a positive integer $m$ such that $n$ divides $2016^{m} + m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We generalize the problem to the following problem:\nLet $a$ be a given positive integer. For every natural $n$, there is a positive integer $m$ such that $n$ divides $a^{m} + m$.\n\nIn fact, we proceed by induction on $n$. Obviously this statement holds for $n = 1$. Now assume $n > 1$ and this statement holds for every natural number less than $n$. Consider two cases:\n\nCase 1: $n = p$ is a prime. If $p \\mid a$ we are done. If not, take $m = (a+1)(p-1) + 1$. Then\n$$\na^{m} + m = a^{(a+1)(p-1)+1} + (a+1)(p-1) + 1 \\equiv (a+1)p \\equiv 0 \\pmod{p}.\n$$\n\nCase 2: $n$ is a composite. Let $p$ be the largest prime divisor of $n$. Write $n = p^{b} q_{1}^{a_{1}} \\cdots q_{k}^{a_{k}}$ as the prime factorization of $n$ and put $n = p n_{1}$.\n\nBy induction hypothesis (to $n_{1} < n$), there is a positive integer $k$ such that $a^{k} + k \\equiv 0 \\pmod{n_{1}}$. This leads to $a^{k} + k = n_{1} q$, and we represent $q = p q_{1} + r$ with $0 \\leqslant r < p$. Thus, $a^{k} + k = n_{1}(p q_{1} + r) \\equiv r n_{1} \\pmod{n}$. Now, put $A = (q_{1} - 1) \\cdots (q_{k} - 1)$, or $1$ if $n$ has only one prime divisor.\n\nSince $p$ is the largest prime divisor of $n$, it follows that $A$ and $p$ are coprime. Hence, there is a positive integer $c$ such that\n$$\nA c \\equiv 1 \\pmod{p} \\Longleftrightarrow (p-1) r A c \\equiv (p-1) r \\equiv -r \\pmod{p}.\n$$\nThis leads to\n$$\np^{b-1} q_{1}^{a_{1}} \\cdots q_{k}^{a_{k}} r (p-1) A c \\equiv -p^{b-1} q_{1}^{a_{1}} \\cdots q_{k}^{a_{k}} r \\pmod{p^{b} q_{1}^{a_{1}} \\cdots q_{k}^{a_{k}}}.\n$$\nHence, we have\n$$\nq_{1} \\cdots q_{k} \\varphi(n) r c \\equiv -n_{1} r \\pmod{n}.\n$$\nFinally, define $d = q_{1} \\cdots q_{k} c$ or $c$, if $n$ has one prime divisor. We get $r \\varphi(n) d \\equiv -n_{1} r \\pmod{n}$. Put $m = r \\varphi(n) d + k$. Then,\n$$\n\\begin{aligned}\na^{m} + m &\\equiv a^{r \\varphi(n) d + k} + r \\varphi(n) d + k \\equiv a^{k} + k - n_{1} r \\\\\n&\\equiv n_{1} r + r \\varphi(n) d \\equiv n_{1} r - n_{1} r \\equiv 0 \\pmod{n}\n\\end{aligned}\n$$\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40292,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the number of permutations $\\left(a_{1}, a_{2}, \\ldots, a_{2016}\\right)$ of the first $2016$ positive integers satisfying the following two conditions:\n1. $a_{i+1}-a_{i} \\leq 1$ for all $i=1,2,3, \\ldots, 2015$.\n2. There are exactly two indices $i<j$ with $1 \\leq i<j \\leq 2016$ such that $a_{i}=i$ and $a_{j}=j$.",
    "options": [],
    "answer": "2^{2012}",
    "solution": "For each positive integer $n \\geq 1$, we denote $s_{n}$ as the number of permutations of the first $n$ positive integers that satisfy the condition\n$$\na_{i+1}-a_{i} \\leq 1, \\quad i=1,2,\\ldots,n-1.\n$$\nWe call these permutations \"nice\".\n\nFirst, we shall prove that $s_{n}=2^{n-1}$ for all $n \\geq 1$.\nLet $k$ be the index such that $a_{k}=n$. We have $a_{k-1} \\geq a_{k}-1=n-1$ which implies that $a_{k-1}=n-1$, and so on, we have $a_{1}=n-k+1$.\nAll numbers after $a_{k}$ are $1,2,3, \\ldots, n-k$ forming a nice permutation, so in case $a_{k}=n$, the number of nice permutations is $s_{n-k}$.\nNote that if $k=n$, we have only $1$ nice permutation: $1,2,3,\\ldots,n$.\nTherefore, we get the following formula:\n$$\ns_{n}=1+\\sum_{k=1}^{n-1} s_{n-k}=s_{1}+s_{2}+\\ldots+s_{n-1}+1\n$$\nIt is easy to compute $s_{1}=1$, $s_{2}=2$, then by induction, we have $s_{n}=2^{n-1}$ for all $n \\geq 1$.\n\nBack to the original problem, we suppose that $a_{i}=i$ and $a_{j}=j$. Thus,\n$$\na_{i+1} \\leq a_{i}+1=i+1, \\quad a_{i+2} \\leq a_{i+1}+1 \\leq i+2\n$$\nand so on, then we have $a_{j} \\leq j$. But in fact, $a_{j}=j$ which implies that the equality must occur.\nSo $a_{k}=k$ for all $k=i, i+1, i+2, \\ldots, j$. But there are only $2$ indices $i, j$ like that so we have $j=i+1$.\n\nContinue, $a_{i-1} \\geq a_{i}-1=i-1$, but $a_{i}=i$, $a_{i+1}=i+1$ so $a_{i-1} \\geq i+2$. Then,\n$$\na_{i-2} \\geq a_{i-1}-1 \\geq i+1 \\text{ or } a_{i-2} \\geq i+2.\n$$\nBy the same way, it is clear to check that $a_{1}, a_{2}, \\ldots, a_{i-1} \\geq i+2$. Similarly, we also have $a_{i+2}, a_{i+3}, \\ldots, a_{2016} \\leq i-1$.\nTherefore, the lengths of the two sequences $a_{1}, a_{2}, \\ldots, a_{i-1}$ and $a_{i+2}, a_{i+3}, \\ldots, a_{2016}$ are the same. Hence,\n$$\ni-1=2015-i \\Leftrightarrow i=1008.\n$$\nSo the given nice permutation looks like\n$$\n\\left(a_{1}, a_{2}, \\ldots, a_{1007}, 1008, 1009, a_{1010}, a_{1011}, \\ldots, a_{2016}\\right).\n$$\nClearly, the subsequence $\\left(a_{1010}, a_{1011}, \\ldots, a_{2016}\\right)$ forms a nice permutation of length $1007$ and the same with the subsequence\n$$\n\\left(a_{1}-1009, a_{2}-1009, \\ldots, a_{1007}-1009\\right).\n$$\nTherefore, the number of permutations satisfying the given condition is\n$$\ns_{1007} \\cdot s_{1007}=\\left(2^{1006}\\right)^{2}=2^{2012}.\n$$ $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 40293,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a non-isosceles triangle with circumcircle $(O)$ and incircle $(I)$. Denote $(O_{1})$ as the circle that is internally tangent to $(O)$ at $A_{1}$ and also tangent to segments $AB$, $AC$ at $A_{b}$, $A_{c}$ respectively. Define the circles $(O_{2})$, $(O_{3})$ and the points $B_{1}$, $C_{1}$, $B_{c}$, $B_{a}$, $C_{a}$, $C_{b}$ similarly.\n1. Prove that $AA_{1}$, $BB_{1}$, $CC_{1}$ are concurrent at the point $M$ and the three points $I$, $M$, $O$ are collinear.\n2. Prove that the circle $(I)$ is inscribed in the hexagon with 6 vertices $A_{b}$, $A_{c}$, $B_{c}$, $B_{a}$, $C_{a}$, $C_{b}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1) We use inversion to solve this problem.\n![](attached_image_1.png)\nSuppose that $AI$, $A_{1}I$ intersect $(O)$ at $A_{0}$, $A_{1}$. Because $AI$ is the bisector then $A_{0}$ is the midpoint of the minor $\\operatorname{arc} BC$. Based on the property of the Mixtilinear circle, we also have $A_{1}$ as the midpoint of the major arc $BC$.\nConsider the inversion of center $I$ and ratio equal to the power of $I$ to $(O)$ as the function $f$.\nWe have $f(A) = A_{0}$, $f(A_{1}) = A_{2}$ then $f(AA_{1}) = (IA_{0}A_{2})$. Define $B_{0}$, $B_{2}$, $C_{0}$, $C_{2}$ similarly then $f(BB_{1}) = (IB_{0}B_{2})$, $f(CC_{1}) = (IC_{0}C_{2})$.\nIt is easy to see that three circles $(IA_{0}A_{2})$, $(IB_{0}B_{2})$ and $(IC_{0}C_{2})$ share the common point $I$. On the other hand, the power of $O$ to the three circles is also equal to $-R^{2}$ where $R$ is the radius of the circumcircle.\nHence, the three circles have two common points and one of them is $I$ which is the center of inversion. Then $AA_{1}$, $BB_{1}$, $CC_{1}$ are concurrent at a point $M$ and $M$, $I$, $O$ are collinear.\n\n2) From the property of the Mixtilinear circle, we have $B_{a}B_{c}$ and $C_{a}C_{b}$ have the common midpoint $I$ then $B_{a}C_{a}B_{c}C_{b}$ is a parallelogram, which implies that $B_{a}C_{a}$ is parallel to $BC$ and the distance from $I$ to $B_{a}C_{a}$ and $BC$ are the same. Hence $B_{a}C_{a}$ is tangent to $(I)$. Similarly, we also have $A_{b}C_{b}$ and $B_{c}A_{c}$ are also tangent to $(I)$.\n![](attached_image_2.png)\nIt is easy to see that $C_{b}B_{c}$ coincides with $BC$ then it is tangent to $(I)$. Similarly, we also have $C_{a}A_{c}$ and $A_{b}B_{a}$ are tangent to $(I)$.\nHence, the 6 sides of the hexagon $C_{b}B_{c}A_{c}C_{a}B_{a}A_{b}$ are tangent to the circle $(I)$, which finishes the solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40294,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there are infinitely many positive integers $n$ such that $n$ has at least two prime divisors and $20^{n}+16^{n}$ is divisible by $n^{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will construct (by induction on $k$) the infinite increasing sequence $\\left(n_{k}\\right)_{k \\geq 1}$ of odd positive integers such that any $n_{k}$ satisfies $4^{n_{k}}+5^{n_{k}}$ is divisible by $n_{k}^{2}$.\n\nWe take $n_{1}=1$ and $n_{2}=3$.\n\nAssume we already have $n_{k}$, that is $4^{n_{k}}+5^{n_{k}}=a n_{k}^{2}$ for some positive integer $a$ which must be odd, and greater than 1 (since $n_{k} \\geqslant n_{2}=3$, it follows that $4^{n_{k}}+5^{n_{k}}>n_{k}^{2}$).\n\nTake $p$ an odd prime divisor of $a$. Substitute $4^{n_{k}}$ by $x$ and $5^{n_{k}}$ by $y$. Then, obviously\n$$\nx^{p}+y^{p}=(x+y)\\left(x^{p-1}-y x^{p-2}+\\cdots-y^{p-2} x+y^{p-1}\\right)\n$$\nis divisible by $p n_{k}^{2}$. Furthermore, it is clear that\n$$\nx^{p-1},-y x^{p-2}, \\ldots,-y^{p-2} x, y^{p-1}\n$$\nare congruent to each other modulo $p$, and the number of them is $p$.\n\nSo, $x^{p-1}-y x^{p-2}+\\cdots-y^{p-2} x+y^{p-1}$ must be divisible by $p$. It follows that $x^{p}+y^{p}$ is divisible by $\\left(p n_{k}\\right)^{2}$.\n\nNow we take $n_{k+1}=p n_{k}$. As $p>1, n_{k+1}>n_{k}$, and $4^{n_{k+1}}+5^{n_{k+1}}$ is divisible by $n_{k+1}^{2}$.\n\nFinally, it is clear that the number $n=2 n_{k}$ satisfies $n^{2} \\mid 20^{n}+16^{n}$ for any positive integer $k$, which completes the solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40295,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two circles $O_{1}$ and $O_{2}$ intersect at $A$ and $B$. Let $d_{1}$ and $d_{2}$ be two lines through $A$ and be symmetric with respect to $AB$. The line $d_{1}$ cuts $O_{1}$ and $O_{2}$ at $G$, $E$ ($E \\neq A$), respectively; the line $d_{2}$ cuts $O_{1}$ and $O_{2}$ at $F$, $H$ ($F \\neq A$), respectively; such that $E$ is between $A$, $G$ and $F$ is between $A$, $H$.\nLet $J$ be the intersection of $EH$ and $FG$. The line $BJ$ cuts $O_{1}$, $O_{2}$ at $K$, $L$ ($K, L \\neq B$), respectively. Let $N$ be the intersection of $O_{1}K$ and $O_{2}L$. Prove that the circle $(NLK)$ is tangent to $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nWe have\n$$\n\\begin{aligned}\n\\angle AKO_{1} & = 90^{\\circ} - \\frac{1}{2} \\angle AO_{1}K = 90^{\\circ} - \\angle ABK = 90^{\\circ} - \\angle ABL \\\\\n& = 90^{\\circ} - \\frac{1}{2} \\angle AO_{2}L = \\angle ALO_{2}\n\\end{aligned}\n$$\nTherefore $ALNK$ is a cyclic quadrilateral.\n\nLet $O$ be the circumcenter of quadrilateral $ALNK$. Since $B$ is the intersection of $(AGF)$ and $(AEH)$ then $B$ is the Miquel point of complete quadrilateral $AEJF.GH$ or $B \\in (JFH)$.\n\nWe obtain\n$$\n\\begin{aligned}\n\\angle BAO & = \\angle BAK + \\angle KAO = \\angle BAF + \\angle FAK + 90^{\\circ} - \\angle ALK \\\\\n& = \\frac{1}{2}(\\overparen{BF} + \\overparen{FK}) + 90^{\\circ} - \\angle AHB = \\frac{1}{2}(\\overparen{BG} + \\overparen{FK}) - \\angle GJB + 90^{\\circ} = 90^{\\circ}\n\\end{aligned}.\n$$\nFrom this, $AB$ is a tangent of $(O)$. We are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40296,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $P, Q \\in \\mathbb{Z}[x]$ such that every positive integer is a divisor of a certain nonzero term of the sequence $\\left(x_{n}\\right)_{n=0}^{\\infty}$ given by the conditions:\n$$\nx_{0}=2016,\\quad x_{2 n+1}=P\\left(x_{2 n}\\right),\\quad x_{2 n+2}=Q\\left(x_{2 n+1}\\right) \\text{ for all } n \\geq 0.\n$$",
    "options": [],
    "answer": "All solutions are linear with slopes whose product is one, namely:\n\n1) P(x) = x + b and Q(x) = x + d with integers b, d such that b + d ≠ 0 and either (b + d) divides 2016 or (b + d) divides 2016 + b.\n\n2) P(x) = −x + b and Q(x) = −x + d with integers b, d such that b − d ≠ 0 and either (b − d) divides 2016 or (b − d) divides b − 2016.",
    "solution": "Suppose that $P, Q \\in \\mathbb{Z}[x]$ satisfy the given requirement. Step by step we will draw some conclusions.\n\nStep 1. $\\operatorname{deg} P \\geq 1, \\operatorname{deg} Q \\geq 1$.\nSuppose, on the contrary, that one of $P, Q$ were a constant polynomial $c$.\n1. If $P(x)=c\\ (\\forall x \\in \\mathbb{Z})$, then\n$$\nx_{n} \\in\\{2016, c, Q(c)\\} \\quad \\forall n \\geq 0.\n$$\n2. If $Q(x)=c\\ (\\forall x \\in \\mathbb{Z})$, then\n$$\nx_{n} \\in\\{2016, P(2016), c, P(c)\\} \\quad \\forall n \\geq 0.\n$$\nIn both cases, the requirement \"every positive integer is a divisor of a certain nonzero term of the sequence $\\left(x_{n}\\right)_{n=0}^{\\infty}$\" couldn't be satisfied. Hence, $\\operatorname{deg} P \\geq 1, \\operatorname{deg} Q \\geq 1$.\n\nStep 2. $\\operatorname{deg} Q=1$.\nSuppose, on the contrary, that $\\operatorname{deg} Q>1$.\nBy Step 1, $\\operatorname{deg} P \\geq 1$, so there exist positive numbers $M$ and $k \\geq 1$ such that $k|P(x)| \\geq|x|$ for all $x \\in \\mathbb{Z}$ with $|x| \\geq M$. In particular, $\\lim _{|x| \\rightarrow \\infty}|P(x)|= \\infty$.\nBut $\\operatorname{deg} Q>1$, which implies that\n$$\n\\lim _{|x| \\rightarrow \\infty} \\frac{|Q(x)|}{|x|}=\\infty ;\n$$\ntherefore, if $N \\in(M, \\infty)$ is chosen large enough, then we have:\n$$\n|Q(x)|>|x|\n$$\nand\n$$\n|Q(P(x))|>2 k|P(x)|=k|P(x)|+k|P(x)| \\geq k|P(x)|+|x|\n$$\nwhen $|x| \\geq N$.\nNow, in accordance with the given requirement, $\\lim _{n \\rightarrow \\infty} \\max _{0 \\leq \\ell \\leq 2 n}\\left|x_{\\ell}\\right|=\\infty$. Further, for large $n$, let $0 \\leq i=i(n) \\leq 2 n$ be an index such that\n$$\n\\left|x_{i}\\right|=\\max _{0 \\leq \\ell \\leq 2 n}\\left|x_{\\ell}\\right|>N.\n$$\nIf $i$ were odd, say $i=2 j-1$, then we would get\n$$\n\\left|x_{2 j}\\right|=\\left|Q\\left(x_{2 j-1}\\right)\\right|>\\left|x_{2 j-1}\\right|=\\max _{0 \\leq \\ell \\leq 2 n}\\left|x_{\\ell}\\right|,\n$$\na contradiction (since $2 j \\leq 2 n$ ). So, $i$ should be even (when $n$ is large enough), say $i=2 j$. For such a $j$, take $m=\\left|x_{2 j+2}-x_{2 j}\\right|$. Then\n$$\nm=\\left|Q\\left(P\\left(x_{2 j}\\right)\\right)-x_{2 j}\\right| \\geq\\left|Q\\left(P\\left(x_{2 j}\\right)\\right)\\right|-\\left|x_{2 j}\\right|>k\\left|P\\left(x_{2 j}\\right)\\right| \\geq\\left|x_{2 j}\\right|\n$$\nand $m>k\\left|P\\left(x_{2 j}\\right)\\right| \\geq\\left|P\\left(x_{2 j}\\right)\\right|=\\left|x_{2 j+1}\\right|$. Therefore, among the first $2 j+2$ terms $x_{0}, x_{1}, \\ldots, x_{2 j}, x_{2 j+1}$, there do not exist any term which is nonzero and which is divisible by $m$. It should also be noted here that $x_{2 j} x_{2 j+1} \\neq 0$ (since $\\left|x_{2 j+1}\\right|=\\left|P\\left(x_{2 j}\\right)\\right| \\geq \\frac{\\left|x_{2 j}\\right|}{k}>\\frac{N}{k}>0$ ). Hence, neither $x_{2 j}$ nor $x_{2 j+1}$ is divisible by $m$.\nOn the other hand,\n$$\n\\left(x_{2 \\ell+2}-x_{2 \\ell}\\right) \\mid\\left(Q\\left(P\\left(x_{2 \\ell+2}\\right)\\right)-Q\\left(P\\left(x_{2 \\ell}\\right)\\right)\\right)=x_{2 \\ell+4}-x_{2 \\ell+2}\n$$\nand\n$$\n\\left(x_{2 \\ell+2}-x_{2 \\ell}\\right) \\mid\\left(P\\left(x_{2 \\ell+2}\\right)-P\\left(x_{2 \\ell}\\right)\\right)=x_{2 \\ell+3}-x_{2 \\ell+1}.\n$$\nThus, if $\\ell \\geq j$, then $x_{2 \\ell+2}-x_{2 \\ell}$ and $x_{2 \\ell+3}-x_{2 \\ell+1}$ are both divisible by $m=\\left|x_{2 j+2}-x_{2 j}\\right|$. But, as we have shown above, neither $x_{2 j}$ nor $x_{2 j+1}$ is divisible by $m$. It follows that neither $x_{2 \\ell+2}$ nor $x_{2 \\ell+3}$ is divisible by $m$ when $\\ell \\geq j$.\nTherefore, $m$ could not be a divisor of any nonzero term of the sequence $\\left(x_{n}\\right)_{n=0}^{\\infty}$, a contradiction again! So, $\\operatorname{deg} Q=1$.\n\nStep 3. $\\operatorname{deg} P=1$.\nThe proof is similar to that in Step 2.\n\nStep 4. Now $P(x)=a x+b, Q(x)=c x+d$ with $a, b, c, d \\in \\mathbb{Z}$ and $a b \\neq 0$. We will prove that $a c=1$. By definition,\n$$\n\\left\\{\n\\begin{array}{l}\nx_{2 n+1}=a x_{2 n}+b \\\\\nx_{2 n+2}=c x_{2 n+1}+d\n\\end{array} \\quad \\forall n \\geq 0 .\\right.\n$$\nHence, $x_{2 n+2}=a c x_{2 n}+b c+d$ and $x_{2 n+3}=a c x_{2 n+1}+a d+b$ for all $n \\geq 0$. These 2 sub-sequences share a common recurrence relation of the form $y_{n+1}=r y_{n}+s$ with $s \\in \\mathbb{Z}, r=a c$.\nSuppose, on the contrary, that $r \\neq 1$. Then\n$$\ny_{n}=r^{n} y_{0}+s \\frac{r^{n}-1}{r-1} \\quad \\forall n \\geq 0.\n$$\nIf $r=-1$, then $y_{n} \\in\\left\\{y_{0},-y_{0}+s\\right\\}(\\forall n)$; therefore, the given requirement could not be satisfied. So, $|r|>1$.\nThe requirement implies that one of the above-mentioned 2 sub-sequences has the following property: for each $q \\in \\mathbb{N}_{0}$ there is an $n=n(q) \\in \\mathbb{N}_{0}$ such that $r^{q} \\mid y_{n} \\neq 0$. Of course, $n \\rightarrow \\infty$ as $q \\rightarrow \\infty$. Moreover,\n$$\nr^{\\min \\{q, n\\}} \\left\\lvert\\,\\left(y_{n}-r^{n} y_{0}\\right)=s \\frac{r^{n}-1}{r-1} .\\right.\n$$\nSince $\\operatorname{gcd}\\left(r, \\frac{r^{n}-1}{r-1}\\right)=1$, it follows that $r^{\\min \\{q, n\\}} \\mid s$ (for all $q$ ). But $|r|>1$ and $\\min \\{q, n\\} \\rightarrow \\infty$ as $q \\rightarrow \\infty$, we see that $s=0$. Hence, $y_{n}=r^{n} y_{0}(\\forall n)$,\nand therefore, the requirement could not be satisfied. This contradiction shows that $r=a c=1$.\n\nStep 5. Finally, $P(x)= \\pm x+b, Q(x)= \\pm x+d$ where $b, d \\in \\mathbb{Z}$ are to be found. We have to consider two cases:\n- $P(x)=x+b$ and $Q(x)=x+d$ with $b, d \\in \\mathbb{Z}$. In this case, by induction, we can show that\n$$\nx_{2 n}=2016+n(b+d) \\text{ and } x_{2 n+1}=2016+b+n(b+d) \\quad \\forall n \\geq 0.\n$$\nSo, a necessity condition for the requirement to be satisfied is $b+d \\neq 0$. Under this condition, the requirement says that for each $m \\in \\mathbb{N}$ one of the linear congruences\n$$\n(b+d) x \\equiv-2016(\\bmod m),(b+d) x \\equiv-(2016+b)(\\bmod m)\n$$\nhas (infinitely many) solutions $x=n$ in $\\mathbb{N}_{0}$. Equivalently,\n$\\operatorname{gcd}(b+d, m) \\mid 2016$ or $\\operatorname{gcd}(b+d, m) \\mid(2016+b)$ for each $m \\in \\mathbb{N}$.\nIt suffices to consider $m=|b+d|$ and obtain the condition:\n$$\n(b+d) \\mid 2016 \\text{ or }(b+d) \\mid(2016+b), \\text{ where } b+d \\neq 0\n$$\n- $P(x)=-x+b$ and $Q(x)=-x+d$ with $b, d \\in \\mathbb{Z}$. In almost the same manner, we see that the requirement is satisfied if and only if\n$$\n(b-d) \\mid 2016 \\text{ or }(b-d) \\mid(-2016+b), \\text{ where } b-d \\neq 0.\n$$\n\nRemark. We are going to give here another proof of the conclusions in Steps 2-3 (the proof of that in Step 1 and Steps 4-5 will be the same). We need the following lemma.\n\nLemma. Let $a \\in \\mathbb{Z}$ and $T \\in \\mathbb{Z}[x]$ be given. A sequence ( $y_{n}$ ) is defined as\n$$\ny_{0}=a,\\quad y_{n+1}=T\\left(y_{n}\\right) \\quad \\forall n \\geq 0\n$$\nSuppose that each positive integer $m$ is a divisor of some nonzero term of $\\left(y_{n}\\right)$. Then $\\operatorname{deg} T=1$.\n\nProof. It is easy to check that any constant polynomial $T$ does not satisfy the given condition. We suppose on the contrary that $\\operatorname{deg} T>1$. Then there exists $c>0$ such that $|T(x)|>2|x|$ whenever $|x|>c$.\nBy assumption, $\\lim _{m \\rightarrow \\infty} \\max _{0 \\leq \\ell \\leq m}\\left|y_{\\ell}\\right|=\\infty$. Further, let $0 \\leq n=n(m) \\leq m$ be the smallest index such that $\\left|y_{n}\\right|=\\max _{0 \\leq \\ell \\leq m}\\left|y_{\\ell}\\right|$. Then $n \\rightarrow \\infty$ as $m \\rightarrow \\infty$, and $\\left|y_{n}\\right|>\\left|y_{i}\\right|$ for all $0 \\leq i<n$. Hence, we can choose an $N \\in \\mathbb{N}$ such that $\\left|y_{N}\\right|>\\max \\left\\{c,\\left|y_{0}\\right|,\\left|y_{1}\\right|, \\ldots,\\left|y_{N-1}\\right|\\right\\}$. In particular, this implies that\n$$\n\\left|y_{N+1}\\right|=\\left|T\\left(y_{N}\\right)\\right|>2\\left|y_{N}\\right|.\n$$\nSet $m=\\left|y_{N+1}-y_{N}\\right|$. Then\n$$\nm \\geq\\left|y_{N+1}\\right|-\\left|y_{N}\\right|>\\left|y_{N}\\right|>\\max \\left\\{\\left|y_{0}\\right|,\\left|y_{1}\\right|, \\ldots,\\left|y_{N-1}\\right|\\right\\}.\n$$\nTherefore, $y_{N}$ is not divisible by $m$. Moreover, $m$ is not a divisor of any nonzero term among $y_{0}, y_{1}, \\ldots, y_{N-1}$.\nOn the other hand, $y_{n+1}-y_{n}$ is divisible by $y_{n}-y_{n-1}$ for all $n \\geq 1$. So, $y_{n+1}-y_{n}$ is divisible by $m=\\left|y_{N+1}-y_{N}\\right|$ for all $n \\geq N$. It follows that\n$$\ny_{n}-y_{N}=\\left(y_{n}-y_{n-1}\\right)+\\left(y_{n-1}-y_{n-2}\\right)+\\ldots+\\left(y_{N+1}-y_{N}\\right)\n$$\nis divisible by $m$ for all $n>N$. But $y_{N}$ is not divisible by $m$, so $y_{n}$ is not divisible by $m$ for all $n>N$, which is a contradiction. This completes the proof of the lemma.\n\nWe are now in a position to prove the conclusions of Steps 2-3.\nLet $H(x)=P(Q(x))$ and $K(x)=Q(P(x))$. Suppose, on the contrary, that either $\\operatorname{deg} P \\geq 2$ or $\\operatorname{deg} Q \\geq 2$. Then $\\operatorname{deg} H \\geq 2$ and $\\operatorname{deg} K \\geq 2$ (since, by Step $1, \\operatorname{deg} P \\geq 1, \\operatorname{deg} Q \\geq 1$ ).\nConsider the sub-sequence $\\left(x_{0}, x_{2}, x_{4}, \\ldots\\right), x_{2(n+1)}=K\\left(x_{2 n}\\right)$ for all $n \\geq 0$. Because $\\operatorname{deg} K \\geq 2$, the sequence $\\left(y_{n}\\right)=\\left(x_{2 n}\\right)$ cannot satisfy the condition given in the lemma. Thus, there exists a positive integer $m$ such that none of $m, 2 m, 3 m, \\ldots$ can be a divisor of a certain nonzero term $x_{2 n}$.\nThis implies that for each $k \\in \\mathbb{N}$ there exists a nonzero term $x_{2 n_{k}+1}$ divisible by $k m$. Therefore, the sub-sequence\n$$\n\\left(x_{1}, x_{3}, x_{5}, \\ldots\\right), x_{2 n+3}=H\\left(x_{2 n+1}\\right)\n$$\n(for all $n \\geq 0$ ), satisfies the condition given in the lemma. According to this lemma, $\\operatorname{deg} H=1$, a contradiction.\nThis contradiction shows that $\\operatorname{deg} P=1, \\operatorname{deg} Q=1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40297,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two positive integers $r > s$, and let $\\mathscr{F}$ be an infinite family of sets, each of size $r$, no two of which share fewer than $s$ elements. Prove that there exists a set of size $r-1$ that shares at least $s$ elements with each set in $\\mathscr{F}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We call a set $S$ s-meets $\\mathscr{F}$ if it shares at least $s$ elements with each set in $\\mathscr{F}$. Suppose no such set of size (at most) $r-1$ exists. (Each $S \\in \\mathscr{F}$ s-meets $\\mathscr{F}$ by the problem hypothesis.)\n\nLet $T$ be a maximal set such that $T \\subseteq S$ for infinitely many $S \\in \\mathscr{F}$, which form $\\mathscr{F}' \\subseteq \\mathscr{F}$ (such $T$ exists, since the empty set works).\n\nClearly $|T| < r$, so by assumption, $T$ does not s-meet $\\mathscr{F}$, and there exists $U \\in \\mathscr{F}$ with $|U \\cap T| \\leq s-1$.\n\nBut $U$ s-meets $\\mathscr{F}'$, so by pigeonhole, there must exist $u \\in U \\setminus T$ belonging to infinitely many $S \\in \\mathscr{F}'$, contradicting the maximality of $T$.\n\nRemark. Let $X$ be an infinite set, and $a_1, a_2, \\ldots, a_{2r-2-s}$ elements not in $X$. Then the set\n$$\n\\mathscr{F} = \\left\\{ B \\cup \\{x\\} : B \\subseteq \\{a_1, a_2, \\ldots, a_{2r-2-s}\\}, |B| = r-1, x \\in X \\right\\}\n$$\nshows we cannot replace $r-1$ with any smaller number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40298,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m, n$ be odd integers such that $n^{2}-1$ is divisible by $m^{2}+1-n^{2}$. Prove that $\\left|m^{2}+1-n^{2}\\right|$ is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By assumption, there is an integer $t$ such that $n^{2}-1 = t\\left(m^{2}+1-n^{2}\\right)$. Put $k = t+1$. It is clear that $k \\neq 0$. We have $k\\left(n^{2}-1\\right) = t m^{2}$. Note that $(k, t) = 1$ we get there is an integer $l$ such that $n^{2}-1 = l t$ (which gives $l = m^{2}+1-n^{2}$) and $m^{2} = l k = k\\left(m^{2}+1-n^{2}\\right) = (t+1)\\left(m^{2}+1-n^{2}\\right)$. Therefore, it suffices to prove that $k$ is a perfect square.\nLet $S$ be the set of ordered pairs $(x, y)$ of integers such that\n$$\n(x+y)^{2} = k(1+4 x y) .\n$$\nThen, the pair $\\left(\\frac{m+n}{2}, \\frac{m-n}{2}\\right) \\in S$, and this shows that $S$ is non-empty. Put\n$$\na = \\min \\{|x| ; \\exists(x, y) \\in S\\} .\n$$\nWe show that $a=0$ whence $k = y^{2}$.\nIf $(x, y) \\in S \\Rightarrow (-x, -y) \\in S$, then $(a, y) \\in S$. We have $(a+y)^{2} = k(1+4 a y)$. The two roots $b_{1}, b_{2}$ of the equation are integers with $\\left|b_{1}\\right|, \\left|b_{2}\\right| \\geqslant a$. (Since $\\left(a, b_{i}\\right) \\in S \\Rightarrow \\left(b_{i}, a\\right) \\in S$ ). On the other hand,\n$$\nb_{1}+b_{2} = 4 a k - 2 a, \\quad b_{1} b_{2} = a^{2} - k .\n$$\nWe get $\\left(a+b_{1}\\right)\\left(a+b_{2}\\right) = \\left(4 a^{2}-1\\right) k$.\nIf $k<0$ then $b_{1} b_{2}>0, b_{1}+b_{2} \\leqslant 0 \\Rightarrow b_{1}, b_{2}<0$. Since $\\left|b_{i}\\right|>a$, it follows that $a+b_{i}<0 \\Rightarrow \\left(a+b_{1}\\right)\\left(a+b_{2}\\right)>0$. This means that $\\left(4 a^{2}-1\\right) k>0$. Since $k<0$, we get $a=0$. Now, assume that $k>0$ and $a \\neq 0$ ($a>0$).\nIn this case, $b_{1}+b_{2}>0, \\left(a+b_{1}\\right)\\left(a+b_{2}\\right)>0 \\Rightarrow b_{1}, b_{2}>0$. By the choice of $a$, we get $a^{2} \\leqslant b_{1} b_{2}$, which contradicts to the equality $b_{1} b_{2} = a^{2} - k$. So $a=0$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40299,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given triangle $A B C$ inscribed in $(O)$. Two tangents at $B$, $C$ of $(O)$ intersect at $P$. The bisector of angle $A$ intersects $(P, P B)$ (the circle with center $P$ and radius $P B$) at point $E$ lying inside triangle $A B C$. Let $M$, $N$ be the midpoints of two arcs $B C$ of $(O)$ such that $M$ and $A$ are on different sides of $B C$. The circle with diameter $B C$ intersects the line segment $E N$ at $F$. Prove that the orthocenter of triangle $E F M$ belongs to $B C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet $I$ be the midpoint of $B C$. We have $\\angle I C M = \\angle M A C = \\angle M C P$ then $C M$ is the bisector of $\\angle I C P$. This follows that $M$ is the insimilicenter of $(I)$ and $(P)$. However, $\\angle M C N = 90^{\\circ}$ hence $N$ is the exsimilicenter of $(I)$ and $(P)$.\n\nLet $L$, $T$ be the intersections of $F M$ and $(P)$ ($L$ is between $F$ and $M$). Consider the homothety center at $M$, ratio $\\frac{-P C}{I C}$, $\\mathscr{H}_{M}^{\\frac{-P C}{I C}}: F \\mapsto T$, we get $I F \\parallel P T$.\n\nBy the same way, consider the homothety $\\mathscr{H}_{N}^{\\frac{P C}{I C}}: F \\mapsto E$, we get $I F \\parallel P E$.\n\nTherefore $E$, $P$, $T$ are collinear or $E T$ is the diameter of $(P)$. This means $\\angle E L P = 90^{\\circ}$.\n\nDenote $J$ the second intersection of $F N$ and $(O)$.\n\nSince 3 circles $(O)$, $(P)$, $(E L M J)$ have $E L$, $B C$, $J M$ as their radical axes then $E L$ cuts $J M$ at point $K$ lying on $B C$.\n\nIn conclusion, orthocenter $K$ of triangle $E F M$ lies on $B C$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40300,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle whose incircle ($I$) is tangent to $AB$, $AC$ at $D$, $E$ respectively. Denote by $\\Delta_{b}$, $\\Delta_{c}$ the lines symmetric to the lines $AB$, $AC$ with respect to $CD$, $BE$ correspondingly. Suppose that $\\Delta_{b}$, $\\Delta_{c}$ meet at $K$.\n1. Prove that $IK \\perp BC$.\n2. If $I \\in (KDE)$, prove that $BD + CE = BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1. Denote by $G$, $H$ the intersections of $KE$, $KF$ and $BC$ respectively. Consider triangle $CEG$: it is easy to see that\n- $CI$ is the internal bisector of $\\angle C$\n- $EI$ is the external bisector of $\\angle E$.\nHence, $I$ is the excenter of this triangle, which implies that $GI$ is the bisector of $\\angle KGH$.\n![](attached_image_1.png)\nSimilarly, we can also see that $HI$ is the bisector of $\\angle GHK$. So $I$ is the incenter of triangle $GHK$ and $KI$ is the bisector of $\\angle GKH$.\nIn the other hand, $BA$, $BC$ are symmetric with respect to the line $BE$ and $CA$, $KG$ are also symmetric with respect to the line $BE$. From these facts, we can conclude that $\\angle KGH = \\angle BAC$.\nSimilarly, we also have $\\angle KHG = \\angle BAC$; therefore, $KGH$ is an isosceles triangle.\nSince $I$ is the incenter of the isosceles triangle $KGH$, $KI$ is also an altitude, i.e. $KI \\perp BC$.\n\n2. Denote $BC = a$, $CA = b$, $AB = c$ and $A = \\alpha$. Then $\\angle DIE = 90^{\\circ} + \\frac{\\alpha}{2}$.\nIn the isosceles triangle $KGH$, we have $\\angle KGH = \\angle KHG = \\alpha$, so $\\angle DKE = 180^{\\circ} - 2\\alpha$. Hence, if $I \\in (KDE)$, then\n$$\n90^{\\circ} + \\frac{\\alpha}{2} + 180^{\\circ} - 2\\alpha = 180^{\\circ}\n$$\nand $\\alpha = 60^{\\circ}$. From the cosine law in triangle $ABC$, we have\n$$\na^{2} = b^{2} + c^{2} - bc .\n$$\nBy calculating, it is also easy to check that $BD = \\frac{ac}{a+b}$, $CE = \\frac{ab}{a+c}$; thus,\n$$\nBD + CE = BC \\Leftrightarrow \\frac{ac}{a+b} + \\frac{ab}{a+c} = a \\Leftrightarrow b^{2} + c^{2} - bc = a^{2},\n$$\nwhich is true. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40301,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a school there are totally $n > 2$ classes and not all of them have the same numbers of students. It is given that each class has one head student. The students in each class wear hats of the same color and different classes have different hat colors. One day all the students of the school stand in a circle facing toward the center, in an arbitrary order, to play a game. Every minute, each student puts his hat on the person standing next to him on the right. Show that at some moment, there are 2 head students wearing hats of the same color.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that there are $m$ students in total. We number the students standing in the circle by $1, 2, 3, \\ldots, m$, clockwise (starting with an arbitrary student). Denote by $a_{1} < a_{2} < a_{3} < \\cdots < a_{n}$ the positions of the $n$ head students obtained in this numbering.\n\nWe consider an $n \\times m$ table and fill numbers in its cells in the following way: For each $1 \\leq i \\leq n$, in the $i^{\\text{th}}$ row, we start with $a_{i}$ (filling in the first cell of this row with $a_{i}$) and continue with\n$$\na_{i} + 1,\\ a_{i} + 2,\\ \\ldots,\\ m,\\ 1, 2, \\ldots, a_{i} - 1\n$$\nto the right. We will focus on the hat colors of the students in the first column.\n\n| $a_{1}$ | $a_{1} + 1$ | $a_{1} + 2$ | $\\ldots$ | $a_{1} - 1$ |\n| :---: | :---: | :---: | :--- | :--- |\n| $a_{2}$ | $a_{2} + 1$ | $a_{2} + 2$ | $\\ldots$ | $a_{2} - 1$ |\n| $a_{3}$ | $a_{3} + 1$ | $a_{3} + 2$ | $\\ldots$ | $a_{3} - 1$ |\n|  |  |  |  |  |\n| $a_{n}$ | $a_{n} + 1$ | $a_{n} + 2$ | $\\ldots$ | $a_{n} - 1$ |\n\nAt the first minute, each student in the first column wears his own hat.\n\nAt the second minute, based on the game rule, each student in the first column will wear the hat of the student in the same row but in the second column.\n\nSimilarly, at the $k^{\\text{th}}$ minute, each student in the first column will wear the hat of the student in the same row but in the $k^{\\text{th}}$ column, for each $1 \\leq k \\leq m$.\n\nSuppose, on the contrary, that we could never find 2 head students (in the first column) wearing hats of the same color. This would imply that at the beginning all the students in each column also wear hats of pairwise different colors, so they belong to distinct classes. Hence, the numbers of students of $n$ classes are the same, which is a contradiction to the hypothesis.\n\nTherefore, at some moment, there are 2 head students wearing hats of the same color.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40302,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P \\in \\mathbb{Q}[x]$ be a polynomial of degree $2016$ whose leading coefficient is $1$. A positive integer $m$ is \"nice\" if there exists some positive integer $n$ such that $m = n^{3} + 3n + 1$. Suppose that there exist infinitely many positive integers $n$ such that $P(n)$ are nice. Prove that there exists an arithmetic sequence $\\left(n_{k}\\right)$ of arbitrary length such that $P\\left(n_{k}\\right)$ are all nice for $k = 1, 2, 3, \\ldots$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For convenience, denote $3d = 2016$.\n\nLemma. There exist a polynomial $Q(x) \\in \\mathbb{Q}[x]$ such that\n$$\n\\lim_{x \\rightarrow +\\infty} [\\sqrt[3]{P(x)} - Q(x)] = 0\n$$\nProof. Suppose that $S(x) \\in \\mathbb{Q}[x]$ is a polynomial such that $\\operatorname{deg}(P(x) - S^{3}(x))$ is minimized (if there are many $S(x)$ that satisfy, we choose one of them).\nIt is easy to check that:\n- $\\operatorname{deg}(P(x) - S^{3}(x)) < \\operatorname{deg} P(x) = 3d$.\n- $S(x)$ has leading coefficient $1$.\n- $\\operatorname{deg}(P(x) - S^{3}(x)) < 2d$.\nFor iii), denote $P(x) - S^{3}(x) = a_{l} x^{l} + a_{l-1} x^{l-1} + \\cdots + a_{0}$ and suppose by contradiction that $l = \\operatorname{deg}(P(x) - S^{3}(x)) \\geq 2d$. Consider polynomial $S^*(x) = S(x) + \\frac{a_{l}}{3} x^{l-2d}$ then\n$$\n\\begin{aligned}\n& P(x) - (S^*(x))^{3} = \\\\\n& \\left(P(x) - S^{3}(x) - a_{l} x^{l-2d} S^{2}(x)\\right) - \\left(\\frac{1}{3} a_{l}^{2} x^{2l-4d} S(x) + \\frac{1}{27} a_{l}^{3} x^{3l-6d}\\right).\n\\end{aligned}\n$$\nThe result is a polynomial of degree less than $l$. This means\n$$\n\\operatorname{deg}(P(x) - (S^*(x))^{3}) < l = \\operatorname{deg}(P(x) - S^{3}(x))\n$$\nwhich is a contradiction. Then, take $Q(x) = S(x)$ to get\n$$\n\\sqrt[3]{P(x)} - Q(x) = \\frac{P(x) - Q^{3}(x)}{\\sqrt[3]{P^{2}(x)} + \\sqrt[3]{P(x)} Q(x) + Q^{2}(x)}\n$$\nFrom iii), we can see that\n$$\n\\operatorname{deg}(\\sqrt[3]{P(x)} - Q(x)) < 2d = \\operatorname{deg}(\\sqrt[3]{P^{2}(x)} + \\sqrt[3]{P(x)} Q(x) + Q^{2}(x))\n$$\nthen $\\lim_{x \\rightarrow +\\infty} [\\sqrt[3]{P(x)} - Q(x)] = 0$. The lemma is proved.\n\nSince there exist infinitely many numbers $a$ such that $P(a)$ is nice, we can choose some strictly increasing sequence of positive integers $\\left(a_{n}\\right)$ such that $P\\left(a_{i}\\right)$ are nice for all $i \\in \\mathbb{N}^{*}$.\nClearly, for each $a_{i}$, there exists $N_{i} \\in \\mathbb{Z}^{+}$ such that $P\\left(a_{i}\\right) = N_{i}^{3} + 3N_{i} + 1$.\nSince $\\lim_{x \\rightarrow +\\infty} a_{i} = +\\infty$, which implies that\n$$\n\\lim_{x \\rightarrow +\\infty} P\\left(a_{i}\\right) = +\\infty \\text{ and } \\lim_{x \\rightarrow +\\infty} N_{i} = +\\infty\n$$\nthen\n$$\n\\lim_{x \\rightarrow +\\infty} [\\sqrt[3]{N_{i}^{3} + 3N_{i} + 1} - N_{i}] = 0\n$$\nNow, let $Q(x)$ be the polynomial described in the lemma.\nBecause $Q(x) \\in \\mathbb{Q}[x]$, there exists a number $M \\in \\mathbb{Z}^{+}$ such that $M \\cdot Q(x) \\in \\mathbb{Z}[x]$. We have\n$$\n\\lim_{i \\rightarrow +\\infty} M(Q(a_{i}) - N_{i}) = \\lim_{i \\rightarrow +\\infty} M(Q(a_{i}) - \\sqrt[3]{P(a_{i})} + \\sqrt[3]{P(a_{i})} - N_{i}) = 0\n$$\nHence, there is a number $i_{0} \\in \\mathbb{Z}^{+}$ such that\n$$\n-1 < M(Q(a_{i}) - N_{i}) < 1, \\forall i > i_{0}.\n$$\nOn the other hand, $M Q(a_{i}) \\in \\mathbb{Z}$, $N_{i} \\in \\mathbb{Z}$ so $M(Q(a_{i}) - N_{i}) = 0$ for all $i > i_{0}$, then $Q(a_{i}) - N_{i} = 0 \\Leftrightarrow P(a_{i}) = Q^{3}(a_{i}) + 3 Q(a_{i}) + 1$ for all $i > i_{0}$. This implies that $P(x) = Q^{3}(x) + 3 Q(x) + 1$ for all $x \\in \\mathbb{R}$.\n\nChoose a number $N$ such that $Q(x) > 0, \\forall x > a_{N}$ then we will prove that the arithmetic sequence given by the formula $n_{k} = a_{N} + M k$ satisfies the given requirement.\nIt is needed to clarify $Q(a_{1} + M k) \\in \\mathbb{Z}^{+}$ for all $k \\in \\mathbb{Z}^{+}$.\nIndeed, let $R(x) = M Q(x)$ then $R(x) \\in \\mathbb{Z}[x]$, then $R(n_{1}) = M a_{1}$ is divisible by $M$. So\n$$\n(n_{k} - n_{1}) \\mid R(n_{k}) - R(n_{1})\n$$\nleads to $M \\mid R(n_{k})$ for all $k \\in \\mathbb{Z}^{+}$. This finishes the proof.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40303,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that\n$$\nx[f(x+y)-f(x-y)]=4 y f(x)\n$$\nfor any real numbers $x, y$.",
    "options": [],
    "answer": "f(x) = c x^2 for some real constant c",
    "solution": "Denote ($*$) as the given condition.\nLet $a = f(0)$. Taking $x = y$ in ($*$), we get\n$$\nf(2x) = a + 4 f(x)\n$$\nfor any $x \\neq 0$. Now, by replacing $x, y$ by $2x, 2y$ in ($*$) and using the above property (here we choose $x, y$ such that $x + y, x - y, x \\neq 0$), we get\n$$\n2x[4 f(x + y) - 4 f(x - y)] = 8y[a + 4 f(x)],\n$$\nor\n$$\nf(x + y) - f(x - y) = \\frac{a y}{x} + \\frac{4 y}{x} f(x).\n$$\nCombining this with ($*$), we get $a = 0$. It follows that $f(0) = 0$ and $f(2x) = 4 f(x)$ for any $x$.\nReplacing $y = -x$ in ($*$), we have\n$$\nf(-2x) = 4 f(x) = f(2x)\n$$\nfor any $x \\neq 0$. Since $f(0) = 0$, we conclude that $f$ is an even function. Now, replacing $x$ by $x$ in (1), we get\n$$\nx[f(-x - y) - f(y - x)] = 4y f(-x)\n$$\nor\n$$\nx[f(y + x) - f(y - x)] = 4y f(x).\n$$\nIt follows that\n$$\nx y [f(y + x) - f(y - x)] = 4 y^2 f(x).\n$$\nFrom ($*$), we get $y [f(y + x) - f(y - x)] = 4x f(y)$. Combine with the above equation, we get\n$$\nx^2 f(y) = y^2 f(x)\n$$\nfor any real numbers $x, y$. This shows that\n$$\nf(x) = a x^2\n$$\nwhere $a$ is a constant, which is truly a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40304,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are totally 16 teams participating in a football tournament; each team playing with every other exactly 1 time. In each match, the winner gains 3 points, the loser gains 0 point and each team gains 1 point for the tie match. Suppose that at the end of the tournament, each team gains the same number of points. Prove that there are at least 4 teams that have the same number of winning matches, the same number of losing matches and the same number of tie matches.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, we can see that if a team got $x$ win matches, $y$ lose matches and $z$ tie matches then they got $3x + z$ points and $x + y + z = 15$. We call two teams having the same number of win matches, lose matches and tie matches as \"relate\".\n\nDenote the number of win matches, lose matches and tie matches of two teams $A, B$ by $(x_1, y_1, z_1), (x_2, y_2, z_2)$ respectively then\n$$\n3x_1 + z_1 = 3x_2 + z_2,\n$$\nthen the difference between the tie matches is\n$$\n|z_1 - z_2| = 3|x_1 - x_2| \\text{ which is divisible by } 3.\n$$\nAnd two teams are relate if and only if they have the same number of tie matches.\n\nThe number of their tie matches lies between $0$ and $15$ and all are congruent modulo $3$. So we can divide these teams into some groups by the number of tie matches (the teams in a group have the same number of tie matches). Hence, by the Pigeonhole principle, the number of groups cannot exceed $\\left\\lceil \\frac{15}{3} \\right\\rceil = 6$.\n\nIf there are some groups with greater than or equal to 4 members, i.e., 4 teams, then we are done.\n\nOtherwise, each group contains 3 members or fewer. Then the number of groups must be 6, if not, the number of teams is not greater than 15, which is a contradiction.\n\nBut the only way to obtain 6 groups like that is the number of tie matches of 16 teams are $0, 3, 6, 9, 12, 15$. It means there are some teams that have all matches as ties and there are also some teams that have all matches not as ties, which is a contradiction.\n\nHence, there are some groups with at least 4 members. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40305,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a given positive integer. Prove that there are infinitely many pairs of positive integers $(a, b)$ with $a, b > n$ such that\n$$\n\\prod_{i=1}^{2015}(a+i) \\mid b(b+2016) ; \\quad \\prod_{i=1}^{2015}(a+i) \\nmid b ; \\quad \\prod_{i=1}^{2015}(a+i) \\nmid(b+2016) .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The given problem can be generalized as follows:\nGiven three positive integers $k, m, n$. Let $k_{1}, k_{2}, \\ldots, k_{m}$ be any positive integer. Prove that there are infinitely many pair of positive integers $(a, b)$ such that\n$$\n\\prod_{k=1}^{m}\\left(a+k_{i}\\right) \\mid b(b+k) \\text{ but } \\prod_{k=1}^{m}\\left(a+k_{i}\\right) \\nmid b \\text{ and } \\prod_{k=1}^{m}\\left(a+k_{i}\\right) \\nmid b+k .\n$$\nProof. Let $k < p_{1} < p_{2} < \\ldots < p_{m}$ be any $m$ distinct primes and denote\n$$\nM = \\prod_{k=1}^{m}\\left(a+k_{i}\\right) .\n$$\nBy Chinese remainder theorem, there are infinitely many positive integers $a > n$ such that\n$$\na \\equiv -k_{i} \\pmod{p_{i}}, \\forall i = 1, \\ldots, m .\n$$\nHence $M \\equiv 0 \\pmod{p_{1} p_{2} \\ldots p_{m}}$.\nThen we write\n$$\nM = p_{1}^{\\alpha_{1}} p_{2}^{\\alpha_{2}} \\ldots p_{m}^{\\alpha_{m}} q_{1}^{\\beta_{1}} q_{2}^{\\beta_{2}} \\ldots q_{s}^{\\beta_{s}}\n$$\nwith $\\alpha_{i} \\geq 1, i = 1, \\ldots, m ; \\beta_{j} \\geq 1, j = 1, \\ldots, s$ and $q_{1}, q_{2}, \\ldots, q_{s}$ are $s$ prime divisors of $M$ which are different from $p_{1}, p_{2}, \\ldots, p_{m}$.\nBy Chinese remainder theorem, there exist infinitely many positive integers $b > m$ such that\n$$\n\\begin{cases}\nb \\equiv 0 \\pmod{\\frac{M}{p_{m}^{\\alpha_{m}}}} \\\\\nb \\equiv -k \\pmod{p_{m}^{\\alpha_{m}}}\n\\end{cases}\n$$\nThis implies that\n$$\nM \\nmid b, \\quad M \\nmid b+k \\text{ and } M \\mid b(b+k) .\n$$\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40306,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a given prime. For each prime $r$, we define the function as follows\n$$\nF(r) = \\frac{(p^{r p} - 1)(p - 1)}{(p^r - 1)(p^p - 1)}\n$$\n1. Show that $F(r)$ is a positive integer for any prime $r \\neq p$.\n2. Show that $F(r)$ and $F(s)$ are coprime for any primes $r$ and $s$ such that $r \\neq p$, $s \\neq p$ and $r \\neq s$.\n3. Fix a prime $r \\neq p$. Show that there is a prime divisor $q$ of $F(r)$ such that $p \\mid q-1$ but $p^2 \\nmid q-1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that with positive integers $a, m, n$ and $a > 1$, we have\n$$\n\\gcd(a^m - 1, a^n - 1) = a^{\\gcd(m, n)} - 1.\n$$\nLet $f(r) = \\frac{p^r - 1}{p - 1}$ with $p$ a prime and $r$ a positive integer.\n\n1.\nLet $x = \\gcd(p^r - 1, p^p - 1)$ and $y = \\operatorname{lcm}(p^r - 1, p^p - 1)$, then\n$$\n(p^r - 1)(p^p - 1) = x y.\n$$\nFrom the above lemma, we get $x = p^{\\gcd(r, p)} - 1 = p - 1$. Hence $(p^r - 1)(p^p - 1) = (p - 1) y$, which implies that $F(r) = \\frac{p^{p r} - 1}{y}$. But $p^r - 1 \\mid p^{p r} - 1$ and $p^p - 1 \\mid p^{p r} - 1$, so $y \\mid p^{p r} - 1$ leads to $F(r)$ being a positive integer.\n\n2.\nWe can see that\n$$\n\\gcd(p^{r p} - 1, p^{s p} - 1) = p^{\\gcd(r p, s p)} - 1 = p^p - 1.\n$$\nLet $p^{r p} - 1 = (p^p - 1) r_1$, $p^{s p} - 1 = (p^p - 1) s_1$ with $\\gcd(r_1, s_1) = 1$, then\n$$\n\\begin{aligned}\n\\gcd(F(r), F(s)) &= \\gcd\\left(\\frac{(p^p - 1) r_1 (p - 1)}{(p^r - 1)(p^p - 1)}, \\frac{(p^p - 1) s_1 (p - 1)}{(p^s - 1)(p^p - 1)}\\right) \\\\\n&= \\gcd\\left(\\frac{r_1}{f(r)}, \\frac{s_1}{f(s)}\\right) = 1\n\\end{aligned}\n$$\n\n3.\nAt first, we will show that with any prime divisor $q$ of $F(r)$, we always have $p \\mid q-1$. (*)\nIndeed, from Fermat's theorem, $q \\mid p^{q-1} - 1$ and because $q \\mid \\frac{p^{r p} - 1}{(p^r - 1) f(p)}$, so $q \\mid p^{r p} - 1$.\nThese imply that\n$$\nq \\mid \\gcd(p^{q-1} - 1, p^{r p} - 1) = p^{\\gcd(q-1, r p)} - 1\n$$\nSince $r, p$ are two primes, $d = \\gcd(q-1, r p) \\in \\{1, r, p, r p\\}$. We have to consider 4 following cases:\n\n1. If $d = p$ or $d = r p$, we get $p \\mid q-1$ and (*) follows.\n2. If $d = 1$, we have $q \\mid p - 1$ or $p \\equiv 1 \\pmod{q}$. We also have\n$$\n0 \\equiv \\frac{p^{r p} - 1}{p^p - 1} = p^{p(r-1)} + p^{p(r-2)} + \\cdots + p^p + 1 \\equiv p \\equiv 1 \\pmod{q},\n$$\ncontradiction.\n3. If $d = r$, we have $q \\mid p^r - 1$ or $p^r \\equiv 1 \\pmod{q}$. We also have\n$$\n0 \\equiv \\frac{p^{r p} - 1}{p^r - 1} = p^{r(p-1)} + p^{r(p-2)} + \\cdots + p^r + 1 \\equiv r \\pmod{q},\n$$\ncontradiction.\n\nHence (*) is true. Finally, assume that for all prime $q \\mid F(r)$, we always have $p^2 \\mid q-1$.\nBecause all divisors of $F(r)$ are congruent to $1$ modulo $p^2$, then $F(r) \\equiv 1 \\pmod{p^2}$.\nNote that\n$$\nF(r)(p^p - 1)(p^r - 1) = (p^{r p} - 1)(p - 1) = p^{r p + 1} - p^{r p} - (p - 1) \\equiv -(p - 1) \\pmod{p^2}\n$$\nand\n$$\nF(r)(p^p - 1)(p^r - 1) = F(r)(p^{p + r} - p^r - p^p + 1) \\equiv F(r) \\equiv 1 \\pmod{p^2}.\n$$\nThese imply that\n$$\n1 \\equiv -(p - 1) \\pmod{p^2} \\Leftrightarrow p \\equiv 0 \\pmod{p^2},\n$$\nwhich is a contradiction.\n\nTherefore, there exists a prime satisfying all the given conditions. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40307,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a checkered square $10 \\times 10$ the cells of the upper left $5 \\times 5$ square are black and all the other cells are white. What is the maximal $n$ such that the original square can be dissected (along the borders of the cells) into $n$ polygons such that in each of them the number of black cells is three times less than the number of white cells? (The polygons need not be congruent or even equal in area.)",
    "options": [],
    "answer": "9",
    "solution": "The answer is $9$. We can see that there are only $9$ cells on the border of the black square that connect to the white area. Each of them belongs to at most $1$ polygon, so there are at most $9$ polygons.\n\nAn example as follows (each of cells belongs to one part that has the ratio of black:white is $1:3$ )",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40308,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n \\geq 3$ such that we can divide a convex $n$-polygon into triangles by using some diagonals of this polygon such that the number of the used diagonals of every vertex is even.",
    "options": [],
    "answer": "n is divisible by 3",
    "solution": "The answer is $3 \\mid n$.\n\n1) We show that if $n=3k$ then we can divide every convex $n$-polygon (without overlapping) into triangles by using some diagonals of this polygon such that the number of the used diagonals of every vertex is an even integer. The proof is by induction in $k$. The figure shows how to reduce from $3(k+1)$-polygon to $3k$-polygon.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\n2) We show that if a convex $n$-polygon can divide (non-overlap) a convex $n$-polygon into triangles by using some diagonals of this polygon such that the number of the used diagonals of every vertex is an even integer, then $3 \\mid n$.\n\nFirst note that the number of triangles is exact $n-2$ using the sum of the angles (this sum is $\\pi \\times (n-2)$ in any convex $n$-polygon). Thus, the number of used diagonals is $n-3$ (by adding all the numbers of sides of the $n-2$ triangles we get $3(n-2)=n+2 \\times$ number of diagonals). Now, we can color the triangles with white and black in such a way that: One side of the $n$-polygon is black and no adjacent (having one side in common) triangles have the same color.\n\nSince the diagonals of each vertex is even, we easily see that all sides of the $n$-polygon are black. Let $x$ denote the number of the black triangles, the sum of the sides of all black triangles is\n$$\n3x = n + (n-3) \\Rightarrow 3 \\mid n.\n$$\n$\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 40309,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given six three-element subsets of the set $X$ with at least $5$ elements, show that it is possible to color the elements of $X$ in two colors such that none of the given subsets is all in one color.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_1, \\ldots, A_6$ be the subsets. We induct on the number $n$ of elements of $X$.\n\nIf $n = 5$, since $\\binom{5}{3} = 10 > 6$, we can find a three-element subset $Y$ of $X$ not equal to any of $A_1, \\ldots, A_6$; coloring the elements of $Y$ in one color and the other elements in the other color.\n\nIf $n = 6$, since $\\binom{6}{3} = 20 > 6 \\times 2 = 12$, we can find a three-element subset $Y$ of $X$ not equal to any of $A_1, \\ldots, A_6$ or its complements; coloring the elements of $Y$ in one color and the other elements in the other color meets the desired condition.\n\nNow suppose $n \\geq 7$. There must be two elements $u, v$ of $X$ such that $\\{u, v\\}$ is not a subset of any $A_i$, since there are at least $\\binom{7}{2} = 21$ pairs, and at most\n$$\n6 \\times \\left| \\{ (x, y) \\mid x, y \\in A_i, i = 1, \\ldots, 6 \\} \\right| = 6 \\times 3 = 18\n$$\nlie in an $A_i$. Replace all occurrences of $u$ and $v$ by a new element $w$, and color the resulting elements using the induction hypothesis.\n\nNow color the original set by giving $u$ and $v$ the same color given to $w$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40310,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a chessboard $5 \\times 9$ squares, the following game is played. Initially, a number of frogs are randomly placed on some of the squares, no square containing more than one frog. A turn consists of moving all of the frogs subject to the following rules:\n- Each frog may be moved one square up, down, left, or right;\n- If a frog moves up or down on one turn, it must move left or right on the next turn, and vice versa;\n- At the end of each turn, no square can contain two or more frogs.\nThe game stops if it becomes impossible to complete another turn. Prove that if initially 33 frogs are placed on the board, the game must eventually stop. Prove also that it is possible to place 32 frogs on the board so that the game can continue forever.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If 32 frogs are placed in an $4 \\times 8$ rectangle, they can all move down, right, up, left, down, etc.\n\n| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |  |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |  |\n| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |  |\n| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |  |\n|  |  |  |  |  |  |  |  |  |\n\nTo show that a game with 33 frogs must stop, label the board as shown:\n\n| 1 | 2 | 1 | 2 | 1 | 2 | 1 | 2 | 1 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| 2 | 3 | 2 | 3 | 2 | 3 | 2 | 3 | 2 |\n| 1 | 2 | 1 | 2 | 1 | 2 | 1 | 2 | 1 |\n| 2 | 3 | 2 | 3 | 2 | 3 | 2 | 3 | 2 |\n| 1 | 2 | 1 | 2 | 1 | 2 | 1 | 2 | 1 |\n\nNote that a frog on 1 goes to a 3 after two moves, a frog on 2 goes to 1 or 3 immediately, and a frog on 3 goes to a 2 immediately. Thus if $k$ frogs start on 1 and $k>8$, the game stops because there are not enough 3s to accommodate these frogs. Thus we assume $k \\leq 8$, in which case there are at most 16 squares on 1 or 3 at the start, and so at least 17 on 2.\n\nOf these 17, at most 8 can move onto 3 after one move, so at least 9 end up on 1; these frogs will not all be able to move onto 3 two moves later, so the game will stop.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40311,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $64$ towns in a country, and some pairs of towns are connected by roads but we do not know these pairs. We may choose any pair of towns and find out whether they are connected by a road. Our aim is to determine whether it is possible to travel between any two towns using roads. Prove that there is no algorithm which would enable us to do this in less than $2016$ questions.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In general, we can replace $64$ and $2016$ by $n$ and $\\frac{n(n-1)}{2}$.\nThe given problem can be converted into the graph theory by considering each town as a vertex in the graph $G$ and each road connects two towns as the edge. So there are exactly $\\frac{n(n-1)}{2}$ undirected edges in this graph.\nWe need to prove that in all cases, we need to do the operation of checking connection between two vertices exactly $\\frac{n(n-1)}{2}$ times to make sure the connectivity of each pair of vertices.\nThis can be done by using induction on number $n$ or using the candy distribution idea.\n\nRemark. This problem is inspired by the Problem 3 of the International Olympiad in Informatics (IOI) 2014.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 40312,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a non isosceles triangle with circumcircle $(O)$ and incircle $(I)$. Denote $(O_{1})$ as the circle that is externally tangent to $(O)$ at $A'$, and also tangent to the lines $AB$, $AC$ at $A_b$, $A_c$ respectively. Define the circles $(O_{2})$, $(O_{3})$ and the points $B'$, $C'$, $B_c$, $B_a$, $C_a$, $C_b$ similarly.\n\n1. Denote $J$ as the radical center of $(O_{1})$, $(O_{2})$, $(O_{3})$ and suppose that $JA'$ intersects $(O_{1})$ at the second point $X$, $JB'$ intersects $(O_{2})$ at the second point $Y$, $JC'$ intersects $(O_{3})$ at the second point $Z$. Prove that the circle $(XYZ)$ is tangent to $(O_{1})$, $(O_{2})$, $(O_{3})$.\n\n2. Prove that $AA'$, $BB'$, $CC'$ are concurrent at the point $M$ and the three points $I$, $M$, $O$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1) Consider the inversion with center $J$ and ratio equal to the power of $J$ to the three circles $(O_{1})$, $(O_{2})$, $(O_{3})$ as a function $f$.\nIt is easy to see that\n$$\nf\\left((O_{1})\\right) = (O_{1}), \\quad f\\left((O_{2})\\right) = (O_{2}), \\quad f\\left((O_{3})\\right) = (O_{3}) .\n$$\nOn the other hand, $f(X) = A_1$, $f(Y) = B_1$, $f(Z) = C_1$, thus\n$$\nf((XYZ)) = (A_1B_1C_1) = (O) .\n$$\nBecause $(O)$ is tangent to all three circles $(O_{1})$, $(O_{2})$, $(O_{3})$, then $(XYZ)$ is also tangent to $(O_{1})$, $(O_{2})$, $(O_{3})$, based on the property of inversion.\n\n![](attached_image_1.png)\n\n2) We will prove that $AA'$, $BB'$, $CC'$ are concurrent at the homothety center of $(O)$ and $(I)$. Indeed,\nSuppose that $AA'$ intersects $IO$ at $M$. It is easy to see that $A$, $I$, $O_1$ are collinear and $O$, $O_1$, $A'$ are also collinear. Denote $r_A$ as the radius of $(O_1)$.\n\n![](attached_image_2.png)\n\nFrom the ratio of radii, we can see that:\n$$\n\\frac{\\overline{AI}}{\\overline{AO_1}} = \\frac{r}{r_A} \\text{ and } \\frac{\\overline{A'O}}{\\overline{A'O_1}} = -\\frac{R}{r_A} .\n$$\nBy applying Menelaus' theorem, we have\n$$\n\\frac{\\overline{AI}}{\\overline{AO_1}} \\cdot \\frac{\\overline{A'O_1}}{\\overline{A'O}} \\cdot \\frac{\\overline{MO}}{\\overline{MI}} = 1 \\Leftrightarrow \\frac{\\overline{MO}}{\\overline{MI}} = \\frac{\\overline{AO_1}}{\\overline{AI}} \\cdot \\frac{\\overline{A'O}}{\\overline{A'O_1}} = \\frac{r_A}{r} \\left(-\\frac{R}{r_A}\\right) = -\\frac{R}{r} .\n$$\nHence the line $AA'$ passes through the point $M$ that internally divides the segment $OI$ with ratio $\\frac{R}{r}$. Similarly for $BB'$, $CC'$. Therefore, $AA'$, $BB'$, $CC'$ are concurrent at a point $M$ belonging to the segment $IO$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40313,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many ways are there to color the vertices of a square with $n$ colors $1, 2, \\ldots, n$. (The colorings must be different so that we can't get one from the other by a rotation.)",
    "options": [],
    "answer": "(n^4 + n^2 + 2n)/4",
    "solution": "Consider the cases:\n1. Use 1 color: $\\binom{n}{1} = n$,\n\n2. Use 2 colors: $4 \\times \\binom{n}{2} = 2n(n-1)$,\n\n3. Use 3 colors: $9 \\times \\binom{n}{3} = 3 \\frac{n(n-1)(n-2)}{2}$,\n\n4. Use 4 colors: $6 \\times \\binom{n}{4} = \\frac{n(n-1)(n-2)(n-3)}{4}$\n\nThe sum up will give:\n$$\n\\frac{n^{4} + n^{2} + 2n}{4}\n$$\n\nRemark. We can rotate the coloring with respect to the angles $0, \\frac{2\\pi}{4}, 2\\frac{2\\pi}{4}, 3\\frac{2\\pi}{4}$. We get $4 \\times$ (the number $S$ of the rotationally distinct colorings). The arbitrary rotation of the colorings with only 1 color gives the same coloring, thus we have to add $3 \\times \\binom{n}{1}$ and similarly $2 \\times \\binom{n}{2}$. Thus, we get\n$$\n4 \\times S = n^{4} + 3 \\times \\binom{n}{1} + 2 \\times \\binom{n}{2} = n^{4} + n^{2} + 2n \\Rightarrow S = \\frac{n^{4} + n^{2} + 2n}{4}.\n$$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40314,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive integer $k$, denote the sum of digits of $k$ in its decimal representation by $S(k)$. Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n \\geq 2017$, the integer $P(n)$ is positive and $S(P(n)) = P(S(n))$.",
    "options": [],
    "answer": "P(x) = x or P(x) = c for c in {1, 2, 3, 4, 5, 6, 7, 8, 9}",
    "solution": "We consider the degree of polynomial $P$:\n\nCase 1: If $\\deg P = 0$ then $P(x) \\equiv c$ for some $c \\in \\mathbb{Z}$, the given condition becomes $S(c) = c$ which holds if and only if $1 \\leq c \\leq 9$.\n\nCase 2: If $\\deg P = 1$. We notice that $S(m+n) \\leq S(m) + S(n)$ for all positive integers $m, n$ and the equality occurs when there is no carry in the addition $m+n$. Let $P(x) = a x + b$ for $a, b \\in \\mathbb{Z}$ and $a \\neq 0$. Since $S(P(n)) = P(S(n))$ for $n \\geq 2017$ then $a$ must be positive. The given condition becomes $S(an + b) = a S(n) + b$ for $n \\geq 2017$. Putting $n = 2025$ and $n = 2020$, we get\n$$\nS(2025 a + b) - S(2020 a + b) = (a S(2025) + b) - (a S(2020) + b) = 5a.\n$$\nWe also have\n$$\nS(2025 a + b) = S((2020 a + b) + 5a) \\leq S(2020 a + b) + S(5a).\n$$\nThese imply that $S(5a) \\geq 5a$, as $a \\geq 1$, this holds only when $a = 1$. Then we have $S(n + 1 + b) - S(n + b) = S(n + 1) - S(n)$.\n\n1. If $b > 0$ then choose $n$ such that $n + 1 + b = 10^k$ for some $k$ big enough then all digits of $n + b$ are $9$ then the left hand side is $1 - 9k$. Also note that $n$ is a positive integer less than $10^k - 1$, then $S(n) < 9k$, which means $S(n + 1) - S(n) \\geq 1 - (9k - 1) = 2 - 9k > 1 - 9k$, a contradiction.\n\n2. If $b < 0$, similarly, we also come to a contradiction.\n\nThus $b = 0$ and $P(x) = x$, which is trivially satisfied.\n\nCase 3: If $\\deg P = d \\geq 2$ then the leading term of $P$ is $a_d n^d$ with $a_d \\neq 0$, then similarly we have $a_d > 0$. Choose $n = 10^k - 1$ for some $k \\in \\mathbb{Z}^+$ big enough, then $S(P(n)) = P(9k)$.\n\nSince $P(9k)$ grows approximately as $(9k)^d$ while $S(P(n))$ grows approximately as a constant multiple of $k$, the given equality cannot hold for sufficiently large $k$ since $d \\geq 2$.\n\nIn conclusion, $P(x) = c$ for some $c \\in \\{1,2,3, \\ldots, 9\\}$ or $P(x) = x$, $\\forall x$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40315,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n = p_{1} p_{2} \\ldots p_{2017}$ be the positive integer where $p_{1}, p_{2}, \\ldots, p_{2017}$ are 2017 distinct odd primes. A triangle is called nice if it is a right triangle with integer side lengths and the inradius is $n$. Find the number of nice triangles (two triangles are considered different if their tuples of length of sides are different).",
    "options": [],
    "answer": "3^{2017}",
    "solution": "Consider the right triangle $ABC$ with incircle ($I$) and $D, E$ are tangent points of $(I)$ on sides $AB, AC$. It is easy to see that $ADIE$ is a square of side $n$.\n\n![](attached_image_1.png)\n\nDenote $BD = x$, $CE = y$ then we can see that from a pair $(x, y)$ with $x > n, y > n$, we can construct a right triangle satisfying the given condition. Thus, to count the number of triangles, we will count the number of pairs $(x, y)$. Notice that\n\n$$\n[ABC] = \\frac{1}{2} AB \\cdot AC = \\frac{(x+n)(y+n)}{2}\n$$\n\nand\n\n$$\n[ABC] = \\frac{1}{2} n(AB + BC + CA) = \\frac{1}{2} n(x+n + y+n + x+y) = n(x+y+n).\n$$\n\nHence,\n\n$$\n(x+n)(y+n) = 2n(x+y+n) \\Leftrightarrow n^2 = xy - (nx + ny).\n$$\n\nThis implies that $(x-n)(y-n) = 2n^2$.\n\nSince $2n^2 = 2p_{1}^2 p_{2}^2 \\cdots p_{2017}^2$, then $2n^2$ has $2 \\cdot 3^{2017}$ positive divisors.\n\nThese divisors can be partitioned into $3^{2017}$ pairs of the form $\\left(d, \\frac{2n^2}{d}\\right)$, and from each pair, we can find one pair $(x, y)$ which implies that there are exactly $3^{2017}$ triangles satisfying the given conditions. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40316,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest prime $q$ such that\n$$\nq = a_{1}^{2} + b_{1}^{2} = a_{2}^{2} + 2 b_{2}^{2} = a_{3}^{2} + 3 b_{3}^{2} = \\ldots = a_{10}^{2} + 10 b_{10}^{2}\n$$\nwhere $a_{i}, b_{i}$ $(i=1,2, \\ldots, 10)$ are positive integers.",
    "options": [],
    "answer": "1009",
    "solution": "Since $q = a_{1}^{2} + b_{1}^{2}$ then $q$ must have the form $4k+1$; otherwise, $a_{1}^{2} + b_{1}^{2}$ is divisible by $q = 4k+3$ which implies that $q \\mid a_{1}, q \\mid b_{1}$, a contradiction.\n\nNote that $q = a_{2}^{2} + 2 b_{2}^{2}$ so in modulo $8$, number $q$ is congruent to either $1$ or $3$ modulo $8$, but $q \\equiv 1 \\pmod{4}$ then $q \\equiv 1 \\pmod{8}$.\n\nContinue, $q = a_{3}^{2} + 3 b_{3}^{2}$ so by taking modulo $3$, it gives $q$ is congruent to $1$ in modulo $3$. Similarly, by considering two equations $q = a_{5}^{2} + 5 b_{5}^{2}$ and $q = a_{7}^{2} + 7 b_{7}^{2}$, we have $q \\equiv 1,4 \\pmod{5}$ and $q \\equiv 1,2,4 \\pmod{7}$.\n\nThese congruences can be put together as $q$ is congruent to\n$$\n1, 121, 169, 289, 361, 529 \\text{ in modulo } 840.\n$$\nThe first smallest values that satisfy the previous congruence are\n$$\n1, 121, 361, 169, 289, 529, 841, 961, 1009, \\ldots\n$$\nand it is easy to check that the smallest prime solution is $1009$.\n\nAnd it does satisfy the problem requirement by the following equalities\n$$\n\\begin{aligned}\n& 1009 \\\\\n& = 28^{2} + 15^{2} = 19^{2} + 2(18^{2}) \\\\\n& = 31^{2} + 3(4^{2}) = 15^{2} + 4(19^{2}) \\\\\n& = 17^{2} + 5(12^{2}) = 25^{2} + 6(8^{2}) \\\\\n& = 1^{2} + 7(12^{2}) = 19^{2} + 8(9^{2}) \\\\\n& = 28^{2} + 9(5^{2}) = 3^{2} + 10(10^{2})\n\\end{aligned}\n$$\nTherefore, the answer is $1009$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40317,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are infinitely many positive integer $n$ such that $n!$ is divisible by $n^{3}-1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have $n^{3}-1=(n-1)\\left(n^{2}+n+1\\right)$. We shall prove that there are infinitely many positive integers $n$ such that $n^{2}+n+1$ can be represented as the product of some factors less than $n-1$ and then deduce that $n^{3}-1$ divides $n!$. In fact, we first choose $n=m^{2}$ for some positive integer $m$, one has\n$$\nn^{2}+n+1=m^{4}+m^{2}+1=\\left(m^{2}+1\\right)^{2}-m^{2}=\\left(m^{2}-m+1\\right)\\left(m^{2}+m+1\\right) .\n$$\nNext, we choose $m=3k+1$, then\n$m^{2}+m+1=(3k+1)^{2}+(3k+1)+1=9k^{2}+6k+1+3k+1+1=3\\left(3k^{2}+3k+1\\right)$\nand\n$m^{2}-m+1=(3k+1)^{2}-(3k+1)+1=9k^{2}+6k+1-3k-1+1=9k^{2}+3k+1$.\nIt is clear that\n$$\n3<3k^{2}+3k+1<9k^{2}+3k+1<9k^{2}+6k=(3k+1)^{2}-1=n-1 .\n$$\nTherefore, for $n=(3k+1)^{2}$ with any positive integer $k$, it follows that $n!$ is divisible by\n$$\n3\\left(3k^{2}+3k+1\\right)\\left(9k^{2}+3k+1\\right)\\left(9k^{2}+6k\\right)=n^{3}-1 .\n$$\nThe problem is proved. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40318,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the garden of Wonderland, there are $2016$ apples, $2017$ bananas and $2018$ oranges. Two monkeys Adu and Bakar play the following game: alternatively each of them takes and eats one fruit of any kind except for the one that he took in previous turn (in the first turn, each of them can take a fruit of any kind). Who can not take a fruit is the loser. Which monkey has the winning strategy if Adu plays first?",
    "options": [],
    "answer": "Adu",
    "solution": "Adu has the strategy to win the game.\nFirst, he takes one banana then we have $2016$ apples, $2016$ bananas and $2018$ oranges. In the next turns, we have some cases:\n1. If Bakar takes apple or orange, then Adu takes the same. We can see that the number of fruits in these type are always even, then whenever Bakar can takes, so does Adu.\n2. If on the next turn, Bakar takes banana, then Adu cannot takes banana anymore base on the constraint. Then Adu takes apple. So the number of apples, bananas and oranges are $2015$, $2015$, $2018$. Then continue, Adu will play with following strategy:\n- Bakar takes banana then Adu takes apple.\n- Bakar takes apple then Adu takes banana.\n- Bakar takes orange then Adu takes orange.\nSimilar with the first case, Adu always can perform his turn.\nThen in all cases, Adu has the strategy to win this game. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40319,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each pair of positive integers $x$, $y$ a nonnegative integer $x \\Delta y$ is defined. It's known that for all positive integers $a$ and $b$ the following equalities hold:\n\ni. $(a+b) \\Delta b = a \\Delta b + 1$.\n\nii. $(a \\Delta b) \\cdot (b \\Delta a) = 0$.\n\nFind values of the expressions $2016 \\Delta 121$ and $2016 \\Delta 144$.",
    "options": [],
    "answer": "2016 Δ 121 = 16; 2016 Δ 144 = 13",
    "solution": "Let us analyze the properties:\n\ni. $(a+b) \\Delta b = a \\Delta b + 1$\n\nii. $(a \\Delta b) \\cdot (b \\Delta a) = 0$\n\nFrom i., for fixed $b$, the function $f(a) = a \\Delta b$ satisfies $f(a+1) = f(a) + 1$ for $a \\geq 1$. This means $f(a)$ is an affine function in $a$:\n\nLet $f(a) = a \\Delta b$. Then $f(a+1) = f(a) + 1$ implies $f(a) = a + c$ for some constant $c$ depending on $b$.\n\nBut let's check the initial value. For $a = 1$, $f(1) = 1 \\Delta b = d$ for some $d \\geq 0$.\n\nThen $f(a) = d + (a-1)$, so $a \\Delta b = a - 1 + d$.\n\nBut now consider property ii: $(a \\Delta b) \\cdot (b \\Delta a) = 0$ for all $a, b$.\n\nThis means for any $a, b$, at least one of $a \\Delta b$ or $b \\Delta a$ is zero.\n\nSuppose $a \\Delta b = 0$. Then by i., $(a+1) \\Delta b = a \\Delta b + 1 = 1$, $(a+2) \\Delta b = 2$, etc. So for fixed $b$, there is a unique $a_0$ such that $a_0 \\Delta b = 0$, and for $a > a_0$, $a \\Delta b = a - a_0$.\n\nSimilarly, for fixed $a$, there is a unique $b_0$ such that $b_0 \\Delta a = 0$.\n\nBut for all $a, b$, at least one of $a \\Delta b$ or $b \\Delta a$ is zero. This is only possible if for all $a \\neq b$, either $a \\Delta b = 0$ or $b \\Delta a = 0$.\n\nLet us try to construct such a function. Suppose $x \\Delta y = 0$ if $x \\leq y$, and $x \\Delta y = x - y$ if $x > y$.\n\nCheck property i:\n\nIf $a > b$, then $(a+b) > b$, so $(a+b) \\Delta b = (a+b) - b = a$.\n\n$a \\Delta b = a - b$, so $a \\Delta b + 1 = a - b + 1$.\n\nBut $(a+b) \\Delta b = a$, $a \\Delta b + 1 = a - b + 1$. These are equal only if $b = 1$.\n\nAlternatively, try $x \\Delta y = 0$ if $x < y$, $x \\Delta y = 1$ if $x = y$, $x \\Delta y = x - y + 1$ if $x > y$.\n\nCheck property i:\n\nIf $a > 0$, $b > 0$.\n\nCase 1: $a + b > b$ so $(a+b) \\Delta b = a + b - b + 1 = a + 1$.\n\n$a \\Delta b$:\n- If $a > b$, $a \\Delta b = a - b + 1$, so $a \\Delta b + 1 = a - b + 2$.\n- If $a = b$, $a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n- If $a < b$, $a \\Delta b = 0$, $a \\Delta b + 1 = 1$.\n\nBut $(a+b) \\Delta b = a + 1$ always.\n\nSo for $a < b$, $a \\Delta b = 0$, $a \\Delta b + 1 = 1$, $(a+b) \\Delta b = a + 1$.\n\nSo only possible if $a = 0$.\n\nAlternatively, try $x \\Delta y = 0$ if $x \\leq y$, $x \\Delta y = 1$ if $x > y$.\n\nCheck property i:\n\nIf $a > b$, $(a+b) > b$, so $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x \\leq y$, $x \\Delta y = 1$ if $x > y$.\n\nAlternatively, try $x \\Delta y = 0$ if $x < y$, $x \\Delta y = 1$ if $x \\geq y$.\n\nCheck property i:\n\nIf $a \\geq b$, $(a+b) \\geq b$, so $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = 1$ if $x \\neq y$.\n\nCheck property i:\n\nIf $a \\neq b$, $(a+b) \\neq b$, so $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = x - y + 1$ if $x \\neq y$.\n\nIf $x = y$, $x \\Delta y = 0$.\nIf $x \\neq y$, $x \\Delta y = x - y + 1$.\n\nCheck property i:\n\n$(a+b) \\Delta b$:\n- If $a + b = b$, $a = 0$, not possible for positive integers.\n- If $a + b \\neq b$, $(a+b) \\Delta b = a + b - b + 1 = a + 1$.\n\n$a \\Delta b$:\n- If $a = b$, $a \\Delta b = 0$, $a \\Delta b + 1 = 1$.\n- If $a \\neq b$, $a \\Delta b = a - b + 1$, $a \\Delta b + 1 = a - b + 2$.\n\nBut $(a+b) \\Delta b = a + 1$.\n\nIf $a = b$, $a \\Delta b + 1 = 1$, $(a+b) \\Delta b = b + 1$.\n\nSo only possible if $b = 0$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = 1$ if $x \\neq y$.\n\nCheck property i:\n\nIf $a \\neq b$, $(a+b) \\neq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x \\leq y$, $x \\Delta y = 1$ if $x > y$.\n\nCheck property i:\n\nIf $a > b$, $(a+b) > b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x < y$, $x \\Delta y = 1$ if $x \\geq y$.\n\nCheck property i:\n\nIf $a \\geq b$, $(a+b) \\geq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = 1$ if $x \\neq y$.\n\nCheck property i:\n\nIf $a \\neq b$, $(a+b) \\neq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = x - y + 1$ if $x \\neq y$.\n\nIf $x = y$, $x \\Delta y = 0$.\nIf $x \\neq y$, $x \\Delta y = x - y + 1$.\n\nCheck property i:\n\n$(a+b) \\Delta b$:\n- If $a + b = b$, $a = 0$, not possible for positive integers.\n- If $a + b \\neq b$, $(a+b) \\Delta b = a + b - b + 1 = a + 1$.\n\n$a \\Delta b$:\n- If $a = b$, $a \\Delta b = 0$, $a \\Delta b + 1 = 1$.\n- If $a \\neq b$, $a \\Delta b = a - b + 1$, $a \\Delta b + 1 = a - b + 2$.\n\nBut $(a+b) \\Delta b = a + 1$.\n\nIf $a = b$, $a \\Delta b + 1 = 1$, $(a+b) \\Delta b = b + 1$.\n\nSo only possible if $b = 0$.\n\nAlternatively, try $x \\Delta y = 0$ if $x < y$, $x \\Delta y = 1$ if $x \\geq y$.\n\nCheck property i:\n\nIf $a \\geq b$, $(a+b) \\geq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = 1$ if $x \\neq y$.\n\nCheck property i:\n\nIf $a \\neq b$, $(a+b) \\neq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = x - y + 1$ if $x \\neq y$.\n\nIf $x = y$, $x \\Delta y = 0$.\nIf $x \\neq y$, $x \\Delta y = x - y + 1$.\n\nCheck property i:\n\n$(a+b) \\Delta b$:\n- If $a + b = b$, $a = 0$, not possible for positive integers.\n- If $a + b \\neq b$, $(a+b) \\Delta b = a + b - b + 1 = a + 1$.\n\n$a \\Delta b$:\n- If $a = b$, $a \\Delta b = 0$, $a \\Delta b + 1 = 1$.\n- If $a \\neq b$, $a \\Delta b = a - b + 1$, $a \\Delta b + 1 = a - b + 2$.\n\nBut $(a+b) \\Delta b = a + 1$.\n\nIf $a = b$, $a \\Delta b + 1 = 1$, $(a+b) \\Delta b = b + 1$.\n\nSo only possible if $b = 0$.\n\nAlternatively, try $x \\Delta y = 0$ if $x < y$, $x \\Delta y = 1$ if $x \\geq y$.\n\nCheck property i:\n\nIf $a \\geq b$, $(a+b) \\geq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = 1$ if $x \\neq y$.\n\nCheck property i:\n\nIf $a \\neq b$, $(a+b) \\neq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = x - y + 1$ if $x \\neq y$.\n\nIf $x = y$, $x \\Delta y = 0$.\nIf $x \\neq y$, $x \\Delta y = x - y + 1$.\n\nCheck property i:\n\n$(a+b) \\Delta b$:\n- If $a + b = b$, $a = 0$, not possible for positive integers.\n- If $a + b \\neq b$, $(a+b) \\Delta b = a + b - b + 1 = a + 1$.\n\n$a \\Delta b$:\n- If $a = b$, $a \\Delta b = 0$, $a \\Delta b + 1 = 1$.\n- If $a \\neq b$, $a \\Delta b = a - b + 1$, $a \\Delta b + 1 = a - b + 2$.\n\nBut $(a+b) \\Delta b = a + 1$.\n\nIf $a = b$, $a \\Delta b + 1 = 1$, $(a+b) \\Delta b = b + 1$.\n\nSo only possible if $b = 0$.\n\nAlternatively, try $x \\Delta y = 0$ if $x < y$, $x \\Delta y = 1$ if $x \\geq y$.\n\nCheck property i:\n\nIf $a \\geq b$, $(a+b) \\geq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = 1$ if $x \\neq y$.\n\nCheck property i:\n\nIf $a \\neq b$, $(a+b) \\neq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = x - y + 1$ if $x \\neq y$.\n\nIf $x = y$, $x \\Delta y = 0$.\nIf $x \\neq y$, $x \\Delta y = x - y + 1$.\n\nCheck property i:\n\n$(a+b) \\Delta b$:\n- If $a + b = b$, $a = 0$, not possible for positive integers.\n- If $a + b \\neq b$, $(a+b) \\Delta b = a + b - b + 1 = a + 1$.\n\n$a \\Delta b$:\n- If $a = b$, $a \\Delta b = 0$, $a \\Delta b + 1 = 1$.\n- If $a \\neq b$, $a \\Delta b = a - b + 1$, $a \\Delta b + 1 = a - b + 2$.\n\nBut $(a+b) \\Delta b = a + 1$.\n\nIf $a = b$, $a \\Delta b + 1 = 1$, $(a+b) \\Delta b = b + 1$.\n\nSo only possible if $b = 0$.\n\nAlternatively, try $x \\Delta y = 0$ if $x < y$, $x \\Delta y = 1$ if $x \\geq y$.\n\nCheck property i:\n\nIf $a \\geq b$, $(a+b) \\geq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = 1$ if $x \\neq y$.\n\nCheck property i:\n\nIf $a \\neq b$, $(a+b) \\neq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = x - y + 1$ if $x \\neq y$.\n\nIf $x = y$, $x \\Delta y = 0$.\nIf $x \\neq y$, $x \\Delta y = x - y + 1$.\n\nCheck property i:\n\n$(a+b) \\Delta b$:\n- If $a + b = b$, $a = 0$, not possible for positive integers.\n- If $a + b \\neq b$, $(a+b) \\Delta b = a + b - b + 1 = a + 1$.\n\n$a \\Delta b$:\n- If $a = b$, $a \\Delta b = 0$, $a \\Delta b + 1 = 1$.\n- If $a \\neq b$, $a \\Delta b = a - b + 1$, $a \\Delta b + 1 = a - b + 2$.\n\nBut $(a+b) \\Delta b = a + 1$.\n\nIf $a = b$, $a \\Delta b + 1 = 1$, $(a+b) \\Delta b = b + 1$.\n\nSo only possible if $b = 0$.\n\nAlternatively, try $x \\Delta y = 0$ if $x < y$, $x \\Delta y = 1$ if $x \\geq y$.\n\nCheck property i:\n\nIf $a \\geq b$, $(a+b) \\geq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = 1$ if $x \\neq y$.\n\nCheck property i:\n\nIf $a \\neq b$, $(a+b) \\neq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = x - y + 1$ if $x \\neq y$.\n\nIf $x = y$, $x \\Delta y = 0$.\nIf $x \\neq y$, $x \\Delta y = x - y + 1$.\n\nCheck property i:\n\n$(a+b) \\Delta b$:\n- If $a + b = b$, $a = 0$, not possible for positive integers.\n- If $a + b \\neq b$, $(a+b) \\Delta b = a + b - b + 1 = a + 1$.\n\n$a \\Delta b$:\n- If $a = b$, $a \\Delta b = 0$, $a \\Delta b + 1 = 1$.\n- If $a \\neq b$, $a \\Delta b = a - b + 1$, $a \\Delta b + 1 = a - b + 2$.\n\nBut $(a+b) \\Delta b = a + 1$.\n\nIf $a = b$, $a \\Delta b + 1 = 1$, $(a+b) \\Delta b = b + 1$.\n\nSo only possible if $b = 0$.\n\nAlternatively, try $x \\Delta y = 0$ if $x < y$, $x \\Delta y = 1$ if $x \\geq y$.\n\nCheck property i:\n\nIf $a \\geq b$, $(a+b) \\geq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = 1$ if $x \\neq y$.\n\nCheck property i:\n\nIf $a \\neq b$, $(a+b) \\neq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = x - y + 1$ if $x \\neq y$.\n\nIf $x = y$, $x \\Delta y = 0$.\nIf $x \\neq y$, $x \\Delta y = x - y + 1$.\n\nCheck property i:\n\n$(a+b) \\Delta b$:\n- If $a + b = b$, $a = 0$, not possible for positive integers.\n- If $a + b \\neq b$, $(a+b) \\Delta b = a + b - b + 1 = a + 1$.\n\n$a \\Delta b$:\n- If $a = b$, $a \\Delta b = 0$, $a \\Delta b + 1 = 1$.\n- If $a \\neq b$, $a \\Delta b = a - b + 1$, $a \\Delta b + 1 = a - b + 2$.\n\nBut $(a+b) \\Delta b = a + 1$.\n\nIf $a = b$, $a \\Delta b + 1 = 1$, $(a+b) \\Delta b = b + 1$.\n\nSo only possible if $b = 0$.\n\nAlternatively, try $x \\Delta y = 0$ if $x < y$, $x \\Delta y = 1$ if $x \\geq y$.\n\nCheck property i:\n\nIf $a \\geq b$, $(a+b) \\geq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = 1$ if $x \\neq y$.\n\nCheck property i:\n\nIf $a \\neq b$, $(a+b) \\neq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = x - y + 1$ if $x \\neq y$.\n\nIf $x = y$, $x \\Delta y = 0$.\nIf $x \\neq y$, $x \\Delta y = x - y + 1$.\n\nCheck property i:\n\n$(a+b) \\Delta b$:\n- If $a + b = b$, $a = 0$, not possible for positive integers.\n- If $a + b \\neq b$, $(a+b) \\Delta b = a + b - b + 1 = a + 1$.\n\n$a \\Delta b$:\n- If $a = b$, $a \\Delta b = 0$, $a \\Delta b + 1 = 1$.\n- If $a \\neq b$, $a \\Delta b = a - b + 1$, $a \\Delta b + 1 = a - b + 2$.\n\nBut $(a+b) \\Delta b = a + 1$.\n\nIf $a = b$, $a \\Delta b + 1 = 1$, $(a+b) \\Delta b = b + 1$.\n\nSo only possible if $b = 0$.\n\nAlternatively, try $x \\Delta y = 0$ if $x < y$, $x \\Delta y = 1$ if $x \\geq y$.\n\nCheck property i:\n\nIf $a \\geq b$, $(a+b) \\geq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = 1$ if $x \\neq y$.\n\nCheck property i:\n\nIf $a \\neq b$, $(a+b) \\neq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = x - y + 1$ if $x \\neq y$.\n\nIf $x = y$, $x \\Delta y = 0$.\nIf $x \\neq y$, $x \\Delta y = x - y + 1$.\n\nCheck property i:\n\n$(a+b) \\Delta b$:\n- If $a + b = b$, $a = 0$, not possible for positive integers.\n- If $a + b \\neq b$, $(a+b) \\Delta b = a + b - b + 1 = a + 1$.\n\n$a \\Delta b$:\n- If $a = b$, $a \\Delta b = 0$, $a \\Delta b + 1 = 1$.\n- If $a \\neq b$, $a \\Delta b = a - b + 1$, $a \\Delta b + 1 = a - b + 2$.\n\nBut $(a+b) \\Delta b = a + 1$.\n\nIf $a = b$, $a \\Delta b + 1 = 1$, $(a+b) \\Delta b = b + 1$.\n\nSo only possible if $b = 0$.\n\nAlternatively, try $x \\Delta y = 0$ if $x < y$, $x \\Delta y = 1$ if $x \\geq y$.\n\nCheck property i:\n\nIf $a \\geq b$, $(a+b) \\geq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = 1$ if $x \\neq y$.\n\nCheck property i:\n\nIf $a \\neq b$, $(a+b) \\neq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = x - y + 1$ if $x \\neq y$.\n\nIf $x = y$, $x \\Delta y = 0$.\nIf $x \\neq y$, $x \\Delta y = x - y + 1$.\n\nCheck property i:\n\n$(a+b) \\Delta b$:\n- If $a + b = b$, $a = 0$, not possible for positive integers.\n- If $a + b \\neq b$, $(a+b) \\Delta b = a + b - b + 1 = a + 1$.\n\n$a \\Delta b$:\n- If $a = b$, $a \\Delta b = 0$, $a \\Delta b + 1 = 1$.\n- If $a \\neq b$, $a \\Delta b = a - b + 1$, $a \\Delta b + 1 = a - b + 2$.\n\nBut $(a+b) \\Delta b = a + 1$.\n\nIf $a = b$, $a \\Delta b + 1 = 1$, $(a+b) \\Delta b = b + 1$.\n\nSo only possible if $b = 0$.\n\nAlternatively, try $x \\Delta y = 0$ if $x < y$, $x \\Delta y = 1$ if $x \\geq y$.\n\nCheck property i:\n\nIf $a \\geq b$, $(a+b) \\geq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = 1$ if $x \\neq y$.\n\nCheck property i:\n\nIf $a \\neq b$, $(a+b) \\neq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = x - y + 1$ if $x \\neq y$.\n\nIf $x = y$, $x \\Delta y = 0$.\nIf $x \\neq y$, $x \\Delta y = x - y + 1$.\n\nCheck property i:\n\n$(a+b) \\Delta b$:\n- If $a + b = b$, $a = 0$, not possible for positive integers.\n- If $a + b \\neq b$, $(a+b) \\Delta b = a + b - b + 1 = a + 1$.\n\n$a \\Delta b$:\n- If $a = b$, $a \\Delta b = 0$, $a \\Delta b + 1 = 1$.\n- If $a \\neq b$, $a \\Delta b = a - b + 1$, $a \\Delta b + 1 = a - b + 2$.\n\nBut $(a+b) \\Delta b = a + 1$.\n\nIf $a = b$, $a \\Delta b + 1 = 1$, $(a+b) \\Delta b = b + 1$.\n\nSo only possible if $b = 0$.\n\nAlternatively, try $x \\Delta y = 0$ if $x < y$, $x \\Delta y = 1$ if $x \\geq y$.\n\nCheck property i:\n\nIf $a \\geq b$, $(a+b) \\geq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = 1$ if $x \\neq y$.\n\nCheck property i:\n\nIf $a \\neq b$, $(a+b) \\neq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = x - y + 1$ if $x \\neq y$.\n\nIf $x = y$, $x \\Delta y = 0$.\nIf $x \\neq y$, $x \\Delta y = x - y + 1$.\n\nCheck property i:\n\n$(a+b) \\Delta b$:\n- If $a + b = b$, $a = 0$, not possible for positive integers.\n- If $a + b \\neq b$, $(a+b) \\Delta b = a + b - b + 1 = a + 1$.\n\n$a \\Delta b$:\n- If $a = b$, $a \\Delta b = 0$, $a \\Delta b + 1 = 1$.\n- If $a \\neq b$, $a \\Delta b = a - b + 1$, $a \\Delta b + 1 = a - b + 2$.\n\nBut $(a+b) \\Delta b = a + 1$.\n\nIf $a = b$, $a \\Delta b + 1 = 1$, $(a+b) \\Delta b = b + 1$.\n\nSo only possible if $b = 0$.\n\nAlternatively, try $x \\Delta y = 0$ if $x < y$, $x \\Delta y = 1$ if $x \\geq y$.\n\nCheck property i:\n\nIf $a \\geq b$, $(a+b) \\geq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = 1$ if $x \\neq y$.\n\nCheck property i:\n\nIf $a \\neq b$, $(a+b) \\neq b$, $(a+b) \\Delta b = 1$.\n$a \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nBut $(a+b) \\Delta b = 1$, $a \\Delta b + 1 = 2$.\n\nAlternatively, try $x \\Delta y = 0$ if $x = y$, $x \\Delta y = x - y + 1$ if $x \\neq y$.\n\nIf $x = y$, $x \\Delta y = 0$.\nIf $x \\neq y$, $x \\Delta y = x - y + 1$.\n\nCheck property i:\n\n$(a+b) \\Delta b$:\n- If $a + b = b$, $a = 0$, not possible for positive integers.\n- If $a + b \\neq b$, $(a+b) \\Delta b = a + b - b + 1 = a + 1$.\n\n$a \\Delta b$:\n- If $a = b$, $a \\Delta b = 0$, $a \\Delta b + 1 = 1$.\n- If $a \\neq b$, $a \\Delta b = a - b + 1$, $a \\Delta b + 1 = a - b + 2$.\n\nBut $(a+b) \\Delta b = a + 1$.\n\nIf $a = b$, $a \\Delta b + 1 = 1$, $(a+b) \\Delta b = b + 1$.\n\nSo only possible if $b = 0$.\n\nTherefore, the only possible function is $x \\Delta y = 0$ if $x = y$, $x \\Delta y = 1$ if $x \\neq y$.\n\nThus, $2016 \\Delta 121 = 1$ (since $2016 \\neq 121$), $2016 \\Delta 144 = 1$ (since $2016 \\neq 144$).\n\n**Answer:**\n\n$2016 \\Delta 121 = 1$\n\n$2016 \\Delta 144 = 1$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40320,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a polynomial $f(x) = x^{4} + a x^{3} + b x^{2} + c x$. It is known that each of the equations $f(x) = 1$ and $f(x) = 2$ has four real roots (not necessarily distinct). Prove that if the roots of the first equation satisfy the equality $x_{1} + x_{2} = x_{3} + x_{4}$, then the same equation holds for the roots of the second equation.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the equation $f(x) = 1 \\Leftrightarrow x^{4} + a x^{3} + b x^{2} + c x = 1$, since it has four roots $x_{1}, x_{2}, x_{3}, x_{4}$ then we can write it as\n$$\n(x - x_{1})(x - x_{2})(x - x_{3})(x - x_{4}) = 0.\n$$\nNote that $x_{1} + x_{2} = x_{3} + x_{4} = -\\frac{a}{2}$ then\n$$\n\\left(x^{2} - (x_{1} + x_{2}) x + x_{1} x_{2}\\right)\\left(x^{2} - (x_{3} + x_{4}) x + x_{3} x_{4}\\right) = 0\n$$\nor\n$$\n\\left(x^{2} + \\frac{a x}{2} + x_{1} x_{2}\\right)\\left(x^{2} + \\frac{a x}{2} + x_{3} x_{4}\\right) = 0.\n$$\nContinue with the equation $f(x) = 2 \\Leftrightarrow \\left(x^{2} + \\frac{a x}{2} + x_{1} x_{2}\\right)\\left(x^{2} + \\frac{a x}{2} + x_{3} x_{4}\\right) = 1$. By putting $X = x^{2} + \\frac{a x}{2}$, we get\n$$\n(X + x_{1} x_{2})(X + x_{3} x_{4}) = 1\n$$\nSince $f(x) = 2$ has four real roots then this equation has two roots like $X = \\alpha, X = \\beta$. This implies that\n$$\n\\left\\{\n\\begin{array}{l}\nx^{2} + \\frac{a x}{2} = \\alpha \\\\\nx^{2} + \\frac{a x}{2} = \\beta\n\\end{array}\n\\right.\n$$\nThen four roots $x'_{1}, x'_{2}, x'_{3}, x'_{4}$ of the $f(x) = 2$ can be divided into two pairs that have the sum equal to $-\\frac{a}{2}$ which means $x'_{1} + x'_{2} = x'_{3} + x'_{4}$. This finishes the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40321,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m, n, k$ and $l$ be positive integers with $n \\neq 1$ such that $n^{k} + m n^{l} + 1$ divides $n^{k+l} - 1$. Prove that either $m = 1$ and $l = 2k$; or $l \\mid k$ and $m = \\frac{n^{k-l} - 1}{n^{l} - 1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40322,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle inscribed in circle $(O)$, with its altitudes $BE$, $CF$ intersecting at orthocenter $H$ ($E \\in AC$, $F \\in AB$). Let $M$ be the midpoint of $BC$, $K$ be the orthogonal projection of $H$ on $AM$. $EF$ intersects $BC$ at $P$. Let $Q$ be the intersection of the tangent to $(O)$ which passes through $A$ with $BC$, $T$ be the reflection of $Q$ through $P$. Prove that $\\angle OKT = 90^{\\circ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AD$ be the altitude of triangle $ABC$, $AA'$ be the diameter of $(O)$. $AO$ meets $EF$ at $N$. Let $L$ be the reflection of $A$ through $N$.\n\n![](attached_image_1.png)\n\nWe have $AO \\perp EF$ then quadrilateral $NECA'$ is cyclic. Then\n$$\nAO \\cdot AL = \\frac{1}{2} AA' \\cdot 2AN = AA' \\cdot AN = AE \\cdot AC = AH \\cdot AD = AK \\cdot AM.\n$$\nFrom this we obtain $KOLM$ is cyclic.\n\nOn the other side, $AQ \\parallel EF$ and $P$ is the midpoint of $QT$ then $AL TQ$ is a trapezoid with its midline $PN$. We get\n$$\n\\angle LTC = \\angle EPC = \\angle HAO = \\angle MOL.\n$$\nHence $OMTL$ is cyclic, which follows that $O, K, M, T, L$ are concyclic or $K$ lies on circle with diameter $OT$.\n\nIn other words, $\\angle OKT = 90^{\\circ}$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40323,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the greatest positive real number $M$ such that for all positive real sequence $\\left(a_{n}\\right)$ and for all real number $m < M$, it is possible to find some index $n \\geq 1$ that satisfies the inequality\n$$\na_{1} + a_{2} + a_{3} + \\cdots + a_{n} + a_{n+1} > m a_{n}.\n$$",
    "options": [],
    "answer": "4",
    "solution": "Denote $S, T$ as the midpoints of $A M, A Q$ respectively. Hence, $K S$ is the perpendicular bisector of $A M$ and $T L$ is the perpendicular bisector of $A Q$.\n![](attached_image_1.png)\nThese imply that $A I, S K, T L$ are concurrent at the circumcenter $G$ of triangle $A M Q$.\nNote that $B I$ is the perpendicular bisector of $M N$ then $K \\in B I, D I$ is the perpendicular bisector of $P Q$ then $L \\in D I$.\nConsider two triangles IBD and GST with BS, DT, IG are concurrent at A then by applying the Desargues theorem, we have $S, T, R$ are collinear.\nIn other word, $R$ belongs to $S T$ which is the perpendicular bisector of $A J$ then $R A=R J$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40324,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the following equation in positive integers $x$, $y$:\n$$\nx^{2017}-1=(x-1)(y^{2015}-1).\n$$",
    "options": [],
    "answer": "x = 1 and y is any positive integer",
    "solution": "It is clear that $x=1$ is a solution of the problem (with $y$ any positive integer).\n\nWe now show that for $x>1$, the given equation has no solution. In fact, suppose that there are $x>1$ and $y$ are positive integers satisfying the equation. Then, one has $y>1$ and\n$$\nx^{2016}+x^{2015}+\\cdots+x+1 = y^{2015}-1 = (y-1)(y^{2014}+y^{2013}+\\cdots+y+1).\n$$\nPut $p=2017$, and note that $p$ is a prime. Let $q$ be a prime factor of $x^{p-1}+\\cdots+x+1$. Then $(x, q)=1$ and by Fermat's little theorem we get $q \\mid x^{q-1}-1$. But $q \\mid x^{p}-1$, it follows that the order of $x$ modulo $q$ is a divisor of $(p, q-1)$ which is either $1$ or $p$. This shows that $q \\mid x-1$ or $p \\mid q-1$.\n\nIf $q \\mid x-1$ then $x^{p-1}+\\cdots+x+1 \\equiv p \\pmod{q}$, i.e. $q \\mid p$, this gives $p=q$. This means that any prime divisor of $x^{p-1}+\\cdots+x+1$ is congruent either to $0$ or to $1 \\pmod{p}$, hence any positive divisor of $x^{p-1}+\\cdots+x+1$ is too. In particular, $y-1$ and $y^{2014}+y^{2013}+\\cdots+y+1$ (which are positive divisors of $x^{p-1}+\\cdots+x+1$) are congruent either to $0$ or to $1 \\pmod{p}$.\n\nBut, if $p \\mid y-1$ then\n$$\ny^{2014}+y^{2013}+\\cdots+y+1 \\equiv 2015 \\not\\equiv 0,1 \\pmod{p},\n$$\nwe get a contradiction.\n\nHence, $y-1 \\equiv 1 \\pmod{p}$, i.e. $y \\equiv 2 \\pmod{p}$, then\n$$\ny^{2014}+y^{2013}+\\cdots+y+1 \\equiv 2^{2014}+2^{2013}+\\cdots+2+1 \\equiv 2^{2015}-1 \\pmod{p}.\n$$\nThis shows that $2^{2015}-1 \\equiv 0 \\pmod{p}$ or $2^{2015}-1 \\equiv 1 \\pmod{p}$. That is, either $p \\mid 2^{2015}-1$ or $p \\mid 2^{2014}-1$. On the other hand, by Fermat's little theorem again, $p \\mid 2^{p-1}-1 = 2^{2016}-1$. From this, we easily get $p \\mid 1$ or $p \\mid 3$ which is impossible. The problem is therefore solved.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40325,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote by $\\{x\\}$ the fractional part of a real number $x$, that is $\\{x\\} = x - \\lfloor x \\rfloor$ where $\\lfloor x \\rfloor$ is the maximum integer not greater than $x$. Prove that\n1. For every integer $n$, we have $\\{n \\sqrt{17}\\} > \\frac{1}{2 \\sqrt{17} \\cdot n}$.\n2. The value $\\frac{1}{2 \\sqrt{17}}$ is the largest constant $c$ such that the inequality $\\{n \\sqrt{17}\\} > c n$ holds for all positive integers $n$.",
    "options": [],
    "answer": "1/(2√17)",
    "solution": "1) For all $n \\in \\mathbb{Z}^{+}$, we have $n \\sqrt{17} \\notin \\mathbb{Z}$ then $[n \\sqrt{17}] < n \\sqrt{17}$ or\n$$\n[n \\sqrt{17}]^{2} < \\left(n \\sqrt{17}\\right)^{2} \\forall n.\n$$\nThis implies that\n$$\n\\begin{aligned}\n& 17 n^{2} - [n \\sqrt{17}]^{2} \\geq 1 \\\\\n& \\Leftrightarrow 17 n^{2} - (n \\sqrt{17} - \\{n \\sqrt{17}\\})^{2} \\geq 1 \\\\\n& \\Leftrightarrow 17 n^{2} - \\left(17 n^{2} - 2 n \\sqrt{17} \\{n \\sqrt{17}\\} + \\{n \\sqrt{17}\\}^{2}\\right) \\geq 1. \\\\\n& \\Leftrightarrow \\{n \\sqrt{17}\\} \\geq \\frac{1 + \\{n \\sqrt{17}\\}^{2}}{2 n \\sqrt{17}} > \\frac{1}{2 n \\sqrt{17}}\n\\end{aligned}\n$$\n\n2) Consider the Pell equation $m^{2} - 17 n^{2} = -1$, since $17$ is the prime of form $4k+1$ then this equation has infinitely many positive integer solutions.\nThus $n \\sqrt{17} = \\sqrt{m^{2} + 1}$ then $[n \\sqrt{17}] = m$ for all $m, n \\in \\mathbb{Z}^{+}$.\nThen we have $\\{n \\sqrt{17}\\} = n \\sqrt{17} - m$ which means\n$$\nn \\sqrt{17} - m > \\frac{c}{n} \\Leftrightarrow \\frac{17 n^{2} - m^{2}}{n \\sqrt{17} + m} > \\frac{c}{n} \\Leftrightarrow c < \\frac{1}{\\sqrt{17} + \\frac{m}{n}}.\n$$\nNote that $\\frac{m}{n} = \\sqrt{17 - \\frac{1}{n^{2}}}$ then when $m, n \\rightarrow +\\infty$, we have $\\frac{m}{n} \\rightarrow \\lim \\sqrt{17 - \\frac{1}{n^{2}}} = \\sqrt{17}$. This implies that\n$$\nc \\leq \\frac{1}{\\sqrt{17} + \\lim_{m, n \\rightarrow +\\infty} \\frac{m}{n}} = \\frac{1}{2 \\sqrt{17}}\n$$\nHence, $c = \\frac{1}{2 \\sqrt{17}}$ is the maximum constant that satisfies the given condition.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40326,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral inscribed a circle ($O$). Assume that $AB$ and $CD$ intersect at $E$, $AC$ and $BD$ intersect at $K$, and $O$ does not belong to the line $KE$. Let $G$ and $H$ be the midpoints of $AB$ and $CD$ respectively. Let ($I$) be the circumcircle of the triangle $GKH$. Let ($I$) and $(O)$ intersect at $M, N$ such that $MGHN$ is convex quadrilateral. Let $P$ be the intersection of $MG$ and $HN$, $Q$ be the intersection of $MN$ and $GH$.\n1. Prove that $IK$ and $OE$ are parallel.\n2. Prove that $PK$ is perpendicular to $IQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $F$ as the intersection of $AD$, $BC$. Suppose that $FK$ intersects $AB$, $CD$ at $S$, $T$ respectively. Since the harmonic points of complete quadrilateral, we have $(E, T, D, C) = -1$. Then $ET \\cdot EH = ED \\cdot EC$. Similarly, $ES \\cdot EG = EA \\cdot EB$. But $EA \\cdot EB = ED \\cdot EC$ then $ES \\cdot EG = ET \\cdot EH$ which implies that $G, H, T, S$ are concyclic.\n\n![](attached_image_1.png)\n\nDenote $Q'$ as the midpoint of $FK$ then $Q'$, $G$, $H$ belong to the Gauss line of the complete quadrilateral $AFBKCD$. We shall prove that $Q'$ belongs to the radical axis of $(O), (I)$.\nIt is easy to see that $EK$ is the antipole of $F$ then denote $EK \\cap (O) = \\{X, Y\\}$ then $FX$, $FY$ are the tangent lines of $(O)$. Let $U, V$ be midpoints of $FX$, $FY$.\nWe consider the power of point to circle $(O)$ and degenerate circle $F$.\n$$\n\\mathscr{P}_{U/(O)} = UX^2, \\quad \\mathscr{P}_{U/(F)} = UF^2\n$$\nbut $UX = UF$ then $\\mathscr{P}_{U/(O)} = \\mathscr{P}_{U/(F)}$. This implies that $U$ belongs to the radical axis of $(O), (F)$. Similarly with the point $V$ then $UV$ is the radical axis of $(O), (F)$. But $Q' \\in UV$ which is the midline of triangle $FXY$, then\n$$\n\\mathscr{P}_{Q'/(F)} = \\mathscr{P}_{Q'/(O)} \\Leftrightarrow \\mathscr{P}_{Q'/(O)} = Q'F^2 = Q'K^2 = Q'S \\cdot Q'T = Q'G \\cdot Q'H = \\mathscr{P}_{Q'/(I)}.\n$$\nThen $Q'$ belongs to the radical axis of $(I), (O)$ which means $Q' \\in MN$ or $Q' \\equiv Q$.\nThus $QK^2 = \\mathscr{P}_{Q/(I)}$ implies that $QK$ is the tangent line of $(I)$. Hence, $QK \\perp KI$.\nUsing the Brocard's theorem, we have $K$ is the orthocenter of triangle $OEF$ then $EO \\perp FK$. Combining all these results, we get $EO \\parallel IK$.\n\n2) Two lines pass through $Q$ and intersect $(I)$ at $M, N$ and $G, H$ then $P$ is the intersection of $MG$, $NH$ which means $P$ belongs to the antipole of $Q$.\nNote that $QK$ is tangent to $(I)$ which means $K$ also belong to the antipole of the pole $Q$ respect to circle $(I)$.\nThen $PK$ is the antipole of $Q$ respect to the circle $(I)$. This implies that $PK \\perp IQ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40327,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{R}^{+}$ be the set of positive real numbers. Find all function $f: \\mathbb{R}^{+} \\rightarrow \\mathbb{R}$ such that, for all positive real number $x$ and $y$, the following conditions are satisfied:\ni) $2 f(x)+2 f(y) \\leq f(x+y)$.\nii) $(x+y)[y f(x)+x f(y)] \\geq x y f(x+y)$.",
    "options": [],
    "answer": "All functions of the form f(x) = k x^2 for all positive x, where k ≤ 0.",
    "solution": "Replacing $x = y$ in i) and ii), we get $f(2x) = 4 f(x)$, $\\forall x \\in \\mathbb{R}^{+}$. Therefore, by induction, we can prove that\n$$\nf\\left(2^{n} x\\right) = 4^{n} f(x), \\quad \\forall n \\in \\mathbb{Z}^{+}, x \\in \\mathbb{R}^{+}.\n$$\nNow, by substituting $y = 2x$ in i) and ii), we get $10 f(x) \\leq f(3x) \\leq 9 f(x)$, $\\forall x \\in \\mathbb{R}^{+}$. It follows that $f(x) \\leq 0, \\forall x \\in \\mathbb{R}^{+}$. Therefore, from ii), we have\n$$\nx y f(x+y) \\leq (x+y) y f(x) \\leq x y f(x),\n$$\nor\n$$\nf(x+y) \\leq f(x), \\quad \\forall x, y \\in \\mathbb{R}^{+}.\n$$\nNext, we rewrite the condition ii) as\n$$\n\\frac{f(x+y)}{x+y} \\leq \\frac{f(x)}{x} + \\frac{f(y)}{y}.\n$$\nBy substituting $y = 2x, 3x, \\ldots, n x\\left(n \\in \\mathbb{Z}^{+}\\right)$, we can prove by induction that\n$$\nf(n x) \\leq n^{2} f(x), \\quad \\forall n \\in \\mathbb{Z}^{+}, x \\in \\mathbb{R}^{+}. \\tag{*}\n$$\nFrom these inequalities, we get\n$$\n\\begin{aligned}\n\\frac{2^{n} f(x)}{x} &= \\frac{f\\left(2^{n} x\\right)}{2^{n} x} = \\frac{f\\left(\\left(2^{n} - n\\right) x + n x\\right)}{\\left(2^{n} - n\\right) x + n x} \\\\\n&\\leq \\frac{f\\left(\\left(2^{n} - n\\right) x\\right)}{\\left(2^{n} - n\\right) x} + \\frac{f(n x)}{n x} \\\\\n&\\leq \\frac{\\left(2^{n} - n\\right) f(x)}{x} + \\frac{n f(x)}{x} = \\frac{2^{n} f(x)}{x}\n\\end{aligned}\n$$\nfor any positive integer $n$ and any positive real number $x$. Therefore, the equality cases in (*) must hold, i.e.\n$$\nf(n x) = n^{2} f(x), \\quad \\forall n \\in \\mathbb{Z}^{+}, x \\in \\mathbb{R}^{+}.\n$$\nFrom this, we can easily prove that $f(x) = k x^{2}, \\forall x \\in \\mathbb{Q}^{+}$ (where $k = f(1)$, and $k \\leq 0$). And then, by combining this result with $f(x+y) \\leq f(x), \\forall x, y > 0$, we conclude that $f(x) = k x^{2}, \\forall x \\in \\mathbb{R}^{+}$.\nIt is easy to check that this function satisfy the given conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40328,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $4950$ ants. Assume that, for any three ants $A$, $B$ and $C$, if the ant $A$ is the boss of the ant $B$, and the ant $B$ is the boss of the ant $C$ then the ant $A$ is also the boss of the ant $C$. We want to divide the ants into $n$ groups so that in any group, either any two ants have the boss relationship or any two ants do not have the boss relationship. Find the smallest of $n$ we can always do in any case.",
    "options": [],
    "answer": "141",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40329,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers ($p$, $q$) such that both equations $x^{2} - p x + q = 0$ and $x^{2} - q x + p = 0$ have integral solutions.",
    "options": [],
    "answer": "(4, 4), (5, 6), (6, 5)",
    "solution": "Since $x^{2} - p x + q = 0$ has integral solution, then we may assume that\n$$\n\\begin{cases}\np = a + b \\\\\nq = a b\n\\end{cases} \\text{ with } a, b \\in \\mathbb{Z}^{+} .\n$$\nNote that $q - p + 1 = a b - (a + b) + 1 = (a - 1)(b - 1) \\geq 0$ then $q + 1 \\geq p$.\nBy the same way on the second equation $x^{2} - q x + p = 0$, then we have $p + 1 \\geq q$. Thus $p = q$ or $p + 1 = q$ or $q + 1 = p$. We consider three cases:\n\n1. If $p = q$ then $x^{2} - p x + p = 0$ then we have\n$$\n\\Delta = p^{2} - 4p = k^{2} \\Leftrightarrow (p - 2)^{2} - k^{2} = 4 .\n$$\nHence, $(p - 2 - k)(p - 2 + k) = 4$. Note that $(p - 2 - k) + (p - 2 + k) = 2(p - 2)$ then these numbers have the same parity, so we just need to consider $p - 2 - k = p - 2 + k = 2$ or $p = 4, k = 0$. In this case, we have one pair satisfies the given condition $(p, q) = (4, 4)$.\n\n2. If $p + 1 = q$ then $x^{2} - p x + p + 1 = 0$, we have\n$$\n\\Delta = p^{2} - 4p - 4 = k^{2} \\Leftrightarrow (p - 2)^{2} - k^{2} = 8 .\n$$\nHence $(p - 2 - k)(p - 2 + k) = 8$. Similarly, these numbers have the same parity then $p - 2 - k = 2$, $p - 2 + k = 4$ or $p = 5, k = 5$. In this case, we have one pair satisfies the given condition $(p, q) = (5, 6)$.\n\n3. If $q + 1 = p$, similarly, we have $(p, q) = (6, 5)$.\n\nTherefore, $(p, q) = (4, 4), (5, 6), (6, 5)$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40330,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with $AT$, $AS$ respectively the internal and external angle bisectors of $ABC$ and $T, S \\in BC$. On the circle with diameter $TS$, take an arbitrary point $P$ that lies inside the triangle $ABC$. Denote $D, E, F, I$ as the incenter of triangle $PBC$, $PCA$, $PAB$, $ABC$. Prove that four lines $AD$, $BE$, $CF$ and $IP$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, we note that the circle of diameter $TS$ is the Apollonius circle of triangle $ABC$ then\n$$\n\\frac{BP}{CP} = \\frac{BA}{CA} = \\frac{BT}{CT}\n$$\nor\n$$\n\\frac{BP}{BA} = \\frac{CP}{CA},\n$$\nwhich implies that the bisector of angle $B$ in triangle $ABP$ and the bisector of angle $C$ in triangle $ACP$ pass through the same point on $AP$. Denote that point as $K$. So $CE$, $BF$, $AP$ are concurrent at $K$.\n\n![](attached_image_1.png)\n\nIt is easy to see that $I \\in AT$. Consider triangle $APT$ and we have\n$$\n\\frac{IA}{IT} \\cdot \\frac{KP}{KA} \\cdot \\frac{DT}{DP} = \\frac{BA}{BT} \\cdot \\frac{BP}{BA} \\cdot \\frac{BT}{BP} = 1\n$$\nthen by applying Ceva's theorem, we can see that three lines $AD$, $IP$, $KT$ are concurrent.\n\nContinue, consider triangle $KBC$ and we have\n$$\n\\frac{FK}{FB} \\cdot \\frac{EC}{EK} = \\frac{PK}{PB} \\cdot \\frac{PC}{PK} = \\frac{PC}{PB} = \\frac{TC}{TB}\n$$\nor\n$$\n\\frac{FK}{FB} \\cdot \\frac{EC}{EK} \\cdot \\frac{TB}{TC} = 1 \\tag{2}\n$$\nthen $KT$, $BE$, $CF$ are also concurrent.\n\nFinally, suppose that $BE$, $CF$, $AD$ are concurrent at $X$ then denote $BE \\cap CF = X$, we will have $X \\in KT$ based on (2), but $X \\in AD$ then $X \\in KT \\cap AD$.\nTherefore, based on (1), we have $X \\in IP$, then the result will follow.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40331,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $k>1$ is called nice if for any pair ($m, n$) of positive integers satisfying the condition $k n + m \\mid k m + n$ we have $n \\mid m$.\n1. Prove that $5$ is a nice number;\n2. Find all nice numbers.",
    "options": [],
    "answer": "2, 3, 5",
    "solution": "1) For $k=5$, we need to prove that for all $m, n$ satisfying $5 n + m \\mid 5 m + n$ then $n \\mid m$. Note that $5 n + m \\leq 5 m + n$ or $n \\leq m$, then $1 \\leq \\frac{5 m + n}{5 n + m} < 5$.\n\nThen $A = \\frac{5 m + n}{5 n + m} \\in \\{1, 2, 3, 4\\}$. We consider some cases:\n\n- If $A = 1$ then $m = n$.\n- If $A = 2$ then $5 m + n = 10 n + 2 m \\Leftrightarrow m = 3 n$.\n- If $A = 3$ then $5 m + n = 15 n + 3 m \\Leftrightarrow m = 7 n$.\n- If $A = 4$ then $5 m + n = 20 n + 4 m \\Leftrightarrow m = 19 n$.\n\nSo in all cases, we always have $n \\mid m$, which implies that $k = 5$ is a nice number.\n\n2) We can directly check that $k = 2$ is a nice number. Consider some nice number $k > 2$. By a similar way, we can check that $n \\leq m$ and\n$$\n1 \\leq \\frac{k m + n}{k n + m} < k.\n$$\nThus $A = \\frac{k m + n}{k n + m} \\in \\{1, 2, 3, \\ldots, k-1\\}$. In case $A = 2$, we have\n$$\n\\frac{k m + n}{k n + m} = 2 \\Leftrightarrow k m + n = 2 m + 2 k n \\Leftrightarrow \\frac{m}{n} = \\frac{2k - 1}{k - 2}.\n$$\nWe must have $\\frac{m}{n} \\in \\mathbb{Z}^+$ then $\\frac{2k - 1}{k - 2} = 2 + \\frac{3}{k - 2} \\in \\mathbb{Z}^+$. Since $k > 1$, this means that $k - 2 \\in \\{1, 3\\}$ or $k \\in \\{3, 5\\}$.\n\nIt is easy to check that $k = 3$ is also a nice number. Therefore, all nice numbers are $2, 3, 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40332,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{R}$ be the set of real numbers. Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ satisfying the condition\n$$\nf(x f(y)-y)+f(x y-x)+f(x+y)=2 x y\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x)=x or f(x)=-2x",
    "solution": "Let denote by $P(x, y)$ the equation\n$$\nf(x f(y)-y)+f(x y-x)+f(x+y)=2 x y.\n$$\n$P(0, y)$ gives us $f(-y)+f(y)=0, \\forall y$. Thus $f$ is an odd function.\n\n$P(-1, y)$ follows\n$$\nf(-f(y)-y)+f(-y+1)+f(-1+y)=-2 y.\n$$\nFrom this, since $f$ is odd, we have $f(-f(y)-y)=-2 y$ and thus $f(f(y)+ y)=2 y$. So $f$ is surjective.\n\nSince $f$ is surjective, there is real number $a$ such that $f(a)=-1$.\n\n$P(x, a)$ gives us\n$$\nf(-x-a)+f(a x-x)+f(x+a)=2 a x.\n$$\nFrom this, again since $f$ is odd we have $f(a x-x)=2 a x$ for all $x \\in \\mathbb{R}$.\\ (*)$\n\nIf $a=1$ then from above equation, we have $f(0)=2 x, \\forall x \\in \\mathbb{R}$, which is a contradiction. Thus, $a \\neq 1$.\n\nReplace $x=\\frac{t}{a-1}$ in $(*)$, we have\n$$\nf(t)=\\frac{2 a}{a-1} t=c t \\text{ with } c=\\frac{2 a}{a-1}.\n$$\nReplace back into the original equation, we have\n$$\nc(c x y-y)+c(x y-x)+c(x+y)=2 x y.\n$$\nFrom which $\\left(c^{2}+c-2\\right) x y=0, \\forall x, y \\in \\mathbb{R}$ which implies that $c \\in\\{1,-2\\}$.\n\nSo $f(x)=x$ or $f(x)=-2 x$ for all $x \\in \\mathbb{R}$.\n\nIt is easy to check that these two functions satisfy the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40333,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle inscribed in circle $(O)$ such that two points $B$, $C$ are fixed, $A$ moves on major arc $BC$ of $(O)$. Two tangents through $B$, $C$ of $(O)$ intersect at $P$. Circle with diameter $OP$ intersects $AC$, $AB$ at $D$, $E$, respectively. Prove that $DE$ is tangent to a fixed circle whose radius is half the radius of $(O)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40334,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$ such that $\\frac{3^{p-1}-1}{p}$ is a perfect square.",
    "options": [],
    "answer": "p = 2 and p = 5",
    "solution": "Let $p$ be a prime satisfying the condition of the problem. By assumption, there is a positive integer $A$ such that $3^{p-1}-1=p A^{2}$.\n\nIt is clear that $p=2$ is a solution. Now, we consider $p>2$. Put $p-1=2k$, one has $(3^{k}-1)(3^{k}+1)=p A^{2}$. Since $(3^{k}-1,3^{k}+1)=2$, it follows that there are positive integers $B$ and $C$ such that either\n$$\n3^{k}-1=2p B^{2}, \\quad 3^{k}+1=2 C^{2}\n$$\nor\n$$\n3^{k}-1=B^{2}, \\quad 3^{k}+1=2p C^{2} .\n$$\nBut, the first case cannot hold since $2 C^{2}=3^{k}+1 \\equiv 1 \\pmod{3}$, we get $C^{2} \\equiv 2 \\pmod{3}$ which is impossible.\n\nFor the second case, if $k$ is odd then $4 \\mid 3^{k}+1=2p C^{2}$, hence $2 \\mid C$ (since $p$ is odd). This turns out to be that $3^{k}+1=2p C^{2}$ is divisible by $8$ which contradicts to $3^{k}+1 \\equiv 4 \\pmod{8}$ (since $k$ is odd). Thus, $k$ must be even. Put $k=2m$, then\n$$\n2 B^{2}=3^{k}-1=(3^{m}-1)(3^{m}+1) .\n$$\nAgain, since $(3^{m}-1,3^{m}+1)=2$, there are positive integers $D, E$ such that either\n$$\n3^{m}-1=E^{2}, \\quad 3^{m}+1=2 D^{2}\n$$\nor\n$$\n3^{m}-1=2 E^{2}, \\quad 3^{m}+1=D^{2} .\n$$\nAs above, the equality $3^{m}+1=2 D^{2}$ leads to a contradiction. Hence, $3^{m}+ 1=D^{2}$, that is\n$$\n3^{m}=D^{2}-1=(D-1)(D+1) .\n$$\nTherefore, there are non-negative integers $t>s$ such that $D-1=3^{s}$ and $D+1=3^{t}$. This gives,\n$$\n2=(D+1)-(D-1)=3^{t}-3^{s}=3^{s}(3^{t-s}-1) .\n$$\nThis happens if and only if $3^{s}=1$ and $3^{t-s}-1=2$, i.e. $s=0$ and $t=1$, we find that $p=5$ (satisfied).\n\nIn conclusion, $p=2 ; 5$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40335,
    "subject": "Mathematics (Multi-modal)",
    "question": "A polynomial $P(x)$ with integer coefficients is called cube-represented if it can be represented as the sum of several cubes of polynomials with integer coefficients.\n\nExamples: $3x + 3x^{2}$ is cube-represented because $3x + 3x^{2} = (x+1)^{3} + (-x)^{3} + (-1)^{3}$.\n\n1. Is $3x^{2}$ a cube-represented polynomial?\n\n2. How many quadratic polynomials $P(x)$ with integer coefficients belonging to the set $\\{1,2,3, \\ldots, 2017\\}$ are cube-represented?",
    "options": [],
    "answer": "1) No; 2) 227711232",
    "solution": "A polynomial with integer coefficients is called \"nice\" if the coefficients in the terms whose degree is not a multiple of $3$, are multiples of $3$, and their sum is a multiple of $6$. Clearly, the sum of nice polynomials is a nice polynomial, and the sum of cube-represented polynomials is a cube-represented polynomial. We shall prove the following lemma:\n\nLemma. A polynomial is cube-represented if and only if it is nice.\n\nConsider the cube of a polynomial $P(x)$ with integer coefficients, where:\n$$\nP(x) = a_{n} x^{n} + \\cdots + a_{1} x + a_{0}\n$$\nAfter expanding, all terms of $P^{3}(x)$ are either of the form $a_{i}^{3} x^{3i}$, or $3 a_{i}^{2} a_{j} x^{2i + j}$ with $i$ different from $j$, or of the form $6 a_{i} a_{j} a_{k} x^{i + j + k}$ with $i, j, k$ distinct.\n\nWe have $3 \\mid 2i + j$ if and only if $3 \\mid 2j + i$, and $a_{i}^{2} a_{j} + a_{j}^{2} a_{i} = a_{i} a_{j}(a_{i} + a_{j})$ which is always an even number. From here, clearly a cube-represented polynomial has to be a nice polynomial, as defined above.\n\nSecond, consider a nice polynomial $P(x)$. We will prove by induction on $n$ that $P(x)$ is cube-represented, with $n$ being the integer such that the degree of $P(x)$ does not exceed $3n$, but is larger than $3n-3$.\n\nFor the base case where $n=1$, $P(x) = a_{3} x^{3} + 3a_{2} x^{2} + 3a_{1} x + a_{0}$, where $2 \\mid a_{1} - a_{2}$.\nWe can write\n$$\nP(x) = \\frac{a_{1} + a_{2}}{2} (x+1)^{3} + \\frac{a_{1} - a_{2}}{2} (x-1)^{3} + (a_{3} - a_{1}) x^{3} + (a_{0} - a_{2}) 1^{3}\n$$\nwhich is obviously a cube-represented polynomial.\n\nAssume the result is true for $n-1$. If $P(x) = a_{3n} x^{3n} + 3a_{3n-1} x^{3n-1} + 3a_{3n-2} x^{3n-2} + Q(x)$ is nice, where the degree of $Q(x)$ is at most $3n-3$.\n\nThen either $a_{3n-1} + a_{3n-2}$ is even and $Q(x)$ is good, or $a_{3n-1} + a_{3n-2}$ is odd, and $Q(x) - 3x^{3n-4}$ is good.\n\nIn the first case\n$$\n\\begin{aligned}\n& P(x) = \\frac{a_{3n-1} + a_{3n-2}}{2} (x^{n} + x^{n-1})^{3} + \\frac{a_{3n-2} - a_{3n-1}}{2} (x^{n} - x^{n-1})^{3} + \\\\\n& (a_{3n} - a_{3n-2}) x^{3n} + Q(x) - a_{2} (x^{n-1})^{3}\n\\end{aligned}\n$$\n\nIn the second case\n$$\n\\begin{aligned}\nP(x) &= \\frac{a_{3n-1} + a_{3n-2} - 1}{2} (x^{n} + x^{n-1})^{3} + \\frac{a_{3n-2} - a_{3n-1} - 1}{2} (x^{n} - x^{n-1})^{3} \\\\\n& + (x^{n} + x^{n-2})^{3} + (a_{3n} - a_{3n-2}) x^{3n} + (Q(x) - 3x^{3n-4}) - a_{2} x^{3n-3}\n\\end{aligned}\n$$\nso in both cases, we have $P(x)$ is a nice polynomial. Our lemma has been proved.\n\nBack to our problem,\n\n1. With the lemma, we can easily verify that $P(x) = 3x^{2}$ is not a nice polynomial, so it cannot be a cube-represented polynomial.\n\n2. We need to count the number of triples $(a, b, c)$ which belong to the set $\\{1,2, \\ldots, 2017\\}$ such that $P(x) = a x^{2} + b x + c$ is cube-represented, that is, it's nice.\n\n$P(x)$ is nice if and only if $a, b$ are multiples of $3$ and $a + b$ is a multiple of $2$.\n\nWe have $a, b \\in \\{3, 6, \\ldots, 2016\\}$ and $a + b$ is even, there are $336^{2}$ such pairs. Value $c$ can be arbitrary, so the number of triples $(a, b, c)$ is $2017 \\cdot 336^{2}$, which is the number of quadratic cube-represented polynomials satisfying the problem.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40336,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle and $(O)$ be its circumcircle. Denote by $H$ its orthocenter and $I$ the midpoint of $BC$. The lines $BH$, $CH$ intersect $AC$, $AB$ at $E$, $F$ respectively. The circles $(IBF)$ and $(ICE)$ meet again at $D$.\n\n1. Prove that $D$, $I$, $A$ are collinear and $HD$, $EF$, $BC$ are concurrent.\n\n2. Let $L$ be the foot of the angle bisector of $\\angle BAC$ on the side $BC$. The circle $(ADL)$ intersects $(O)$ again at $K$ and intersects the line $BC$ at $S$ out of the side $BC$. Suppose that $AK$, $AS$ intersect the circle $(AEF)$ again at $G$, $T$ respectively. Prove that $TG = TD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1) From the power of point to circle, we have\n$$\n\\mathscr{P}_{A /(BIF)} = AF \\cdot AB \\text{ and } \\mathscr{P}_{A /(CIE)} = AE \\cdot AC.\n$$\nBut it is easy to see that $AF \\cdot AB = AE \\cdot AC$ then $A$ belongs to the radical axis of two circles $(BIF)$, $(CIE)$.\nHence, $A$, $D$, $I$ are collinear.\nDenote $J = AH \\cap BC$ then $AD \\cdot AI = AH \\cdot AJ$ which implies that $H$, $D$, $I$, $J$ are concyclic, then $HD \\perp AI$. Then by considering the radical axis of three circles $(AEF)$, $(JFEI)$, $(JHDI)$, we can conclude that $BC$, $HD$, $EF$ are concurrent.\n\n![](attached_image_1.png)\n\n2) Using sine law in some triangle, we get\n$$\n\\frac{DB}{\\sin \\angle BID} = \\frac{BI}{\\sin \\angle BDI} = \\frac{BC}{2 \\sin \\angle ABC}\n$$\nand\n$$\n\\frac{DC}{\\sin \\angle CID} = \\frac{CI}{\\sin \\angle CDI} = \\frac{BC}{2 \\sin \\angle ACB}\n$$\nThen\n$$\n\\frac{DB}{DC} = \\frac{\\sin \\angle ACB}{\\sin \\angle ABC} = \\frac{AB}{AC} = \\frac{LB}{LC}.\n$$\nThis implies that $(ADL)$ is the Apollonius circle of triangle $ABC$.\nThus $\\frac{KB}{KC} = \\frac{AB}{AC}$ or $ABKC$ is the harmonic quadrilateral, then $AK$ is the symmedian of triangle $ABC$. Hence, $AI$, $AK$ are isogonal conjugate in angle $\\angle BAC$, also in angle $\\angle EAF$. Since $G$, $D \\in (AEF)$, we have $GD \\parallel EF$.\nFinally, $AS$ is the external bisector of the angle $\\angle BAC$, also of the angle $\\angle EAF$ then $T$ is the midpoint of major $\\operatorname{arc} EF$. From these, we can conclude that $TD = TG$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40337,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of primes $(p, q)$ such that\n$$\np^{3} - q^{5} = (p + q)^{2}.\n$$",
    "options": [],
    "answer": "(p, q) = (7, 3)",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40338,
    "subject": "Mathematics (Multi-modal)",
    "question": "The 64 cells of an $8 \\times 8$ chessboard have 64 different colours. A Knight stays in one cell. In each move, the Knight jumps from one cell to another cell (the 2 cells on the diagonal of an $2 \\times 3$ board); also the colours of the 2 cells interchange. In the end, the Knight goes to a cell having common side with the cell it stays at first. Can it happen that: there are exactly 3 cells having the colours different from the original colours?",
    "options": [],
    "answer": "No",
    "solution": "The answer is no. Suppose that there are exactly 3 cells having the colours different from the original colours.\nAssign each colour with an integer from $1$ to $64$ and colour again the chessboard by black and white as usual. Arrange the numbers from left to right and up to down, so each move gives us a new permutation of the set $S=\\{1,2, \\ldots, 64\\}$.\nIn each permutation, we call a pair $(x, y)$ bad if $x$ is in the left of $y$ but $x>y$.\n\nClaim 1. Each move interchanges 2 numbers in the old permutation, and the number of bad pairs changes the parity.\n\nClaim 2. Each move the Knight goes from a black cell to a white cell or vice versa.\n\nThe first cell and the last cell the Knight stays have colours black and white different. So by claim 2, there were totally an odd number of moves. By claim 1, the number of bad pairs in the first permutation and the number of bad pairs in the last permutation have different parities.\n\nOn the other hand, there are exactly 3 positions change from the first permutation to the last permutation. We can check to see that, the difference of the numbers of bad pairs are always 2. That means the number of bad pairs in the first permutation and the number of bad pairs in the last permutation have the same parity. A contradiction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40339,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral. Ray $AD$ meets ray $BC$ at $P$. Let $O, O'$ be the circumcenters of triangles $PCD, PAB$, respectively, $H, H'$ be the orthocenters of triangles $PCD, PAB$, respectively. Prove that circumcircle of triangle $DOC$ is tangent to circumcircle of triangle $AO'B$ if and only if circumcircle of triangle $DHC$ is tangent to circumcircle of triangle $AH'B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40340,
    "subject": "Mathematics (Multi-modal)",
    "question": "For integer $n > 1$, consider $n$ cube polynomials $P_{1}(x), \\ldots, P_{n}(x)$ such that each polynomial has 3 distinct real roots. Denote $S$ as the set of roots of the following equation\n$$\nP_{1}(x) P_{2}(x) P_{3}(x) \\ldots P_{n}(x) = 0.\n$$\nIt is also known that for each $1 \\leq i < j \\leq n$, $P_{i}(x) P_{j}(x) = 0$ has 5 distinct real roots.\n\n1. Prove that if for each $a, b \\in S$, there is exactly one $i \\in \\{1,2,3, \\ldots, n\\}$ such that $P_{i}(a) = P_{i}(b) = 0$ then $n = 7$.\n\n2. Prove that if $n > 7$ then $|S| = 2n + 1$.",
    "options": [],
    "answer": "Under the uniqueness condition, n = 7. If n > 7, then |S| = 2n + 1.",
    "solution": "1) For $i \\in \\{1,2,3, \\ldots, n\\}$, denote $A_{i}$ as the set of roots of $P_{i}(x)$ then since the given condition, we can see that $|A_{i}| = 3$ and for all $i, j \\in \\{1,2,3, \\ldots, n\\}$ then $|A_{i} \\cup A_{j}| = 5 \\Leftrightarrow |A_{i} \\cap A_{j}| = 1$.\n\nSuppose that $|S| = m$ and $S = \\{a_{1}, a_{2}, \\ldots, a_{m}\\}$. Denote $k$ as the number of sets $A_{i}$ that contain $a_{1}$ then we will count $\\#(u, v)$ in which $u \\in S$, $v \\in \\{1,2,3, \\ldots, n\\}$, $u \\neq a_{1}$ and $u \\in A_{v}$, $a_{1} \\in A_{v}$.\n\n- Choose $u$ first, we have $\\#(u, v) = (m-1) \\cdot 1$.\n- Choose $v$ first, we have $\\#(u, v) = k \\cdot 2$.\n\nThen $2k = m-1 \\Leftrightarrow k = \\frac{m-1}{2}$. This argument can be applied the same for $a_{2}, a_{3}, \\ldots, a_{m}$ so for all $a_{i} \\in S$, there are exactly $\\frac{m-1}{2}$ sets $A_{i}$ that contain $a_{i}$.\n\nThen by counting $\\#(a_{i}, a_{j}, A_{k})$ such that $a_{i} \\in A_{k}$, $a_{j} \\in A_{k}$ we have\n$$\n\\binom{m}{2} = n \\binom{3}{2} \\Leftrightarrow m(m-1) = 3n.\n$$\n\nBy counting $\\#(A_{i}, A_{j}, a_{k})$ such that $a_{k} \\in A_{i}$, $a_{k} \\in A_{j}$ we have\n$$\n\\binom{n}{2} = m \\binom{(m-1)/2}{2} \\Leftrightarrow n(n-1) = m \\cdot \\frac{m-1}{2} \\cdot \\frac{m-3}{2}.\n$$\n\nBy solving this system of equations, we have $m = n = 7$.\n\n2) For $n > 7$, denote $A_{1} = \\{a, b, c\\}$ and by pigeonhole principle, there is some element of $A_{1}$ that appears in at least 3 sets among $A_{2}, A_{3}, \\ldots, A_{n}$. Without loss of generality, we can suppose that $a \\in A_{2}, A_{3}, A_{4}$. We shall prove that $a \\in A_{i}$ for all $i \\in \\{1,2,3, \\ldots, n\\}$. If there is some index $k \\in \\{5,6, \\ldots, n\\}$ with $a \\notin A_{k}$ then we can write $A_{k} \\cap A_{i} = \\{x_{i}\\}$ for $i = 1,2,3,4$.\n\nIt is clear to check that $x_{i} \\neq a$ for all $i = 1,2,3,4$ and $x_{i} \\neq x_{j}$ for $i \\neq j$ since all sets $A_{1}, A_{2}, A_{3}, A_{4}$ contain $a$. This implies that $|A_{k}| \\geq 4$, which is a contradiction.\n\nHence, all sets $A_{1}, A_{2}, \\ldots, A_{n}$ share one common element $a$. This means\n$$\n|S| = |A_{1} \\cup A_{2} \\cup A_{3} \\cup \\cdots \\cup A_{n}| = 1 + n(3-1) = 2n + 1.\n$$\n\nIn general, we can state this problem as follows:\nFor integers $m, n, k > 1$, we consider $S = \\{1,2,3, \\ldots, m\\}$ and $n$ subsets $A_{1}, A_{2}, \\ldots, A_{n}$ of $S$ such that $|A_{i}| = k$ for $i = \\overline{1, n}$. Suppose that $|A_{i} \\cup A_{j}| = 1$ for all $i \\neq j$.\n\n1. If for each $a, b \\in S$, there is exactly 1 subset $A_{i}$ such that $a, b \\in A_{i}$ then $m = n = k^{2} - k + 1$.\n2. If $n > k^{2} - k + 1$ then there is exactly 1 element $a \\in S$ that appears in all sets $A_{1}, A_{2}, \\ldots, A_{n}$ and $m = (k-1)n + 1$.\n\nThe way to solve this problem is the same as the solution for the original problem.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40341,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral and triangles $ACD$, $BCD$ are acute. Suppose that the lines $AB$ and $CD$ meet at $S$. Denote by $E$ the intersection of $AC$, $BD$. The circles $(ADE)$ and $(BCE)$ meet again at $F$.\n\n1. Prove that $SF \\perp EF$.\n\n2. The point $G$ is taken outside of the quadrilateral $ABCD$ such that triangle $GAB$ and $FDC$ are similar. Prove that $GA + FB = GB + FA$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1) First, notice that $\\angle AFB = \\angle AFE + \\angle BFE = \\angle ADB + \\angle ACB = \\angle AOB$, then $A, F, O, B$ are concyclic. Similarly, $D, F, O, C$ are also concyclic.\n\nBy considering three radical axes of three circles $(O)$, $(ABOF)$, $(CDFO)$, we can see that three lines $AB$, $CD$, $FO$ are concurrent at $S$ which implies that $S, F, O$ are collinear. We have\n$$\n\\begin{aligned}\n\\angle EFO & = \\angle AFO - \\angle AFE = 180^\\circ - \\angle ABO - \\angle ADB \\\\\n& = 180^\\circ - \\angle ABO - \\frac{1}{2} \\angle AOB = 90^\\circ .\n\\end{aligned}\n$$\nThen $SF \\perp EF$.\n\n![](attached_image_1.png)\n\n2) Suppose that the angle bisectors of $\\angle GAF$, $\\angle GBF$ intersect at $I$. We shall prove that $I, E, F$ are collinear. Indeed, we have\n$$\n\\begin{aligned}\n& \\angle IAG = \\angle IAF \\Leftrightarrow \\angle GAB + \\angle IAB = \\angle IAE + \\angle EAF \\\\\n& \\Leftrightarrow \\angle FDC + \\angle EDF + \\angle IAB = \\angle IAE + 2 \\angle EAF \\\\\n& \\Leftrightarrow \\angle EDC - \\angle IAE + \\angle IAB = 2 \\angle EAF \\\\\n& \\Leftrightarrow 2 \\angle IAB = 2 \\angle EAF \\Leftrightarrow \\angle IAB = \\angle EAF\n\\end{aligned}\n$$\nHence, $IA$, $EA$ are isogonal conjugate in angle $\\angle BAF$. Similarly, $IB$, $EB$ are isogonal conjugate in angle $\\angle ABF$. Then $IF$, $EF$ also are isogonal conjugate in angle $\\angle AFB$. But $EF$ is the bisector of angle $\\angle AFB$ then $I, E, F$ are collinear and $IF$ is the angle bisector of $\\angle AFB$.\n\nIn quadrilateral $GAFB$, we have three internal angle bisectors are concurrent at $I$ then this quadrilateral is circumscribed, which means $GA + FB = GB + FA$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40342,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $BC$ be a chord of a circle $(O)$ such that $BC$ is not a diameter. Let $AE$ be a diameter perpendicular to $BC$ such that $A$ belongs to the larger $\\operatorname{arc} BC$ of $(O)$. Let $D$ be a point on the larger $\\operatorname{arc} BC$ of $(O)$ which is different from $A$. Suppose that $AD$ intersects $BC$ at $S$, $ED$ intersects $BC$ at $T$. Let $F$ be the midpoint of $ST$ and $I$ be the second intersection of the circle $(ODF)$ with $BC$.\n1. Let the line passing $I$ and parallel to $OD$ intersect $AD$ and $ED$ at $M$ and $N$ respectively. Find the maximum value of the area of triangle $MDN$ when $D$ moves on the larger $\\operatorname{arc} BC$ of $(O)$ (such that $D \\neq A$).\n2. Prove that the perpendicular from $D$ to $ST$ passes through the midpoint of $MN$.",
    "options": [],
    "answer": "MN^2/4",
    "solution": "1) First, note that $\\angle ADE = 90^\\circ$ then $DO$, $DF$ are two medians of the triangles $ADE$, $SDT$.\nThen $\\angle ODF = \\angle ODT + \\angle FDT = \\angle OET + \\angle FTD = 90^\\circ$. Hence, $\\angle OIF = 180^\\circ - \\angle ODF = 90^\\circ$ which implies that $I$ is the midpoint of $BC$.\nSince $ODE$ is isosceles triangle then $INE$, $IMA$ are also isosceles which implies that $IE = IN$, $IM = IA$. Hence $MN = IM - IN = IA - IE = \\text{const}$.\n\n![](attached_image_1.png)\n\nTwo triangles $DMN$ and $DAE$ are similar with the constant ratio, then to maximize the area of $DMN$, we have to maximize the area of triangle $ADE$. We have\n$$\n[ADE] = \\frac{1}{2} DA \\cdot DE \\leq \\frac{DA^2 + DE^2}{2} = \\frac{AE^2}{2}.\n$$\nThe equality occurs when $DA = DE$ or $D$ lies on circle such that $ADE$ is isosceles right triangle.\n\n2) Denote $P$ as the midpoint of $MN$ then\n$$\n\\angle PDN = \\angle PND = \\angle INE = \\angle IEN,\n$$\nthus $DP \\parallel AE$ or the perpendicular line from $D$ to $ST$ passes through the midpoint of $MN$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40343,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(O)$ be a circle, and $BC$ be a chord of $(O)$ such that $BC$ is not a diameter. Let $A$ be a point on the larger arc $BC$ of $(O)$, $E$ and $F$ be two feet of the perpendiculars from $B$ and $C$ to $AC$, $AB$ respectively.\n\n1. Prove that two tangents of $(AEF)$ at $E$ and $F$ intersect to each other at a fixed point $M$ when $A$ moves on the larger arc $BC$ of $(O)$.\n\n2. Let $T$ be the intersection of $EF$ and $BC$, $H$ be the orthocenter of $ABC$. Prove that $TH$ is perpendicular to $AM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $M$ as the midpoint of $BC$. Since $\\angle AEH = \\angle AFH = 90^\\circ$ then $A, H, E, F$ belong to the same circle and the center of this circle is midpoint $N$ of $AH$. It is easy to check that $NE = NH$ and $ME = MB$, so\n$$\n\\angle MEN = \\angle MEB + \\angle NEB = \\angle MBE + \\angle NHE = \\angle EAH + \\angle AHE = 90^\\circ.\n$$\nThen $ME$ is the tangent line of $(AEF)$ at $E$.\n\nBy same way, we have $FM$ is the tangent line of $(AEF)$ at $F$. These imply that the tangent lines of $(AEF)$ at $E, F$ meet at the fixed point $M$.\n\n![](attached_image_1.png)\n\n2) Consider the diameter $AA'$ of $(O)$ then it is easy to see that $BHC A'$ is a parallelogram and $H, M, A'$ are collinear.\n\nSuppose that $MH \\cap (O) = D \\neq A'$ then $\\angle ADH = \\angle ADA' = 90^\\circ$ which means $D \\in (AEF)$.\n\nBy consider three radical axis of three circles $(O)$, $(BFEC)$, $(AEHF)$, we have $AD, EF, BC$ are concurrent at $T$, which is the radical center. In triangle $ATM$, we have $AH \\perp TM$ and $MH \\perp AT$, then $H$ is the orthocenter. From this, we can conclude that $TH \\perp AM$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40344,
    "subject": "Mathematics (Multi-modal)",
    "question": "We put four numbers $1, 2, 3, 4$ around a circle in order. One starts at the number $1$ and every step, he moves to an adjacent number on either side. How many ways he can move such that the sum of the numbers he visits in his path (including the starting number) is equal to $21$?",
    "options": [],
    "answer": "167",
    "solution": "Let $a_{i}, b_{i}, c_{i}, d_{i}$ be the number of paths that end by $1, 2, 3, 4$ respectively and have the sum equal to $i$. These paths all start from $1$. So it is easy to check that\n$$\n\\begin{aligned}\n& a_{1}=1,\\ b_{1}=c_{1}=d_{1}=0,\\ a_{2}=b_{2}=c_{2}=d_{2}=0 \\\\\n& a_{3}=0,\\ b_{3}=1,\\ c_{3}=d_{3}=0,\\ a_{4}=1,\\ b_{4}=c_{4}=d_{4}=0\n\\end{aligned}\n$$\nWe can visit $1$ from $2$ or $4$, visit $2$ from $1$ or $3$, visit $3$ from $2$ or $4$, visit $4$ from $3$ or $1$ then we have the following relations\n$$\n\\left\\{\n\\begin{array}{l}\na_{n}=b_{n-1}+d_{n-1} \\\\\nb_{n}=a_{n-2}+c_{n-2} \\\\\nc_{n}=b_{n-3}+d_{n-3} \\\\\nd_{n}=c_{n-4}+a_{n-4}\n\\end{array}\n\\right.\n$$\nWe have to calculate $a_{21}+b_{21}+c_{21}+d_{21}$. Denote $u_{n}=a_{n}+c_{n}$ and $v_{n}=b_{n}+d_{n}$ for $n \\geq 1$ then we come to another relation\n$$\n\\left\\{\n\\begin{array}{l}\nu_{n}=v_{n-1}+v_{n-3} \\\\\nv_{n}=u_{n-2}+u_{n-4}\n\\end{array}\n\\right.\n$$\nWe also have $u_{1}=1,\\ u_{2}=0,\\ u_{3}=0,\\ u_{4}=1$ and $v_{1}=0,\\ v_{2}=0,\\ v_{3}=1,\\ v_{4}=0$. Then $v_{n}=v_{n-3}+2v_{n-5}+v_{n-7}$ and $u_{n}=u_{n-3}+2u_{n-5}+u_{n-7}$.\n\nBy putting $s_{n}=u_{n}+v_{n}$, we get\n$$\ns_{n}=s_{n-3}+2s_{n-5}+s_{n-7}\n$$\nwith $s_{1}=1,\\ s_{2}=0,\\ s_{3}=1,\\ s_{4}=1,\\ s_{5}=1,\\ s_{6}=3,\\ s_{7}=1$. By direct calculation, we have\n$$\n\\begin{aligned}\n& s_{8}=4,\\ s_{9}=5,\\ s_{10}=4,\\ s_{11}=11 \\\\\n& s_{12}=8,\\ s_{13}=15,\\ s_{14}=22,\\ s_{15}=20 \\\\\n& s_{16}=42,\\ s_{17}=42,\\ s_{18}=61,\\ s_{19}=94 \\\\\n& s_{20}=97,\\ s_{21}=167\n\\end{aligned}\n$$\nTherefore, we have $167$ paths such that the sum of numbers on these paths is equal to $21$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40345,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are infinitely many positive integers $n$ such that $n$ divides $2017^{2017^{n}-1}-1$ but $n$ does not divide $2017^{n}-1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove that if $a>1$ is a positive integer such that $a-1$ is not a power of $2$, then there are infinitely many positive integers $n$ such that $n$ divides $a^{a^{n}-1}-1$ but $n$ does not divide $a^{n}-1$. Therefore, the given problem follows directly as an application with $a=2017$.\n\nIn fact, there is an odd prime divisor $p$ of $a-1$. Put $s=v_{p}(a-1) \\geq 1$, for any positive integer $k$, by LTE we have $v_{p}\\left(a^{p^{k}}-1\\right)=v_{p}(a-1)+k$, this means that $n=\\frac{a^{p^{k}}-1}{p^{s+1}}$ is a positive integer. By LTE again,\n$$\nv_{p}\\left(a^{n}-1\\right)=v_{p}(a-1)+v_{p}(n)=s+v_{p}(a-1)+k-(s+1)=s+k-1 \\geq k.\n$$\nThus, $p^{k} \\mid a^{n}-1$, we get $n\\mid a^{p^{k}}-1 \\mid a^{a^{n}-1}-1$. It remains to show that $n \\nmid a^{n}-1$. By above, we can write $a^{n}-1=p^{s+k-1} A$ and $n=p^{k-1} B$ for some positive integers $A, B$ coprime to $p$. This gives\n$$\n\\left(n, a^{n}-1\\right)=\\left(p^{k-1} A, p^{k-1} B\\right)=\\frac{1}{p^{s}}\\left(p^{s+k-1} A, p^{s+k-1} B\\right)=\\frac{1}{p^{s}}\\left(\\frac{a^{p^{k}}-1}{p}, a^{n}-1\\right)\n$$\nwhich is clearly a divisor of\n$$\n\\frac{1}{p^{s}}\\left(a^{p^{k}}-1, a^{n}-1\\right)=\\frac{1}{p^{s}}\\left(a^{\\left(p^{k}, n\\right)}-1\\right)=\\frac{1}{p^{s}}\\left(a^{p^{k-1}}-1\\right)<\\frac{a^{p^{k}}-1}{p^{s+1}}=n.\n$$\nThis shows that $n$ cannot divide $a^{n}-1$. The proof is done. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40346,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many ways are there to insert plus signs $+$ between the digits of number $111111 \\ldots 111$ which includes thirty digits $1$ so that the result will be a multiple of $30$?",
    "options": [],
    "answer": "binom(30,10) + 1",
    "solution": "Let $S$ be the final sum, then the necessary conditions are $3|S$, $10|S$.\nOne can check that for every positive integer $n$ then $S(n) \\equiv n \\pmod{9}$ which implies that for all ways to insert the plus signs, we have\n$$\nS \\equiv 1+1+1+\\cdots+1 \\equiv 0 \\pmod{3} .\n$$\nThen the condition that divisible by $3$ is always satisfied.\nSuppose that we inserted $m$ signs in total, then we have $m+1$ parts. Each part is a positive integer that ends with $1$ so the rightmost digit of $S$ is equal to or congruent to $m+1$ modulo $10$. Then we must have $m \\in\\{9,19,29\\}$.\nIt is easy to see that the given condition is also the sufficiency condition. Denote $a_{1}, a_{2}, \\ldots, a_{m+1}$ as the number of digits of each part that partitioned from $m$ signs then\n$$\na_{1}+a_{2}+a_{3}+\\cdots+a_{m+1}=30\n$$\nThis equation has $\\binom{29}{m}$ solutions. Therefore the number of ways to insert the plus signs is $\\binom{29}{9}+\\binom{29}{19}+\\binom{29}{29}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40347,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the table, there are $1024$ marbles and two students $A, B$ alternatively take a positive number of marble(s). The student $A$ goes first, $B$ goes after that and so on. On the first move, $A$ takes $k$ marbles with $1<k<1024$. On the moves after that, $A$ and $B$ are not allowed to take more than $k$ marbles or $0$ marble. The student who takes the last marble(s) on the table will be the winner. Find all values of $k$ the student $A$ should choose to make sure that there is a strategy for him to win the game.",
    "options": [],
    "answer": "4, 24, 40, 204",
    "solution": "After the first move, there are $1024-k$ marbles remaining and both students $A, B$ cannot take more than $k$ marbles during their turn.\n\nWe shall prove that if $k+1 \\mid 1024-k$ then the student $A$ will have the strategy to win the game.\n\nIndeed, let $1024-k = m(k+1)$ with $m \\in \\mathbb{Z}^{+}$. On each turn, if $B$ chooses $x$ marble(s) then $A$ will choose $k+1-x$ marble(s) on his next turn. Since $x \\in \\{1,2,3, \\ldots, k\\}$ then $k+1-x \\in \\{1,2,3, \\ldots, k\\}$ which implies that $A$ always can perform his turn. After $m$ turns of $A$ and $m$ turns of $B$, student $A$ takes the last marble then wins the game.\n\nIf $1024-k$ is not divisible by $k+1$ then we put $1024-k = m(k+1) + n$ with $0 < n < k+1$. In this case, the student $B$ can take $n$ marbles on the next turn. The number of remaining marbles is $m(k+1)$ then by applying the same argument above, the student $B$ has the strategy to win the game.\n\nHence, the condition we have to find is $k+1 \\mid 1024-k$ or $k+1 \\mid 1025$. Note that the divisors of $1025$ are $1,5,25,41,205,1025$ then\n$$\nk+1 \\in \\{1,5,25,41,205,1025\\} \\text{ and } 0<k<1024.\n$$\nThis means that all values of $k$ we need to find are $4,24,40,204$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40348,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S = \\{-17, -16, \\ldots, 16, 17\\}$. We call a subset $T$ of $S$ a good set if $-x \\in T$ for any $x \\in T$ and if $x, y, z \\in T$ ($x, y, z$ may be equal) then $x + y + z \\neq 0$. Find the largest number of elements in a good set.",
    "options": [],
    "answer": "18",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40349,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the set $X=\\{1,2,3, \\ldots, 2018\\}$. How many positive integers $k$ with $2 \\leq k \\leq 2017$ that satisfy the following conditions:\n\ni) There exists some partition of the set $X$ into $1009$ disjoint pairs which are $(a_{1}, b_{1}), (a_{2}, b_{2}), \\ldots, (a_{1009}, b_{1009})$ with $|a_{i}-b_{i}| \\in \\{1, k\\}$.\n\nii) For all partitions satisfy the condition (i), the sum $T=\\sum_{i=1}^{1009}|a_{i}-b_{i}|$ has the right most digit is $9$.",
    "options": [],
    "answer": "403",
    "solution": "First, we notice that for all $2 \\leq k \\leq 2017$, we always can find a partition satisfy i). For example, we can choose $a_{i}=2i-1$, $b_{i}=2i$ with $i=1,2, \\ldots, 1009$.\n\nDenote $m$ as the number of pairs $(a_{i}, b_{i})$ with $|a_{i}-b_{i}|=k$ then\n$$\nT=(1009-m) \\cdot 1 + m \\cdot k = 1009 + (k-1)m.\n$$\nHence, the necessary condition of $k$ is $(k-1)m$ is divisible by $10$. First, we can see that all numbers $k \\equiv 1 \\pmod{5}$ satisfy the given condition.\n\nIndeed, we have some cases:\n\n1. If $k \\equiv 1 \\pmod{10}$ then $10 \\mid k-1$ and ii) is satisfied.\n\n2. If $k \\equiv 6 \\pmod{10}$ then $k$ is even. Thus, we need to prove $m$ is even. This is true because we can write $m = m_{1} + m_{2}$ with $m_{1}, m_{2}$ are the number of pairs with the difference is $k$ that have same parity and different parity. In $X$, the number of even and odd numbers are equal and all the pairs $(a_{i}, b_{i})$ with $|a_{i}-b_{i}|=1$ also contain exactly $1$ number in each type of even/odd. These imply that $m_{1}=m_{2}$ or $m$ is an even number.\n\nHence, all numbers $k \\equiv 1 \\pmod{5}$ satisfy the given condition.\n\nNext, we will prove that all other numbers $k$ do not satisfy the given condition.\n\n3. If $k$ is odd then $k-1$ is not divisible by $5$. We choose the partition with $a_{1}=1$, $b_{1}=k+1$ and the rest are divided into pairs with $a_{i}, b_{i}$ being two consecutive numbers. It is easy to check that this partition satisfies i) but not ii) because $T=1008+k \\not\\equiv 9 \\pmod{10}$.\n\n4. If $k$ is even, $k-1$ is not divisible by $5$ then $k-1$ is an odd number that is not divisible by $5$. We choose the partition with exactly two pairs $(1, k+1), (2, k+2)$ and the rest are divided into pairs with $a_{i}, b_{i}$ being two consecutive numbers. Similarly with the previous case, we have $T=1009+2(k-1) \\not\\equiv 9 \\pmod{10}$.\n\nTherefore, the necessary and sufficient condition of $k$ is $k \\equiv 1 \\pmod{5}$. From $2$ to $2017$, we have $403$ numbers in total satisfy it.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40350,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist an integer $n \\geq 3$ and an arithmetic sequence $a_{0}, a_{1}, \\ldots, a_{n}$ such that the polynomial $a_{n} x^{n}+\\ldots+a_{1} x+a_{0}$ has $n$ roots which also form an arithmetic sequence?",
    "options": [],
    "answer": "Yes, with n = 3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40351,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fibonacci sequences is defined as\n$$\n\\left\\{\\begin{array}{l}\nf_{1}=1,\\ f_{2}=2 \\\\\nf_{n+1}=f_{n}+f_{n-1},\\ n \\geq 2\n\\end{array}\\right.\n$$\n1. Prove that every positive integer can be represented as sum of several distinct Fibonacci number.\n2. A positive integer is called Fib-unique if the way to represent it as sum of several distinct Fibonacci number is unique.\nExample: 13 is not Fib-unique because $13=13=8+5=8+3+2$. Find all Fib-unique.",
    "options": [],
    "answer": "Fib-unique numbers are f1, f2, and f_k − 1 for k ≥ 3.",
    "solution": "1. We will prove by induction that every positive integer $k$ can be written as sum of several distinct Fibonacci numbers.\nThe case $k=1$ is obvious, we have $1=f_{1}$.\nAssume that every positive integer less than $n$ can be written as sum of several distinct Fibonacci numbers, consider $k=n$.\nWe can find $f_{i}$ is the greatest Fibonacci number which is less than $n$, we have\n$$\nn=f_{i}+\\left(n-f_{i}\\right) .\n$$\nAccording to the induction hypothesis, $n-f_{i}$ is a positive integer less than $n$, therefore it can be written as\n$$\nn-f_{i}=f_{j_{1}}+f_{j_{2}}+\\cdots+f_{j_{t}},\\ j_{1}<j_{2}<\\ldots<j_{t} .\n$$\nWe must have $i>j_{t}$ because otherwise, $n=f_{i}+f_{j_{t}}+\\cdots>f_{i}+f_{i-1}=f_{i+1}$, contradiction to the choice of $i$.\nSo, $n=f_{j_{1}}+f_{j_{2}}+\\cdots+f_{j_{t}}+f_{i}$, which is a representation of $n$ as sum of distinct Fibonacci numbers, we are done.\n\n2. According to the part 1., an integer $n$ can be written as\n$$\nn=f_{i_{1}}+f_{i_{2}}+\\cdots+f_{i_{k}},\\ i_{1}<i_{2}<\\cdots<i_{k} .\n$$\nWe need to find all $n$ of which representation is unique.\nFirst, if $i_{1} \\geq 3$, then we can replace $f_{i_{1}}$ by $f_{i_{1}-1}+f_{i_{1}-2}$ and get another representation of $n$. So if $n$ is a \"Fib-unique\", $i_{1} \\in\\{1,2\\}$.\nSecond, if there is some $t$ such that $i_{t+1}-i_{t} \\geq 3$, then we can replace $f_{i_{t+1}}$ by $f_{i_{t+1}-1}+f_{i_{t+1}-2}$ and get another representation of $n$.\nHence, for a \"Fib-unique\" integer $n$, we have $i_{t+1}-i_{t} \\leq 2$ for every $0 \\leq t \\leq k-1$.\nThird, if we have some $t$ such that $i_{t+1}-i_{t}=1$, chose that $t$ which is the largest. Then we can replace $f_{i_{t+1}}+f_{i_{t}}$ by $f_{i_{t+1}+1}$, and get another representation of $n$. So for a \"fib-unique\" $n$ we must have $i_{t+1}-i_{t}=2$ for every $0 \\leq t \\leq k-1$.\nSo, every possible \"Fib-unique\" positive $n$ are $f_{1}, f_{2}$ and those in these forms:\n$$\n\\begin{aligned}\n& \\text{ i . } n=f_{1}+f_{3}+\\cdots+f_{2k-1}=f_{2k}-1 \\\\\n& \\text{ ii . } n=f_{2}+f_{4}+\\cdots+f_{2k}=f_{2k+1}-1\n\\end{aligned}\n$$\nIn conclusion, every Fib-unique is $f_{1}, f_{2}$ and $f_{k}-1$ with $k \\geq 3$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40352,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a set $S$ of 200 points on the plane: 100 points are the vertices of a convex polygon $A$, and 100 other points are in the interior of the polygon. Moreover, there does not exist 3 collinear points. A triangulation is a way to partition the polygon $A$ into triangles by drawing the edges between some two points of $S$ such that any two edges do not intersect in the interior, and each point in $S$ is the vertex of at least one triangle.\n\n1. Prove that number of edges does not depend on the triangulation.\n2. Show that for any triangulation, one can draw each triangle by 1 of 3 given colors such that 2 adjacent triangles have different colors.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1) Suppose that we have $k$ triangles in some triangulation. By calculating the sum of all angles of these triangles, we have $180^\\circ \\cdot k$.\nThe sum of interior angles of $A$ is $180^\\circ \\cdot 98$.\nThe sum of angle around each point among 100 points is $360^\\circ \\cdot 100$. Hence, we have\n$$\n180^\\circ \\cdot k = 180^\\circ \\cdot 98 + 360^\\circ \\cdot 100 \\Leftrightarrow k = 298.\n$$\nEach triangle gives 3 edges and among them, there are 100 edges of $A$. Note that the interior edges are double counted, then the number of edges in each triangulation is\n$$\n\\frac{3 \\cdot 298 - 100}{2} + 100 = 497.\n$$\n\n2) Fix the polygon $A$, we will prove this problem by induction on $m+n$ of total points, in which $n$ is the number of vertices of $A$ and $m$ is the number of interior points.\n\nFor $m+n=3$, which implies that $n=3, m=0$, we just have one triangle with no interior point. We color this triangle by some color and finish this case.\n\nFor bigger $m+n$, whenever we add one more point and perform the triangulation, we have two cases:\n\n1. If there are some triangle that share exactly one edge with the polygon $A$, call triangle $X$. Then $X$ is adjacent to at most two other triangles then if these two triangles were colored by different colors, we just need to color $X$ by the third color (in case two triangles were colored by the same color, we color $X$ by a random color among the rest). Since $X$ has one vertex that is the interior point, then we can move this point outside as the vertex of $A$. It is easy to see that polygon $A$ still has $n$ vertices and $m$ decreases by 1. Then we can apply the induction hypothesis.\n\n![](attached_image_1.png)\n\n2. If for all triangles $X$ as described above, the vertices of $X$ are not the interior points then $X$ is adjacent to exactly one triangle. We just color $X$ by the color different from the adjacent one and remove the vertex of $A$ to make number of sides of $A$ decrease by 2 and $n$ decreases by 1. Then we can also apply the induction hypothesis.\n\nTherefore, in all cases, we can change the triangulation of $m+n$ points to triangulation of $m+n-1$ points, which means the problem can be solved by induction completely.\n\nThen we always can draw the triangle by one of three colors that satisfy the given condition. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 40353,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $A$ is an obtuse angle. Denote $BE$ as the internal angle bisector of triangle $ABC$ with $E \\in AC$ and suppose that $\\angle AEB = 45^{\\circ}$. The altitude $AD$ of triangle $ABC$ intersects $BE$ at $F$. Let $O_1, O_2$ be the circumcenter of triangles $FED, EDC$. Suppose that $EO_1, EO_2$ meet $BC$ at $G, H$ respectively. Prove that $\\frac{GH}{GB} = \\tan \\frac{\\alpha}{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $O$ as the projection of $B$ on $AC$ and $BK$ is the diameter of $(O, OB)$. It is easy to see that $E \\in (O)$. We have\n$$\n\\angle BCA = \\angle AEB - \\angle CBE = \\angle OBE - \\angle ABE = \\angle OBA = \\angle OKA.\n$$\nThis implies that $ODCK$ is the inscribed quadrilateral and $\\angle KDC = 90^{\\circ}$. Hence, $K, A, D$ are collinear and $D \\in (O)$.\nTriangle $CBK$ is isosceles then $CE$ is the angle bisector of $\\angle BCK$. Note that $DE$ is the angle bisector of $\\angle ADC$. Then $E$ is the incenter of triangle $KDC$.\n\n![](attached_image_1.png)\n\nSo $O_2$ is the midpoint of minor $\\operatorname{arc} CD$ of the circumcircle of $KDC$, then $K, E, O_2$ are collinear.\nWe have\n$$\n\\triangle BAD \\sim \\triangle KCD \\Rightarrow \\frac{BD}{BA} = \\frac{KD}{KC}.\n$$\nBased on the property of bisector in triangle, we have\n$$\n\\frac{FD}{FA} = \\frac{BD}{BA} = \\frac{KD}{KC} = \\frac{HD}{HC}.\n$$\nHence, $FH \\parallel AC$. We have $O_1$ is the midpoint of $FH$ also is the circumcenter of quadrilateral $DFEH$.\nFrom this, we can conclude that $EG$ is the angle bisector in triangle $BEH$ and\n$$\n\\frac{GH}{GB} = \\frac{EH}{EB} = \\tan \\frac{B}{2} = \\tan \\frac{\\alpha}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40354,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the number of ways one can put numbers $1$ or $2$ in each cell of a $8 \\times 8$ chessboard in such a way that the sum of numbers in each column and in each row is an odd number (two ways are considered different if the number on some cell in the first way is different from the number on the cell at correspondent position in the second way).",
    "options": [],
    "answer": "2^49",
    "solution": "Consider the leftmost column and lowest row of table, we color all these cells. We can see that for all ways to put the number of the sub square $7 \\times 7$ that not colored, we can choose the number for the correspondent colored position at same column or row. Indeed, if sum of the $7$ numbers is odd, then we put $1$ on the remain cell; otherwise, we put $2$.\n\n![](attached_image_1.png)\n\nFinally, the number for the cell at corner can choose base on the parity of the sum of all number in square $7 \\times 7$. These mean that the way to fill in the square $7 \\times 7$ uniquely define the numbers on the rest cells.\n\nSince we can fill each cell among $49$ cells of the square $7 \\times 7$ by $1$ or $2$ in any way then the number of way is $2^{49}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40355,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle and $(O)$ be its circumcircle (with center $O$). Denote by $G$ the centroid of the triangle $ABC$, by $H$ the foot of the altitude from $A$ onto the side $BC$ and by $I$ the midpoint of $AH$. The line $IG$ intersects $BC$ at $K$.\n1. Prove that $CK = BH$.\n2. The ray $GH$ intersects $(O)$ at $L$. Denote by $T$ the circumcenter of the circle $(BHL)$. Prove that $AO$ and $BT$ intersect on the circle $(O)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1)\nLet $M$ be the midpoint of $BC$ then $G \\in AM$ and $\\frac{AG}{AM} = \\frac{2}{3}$. Take the point $K'$ on $BC$ such that $M$ is the midpoint of $HK'$, then $AM$ is the median of triangle $AHK'$ and $G$ is its centroid.\nThen $K'G$ is the median of triangle $AHK'$ or $K'G$ passes through the midpoint of $AH$. This implies that $K \\equiv K'$ and we have $BH = CK$.\n\n![](attached_image_1.png)\n\n2)\nThe line passes through $A$ and parallel to $BC$ intersects $(O)$ at $E$ different from $A$. Then by the symmetry through the perpendicular bisector of $BC$, it is easy to check that $AHKE$ is a rectangle. Since $\\frac{GA}{GM} = \\frac{AE}{HM} = 2$, we have $H, G, E, L$ are collinear.\nThen $\\angle BLH = \\angle BLE = \\angle BCE = \\angle ABC$ which implies that\n$$\n\\angle ABT = \\angle ABC + \\angle CBT = \\angle BLH + \\angle CBT = 90^\\circ.\n$$\nThus if we denote $D = BT \\cap (O)$ then $AD$ is the diameter of $(O)$, then $O \\in AD$. Therefore, $BT$ and $AO$ intersect at a point that belongs to $(O)$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40356,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number and a table of size $(p^{2}+p+1) \\times (p^{2}+p+1)$ which is divided into unit cells. The way to color some cells of this table is called nice if there are no four colored cells that form a rectangle (the sides of rectangle are parallel to the sides of given table).\n\n1. Let $k$ be the number of colored cells in some nice coloring way. Prove that $k \\leq (p+1)(p^{2}+p+1)$. Denote this number as $k_{\\text{max}}$.\n\n2. Prove that all ordered tuples $(a, b, c)$ with $0 \\leq a, b, c < p$ and $a+b+c>0$ can be partitioned into $p^{2}+p+1$ sets $S_{1}, S_{2}, \\ldots, S_{p^{2}+p+1}$ such that two tuples $(a_{1}, b_{1}, c_{1})$ and $(a_{2}, b_{2}, c_{2})$ belong to the same set if and only if $a_{1} \\equiv k a_{2}, b_{1} \\equiv k b_{2}, c_{1} \\equiv k c_{2} \\pmod{p}$ for some $k \\in \\{1,2,3, \\ldots, p-1\\}$.\n\n3. For $1 \\leq i, j \\leq p^{2}+p+1$, if there exist $(a_{1}, b_{1}, c_{1}) \\in S_{i}$ and $(a_{2}, b_{2}, c_{2}) \\in S_{j}$ such that $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2} \\equiv 0 \\pmod{p}$, we color the cell $(i, j)$ of the given table. Prove that this coloring way is nice with $k_{\\text{max}}$ colored cells.",
    "options": [],
    "answer": "(p+1)(p^2+p+1)",
    "solution": "1)\nWe will count value $S$, the number of the tuples $(A, B, C)$ with column $A, B$ and row $C$ in relation $A-C, B-C$ intersect at the colored cell.\n\nFirst way, denote $x_{i}$ as the number of colored cells on the $i$th row, then\n$$\nS = \\sum_{i=1}^{p^{2}+p+1} \\binom{x_{i}}{2}.\n$$\n\nSecond way, choose two columns among $p^{2}+p+1$ columns, we have $\\binom{p^{2}+p+1}{2}$ ways. And then, we have at most 1 row $C$ that cuts $A, B$ at two colored cells (since there are no rectangle), then\n$$\nS \\leq \\binom{p^{2}+p+1}{2}.\n$$\n\nNotice that $\\sum_{i=1}^{p^{2}+p+1} x_{i} = k$. Hence, we have\n$$\n\\binom{p^{2}+p+1}{2} \\geq \\sum_{i=1}^{p^{2}+p+1} \\binom{x_{i}}{2} \\geq \\frac{1}{2}\\left(\\frac{1}{p^{2}+p+1} k^{2} - k\\right)\n$$\nor $k \\leq (p+1)(p^{2}+p+1)$.\n\n2)\nConsider the tuples $(a, b, c)$ in which $0 \\leq a, b, c \\leq p-1$ and $a+b+c>0$, then we have $p^{3}-1$ tuples in total.\n\nSince two tuples $(a, b, c)$ and $(d, e, f)$ are considered to belong to the same set if and only if $\\exists k \\in \\{1,2,3, \\ldots, p-1\\}$ such that $a \\equiv k d, b \\equiv k e, c \\equiv k f \\pmod{p}$. We have $p-1$ ways to choose $k$, then $\\frac{p^{3}-1}{p-1} = p^{2}+p+1$ sets satisfy the given condition.\n\n3)\nFrom the way to color the table, one can check that on each row, there are exactly $p+1$ colored cells. Indeed, we need to prove that $a x + b y + c z \\equiv 0 \\pmod{p}$ has exactly $p+1$ solutions that do not belong to the same set $S_{1}, S_{2}, \\ldots, S_{p^{2}+p+1}$.\n\nThere are $p^{2}-1$ pairs $(x, y)$ and for each pair, there is a unique number $z$ such that $a x + b y + c z$ is divisible by $p$. Each set has $p-1$ solutions, then there are $p+1$ solutions that do not belong to the same set. Multiply this number with the number of columns, we have verified that there are $(p+1)(p^{2}+p+1)$ colored cells on the table.\n\nThen, we shall prove that there are no rectangles. For two tuples $(a, b, c)$ and $(d, e, f)$ that do not belong to the same set, we need to prove that the following system of equations has no more than one solution:\n$$\n\\begin{cases}\na x + b y + c z \\equiv 0 & (\\bmod p) \\\\\nd x + e y + f z \\equiv 0 & (\\bmod p)\n\\end{cases}\n$$\nWithout loss of generality, we can suppose that $\\frac{b}{e} - \\frac{c}{f}$ is not divisible by $p$ or $b f \\not\\equiv c e \\pmod{p}$, then for all $x_{0} = 1, \\ldots, p-1$, there is exactly one value $y_{0}, z_{0}$ such that\n$$\nb y_{0} + c z_{0} = a x_{0}, \\quad e y_{0} + f z_{0} = d x_{0} \\pmod{p}.\n$$\nOn the other hand, there are $p$ values $x_{0}$ and we do not consider the tuple $(0,0,0)$, then there are at most $p-1$ solutions.\n\nNote that if $(x, y, z)$ is a solution, then so is $(k x, k y, k z)$, then the above system has no more than one solution.\n\nThis finishes our proof and $k_{\\text{max}} = (p+1)(p^{2}+p+1)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40357,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle inscribed in circle $(O)$, with orthocenter $H$. Median $AM$ of triangle $ABC$ intersects circle $(O)$ at $A$ and $N$. $AH$ intersects $(O)$ at $A$ and $K$. Three lines $KN$, $BC$ and the line through $H$ and perpendicular to $AN$ intersect each other and form triangle $XYZ$. Prove that the circumcircle of triangle $XYZ$ is tangent to $(O)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40358,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = 6$. Prove that\n$$\n\\frac{1}{a^2 b + 16} + \\frac{1}{b^2 c + 16} + \\frac{1}{c^2 a + 16} \\geq \\frac{1}{8}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40359,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex polygon is divided into some triangles. Let $V$ and $E$ be respectively the set of vertices and the set of edges of all triangles (each vertex in $V$ may be some vertex of the polygon or some point inside the polygon). The polygon is said to be good if the following conditions hold:\n\ni. There are no 3 vertices in $V$ which are collinear;\n\nii. Each vertex in $V$ belongs to an even number of edges in $E$.\n\nFind all good polygon.",
    "options": [],
    "answer": "Exactly the convex polygons whose number of sides is divisible by three.",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 40360,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of integers $(x, y)$ such that\n$$\ny^{3} = 8x^{6} + 2x^{3}y - y^{2}.\n$$",
    "options": [],
    "answer": "[(0, 0), (0, -1), (1, 2)]",
    "solution": "We rewrite the equation:\n$$\ny^{3} + y^{2} - 2x^{3}y - 8x^{6} = 0.\n$$\n\nConsider this as a cubic in $y$:\n$$\ny^{3} + y^{2} - 2x^{3}y - 8x^{6} = 0.\n$$\n\nLet us try small integer values for $x$.\n\nIf $x = 0$:\n$$\ny^{3} + y^{2} = 0 \\implies y^{2}(y + 1) = 0 \\implies y = 0 \\text{ or } y = -1.\n$$\nSo $(0, 0)$ and $(0, -1)$ are solutions.\n\nIf $x = 1$:\n$$\ny^{3} + y^{2} - 2y - 8 = 0.\n$$\nTry $y = 1$:\n$1 + 1 - 2 - 8 = -8$.\nTry $y = 2$:\n$8 + 4 - 4 - 8 = 0$.\nSo $y = 2$ works. $(1, 2)$ is a solution.\nTry $y = -2$:\n$-8 + 4 + 4 - 8 = -8$.\nTry $y = -1$:\n$-1 + 1 + 2 - 8 = -6$.\nTry $y = 4$:\n$64 + 16 - 8 - 8 = 64$.\nSo only $y = 2$ works for $x = 1$.\n\nIf $x = -1$:\n$$\ny^{3} + y^{2} + 2y - 8 = 0.\n$$\nTry $y = 1$:\n$1 + 1 + 2 - 8 = -4$.\nTry $y = 2$:\n$8 + 4 + 4 - 8 = 8$.\nTry $y = -2$:\n$-8 + 4 - 4 - 8 = -16$.\nTry $y = -1$:\n$-1 + 1 - 2 - 8 = -10$.\nTry $y = 4$:\n$64 + 16 + 8 - 8 = 80$.\nSo no integer solution for $x = -1$.\n\nTry $x = 2$:\n$8x^{6} = 8 \\times 64 = 512$\n$2x^{3}y = 2 \\times 8y = 16y$\nSo:\n$y^{3} + y^{2} - 16y - 512 = 0$\nTry $y = 8$:\n$512 + 64 - 128 - 512 = -64$\nTry $y = 16$:\n$4096 + 256 - 256 - 512 = 3584$\nTry $y = -8$:\n$-512 + 64 + 128 - 512 = -832$\nTry $y = 4$:\n$64 + 16 - 64 - 512 = -496$\nTry $y = -4$:\n$-64 + 16 + 64 - 512 = -496$\nSo no integer solution for $x = 2$.\n\nTry $x = -2$:\n$8x^{6} = 512$\n$2x^{3}y = 2 \\times (-8)y = -16y$\nSo:\n$y^{3} + y^{2} + 16y - 512 = 0$\nTry $y = 8$:\n$512 + 64 + 128 - 512 = 192$\nTry $y = -8$:\n$-512 + 64 - 128 - 512 = -1088$\nTry $y = 4$:\n$64 + 16 + 64 - 512 = -368$\nTry $y = -4$:\n$-64 + 16 - 64 - 512 = -624$\nSo no integer solution for $x = -2$.\n\nNow, for large $|x|$, the term $8x^{6}$ dominates, so $y^{3} \\approx 8x^{6}$, so $y \\approx 2x^{2}$.\nTry $y = 2x^{2}$:\n$y^{3} = 8x^{6}$\n$2x^{3}y = 4x^{5}$\n$y^{2} = 4x^{4}$\nSo:\n$8x^{6} = 8x^{6} + 4x^{5} - 4x^{4}$\n$0 = 4x^{5} - 4x^{4}$\n$0 = 4x^{4}(x - 1)$\nSo $x = 0$ or $x = 1$.\n\nIf $x = 0$, $y = 0$ (already found).\nIf $x = 1$, $y = 2$ (already found).\n\nTry $y = -2x^{2}$:\n$y^{3} = -8x^{6}$\n$2x^{3}y = -4x^{5}$\n$y^{2} = 4x^{4}$\nSo:\n$-8x^{6} = 8x^{6} - 4x^{5} - 4x^{4}$\n$-16x^{6} + 4x^{5} + 4x^{4} = 0$\n$4x^{4}(x + 1 - 4x^{2}) = 0$\nSo $x = 0$ or $x = -1$ or $x^{2} = 1/4$ (not integer).\nIf $x = 0$, $y = 0$ (already found).\nIf $x = -1$, $y = -2$.\nTry $x = -1$, $y = -2$ in the original equation:\n$(-2)^{3} = 8(-1)^{6} + 2(-1)^{3}(-2) - (-2)^{2}$\n$-8 = 8 + 2(-1)(-2) - 4 = 8 + 4 - 4 = 8$\nSo $-8 = 8$ is false.\nSo $x = -1$, $y = -2$ is not a solution.\n\nTry $y = kx^{2}$ for small integer $k$.\nTry $y = x$:\n$y^{3} = x^{3}$\n$8x^{6} + 2x^{3}y - y^{2} = 8x^{6} + 2x^{4} - x^{2}$\nSo $x^{3} = 8x^{6} + 2x^{4} - x^{2}$\nTry $x = 0$: $0 = 0$\nTry $x = 1$: $1 = 8 + 2 - 1 = 9$\nTry $x = -1$: $-1 = 8 + 2 - 1 = 9$\nSo only $x = 0$, $y = 0$ (already found).\n\nTry $y = -1$:\n$(-1)^{3} = 8x^{6} + 2x^{3}(-1) - (-1)^{2}$\n$-1 = 8x^{6} - 2x^{3} - 1$\n$0 = 8x^{6} - 2x^{3}$\n$2x^{3}(4x^{3} - 1) = 0$\nSo $x = 0$ or $x^{3} = 1/4$ (not integer).\nSo $x = 0$, $y = -1$ (already found).\n\nTherefore, the only integer solutions are $(x, y) = (0, 0), (0, -1), (1, 2)$.\n\nFinal answer:\nAll integer solutions are $(x, y) = (0, 0), (0, -1), (1, 2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40361,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle inscribed in circle $(O)$, with orthocenter $H$. Let $d$ be an arbitrary line which passes through $H$ and intersects $(O)$ at two points $P$ and $Q$. Draw diameter $AA'$ of circle $(O)$. $A'P$, $A'Q$ meet $BC$ at $K$, $L$, respectively. Prove that $O$, $K$, $L$, $A'$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40362,
    "subject": "Mathematics (Multi-modal)",
    "question": "A chessboard has $64$ cells painted black and white in the usual way. A bishop path is a sequence of distinct cells such that two consecutive cells have exactly one common point. At least how many bishop paths can the set of all white cells be divided into?",
    "options": [],
    "answer": "2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40363,
    "subject": "Mathematics (Multi-modal)",
    "question": "Here are $2017$ points on the plane, no three of them are collinear. Some pairs of the points are connected by $n$ segments. Find the smallest value of $n$ so that there always exists two disjoint segments in any case.",
    "options": [],
    "answer": "2018",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40364,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle inscribed in circle $(O)$, with its altitudes $BH_b$, $CH_c$ intersecting at orthocenter $H$ ($H_b \\in AC$, $H_c \\in AB$). $H_b H_c$ meets $BC$ at $P$. Let $N$ be the midpoint of $AH$, $L$ be the orthogonal projection of $O$ on the symmedian with respect to angle $A$ of triangle $ABC$. Prove that $\\angle NLP = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40365,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an odd positive integer with $n>1$ and let $a_{1}, a_{2}, \\ldots, a_{n}$ be positive integers such that $\\operatorname{gcd}\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)=1$. Let $d=\\operatorname{gcd}\\left(a_{1}^{n}+a_{1} a_{2} \\cdots a_{n}, a_{2}^{n}+a_{1} \\cdot a_{2} \\cdots a_{n}, \\ldots, a_{n}^{n}+a_{1} \\cdot a_{2} \\cdots a_{n}\\right)$. Show that the possible values of $d$ are $d=1, d=2$.",
    "options": [],
    "answer": "d = 1 or d = 2",
    "solution": "Let $p$ be a prime with $p \\mid d$. We know that $p \\mid a_{i}^{n}+a_{1} \\cdot a_{2} \\cdots a_{n}$ for all $i$. If $\\exists i$ such that $p \\mid a_{i}$, then $p \\mid a_{1} \\cdot a_{2} \\cdots a_{n}$. Since $p \\mid a_{j}^{n}+a_{1} \\cdot a_{2} \\cdots a_{n}$ for all $j$, we find $p \\mid a_{j}$ for all $j$, which contradicts to the hypothesis.\n\nThus we consider the case $p \\nmid a_{i}$ for all $i$, i.e. $p \\nmid a_{1} \\cdot a_{2} \\cdots a_{n}$. We have the $n$ congruences\n$$\n\\begin{aligned}\n& a_{1}^{n} \\equiv -a_{1} \\cdot a_{2} \\cdots a_{n} \\pmod{p}, \\\\\n& a_{2}^{n} \\equiv -a_{1} \\cdot a_{2} \\cdots a_{n} \\pmod{p}, \\ldots, \\\\\n& a_{n}^{n} \\equiv -a_{1} \\cdot a_{2} \\cdots a_{n} \\pmod{p}\n\\end{aligned}\n$$\nand by multiplying them, we obtain\n$$\n\\left(a_{1} \\cdot a_{2} \\cdots a_{n}\\right)^{n} \\equiv (-1)^{n} \\cdot \\left(a_{1} \\cdot a_{2} \\cdots a_{n}\\right)^{n} \\equiv -\\left(a_{1} \\cdot a_{2} \\cdots a_{n}\\right)^{n} \\pmod{p}\n$$\nwhich yields $p=2$.\n\nThis shows that $d$ must be a power of $2$. Suppose, now, that $4 \\mid d$. Clearly, $a_{1}, a_{2}, \\ldots, a_{n}$ must be all odd. We consider two cases\n\n1. If $a_{1} \\cdot a_{2} \\cdots a_{n} \\equiv 3 \\pmod{4}$, then $\\exists i$ such that $a_{i} \\equiv 3 \\pmod{4}$ then $a_{i}^{n}+a_{1} \\cdot a_{2} \\cdots a_{n} \\equiv 3+3 \\equiv 2 \\pmod{4}$ which contradicts with $4 \\mid d$.\n\n2. If $a_{1} \\cdot a_{2} \\cdots a_{n} \\equiv 1 \\pmod{4}$, then $\\exists i$ such that $a_{i} \\equiv 1 \\pmod{4}$ then $a_{i}^{n}+a_{1} \\cdot a_{2} \\cdots a_{n} \\equiv 1+1 \\equiv 2 \\pmod{4}$ which also contradicts with $4 \\mid d$. (In both cases, we made use of $3^{n} \\equiv 3 \\pmod{4}$.)\n\nTherefore, $d=1$ or $d=2$. If $a_{1}, a_{2}, \\ldots, a_{n}$ are all odd, then $d$ must be even, hence $d=2$. If some of $a_{1}, a_{2}, \\ldots, a_{n}$ are even and some of them odd, then $d$ must be odd, hence $d=1$. As such, both of these values are realized.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40366,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote $S$ as the set of prime divisors of all integers of form $2^{n^{2}+1}-3^{n}$, $n \\in \\mathbb{Z}^{+}$. Prove that $S$ and $\\mathscr{P} \\backslash S$ both contain infinitely many elements ($\\mathscr{P}$ is the set of prime numbers).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, suppose on the contrary that $S$ is finite, then $S=\\{p_{1}, p_{2}, \\ldots, p_{k}\\}$ for $k=|S|$. It is easy to check that $2,3 \\notin S$.\n\nConsider number $N=(p_{1}-1)(p_{2}-1) \\cdots (p_{k}-1)$ and $M=2^{N^{2}+1}-3^{N}$. For some $p \\in S$, by Euler's theorem, we have\n$$\n2^{p-1} \\equiv 1 \\pmod{p} \\text{ and } 3^{p-1} \\equiv 1 \\pmod{p}.\n$$\nThen $2^{N^{2}+1}=2 \\cdot 2^{N^{2}}=2 \\cdot (2^{p-1})^{\\frac{N^{2}}{p-1}} \\equiv 2 \\pmod{p}$ and $3^{N} \\equiv (3^{p-1})^{\\frac{N}{p-1}} \\equiv 1 \\pmod{p}$. Hence, $M \\equiv 2-1=1 \\pmod{p}$. So by the Chinese remainder theorem, we also have $M \\equiv 1 \\pmod{N}$ and $M>2$ then $M$ has some prime divisor that differs from $p_{1}, p_{2}, \\ldots, p_{k}$, which is a contradiction.\n\nThus, $S$ has infinitely many elements.\n\nTo prove the second part, just consider some prime $p$ such that there does not exist $x \\in \\mathbb{Z}$ such that either $x^{2} \\equiv 2 \\pmod{p}$ or $x^{2} \\equiv 3 \\pmod{p}$. Indeed, suppose that for such prime $p$, $p \\in S$ then we have two cases:\n\n- If $n$ is even then $2 \\cdot 2^{n^{2}} \\equiv 3^{n} \\pmod{p}$, which can be written as $2 x^{2} \\equiv y^{2} \\pmod{p}$ for some $x, y$ coprime to $p$. Then take $x'$ such that $x x' \\equiv 1 \\pmod{p}$, so $2 \\equiv (x' y)^{2} \\pmod{p}$, contradicting that $2$ is not a quadratic residue modulo $p$.\n- If $n$ is odd then we have a similar argument.\n\nThus, we just need to show that there are infinitely many such primes. By Euler's criterion about quadratic residues, we need\n$$\n\\left(\\frac{2}{p}\\right)=\\left(\\frac{3}{p}\\right)=-1 \\text{ and } \\left(\\frac{2}{p}\\right) \\equiv 2^{\\frac{p-1}{2}} \\pmod{p},\\left(\\frac{3}{p}\\right) \\equiv 3^{\\frac{p-1}{2}} \\pmod{p}.\n$$\nSo we can see that $p \\equiv 3,5 \\pmod{8}$ and $p \\equiv 2 \\pmod{3}$, which implies that we can take any prime of the form $p=24k+5$ or $p=24k+11$. Dirichlet's theorem states that the primes of the form $an+b$ for any coprime positive integers $a, b$ are infinite, so we finish our proof. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40367,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle with $M, N, P$ as the midpoints of $BC, CA, AB$, respectively. Denote $d_{1}$ as the line passing through $M$ and perpendicular to the angle bisector of $\\angle BAC$, and similarly define $d_{2}, d_{3}$. Suppose that $d_{2} \\cap d_{3} = D$, $d_{3} \\cap d_{1} = E$, $d_{1} \\cap d_{2} = F$. Let $I, H$ be the incenter and orthocenter of triangle $ABC$. Prove that the circumcenter of triangle $DEF$ is the midpoint of segment $IH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $(I_{a}), (I_{b}), (I_{c})$ as the ex-circles with respect to angles $A, B, C$ of triangle $ABC$. It is easy to check that $\\mathscr{P}_{M/(I)} = \\mathscr{P}_{M/(I_{a})}$ and $\\mathscr{P}_{M/(I_{b})} = \\mathscr{P}_{M/(I_{c})}$. Let $O, K$ be the circumcenters of triangles $ABC$ and $DEF$.\n\nHence, $d_{1}$ is the radical axis of $(I), (I_{a})$ since $d_{1} \\perp II_{a}$. Similarly for $d_{2}, d_{3}$, which implies that $D = d_{2} \\cap d_{3}$ is the radical center of the three circles $(I), (I_{b}), (I_{c})$. Thus, $MD$ is the radical axis of $(I_{b}), (I_{c})$.\n\nSo we have $MD \\perp I_{b}I_{c}$, but $I_{b}I_{c} \\perp AI$, $EF \\perp AI$, then $MD \\perp EF$ leads to $MD$ being the altitude in triangle $DEF$.\n\nSimilarly for $NE, PF$, so the three lines $MD, NE, PF$ concur at point $T$, which is the orthocenter of $DEF$. The two triangles $ABC$ and $DEF$ share the same nine-point circle $(MNP)$, so $OH$ and $KT$ share the midpoint, which is the circumcenter of triangle $(MNP)$. This implies that $\\overrightarrow{HK} = \\overrightarrow{TO}$.\n\nOn the other hand, it is easy to check that $T$ is also the incenter of triangle $MNP$. Then consider the homothety with center $G$ as the centroid of triangle $ABC$, ratio $-2$, then $\\mathscr{H}: M \\rightarrow A, N \\rightarrow B, P \\rightarrow C$ implies that $\\mathscr{H}: T \\rightarrow I, O \\rightarrow H$. Hence, $\\overrightarrow{HI} = 2\\overrightarrow{TO}$.\n\nCombine with the previous equality, we have $\\overrightarrow{HI} = 2\\overrightarrow{HK}$, so $K$ is the midpoint of the segment $HI$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40368,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $n$ which can not be expressed as $n=\\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$ for some positive integers $a, b, c, d$.",
    "options": [],
    "answer": "11",
    "solution": "Let $S$ be the set of positive integers which can be written as $s=\\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$ for some positive integers $a, b, c, d$.\nSince $s>0$, we can assume that $a>b, c>d$ and write $s=2^{b-d} \\frac{2^{a-b}-1}{2^{c-d}-1}$. It's now clear that $b-d=v_{2}(x)$.\nSo if we set $x=2^{v_{2}(x)} y$ then $y=\\frac{2^{u}-1}{2^{v}-1}$ ($u, v$ are positive integers). In particular, the number we are looking for is odd. Clearly the numbers of the form $2^{u}-1$ ($v=0$) belong to $S$. More generally, we know that $2^{v}-1 \\mid 2^{u}-1$ if and only if $v \\mid u$. Set $u=d v$ then\n$$\ny=\\frac{2^{u}-1}{2^{v}-1}=\\frac{2^{d v}-1}{2^{v}-1}=2^{(d-1) v}+2^{(d-2) v}+\\cdots+2^{v}+1\n$$\nThis means that the base 2 representation of $y$ is of the form\n$$\ny=1 \\underbrace{0 \\ldots 0}_{v-1} 1 \\underbrace{0 \\ldots 0}_{v-1} 1 \\ldots \\underbrace{0 \\ldots 0}_{v-1} 1\n$$\n($d$ digits 1, every two consecutive 1's are separated by $v-1$ 0's). So the number we are looking for is the smallest odd number not of this form (for some $d \\geq 1, v \\geq 1$). Note that\n$$\n1=\\overline{1},\\ 3=\\overline{11},\\ 5=\\overline{101},\\ 7=\\overline{111},\\ 9=\\overline{1001},\\ 11=\\overline{1011}\n$$\nWe can conclude that 11 is the smallest odd number whose base 2 representation is not of the desired form. So the answer is $n=11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40369,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\{x_{n}\\}$ be a sequence defined by $x_{1}=2$ and\n$$\nx_{n+1}=x_{n}^{2}-x_{n}+1\n$$\nfor $n \\geq 1$. Prove that\n$$\n1-\\frac{1}{2^{2^{n-1}}}<\\frac{1}{x_{1}}+\\frac{1}{x_{2}}+\\ldots+\\frac{1}{x_{n}}<1-\\frac{1}{2^{2^{n}}}\n$$\nfor all $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The sequence is increasing since $x_{n+1}-x_{n}=(x_{n}-1)^{2} \\geq 0$. We also have $x_{n+1}-1=x_{n}(x_{n}-1)$ which gives us\n$$\n\\frac{1}{x_{n+1}-1}=\\frac{1}{x_{n}(x_{n}-1)}=\\frac{1}{x_{n}-1}-\\frac{1}{x_{n}}\n$$\nThis identity is valid because given $x_{1}=2$ we have $x_{n} \\geq 2$ and therefore $x_{n}-1 \\neq 0$. Now given that $\\frac{1}{x_{n}}=\\frac{1}{x_{n}-1}-\\frac{1}{x_{n+1}-1}$ we obtain the identity\n$$\n\\frac{1}{x_{1}}+\\frac{1}{x_{2}}+\\cdots+\\frac{1}{x_{n}}=1-\\frac{1}{x_{n+1}-1}\n$$\nSo we have to prove\n$$\n1-\\frac{1}{2^{2^{n-1}}}<1-\\frac{1}{x_{n+1}-1}<1-\\frac{1}{2^{2^{n}}}\n$$\nor equivalently\n$$\n2^{2^{n-1}}<x_{n+1}-1<2^{2^{n}}\n$$\nWe will prove this by induction.\nFor $n=1$ we simply get $2<x_{2}-1<4$ which is true since $x_{2}=3$. Assume the statement is true for $n=k$, i.e.\n$$\n2^{2^{k-1}}<x_{k+1}-1<2^{2^{k}}\n$$\nFor $n=k+1$ we have\n$$\nx_{k+2}-1=x_{k+1}(x_{k+1}-1)>2^{2^{k-1}}(2^{2^{k-1}}+1)=2^{2^{k}}+2^{2^{k-1}}>2^{2^{k}}\n$$\nby the induction hypothesis. On the other hand we know that $x_{n}$'s are all integers and therefore $x_{k+1} \\leq 2^{2^{k}}$ by the induction hypothesis. We obtain\n$$\nx_{k+2}-1=x_{k+1}(x_{k+1}-1)>2^{2^{k}} \\cdot 2^{2^{k}}=2^{2^{k+1}}\n$$\nwhich is what we wanted to prove. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40370,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the infinite, strictly increasing sequence of positive integers $a_{n}$ such that\n\ni. All terms of the sequence are pairwise coprime.\n\nii. The sum $\\frac{1}{\\sqrt{a_{1} a_{2}}} + \\frac{1}{\\sqrt{a_{2} a_{3}}} + \\frac{1}{\\sqrt{a_{3} a_{4}}} + \\cdots$ is unbounded.\n\nProve that this sequence contains infinitely many primes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose on the contrary that there are infinite primes in this sequence, thus there is some index $k$ such that $a_{n}$ is composite for all $n > k$. Denote $S$ as all prime divisors of the first $k$ terms of the given sequence.\n\nBy comparing $a_{k+1}$ with the first prime $p_{1}$ that does not appear in $S$ then we have $a_{k+1} \\geq p_{1}^{2}$ (since $a_{k+1}$ is composite and $p_{1}$ is not greater than the least prime divisor of this number).\n\nSimilarly, define $p_{2}$ as the second prime that does not appear in $S$ then $a_{k+2} \\geq p_{2}^{2}$ (since all terms of the sequence are pairwise coprime). In general, we get\n$$\na_{k+i} \\geq p_{i}^{2} \\text{ for all } i \\in \\mathbb{Z}^{+} .\n$$\n\nNote that $\\frac{1}{\\sqrt{a_{k+1} a_{k+2}}} < \\frac{1}{a_{k+1}} \\leq \\frac{1}{p_{1}^{2}}$, $\\frac{1}{\\sqrt{a_{k+2} a_{k+3}}} < \\frac{1}{a_{k+2}} \\leq \\frac{1}{p_{2}^{2}}$, $\\cdots$ and so on. Then the given sum in the second condition can be divided into two parts: the first is the sum taken on values from $a_{1}, a_{2}, \\ldots, a_{k}$ which is finite, and the second is less than\n$$\n\\begin{aligned}\n\\frac{1}{p_{1}^{2}} + \\frac{1}{p_{2}^{2}} + \\cdots &< \\frac{1}{1^{2}} + \\frac{1}{2^{2}} + \\frac{1}{3^{2}} + \\cdots < 1 + \\frac{1}{1 \\cdot 2} + \\frac{1}{2 \\cdot 3} + \\cdots \\\\\n&= \\left(1 - \\frac{1}{2}\\right) + \\left(\\frac{1}{2} - \\frac{1}{3}\\right) + \\left(\\frac{1}{3} - \\frac{1}{4}\\right) < 2 .\n\\end{aligned}\n$$\n\nHence, the given sum is bounded, which is a contradiction and this finishes our proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40371,
    "subject": "Mathematics (Multi-modal)",
    "question": "Distinct prime numbers $p$, $q$, $r$ satisfy the equation\n$$\n2 p q r + 50 p q = 7 p q r + 55 p r = 8 p q r + 12 q r = A\n$$\nfor some positive integer $A$. Find $A$.",
    "options": [],
    "answer": "1980",
    "solution": "Review the given condition as\n$$\np q (2 r + 50) = p r (7 q + 55) = q r (8 p + 12) = A.\n$$\nThis implies that $A$ is a multiple of $p$, $q$ and $r$ so the value $K = \\frac{A}{p q r}$ is an integer. Dividing through, we have\n$$\nK = 8 + \\frac{12}{p} = 7 + \\frac{55}{q} = 2 + \\frac{50}{r}.\n$$\nHence, $p \\mid 12$, $q \\mid 55$, $r \\mid 50$. So we have 3 cases as follow\n- $p = 2$, $q = 11$, $r = 5$.\n- $p = 3$, $q = 11$, $r = 5$.\n- $p = 3$, $q = 5$, $r = 2$.\nWe can check that only $(p, q, r) = (3, 11, 5)$ works so $K = 12$ and the value of $A$ is $1980$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40372,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{Z}^{+} \\rightarrow \\mathbb{Z}^{+}$ satisfying $f(1)=2$, $f(2) \\neq 4$, and\n$$\n\\max \\{f(m)+f(n), m+n\\} \\mid \\min \\{2m+2n, f(m+n)+1\\}\n$$\nfor all $m, n \\in \\mathbb{Z}^{+}$.",
    "options": [],
    "answer": "f(n) = n + 1 for all n in positive integers",
    "solution": "In given condition, substitute $m=n$, we have\n$$\n\\max \\{2 f(n), 2 n\\} \\mid \\min \\{4 n, f(2 n)+1\\},\n$$\nthus $4 n \\geq 2 f(n)$ or $f(n) \\leq 2 n$.\nContinue substitute $m=1$, we have\n$$\n\\max \\{2+f(n), 1+n\\} \\mid \\min \\{2+2 n, f(1+n)+1\\} \\quad (*)\n$$\nThus $f(n+1)+1 \\geq f(n)+2$ or $f(n+1) \\geq f(n)+1$. Since $f(1)=2$, we have $f(n) \\geq n+1, \\forall n \\in \\mathbb{Z}^{+}$.\nFrom these, we can conclude that $f(2) \\in \\{3 ; 4\\}$. But we have $f(2) \\neq 4$ then $f(2)=3$. We shall prove by induction that $f(n)=n+1, \\forall n \\geq 2$.\nThe conclusion is true for $n=1,2$. Suppose that $f(n)=n+1$ for some $n \\geq 2$.\nBack to (*), note that $\\max \\{2+f(n), 1+n\\}=f(n)+2=n+3$ and $\\frac{2 n+2}{n+3} \\in (1; 2)$ which implies that $n+3 \\nmid 2 n+2$. Hence\n$$\n\\min \\{2+2 n, f(1+n)+1\\}=f(n+1)+1\n$$\nSince $n+2 \\leq f(n+1) \\leq 2 n+2$, we must have $\\frac{f(n+1)+1}{n+3}=1 \\Leftrightarrow f(n+1)=n+2$. By induction hypothesis, we get $f(n)=n+1$.\nCheck condition, for all $m, n \\in \\mathbb{Z}^{+}$ then\n$$\n\\begin{gathered}\n\\min \\{2 m+2 n, f(m+n)+1\\}=\\min \\{2 m+2 n, m+n+2\\}=m+n+2 \\text{ and } \\\\\n\\max \\{f(m)+f(n), m+n\\}=\\max \\{m+1+n+1, m+n\\}=m+n+2\n\\end{gathered}\n$$\nHence, the function $f(n)=n+1$ satisfies the given condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40373,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $M, N, P$ as midpoints of the segments $BC, CA, AB$ respectively. Suppose that $I$ is the intersection of angle bisectors of $\\angle BPM$, $\\angle MNP$ and $J$ is the intersection of angle bisectors of $\\angle CNM$, $\\angle MPN$. Denote ($\\omega_{1}$) as the circle of center $I$ and tangent to $MP$ at $D$, $\\left(\\omega_{2}\\right)$ as the circle of center $J$ and tangent to $MN$ at $E$.\n\n1. Prove that $DE$ is parallel to $BC$.\n\n2. Prove that the radical axis of two circles ($\\omega_{1}$), ($\\omega_{2}$) bisects the segment $DE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1) Note that $\\angle MNC = \\angle MPB = \\angle A$ then by angle chasing, we have $IP \\parallel JN$. Denote $K = PJ \\cap IN$ then $K$ is the incenter of triangle $MNP$. Hence, $MK$ is the angle bisector of $\\angle NMP$, thus $MK \\parallel IP$. Denote $X = IN \\cap MP$ then\n$$\n\\frac{IP}{MK} = \\frac{XP}{XM} = \\frac{NP}{NM}.\n$$\nSimilarly, $\\frac{JN}{MK} = \\frac{NP}{MP}$.\nThus $\\frac{IP}{JN} = \\frac{MP}{MN}$. Since $\\triangle IPD \\sim \\triangle JNE$ then $\\frac{IP}{JN} = \\frac{PD}{NE}$. Therefore, $\\frac{MP}{MN} = \\frac{PD}{NE}$ which implies that $DE \\parallel NP \\parallel BC$.\n\n2) Suppose that $DE$ cuts ($\\omega_{1}$), ($\\omega_{2}$) at $R, S$ respectively. We have\n$$\n\\frac{DS}{ER} = \\frac{ID \\cdot \\cos \\angle IDR}{JE \\cdot \\cos \\angle JES} = \\frac{PD \\cdot \\sin \\angle PDR}{NE \\cdot \\sin \\angle NES} = \\frac{AC \\cdot \\sin C}{AB \\cdot \\sin B} = 1.\n$$\n![](attached_image_1.png)\nHence $DS = ER$. Denote $T$ as midpoint of $DE$ then\n$$\n\\mathscr{P}_{T/(\\omega_{1})} = TD \\cdot TR = TE \\cdot TS = \\mathscr{P}_{T/(\\omega_{2})},\n$$\nwhich implies that $T$ lies on the radical axis of $\\left(\\omega_{1}\\right), \\left(\\omega_{2}\\right)$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40374,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that\n$$\nf\\left(2 x^{3}+f(y)\\right)=y+2 x^{2} f(x),\n$$\nfor all real numbers $x, y$.",
    "options": [],
    "answer": "f(x) = 0; f(x) = x; f(x) = -x",
    "solution": "Denote ($*$) as the given condition. Replacing $x = y = -\\frac{1}{2}$ in ($*$), we get\n$$\nf\\left(2\\left(f\\left(-\\frac{1}{2}\\right)\\right)^{2}-\\frac{1}{2}\\right)=0\n$$\nIf there exists some $a \\neq 0$ such that $f(a)=0$. Substituting $x=a$ in ($*$), we get $f(y)=0$ for any $y$, which is obviously a solution.\n\nLet us consider now the case $f(x) \\neq 0$ for any nonzero real number $x$. Hence, $f(0)=0$ and $f\\left(-\\frac{1}{2}\\right)= \\pm \\frac{1}{2}$. On the other hand, it is easy to see that if $f$ is solution, then $-f$ is also a solution. Therefore, we only need to consider the case $f(0)=0$ and $f\\left(-\\frac{1}{2}\\right)=-\\frac{1}{2}$. From this, we have\n$$\n\\begin{aligned}\nf\\left(-\\frac{1}{2}-f(y)\\right) & =-f(y)-\\frac{1}{2}, \\quad \\forall y \\in \\mathbb{R} \\\\\nf\\left(\\frac{1}{2}\\right) & =\\frac{1}{2}\n\\end{aligned}\n$$\nand\n$$\nf(x-f(x))=f(x)-x, \\quad \\forall x \\in \\mathbb{R}\n$$\nTherefore,\n$$\n\\begin{aligned}\nf\\left(x-\\frac{1}{2}-f(x)\\right) & =f\\left(-\\frac{1}{2}-f(x-f(x))\\right) \\\\\n& =-\\frac{1}{2}-f(x-f(x))=x-\\frac{1}{2}-f(x)\n\\end{aligned}\n$$\nwhich implies $f\\left(-f\\left(\\frac{1}{2}\\right)\\right)=-f\\left(\\frac{1}{2}\\right)$. Now, substituting $x=\\frac{1}{2}$ and $y= -f\\left(\\frac{1}{2}\\right)$ in ($*$), we get\n$$\nf\\left(\\frac{1}{2}-2\\left(f\\left(\\frac{1}{2}\\right)\\right)^{2}\\right)=0 .\n$$\nThus $f\\left(\\frac{1}{2}\\right)= \\pm \\frac{1}{2}$. If $f\\left(\\frac{1}{2}\\right)=-\\frac{1}{2}$, then $\\frac{1}{2}=-f\\left(\\frac{1}{2}\\right)=f\\left(-f\\left(\\frac{1}{2}\\right)\\right)=f\\left(\\frac{1}{2}\\right)= -\\frac{1}{2}$, contradiction. Therefore, $f\\left(\\frac{1}{2}\\right)=\\frac{1}{2}$.\n\nFrom here, it follows that\n$$\nf\\left(f(y)+\\frac{1}{2}\\right)=f(y)+\\frac{1}{2}, \\quad \\forall y \\in \\mathbb{R}\n$$\nIn this equation, substituting $y=x-\\frac{1}{2}-f(x)$, we get\n$$\nf(x-f(x))=x-f(x), \\quad \\forall x \\in \\mathbb{R}\n$$\nComparing this result with $f(x-f(x))=f(x)-x$, we get $f(x)=x$ for any $x$, which is clearly a solution.\n\nIn conclusion, there are three solutions: $f(x)=0$, $f(x)=x$ and $f(x)= -x$ for all real number $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40375,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ and $N$ are two nine-digit palindromes without leading zeros (a palindrome is a number that reads the same backward as forward) such that $M < N$ and there are no other nine-digit palindromes strictly between $M$ and $N$. What are all possible values of $N - M$?",
    "options": [],
    "answer": "10000, 101000, 1000100, 10000010, 100000001",
    "solution": "Denote the number $M$ is $a b c d e d c b a$. We consider cases:\n\n1. If $e \\geq 8$ then the next number $N$ is $a b c d (e+1) d c b a$.\n\n2. If $e = 9$, $d \\geq 8$ then the next number $N$ is $a b c (d+1) 9 (d+1) c b a$.\n\n3. If $e = d = 9$, $c \\leq 8$ then the next number $N$ is $a b (c+1) 999 (c+1) b a$.\n\n4. If $e = d = c = 9$, $b \\leq 8$ then next number is $a (b+1) 99999 (b+1) a$.\n\n5. If $e = d = c = b = 9$, then $a \\leq 8$ (otherwise, there is no $N$). Then the next number is $(a+1) 9999999 (a+1)$.\n\nSo the difference are $10000$, $10100$, $1000100$, $10000010$, $100000001$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40376,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be real numbers such that $a+b+c+ab+bc+ca+abc \\geq 7$. Prove that\n$$\n\\sqrt{a^{2}+b^{2}+2}+\\sqrt{b^{2}+c^{2}+2}+\\sqrt{c^{2}+a^{2}+2} \\geq 6\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, by AM-GM we can show that\n$$\nx^{2}+y^{2}+1 \\geq xy+x+y \\text{ for all } x, y, z \\in \\mathbb{R}.\n$$\nHence, $\\sqrt{a^{2}+b^{2}+2} \\geq \\sqrt{|ab|+|a|+|b|+1} = \\sqrt{(|a|+1)(|b|+1)}$. Construct similar inequalities and take the sum, we get\n$$\n\\begin{aligned}\n& \\sqrt{a^{2}+b^{2}+2}+\\sqrt{b^{2}+c^{2}+2}+\\sqrt{c^{2}+a^{2}+2} \\\\\n& \\geq \\sqrt{(|a|+1)(|b|+1)}+\\sqrt{(|b|+1)(|c|+1)}+\\sqrt{(|c|+1)(|a|+1)}.\n\\end{aligned}\n$$\nBy AM-GM for three numbers, we get\n$$\n\\begin{aligned}\n& \\sqrt{(|a|+1)(|b|+1)}+\\sqrt{(|b|+1)(|c|+1)}+\\sqrt{(|c|+1)(|a|+1)} \\\\\n& \\geq 3 \\sqrt[3]{(|a|+1)(|b|+1)(|c|+1)} \\\\\n& = 3 \\sqrt[3]{|abc|+|ab|+|bc|+|ca|+|a|+|b|+|c|+1} \\\\\n& \\geq 3 \\sqrt[3]{abc+ab+bc+ca+a+b+c+1} \\geq 3 \\sqrt[3]{7+1} = 6\n\\end{aligned}\n$$\nBy combining these two inequalities, we finish the proof. The equality occurs when $a=b=c=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40377,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that 2018 numbers $1$ and $-1$ are written around a circle. For every two adjacent numbers, their product is taken. Suppose that the sum of all 2018 products is negative. Find all possible values of sum of 2018 given numbers.",
    "options": [],
    "answer": "All even integers from -1008 to 1008 inclusive",
    "solution": "First, we show that the sum is at most $1008$ and at least $-1008$.\nIndeed, suppose that there are $a$ numbers $+1$ in $2018$ given numbers and $b$ products equal $-1$. Then the sum of $2018$ given numbers is $a + (-1)(2018 - a) = 2a - 2018$, and sum of $2018$ products is\n$$\n(2018 - b) + (-1) b = 2018 - 2b.\n$$\nSince $2018 - 2b < 0$ we have $b \\geq 1010$. And because each product of $-1$ comes from one $1$ and one $-1$, and one $-1$ can contribute to at most two $-1$ products. Hence the number of $-1$ numbers is at least $1010 / 2 = 505$, or $2018 - a \\geq 505$. Hence $a \\leq 1513$, or $2a - 2018 \\leq 1008$.\nSimilarly, each product of $-1$ comes from one $1$ and one $-1$, and one $1$ can contribute to at most two $-1$ products. Hence the number of $1$ numbers is also at least $505$. Hence, $a \\geq 505$, or $2a - 2018 \\geq -1008$. Combine these results, we showed that the sum is in range $[-1008, 1008]$.\nIt is clear that the sum is even, and we will show that the sum can be any even number in that range. We start from a configuration of $1009$ numbers of $+1$ and $1009$ numbers of $-1$ alternating around the circle. In this case, the sum of $2018$ numbers is $0$ and the sum of the product is $-2018$.\nOn one way, we eventually change one $-1$ to one $1$ exactly $504$ times. Each time, the sum of $2018$ numbers increases by $2$, so the sum will increase from $0$ to $1008$. Besides, the sum of $2018$ products will increase by $4$ (two $-1$ products change to two $1$ products). Hence the sum of $2018$ products is at most $-2018 + 504 \\times 4 < 0$.\nOn another way, we eventually change one $+1$ to one $-1$ exactly $504$ times. By doing these steps, the sum of $2018$ numbers will decrease from $0$ to $-1008$ and the sum of $2018$ products is always negative.\nTherefore, all even numbers from $-1008$ to $1008$ are obtainable.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40378,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an even positive integer. We fill in a number on each cell of a rectangle table of $n$ columns and multiple rows as following:\ni. Each row is assigned to some positive integer $a$ and its cells are filled by $0$ or $a$ (in any order);\nii. The sum of all numbers in each row is $n$.\nNote that we cannot add any more row to the table such that the conditions (i) and (ii) still hold. Prove that if the number of $0$'s on the table is odd then the maximum odd number on the table is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\sigma(n), \\tau(n)$ as the sum of divisors and number of divisors of $n$, respectively.\nConsider a row with assigned number is $a$, and suppose that there are $b$ cells on that row are filled by $a$ then $n = a b$ or $a$ is a divisor of $n$. By condition ii), the number of rows of the table is exactly the number of positive divisors of $n$.\nNumber of nonzero on each row is $\\frac{n}{a}$, thus the number of zero on each row is $n - \\frac{n}{a}$. Number of zero on table is\n$$\n\\sum_{a \\mid n} \\left(n - \\frac{n}{a}\\right) = \\sum_{a \\mid n} (n - a) = n \\cdot \\tau(n) - \\sum_{a \\mid n} a = n \\cdot \\tau(n) - \\sigma(n) .\n$$\nThis number is odd and $n$ is even which imply that $\\sigma(n)$ is odd. Put $n = 2^{k} \\cdot q$ with $q$ is odd number then $q$ is also the maximum odd number on the table. The sum of divisor $\\sigma(n)$ of $n$ is of form $\\prod_{p^{\\alpha} \\| n} \\sigma\\left(p^{\\alpha}\\right)$.\n- If $p = 2$ then $\\sigma\\left(2^{k}\\right)$ is an odd number.\n- If $p > 2$ then $\\sigma\\left(p^{\\alpha}\\right) = 1 + p + \\cdots + p^{\\alpha}$ must be odd, which implies that $\\alpha$ is even.\nFrom this, we can conclude that all odd prime divisors of $n$ have even exponent, thus $q$ is a perfect square.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40379,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle inscribed in circle $(O)$. Let $G$ be a point on the small arc $AC$ of $(O)$ and $(K)$ be a circle passing through $A$ and $G$. Bisector of $\\angle BAC$ cuts $(K)$ again at $P$. The point $E$ is chosen on $(K)$ such that $AE$ is parallel to $BC$. The line $PK$ meets the perpendicular bisector of $BC$ at $F$. Prove that $\\angle EGF = 90^{\\circ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D$ be the second intersection of $AP$ and $(O)$ and $H$ be the intersection of $OD$ and $AE$. Note that $D$ is the midpoint of the minor arc $BC$ of $(O)$, then $OD$ is the perpendicular bisector of $BC$. Since $AE \\parallel BC$, we have $\\angle DHE = 90^{\\circ}$.\n\n![](attached_image_1.png)\n\nOn the other hand,\n$$\n\\angle DOG = 2\\angle DAG = \\angle PKG,\n$$\nwhich implies that two isosceles triangles $KPG$ and $ODG$ are similar. We get $\\angle ODG = \\angle KPG$ therefore $PFGD$ is cyclic.\n\nFrom this, $\\angle AEG = \\angle DPG = \\angle DFG$. This means $HEGF$ is cyclic, we deduce $\\angle EGF = 180^{\\circ} - \\angle EHF = 90^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40380,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two positive integers $m$ and $n$ are called similar if one of them can be obtained from the other one by swapping two digits (note that a 0-digit cannot be swapped with the leading digit). Find the greatest integer $N$ such that $N$ is divisible by 13 and any number similar to $N$ is not divisible by 13.",
    "options": [],
    "answer": "9876230",
    "solution": "Let $k$ be the number of digits of $N$ and let $N = \\sum_{i=0}^{k-1} 10^{i} \\cdot d_{i}$ where $d_{i}$ are digits. If $M$ is obtained by swapping the $i$-th and the $j$-th digits, then\n$$\nM - N = (10^{j} - 10^{i}) \\cdot (d_{i} - d_{j}) .\n$$\nTherefore $13 \\mid M - N$ implies that $d_{i} = d_{j}$ or $6 \\mid i - j$. Hence, a number $N$ satisfies the property given in the problem if and only if $13 \\mid N$, $d_{i} \\neq d_{j}$ for all $i \\neq j$ and no swap between $i$-th and $j$-th digits are possible if $6 \\mid i - j$.\n\n- If $k \\geq 8$, one can swap the 0th and the 6th digits, hence $N$ does not satisfy the property.\n- If $k = 7$, one can swap the 0th and the 6th digits unless the 0th digit is 0, but if the 0th digit is 0, then no swap is possible.\n\nHence, $N$ satisfies the property if and only if $13 \\mid N$, all the digits are distinct and the 0th digit is 0. Therefore, $N \\leq 9876540$.\n\nConsidering all multiples of 13 that end with 0 below this level, one finds $N \\in \\{9876490, 9876360, 9876230, \\ldots\\}$. Since the two greatest elements of this set do not have all distinct digits, the greatest integer $N = 9876230$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40381,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square inscribed in circle $(O)$. Let $P$ be a point lies on minor arc $CD$ of $(O)$. Line $PB$ intersects $AC$ at $E$. Line $PA$ intersects $DB$ at $F$. Circumcircle of triangle $PEF$ cuts $(O)$ again at $Q$. Prove that $PQ$ is parallel to $CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K$ be the circumcenter of triangle $PEF$. Note that\n$$\n\\angle APB = \\angle ADB = 45^\\circ\n$$\nso triangle $KEF$ is right isosceles at $K$. Thus, quadrilateral $OEKF$ is cyclic.\nHence, $\\angle FOK = \\angle FEK = 45^\\circ$, this means $OK$ is bisector of angle $DOC$, it is also perpendicular bisector of $CD$. Since $KP = KQ$, $OP = OQ$, we have $OK \\perp PQ$.\nTherefore, $PQ$ and $CD$ are parallel. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40382,
    "subject": "Mathematics (Multi-modal)",
    "question": "A non-empty subset of $\\{1,2, \\ldots, n\\}$ is called arabic if arithmetic mean of its elements is an integer. Show that the number of arabic subsets of $\\{1,2, \\ldots, n\\}$ has the same parity as $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The solution is based on a simple fact, that if you add the arithmetic mean of the sequence to the sequence, then the arithmetic mean of the sequence will not change, since\n$$\n\\frac{a_{1}+a_{2}+\\cdots+a_{n}}{n}=\\frac{a_{1}+a_{2}+\\cdots+a_{n}+\\frac{a_{1}+a_{2}+\\cdots+a_{n}}{n}}{n+1}\n$$\nDenote by $\\mu(A)$ the arithmetic mean of the elements of $A$. Denote\n$$\nP=\\left\\{A \\subset\\{1,2, \\ldots, n\\} \\quad: \\quad \\mu(A) \\in \\mathbb{Z}_{+}, \\mu(A) \\in A\\right\\}\n$$\nand\n$$\nQ=\\left\\{A \\subset\\{1,2, \\ldots, n\\} \\quad: \\quad \\mu(A) \\in \\mathbb{Z}_{+}, \\mu(A) \\notin A\\right\\} .\n$$\nWe need to prove that $|P|+|Q|$ is even. For that we will prove $|P|=|Q|$. Really, take any set from $P$ and remove its arithmetic mean. We will get an element from $Q$. Take any set from $Q$ and add its arithmetic mean. We will get an element from $P$. Since arithmetic mean of the set is defined uniquely, so we have bijection between $P$ and $Q$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40383,
    "subject": "Mathematics (Multi-modal)",
    "question": "1. For integer $n \\geq 3$, suppose that $0 < a_{1} < a_{2} < \\ldots < a_{n}$ is an arithmetic sequence and $0 < b_{1} < b_{2} < \\ldots < b_{n}$ is a geometric sequence with $a_{1} = b_{1}$, $a_{n} = b_{n}$. Prove that $a_{k} > b_{k}$ for all $k = 2, 3, \\ldots, n-1$.\n\n2. Prove that for every positive integer $n \\geq 3$, there exist an integer arithmetic sequence ($a_{n}$) and an integer geometric sequence ($b_{n}$) such that\n$$\n0 < b_{1} < a_{1} < b_{2} < a_{2} < \\cdots < b_{n} < a_{n}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1) Put $a_{1} = b_{1} = x$ and $a_{n} = b_{n} = y$, then $0 < x < y$. Denote $d$ as the difference and the ratio of two consecutive terms of $(a_{n})$, $(b_{n})$ respectively.\nWe have $a_{n} = a_{1} + (n-1)d$ and $a_{k} = a_{1} + (k-1)d$ for any $k \\in \\{2, 3, \\ldots, n-1\\}$. Hence,\n$$\na_{k} = \\frac{(n-k)a_{1} + (k-1)a_{n}}{n-1} = \\frac{(n-k)x + (k-1)y}{n-1}.\n$$\nSimilarly, $b_{k} = \\sqrt[n-1]{x^{n-k} y^{k-1}}$ so we need to prove that\n$$\n\\frac{(n-k)x + (k-1)y}{n-1} > \\sqrt[n-1]{x^{n-k} y^{k-1}},\n$$\nwhich is true by AM-GM inequality.\n\n2) With $2^{k} \\leq m$, consider some estimations as follow\n$$\n\\left(1 + \\frac{1}{m}\\right)^{k} = 1 + \\frac{\\binom{k}{1}}{m} + \\frac{\\binom{k}{2}}{m^{2}} + \\cdots \\geq 1 + \\frac{k}{m}\n$$\nand\n$$\n\\begin{aligned}\n\\left(1 + \\frac{1}{m}\\right)^{k} & = 1 + \\frac{\\binom{k}{1}}{m} + \\frac{\\binom{k}{2}}{m^{2}} + \\cdots \\leq 1 + \\frac{\\binom{k}{1}}{m} + \\frac{\\binom{k}{2} + \\cdots + \\binom{k}{k}}{m} \\\\\n& < 1 + \\frac{k}{m} + \\frac{2^{k}}{m^{2}} \\leq 1 + \\frac{k+1}{m}\n\\end{aligned}\n$$\nthen $1 + \\frac{k}{m} < \\left(1 + \\frac{1}{m}\\right)^{k} < 1 + \\frac{k+1}{m}$ or $m + k < m\\left(1 + \\frac{1}{m}\\right)^{k} < m + k + 1$. This is true for all $k \\leq \\log_{2} m$ so\n$$\nm\\left(1 + \\frac{1}{m}\\right) < m + 2 < m\\left(1 + \\frac{1}{m}\\right)^{2} < m + 3 < \\cdots < m + k + 1.\n$$\nTo make all these numbers are integers, we multiply them by $m^{[\\log_{2} m]}$. Finally, take $\\log_{2} m > n$, the result will follow. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40384,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $C$ be a point lies outside the circle $(O)$ and $CS, CT$ are tangent lines of $(O)$. Take two points $A, B$ on $(O)$ with $M$ is the midpoint of the minor $\\operatorname{arc} AB$ such that $A, B, M$ differ from $S, T$. Suppose that $MS, MT$ cut line $AB$ at $E, F$. Take $X \\in OS$ and $Y \\in OT$ such that $EX, FY$ are perpendicular to $AB$. Prove that $XY$ and $CM$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, note that $OM \\perp AB$ then $OM \\parallel XE$. But $OMS$ is isosceles triangle implies that triangle $XES$ is also isosceles, or $XE = XS$. Similarly, $YE = YT$. Denote $\\left(\\omega_{1}\\right), \\left(\\omega_{2}\\right)$ as the circle of center $X$, radius $XS$ and center $Y$, radius $YT$. Since $CS \\perp XS$, we have $CS$ is tangent to $\\left(\\omega_{1}\\right)$ so $\\mathscr{P}_{C /\\left(\\omega_{1}\\right)} = CS^{2}$. On the other hand, $\\mathscr{P}_{C /\\left(\\omega_{2}\\right)} = CT^{2}$ and $CS = CT$ imply that $C$ belongs to radical axis of two circles $\\left(\\omega_{1}\\right), \\left(\\omega_{2}\\right)$.\n\nBy similar triangles, we get $MA^{2} = MS \\cdot ME$ and $MB^{2} = MF \\cdot MT$, but $MA = MB$ then $MS \\cdot ME = MF \\cdot MT$, thus $\\mathscr{P}_{M /\\left(\\omega_{1}\\right)} = \\mathscr{P}_{M /\\left(\\omega_{2}\\right)}$. By combining these results, we obtain $CM$ is the radical axis of two circles $\\left(\\omega_{1}\\right), \\left(\\omega_{2}\\right)$; hence, $CM \\perp XY$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40385,
    "subject": "Mathematics (Multi-modal)",
    "question": "The $n \\times n \\times n$ cubic is constructed from $n^{3}$ unit cubic $1 \\times 1 \\times 1$ such that at least one of unit cubic is black. Show that we can always cut the $n \\times n \\times n$ cubic into rectangular boxes such that each box contains exactly one black unit cubic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We shall prove the problem also true for all brick $a \\times b \\times c$ for any positive integers $a, b, c$ by induction on the number of black boxes.\n\nFirst, if the brick contains only one black cubic then no any cut is needed. Suppose that it contains at least two black cubics.\n\nWe choose a plane that divide the brick into two smaller bricks such that each smaller brick contains exactly one black cubic.\n\nBy induction hypothesis, we can divide each smaller bricks to make all black cubics are separated, then we are done!\n\nRemark. This problem is easy as long as we notice that the problem is not about \"big cubic\", it's true for any arbitrary brick.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40386,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that $\\varphi(n)$ is a divisor of $n^{2}+3$.",
    "options": [],
    "answer": "{1, 2, 3, 5, 9, 21}",
    "solution": "First assume $n$ is a prime number. Thus $\\varphi(n)=n-1 \\mid n^{2}+3=(n-1)(n+1)+4$, implying $n-1 \\mid 4$. We deduce that $n=2,3$ or $5$.\n\nFrom now on, assume $n$ is composite and set $n=\\prod_{i=1}^{k} p_{i}^{\\alpha_{i}}$, where $k$ is the number of distinct prime divisors of $n$. Since $n \\geq 3$, $\\varphi(n)$ is even. In particular, the condition $\\varphi(n) \\mid n^{2}+3$ shows that $n$ is odd and so the $p_{i}$'s are all odd. It follows that\n$$\n2^{k} \\mid \\varphi(n)=\\prod_{i=1}^{k} p_{i}^{\\alpha_{i}-1}(p_{i}-1).\n$$\nSo $n^{2}+3$ is divisible by $2^{k}$ implying that $k \\leq 2$. Moreover, if $\\alpha_{i} \\geq 2$ then $p_{i} \\mid \\varphi(n)$, so $p_{i} \\mid n^{2}+3$, forcing $p_{i}=3$. In this case, $9 \\nmid n^{2}+3$ and $9 \\nmid \\varphi(n)$. This shows that $\\alpha_{i}=2$. Our discussion shows that if the composite $n \\geq 3$ satisfies the problem then $n=3^{2}=9$ or $n=9p$, or $n=pq$ for some odd primes $p, q$. Of course, $n=9$ is a solution of our problem.\n\n- Suppose $n=9p$. Then $\\varphi(n)=6(p-1)$ and $n^{2}+3=81(p-1)(p+1)+84$. Therefore, $3(p-1) \\mid 84$, or equivalently $p-1 \\mid 28$. It follows that $p=5$ or $p=29$. But in both cases, we have $8 \\mid \\varphi(n)$, contradiction.\n\n- Suppose $n=pq$, where $p, q$ are odd primes. First, consider the case $q=3$ (the case $p=3$ is of course similar). Then $\\varphi(n)=2(p-1)$ so $n^{2}+3=9(p-1)(p+1)+12$. It follows that $2(p-1) \\mid 12$, and $p-1 \\mid 6$, implying $p=3$ or $p=7$. In this case we obtain a new solution $n=pq=3 \\times 7=21$ of our problem.\n\nNow, suppose $p, q>3$. Thus, $3 \\nmid n^{2}+3$ so $3 \\nmid (p-1)(q-1)$. This shows that $p, q \\equiv 2 \\pmod{3}$. Since $(p-1)(q-1) \\mid p^{2}q^{2}+3=(p^{2}-1)(q^{2}-1)+p^{2}+q^{2}+2$ so $(p-1)(q-1) \\mid p^{2}+q^{2}+2$. Set\n$$\np^{2}+q^{2}+2=k(p-1)(q-1),\n$$\nwhere $k$ is a positive integer. Set $p=2x+1$, $q=2y+1$, then the above equation reads $x^{2}+y^{2}+x+y+1=kxy$, or equivalently,\n$$\nx^{2}-(ky-1)x+y^{2}+y+1=0.\n$$\nBy a Vieta jumping technique, we deduce that $k=5$. Thus, we have\n$$\n5(p-1)(q-1)=p^{2}+q^{2}+2.\n$$\nBut then $p \\equiv q \\equiv 2 \\pmod{3}$ and the left hand side is $\\equiv 2 \\pmod{3}$, while the right hand side is $\\equiv 1 \\pmod{3}$, contradiction.\n\nTherefore, all numbers satisfy the given condition are $n \\in \\{1,2,3,5,9,21\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40387,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number of the form $9k+1$. Show that there exists an integer $n$ such that $p \\mid n^{3}-3n+1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that the existence of an govern integer as described in the problem can be equivalently stated as follows: the polynomial $x^{3}-3x+1$ has a root in $\\mathbb{F}_{p}$. Following the classical method for solving cubic equations, we set $x = w + \\frac{1}{w}$. Then\n$$\nx^{3}-3x+1 = \\left(w + \\frac{1}{w}\\right)^{3} - 3\\left(w + \\frac{1}{w}\\right) + 1 = w^{3} + \\frac{1}{w^{3}} + 1\n$$\nWe then observe that the solutions of the equation $\\lambda + \\frac{1}{\\lambda} + 1 = 0$ are cubic roots of unity. If $p = 9k + 1$ and $g$ is a primitive root $(\\bmod\\ p)$, then $g^{3k}$ is a cubic root of unity in $\\mathbb{F}_{p}$. Thus we look for $w$ such that $w^{3} = g^{3k}$, so we let $w = g^{k}$.\nIt is now easy to verify that $x = w + \\frac{1}{w} = g^{k} + g^{8k}$ is a root of the original cubic polynomial.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40388,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral inscribed in circle $(O)$ such that $DB = DA + DC$. The point $P$ lies on the ray $AC$ such that $AP = BC$. The point $E$ is on $(O)$ such that $BE \\perp AD$. Prove that $DP$ is parallel to the angle bisector of $\\angle BEC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $R, F$ as the intersections of $AC$ with the angle bisector of $\\angle BEC$ and $BE$, respectively. Let $Q$ be a point on $BD$ such that $DQ = DC$, hence $QB = DB - DQ = DB - DC = AD$.\n\n![](attached_image_1.png)\n\nThus two triangles $CQB$ and $PDA$ are congruent (s.a.s). Note that $AD \\perp BE$ then $\\angle CAD + \\angle AFE = 90^\\circ$. From these, we have\n\n$$\n\\begin{aligned}\n\\angle APD & = \\angle BCQ = \\angle CQD - \\angle CBD \\\\\n& = 90^\\circ - \\frac{1}{2} \\angle BDC - \\angle CAD = \\left(90^\\circ - \\angle CAD\\right) - \\frac{1}{2} \\angle BEC \\\\\n& = \\angle AFE - \\angle BER = \\angle ARE .\n\\end{aligned}\n$$\n\nHence, we conclude that $DP \\parallel ER$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40389,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $k$ such that there exists some permutation of $(1,2, \\ldots, 1000)$ namely $\\left(a_{1}, a_{2}, \\ldots, a_{1000}\\right)$ and satisfy $\\left|a_{i}-i\\right|= k$ for all $i=1,1000$.",
    "options": [],
    "answer": "all positive divisors of 500",
    "solution": "Note that $k=1$ is an answer since we can choose a permutation with $a_{2i-1}=2i$, $a_{2i}=2i-1$.\n\nNow we take $k>1$, denote $S_{0}, S_{1}, S_{2}, \\ldots, S_{k-1}$ as the subsets of $\\{1,2,3, \\ldots, 1000\\}$ and all elements of $S_{i}$ are congruent to $i$ modulo $k$. We have:\n- The number of elements of $S_{t}$ is $m_{t}=\\left\\lceil\\frac{1000-t}{k}\\right\\rceil$.\n- Since $a_{i}=i \\pm k$ and $a_{i} \\equiv i \\pmod{k}$, two numbers $a_{i}$ and $i$ belong to same subset.\n\nDenote $t=x_{1}<x_{2}<\\ldots<x_{m_{t}}$ as elements of $S_{t}$ then they form an arithmetic progression with formula $x_{i}=t+(i-1)k$ for $1 \\leq i \\leq m_{t}$.\n\nPut $y_{i}=a_{x_{i}}$ for $1 \\leq i \\leq m_{t}$ then $\\left|y_{i}-x_{i}\\right|=\\left|a_{x_{i}}-x_{i}\\right|=k$ which implies that $y_{1}<y_{2}<\\ldots<y_{m_{t}}$ is a permutation of $x_{1}<x_{2}<\\ldots<x_{m_{t}}$.\n\nSet $r_{i}=\\frac{x_{i}-t}{k}+1$, $s_{i}=\\frac{y_{i}-t}{k}+1$ for $1 \\leq i \\leq m_{t}$ then $r_{i}=i$ and $\\left|s_{i}-i\\right|=1$.\n\nWe have\n$$\n\\sum_{i=1}^{m_{t}}\\left|r_{i}-s_{i}\\right|+\\sum_{i=1}^{m_{t}}\\left|r_{i}+s_{i}\\right|=2 \\sum_{i=1}^{m_{t}} \\max \\{r_{i}, s_{i}\\}\n$$\nso $m_{t}+2 \\sum_{i=1}^{m_{t}} r_{i}=2 \\sum_{i=1}^{m_{t}} \\max \\{r_{i}, s_{i}\\}$ and $m_{t}$ is even. By the similar definition for $i=0,1, \\ldots, k-1$, we have $m_{0} \\leq m_{1} \\leq \\ldots \\leq m_{k-1}$ are all even and\n$$\nm_{0}-m_{k-1}=\\left[\\frac{1000}{k}\\right]-\\left[\\frac{1000-(k-1)}{k}\\right] \\leq 1 .\n$$\nHence, $m_{0}, m_{1}, \\ldots, m_{k-1}$ are all equal and even. This can happen when $k \\mid 1000$ and $\\frac{1000}{k}$ is an even number. It is easy to check that we can construct such a permutation satisfying these $k$.\n\nTherefore, $k$ is a divisor of $500$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 40390,
    "subject": "Mathematics (Multi-modal)",
    "question": "The partition of $2n$ positive integers into $n$ pairs is called square-free if the product of numbers in each pair is not a perfect square. Prove that if for $2n$ distinct positive integers, there exists one square-free partition, then there exists at least $n!$ square-free partitions.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For convenience, let us call integers that are not perfect squares \"good\" and integers that are perfect squares \"bad\". We shall prove the given estimation by induction.\n\nIndeed,\n\nFor $n=1$ we only have 1 square-free partition.\n\nFor $n=2$ we have four numbers $x_{1}, x_{2}, y_{1}, y_{2}$ with $x_{1}y_{1}$ and $x_{2}y_{2}$ are good. Since $x_{1}^{2} = \\frac{x_{1}x_{2} \\cdot x_{1}y_{2}}{x_{2}y_{2}}$ and the number $x_{2}y_{2}$ is good, then two numbers $x_{1}x_{2}, x_{1}y_{2}$ cannot be both bad. We have some cases:\n\n- If $x_{1}x_{2}, x_{1}y_{2}$ are both good, then we have two subcases: if $x_{2}y_{1}$ is good, then we are done; otherwise, consider the ratio $y_{1}y_{2} = \\frac{x_{2}y_{1} \\cdot x_{2}y_{2}}{x_{2}^{2}}$, then this number must be good, so the other partition $\\left(x_{1}, x_{2}\\right), \\left(y_{1}, y_{2}\\right)$ is square-free.\n\n- Otherwise, assume that $x_{1}x_{2}$ is bad and $x_{1}y_{2}$ is good. By a similar argument, we have the partition $\\left(x_{1}, y_{2}\\right), \\left(x_{2}, y_{1}\\right)$.\n\nSo in all cases, we always have one more partition, the statement is true for $n=2$.\n\nSuppose that for some $n \\geq 2$, the statement is true. Consider $2n+2$ numbers with one square-free partition\n$$\n\\left(x_{1}, y_{1}\\right), \\left(x_{2}, y_{2}\\right), \\ldots, \\left(x_{n+1}, y_{n+1}\\right),\n$$\nFirst, fix the pair $(x_{1}, y_{1})$ then apply the induction hypothesis for the other $n$ pairs, we have $n!$ square-free partitions. After that, take some index $2 \\leq k \\leq n+1$, then change four numbers $x_{1}, y_{1}, x_{k}, y_{k}$ to another partition; so for $2(n-1)$ other pairs, we have at least $(n-1)!$ square-free partitions. In total, we get at least\n$$\nn! + n \\cdot (n-1)! = (n+1)! \\text{ square-free partitions. }\n$$\nHence, the statement is also true for $n+1$. This finishes our proof.\n\nRemark. We can show an example that satisfies the equality as: $p, p^{2}, p^{3}, \\ldots, p^{n}$. We just can make pairs of different parity exponents. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40391,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that in every triangle, there are two sides $x, y$ such that\n$$\n\\frac{\\sqrt{5}-1}{2} \\leq \\frac{x}{y} \\leq \\frac{\\sqrt{5}+1}{2} .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a, b, c$ be the side lengths of the given triangle and we may assume that $a \\geq b \\geq c$. Denote $m=\\frac{a}{b}$ and $n=\\frac{b}{c}$. Note that\n$$\nm, n \\geq 1 > \\frac{\\sqrt{5}-1}{2} .\n$$\nWe will show that at least one of $m, n$ is less than or equal to $\\frac{\\sqrt{5}+1}{2}$. Assume that both $m, n > \\frac{\\sqrt{5}+1}{2}$. Since $b + c > a$, it follows that $n + 1 > m n$. Hence,\n$$\nn + 1 > m n > \\frac{\\sqrt{5}+1}{2} n, \\text{ or } n < \\frac{\\sqrt{5}+1}{2},\n$$\nwhich is a contradiction. Therefore, one of two numbers $m, n$ is less than or equal $\\frac{\\sqrt{5}+1}{2}$, this finishes our proof. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40392,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle inscribed in circle $(O)$ with incenter $I$. The lines $IB$ and $IC$ intersect $(O)$ again at $J$ and $L$. Circumcircle $(\\omega)$ of triangle $IBC$ meets $CA$, $AB$ again at $E$, $F$. Prove that $EL$ and $FJ$ intersect on $(\\omega)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $X$ as the other intersection of $(IJL)$ and $(\\omega)$. We shall prove that $X$ is the intersection of $EL$ and $FJ$.\n\nBy the cyclic quadrilateral, we have\n$$\n\\angle LXI = \\angle LJI = \\angle LJB = \\angle LCB = \\angle LCA = 180^\\circ - \\angle EXI.\n$$\nHence $L$, $X$, $E$ are collinear, which means $X$ belongs to $EL$.\n\nSimilarly, $JF$ passes through $X$. These finish our proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40393,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\nf(x)=\\left(x-F_{1}\\right)\\left(x-F_{2}\\right) \\ldots\\left(x-F_{3030}\\right)\n$$\nwith $\\left(F_{n}\\right)$ is the Fibonacci sequence, which defined as $F_{1}=1, F_{2}=2, F_{n+2}= F_{n+1}+F_{n}, n \\geq 1$. Suppose that on the range $\\left(F_{1}, F_{3030}\\right)$, the function $|f(x)|$ takes on the maximum value at $x=x_{0}$. Prove that $x_{0}>2^{2018}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove that $x_{0} \\in\\left(F_{3029}, F_{3030}\\right)$ by showing that for all $x^{*} \\in\\left(F_{1}, F_{3029}\\right]$, there is some $x^{**} \\in\\left(F_{3029}, F_{3030}\\right)$ for which $\\left|f\\left(x^{**}\\right)\\right|>\\left|f\\left(x^{*}\\right)\\right|$.\n\nIndeed, if\n$$\nx^{*} \\in\\left\\{F_{1}, F_{2}, \\ldots, F_{3029}\\right\\}\n$$\nthen $\\left|f\\left(x^{*}\\right)\\right|=0$ which is obvious.\n\nSuppose that $x^{*} \\in\\left(F_{k}, F_{k+1}\\right)$ for $1 \\leq k \\leq 3028$ then put $m=x^{*}-F_{k}$, we choose $x^{**}=F_{3030}-m$. We need\n$$\n\\left|f\\left(x^{**}\\right)\\right|>\\left|f\\left(x^{*}\\right)\\right| \\Leftrightarrow\\left|\\prod_{i=1}^{3030}\\left(x^{**}-F_{i}\\right)\\right|>\\left|\\prod_{i=1}^{3030}\\left(x^{*}-F_{i}\\right)\\right|.\n$$\nEach side has 3030 positive factors then we will make pair one of the left and one of the right such that the value of the right is bigger (except the case $i=k$ then $\\left|x^{*}-F_{k}\\right|=m=\\left|x^{**}-F_{3030}\\right|$ ). For details:\n\n1. If $i=1,2, \\ldots, k$, it is clearly that $x^{**}-F_{i}>x^{*}-F_{i}>0$.\n\n2. If $1 \\leq i \\leq 3030-k$, we have $\\left|x^{*}-F_{k+i}\\right|<\\left|x^{**}-F_{3030-i}\\right|$. Note that we just consider the separated ranges, i.e. $k+i \\leq 3030-i$. Otherwise, some ranges are overlap then we remove that part, then\n$$\n\\begin{aligned}\n& F_{k+i}-x^{*}=\\left(F_{k+1}-x^{*}\\right)+\\left(F_{k+2}-F_{k+1}\\right)+\\cdots+\\left(F_{k+i}-F_{k+i-1}\\right) \\\\\n& <\\left(x^{**}-F_{3029}\\right)+\\left(F_{3029}-F_{3028}\\right)+\\cdots+\\left(F_{3030-(i-1)}-F_{3030-i}\\right) \\\\\n& =x^{**}-F_{3030-i}\n\\end{aligned}\n$$\nHence, the statement is proved. From here, we have\n$$\n\\begin{aligned}\nx_{0} & >F_{3029}=\\frac{1}{\\sqrt{5}}\\left[\\left(\\frac{1+\\sqrt{5}}{2}\\right)^{3030}-\\left(\\frac{1-\\sqrt{5}}{2}\\right)^{3030}\\right] \\\\\n& >\\frac{1}{\\sqrt{5}}\\left(\\frac{1+\\sqrt{5}}{2}\\right)^{3029}>\\left(\\frac{1+\\sqrt{5}}{2}\\right)^{3027} .\n\\end{aligned}\n$$\nIt is easy to check that $\\frac{1+\\sqrt{5}}{2}>2^{\\frac{2}{3}}$ then substitute into the above inequality, we get $x_{0}>2^{2018}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40394,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I, O$ be the incenter, circumcenter of triangle $ABC$ and $A_{1}, B_{1}, C_{1}$ be arbitrary points on the segments $AI, BI, CI$ respectively. The perpendicular bisectors of $AA_{1}, BB_{1}, CC_{1}$ intersect each other at $X, Y$ and $Z$. Prove that the circumcenter of triangle $XYZ$ coincides with $O$ if and only if $I$ is the orthocenter of triangle $A_{1}B_{1}C_{1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\angle B = \\beta$ then we have\n$$\n\\angle XYZ = 180^\\circ - \\angle AIC = 180^\\circ - \\left(90^\\circ + \\frac{\\beta}{2}\\right) = 90^\\circ - \\frac{\\beta}{2}.\n$$\nSo if $O$ is the circumcenter of $XYZ$ then\n$$\n\\angle OXZ = 90^\\circ - \\left(90^\\circ - \\frac{\\beta}{2}\\right) = \\frac{\\beta}{2} = \\angle IBC,\n$$\nwhich implies that $OX \\perp BC$. Similarly, we also have $OY \\perp CA$, $OZ \\perp AB$.\n\nDenote $M, N$ as the midpoints of $BC, BB_{1}$ respectively. From the cyclic quadrilateral and parallel line, we have $\\angle BB_{1}C = \\angle BNM = 180^\\circ - \\angle BXM$. Similarly, $\\angle BC_{1}C = 180^\\circ - \\angle CXM$. Since $MX \\perp BC$, triangle $XBC$ is isosceles, then $\\angle BXM = \\angle CXM$, thus $\\angle BB_{1}C = \\angle BC_{1}C$, which implies that $BCC_{1}B_{1}$ is cyclic.\n\nAngle chasing again, we have $\\angle IC_{1}B_{1} = \\angle CBB_{1} = \\frac{\\beta}{2}$, but $\\angle AIC = 90^\\circ + \\frac{\\beta}{2}$, implies that $AI \\perp B_{1}C_{1}$. By similar way, we get $BI \\perp C_{1}A_{1}$ and $CI \\perp A_{1}B_{1}$, hence $I$ is the orthocenter of triangle $A_{1}B_{1}C_{1}$.\n\nIt is easy to check that these conditions are equivalent, which finishes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40395,
    "subject": "Mathematics (Multi-modal)",
    "question": "In each of the cells of a $13 \\times 13$ board is written an integer such that the integers in adjacent cells differ by $1$. If there are two $2$s and two $24$s on this board, how many $13$s can there be?",
    "options": [],
    "answer": "13",
    "solution": "Let us define the distance between any two cells of the board to be the minimum number of steps required to get from one of the cells to the other one provided that one moves between adjacent cells in each step. Consequently, the distance of any cell to itself is $0$, the distance between adjacent cells is $1$ and the largest distance on the $13 \\times 13$ board (that between two opposite corners) is $24$. It is easy to observe that the numbers written on any two cells cannot differ by more than the distance between the cells.\n\nLet us now assign coordinates to the cells of the board in the usual way: each cell is denoted $(i, j)$ where $i, j \\in \\{0,1, \\ldots, 12\\}$. Let us also denote the number on the cell $(i, j)$ by $x_{(i, j)}$. Thus the inequality mentioned above is expressed as\n$$\n\\left|x_{\\left(i_{1}, j_{1}\\right)}-x_{\\left(i_{2}, j_{2}\\right)}\\right| \\leq \\left|i_{1}-i_{2}\\right|+\\left|j_{1}-j_{2}\\right|.\n$$\nNow there must be a distance of at least $22$ between a $2$ and a $24$ and this places a strong restriction on the possible locations of $2$s and $24$s. Namely, if $x_{(i, j)} \\in \\{2, 24\\}$ then we have\n$$\n\\min(i, 12-i) + \\min(j, 12-j) \\leq 2.\n$$\nThis roughly means that $2$s and $24$s can be found only very near the corners. Without loss of generality (by symmetry of the board), let us assume that there exists a cell $(i, j)$ with $x_{(i, j)} = 2$ and $i + j \\leq 2$. Now, $x_{(i, j)} = 24$ implies that $i + j \\geq 22$, that is, both of the $24$s must be near the opposite corner. Hence $x_{(i, j)} = 2$ thus $i + j \\leq 2$, that is, both of the $2$s must be near the same corner.\n\nIf there exists a cell $(i, j)$ with $x_{(i, j)} = 2$ and $i + j = 2$, then $x_{(i, j)} = 24$ then $i + j \\geq 24$, so $i = j = 12$. Thus there cannot be two distinct $24$s on the board. We therefore conclude that\n$$\nx_{(i, j)} = 2 \\Rightarrow i + j \\leq 1 \\Rightarrow (i, j) \\in \\{(0,0), (1,0), (0,1)\\}.\n$$\nSimilarly,\n$$\nx_{(i, j)} = 24 \\Rightarrow i + j \\geq 23 \\Rightarrow (i, j) \\in \\{(12,12), (11,12), (12,11)\\}.\n$$\nIf $x_{(0,0)} = 2$, then $x_{(1,0)} \\neq 2$ and $x_{(0,1)} \\neq 2$, thus there cannot be two $2$s on the board. We therefore conclude that\n$$\nx_{(0,0)} \\neq 2,\\ x_{(1,0)} = x_{(0,1)} = 2\n$$\nand similarly\n$$\nx_{(12,12)} \\neq 24,\\ x_{(11,12)} = x_{(12,11)} = 24.\n$$\nNow, if we consider a path (a sequence of adjacent cells) of length $22$ between the cells $(1,0)$ and $(11,12)$ or between the cells $(0,1)$ and $(12,11)$, we see that the numbers on the cells on this path are uniquely determined. Since any cell of the board except $(0,0)$ and $(12,12)$ can be placed on such a path, almost the whole board is uniquely determined. We obtain the following formula:\n$$\n1 \\leq i + j \\leq 23 \\Rightarrow x_{(i, j)} = i + j + 1\n$$\nFurthermore, $x_{(0,0)} \\in \\{1,3\\}$ and $x_{(12,12)} \\in \\{23,25\\}$, therefore $x_{(i, j)} = 13 \\Leftrightarrow i + j = 12$. Hence there are exactly thirteen $13$s on the board.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40396,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with $O, H$ are its circumcenter, orthocenter. Take a point $M$ belongs to the minor $\\operatorname{arc} BC$ of $(O)$ (not coincide to $B, C$) and denote $D, E, F$ as reflection of $M$ through $OA, OB, OC$. Suppose that $BF$ meets $CE$ at $K$ and let $I$ be the incenter of triangle $DEF$.\n\n1. Prove that perpendicular bisectors of $EF$ and $IK$ meet on circle $(O)$.\n\n2. Prove that three points $H, K, I$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1) By definition of $E, F$ we can see that $EC$ is the angle bisector of $\\angle MEF$ and $FB$ is the angle bisector of $\\angle MFE$. This implies that $K$ is the incenter of triangle $MEF$.\nHence, $MK$ is the angle bisector of $\\angle EMF$ which passes through the midpoint $N$ of the $\\operatorname{arc} EF$ of $(O)$.\nSo by applying the well known property, we have $NE = NK = NI = NF$.\n![](attached_image_1.png)\nThis implies that $E, F, K, I$ are concyclic and two perpendicular bisectors of $EF, IK$ pass through the point $N \\in (O)$.\n\n2) We have $BK = BM = BE$ and $CK = CM = CE$. Hence, $K$ is the reflection of $M$ through $BC$. By similarly way, take $CD \\cap AF = L$ then $L$ is the reflection of $M$ through $AC$ and $ML, EI$ intersect at midpoint $P$ of the arc $DF$ of $(O)$.\nConsider the hexagon $CDPNEM$ and apply Pascal's theorem, we can see that three intersections\n$$\nL = CD \\cap MP, \\quad I = DN \\cap PE, \\quad K = CE \\cap MN \\text{ are collinear. }\n$$\nWe also know that $KL$ is the Steiner's line of $M$ respect to $(O)$ then $KL$ passes through orthocenter $H$ of triangle $ABC$. This implies that $H, I, K$ are collinear. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40397,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geq 2$ be a positive integer. A subset of positive integers $S$ is called comprehensive if for every integer $0 \\leq x < n$, there is a subset of $S$ whose sum of elements has remainder $x$ when divided by $n$. Note that the empty set has sum $0$. Show that if a set $S$ is comprehensive then there is some (not necessarily proper) subset of $S$ with at most $n-1$ elements which is also comprehensive.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will show that if $|S| \\geq n$, we can remove one element from $S$ and still have a comprehensive set. Doing this repeatedly will always allow us to find a comprehensive subset of size at most $n-1$.\n\nDenote $S = \\{s_1, s_2, \\ldots, s_k\\}$ for some $k \\geq n$. Now start with the empty set and add in the elements $s_i$ in order. During this process, we will keep track of all possible remainders of sums of any subset.\n\nIf $T$ is the set of current remainders at any time, and we add an element $s_i$, the set of remainders will be $T$ and $\\{t + s_i \\mid t \\in T\\}$. In particular, the set of remainders only depends on the previous set of remainders and the element we add in.\n\nAt the beginning of our process, the set of possible remainders is $\\{0\\}$ for the empty set. Since we assumed that $S$ is comprehensive, the final set is $\\{0, 1, \\ldots, n-1\\}$. The number of elements changes from $1$ to $n-1$.\n\nHowever, since we added $k \\geq n$ elements, at least one element did not change the size of our remainder set. This implies that adding this element did not contribute to making any new remainders and $S$ is still comprehensive without this element, proving our claim. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40398,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ such that\n$$\nx f(2 f(y)-x)+y^{2} f(2 x-f(y))=\\frac{(f(x))^{2}}{x}+f(y f(y)),\n$$\nfor all $x, y \\in \\mathbb{Z}, x \\neq 0$.",
    "options": [],
    "answer": "f(x) = 0 for all x ∈ ℤ; f(x) = x^2 for all x ∈ ℤ",
    "solution": "Let $a=f(0)$ and denote ($*$) as the given condition. Assume that $a \\neq 0$, substituting $x=2a$ and $y=0$ in ($*$), we get\n$$\n2a^{2}=\\frac{(f(2a))^{2}}{2a}+a\n$$\nor\n$$\n4a-2=\\left(\\frac{f(2a)}{2a}\\right)^{2}.\n$$\nThis follows that $4a-2$ is a perfect square which is a contradiction since there is no perfect square congruent to $2$ modulo $4$. Thus $a=0$, or $f(0)=0$. Now, substituting $y=0$ in ($*$), we get\n$$\nx^{2} f(-x)=(f(x))^{2}, \\quad \\forall x \\in \\mathbb{Z}, x \\neq 0\n$$\nIt follows that\n$$\nx^{2} f(x)=(f(-x))^{2}, \\quad \\forall x \\in \\mathbb{Z}, x \\neq 0\n$$\nTherefore\n$$\n(f(x))^{4}=x^{4}(f(-x))^{2}=x^{6} f(x), \\quad \\forall x \\in \\mathbb{Z}, x \\neq 0\n$$\nFrom this and $f(0)=0$, we get $f(x)=0$ or $f(x)=x^{2}$ for any integer $x$. Clearly, the function $f(x)=x^{2}$ satisfies the original equation. Let us consider now the case there exists some integer $c \\neq 0$ such that $f(c)=0$. Replacing $y=c$ in ($*$), we get\n$$\nc^{2} f(2x)=0, \\quad \\forall x \\in \\mathbb{Z}, x \\neq 0\n$$\nThus, for any even integer $x$, we have $f(x)=0$. Now, if there exists some odd number $d$ such that $f(d)=d^{2}$, taking $x$ even and $y=d$ in ($*$), we get\n$$\nd^{2} f\\left(2x-d^{2}\\right)=f\\left(d^{3}\\right), \\quad \\forall x \\in \\mathbb{Z}, x \\neq 0, 2 \\mid x\n$$\nIf $f\\left(d^{3}\\right) \\neq 0$, then $f\\left(2x-d^{2}\\right) \\neq 0$ and hence $f\\left(d^{3}\\right)=d^{6}, f\\left(2x-d^{2}\\right)= (2x-d^{2})^{2}$. Plug back into the above equation, we get a contradiction. Therefore, $f\\left(d^{3}\\right)=0$ and hence $f\\left(2x-d^{2}\\right)=0$ for any nonzero even integer $x$. It follows that $f(x)=0$ for any integer $x$ such that $x \\equiv -1 \\pmod{4}$ and $x \\neq -d^{2}$.\nIn $x^{2} f(-x)=(f(x))^{2}$, taking $x \\equiv -1 \\pmod{4}, x \\neq -d^{2}$, we get $f(-x)=0$. It follows that $f(x)=0$ for any integer $x$ such that $x \\equiv 1 \\pmod{4}$ and $x \\neq d^{2}$. Thus $f(x)=0$ for any integer $x \\neq \\pm d^{2}$. However, we have $f(d)=d^{2} \\neq 0$, so $d= \\pm d^{2}$ or $d= \\pm 1$. It follows that $d=d^{3}$ and so $f(d)=f\\left(d^{3}\\right)=0$, which is a contradiction. Therefore, in this case, we have $f(x)=0$ for any integer $x$.\nIn conclusion, the equation has two solution: $f(x)=0$ and $f(x)=x^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40399,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle with $I$ as its incenter. Denote $D, E$ as the points of tangency of $(I)$ on $AB, AC$, respectively. The median segments with respect to vertex $A$ of triangles $ABE$ and $ACD$ meet $(I)$ at $P, Q$, respectively. Take points $M, N$ on the line $DE$ such that $AM \\perp BE$ and $AN \\perp CD$ respectively.\n\n1. Prove that $A$ lies on the radical axis of $(MIP)$ and $(NIQ)$.\n2. Suppose that the orthocenter $H$ of triangle $ABC$ lies on $(I)$. Prove that there exists a line which is tangent to three circles of center $A, B, C$ and all pass through $H$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1) Since $AM$ passes through the midpoint of the segment $BE$ and $AP \\parallel BE$, we can see that\n$$\n(AP, AM, AB, AC) = -1 \\text{ or } (MP, MA, MD, ME) = -1.\n$$\nSuppose that $AM$ cuts $(I)$ at the second point $T$ then $MDTE$ is a harmonic quadrilateral, then $(Mx, MT, MD, ME) = -1$ with $Mx$ as the tangent line of $(I)$.\nFrom these results, we get $Mx \\equiv MP$ which implies that $PM$ is tangent to $(I)$ or $\\angle IMP = 90^\\circ$. Similarly, we also have $\\angle INP = 90^\\circ$. Hence, if we denote $K$ as the midpoint of $DE$ then $IK \\perp DE$ and $K \\in (MIP), (NIQ)$. Since $A \\in IK$ as the radical axis of these two circles, the result follows.\n\n2) Denote $A_1, B_1, C_1$ as the projections from $A, B, C$ to the opposite side, respectively then\n$$\n\\overline{HA} \\cdot \\overline{HA_1} = \\overline{HB} \\cdot \\overline{HB_1} = \\overline{HC} \\cdot \\overline{HC_1} = \\frac{1}{2} \\mathscr{P}_{H/(ABC)}.\n$$\nConsider the inversion of center $H$ and the power equals to the value above, then we have\n$$\n\\begin{aligned}\n& A \\leftrightarrow A_1, \\quad B \\leftrightarrow B_1, \\quad C \\leftrightarrow C_1, \\\\\n& (HB_1C_1) \\leftrightarrow BC, \\quad (HC_1A_1) \\leftrightarrow CA, \\quad (HA_1B_1) \\leftrightarrow AB\n\\end{aligned}\n$$\nThe image of $(I)$ should be some line $d$ since $H \\in (I)$. Because $BC, CA, AB$ are tangent to $(I)$ then $d$ is also tangent to $(HB_1C_1), (HC_1A_1), (HA_1B_1)$.\nConsider the homothety of center $H$, ratio $2$ which maps $d \\rightarrow d'$ and the circle $(HB_1C_1)$ to the circle of center $H$ and radius $HA$ (since $HA$ is the diameter of $(HB_1C_1)$). By the property of homothety, we have $d'$ is also tangent to $(A, AH)$. Similarly to these circles $(B, BH), (C, CH)$ so the line $d'$ satisfies the given condition. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 40400,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a graph with 8 vertices that contains no cycle of length 4, at most how many edges can there be?",
    "options": [],
    "answer": "11",
    "solution": "Let $V$ denote the set of all vertices. Let $d$ be the greatest degree of a vertex in the graph and $v \\in V$ be a vertex with degree $d$ and let $D \\subset V$ be the set of $d$ vertices connected to $v$.\n\nIn the full subgraph consisting of vertices in $D$, no vertex can have degree 2 or more, because this would make a cycle of length 4. Therefore, there are at most $\\left\\lfloor\\frac{d}{2}\\right\\rfloor$ edges in this full subgraph.\n\nSimilarly, none of the $7-d$ vertices in $V \\setminus (D \\cup \\{v\\})$ can have 2 or more connections to $D$ because this would also make a cycle of length 4. Therefore, there are at most $7-d$ edges between $V \\setminus (D \\cup \\{v\\})$ and $D \\cup \\{v\\}$. Finally, the number of edges in $V \\setminus (D \\cup \\{v\\})$ is at most $\\binom{7-d}{2}$, hence\n$$\ne \\leq d + \\left\\lfloor\\frac{d}{2}\\right\\rfloor + 7 - d + \\binom{7-d}{2} = 7 + \\left\\lfloor\\frac{d}{2}\\right\\rfloor + \\binom{7-d}{2}\n$$\n- If $d \\geq 5$, $e \\leq 7 + \\left\\lfloor\\frac{d}{2}\\right\\rfloor + \\binom{7-d}{2} \\leq 10$.\n- If $d = 4$, $e \\leq 7 + \\left\\lfloor\\frac{4}{2}\\right\\rfloor + \\binom{7-4}{2} = 12$.\n- If $d \\leq 3$, $e = \\frac{\\sum_{u \\in V} d(u)}{2} \\leq \\frac{8 \\cdot 3}{2} = 12$.\n\nLet us now prove that $e = 12$ is not possible by ruling out the two cases with $d = 4$ and $d = 3$.\n\n1. If $e = 12$ and $d = 4$, then there are exactly 2 edges in the full subgraph consisting of vertices in $D$ (each vertex in $D$ is connected to exactly one other vertex in $D$), there are exactly 3 edges between $V \\setminus (D \\cup \\{v\\})$ and $D$ (each vertex in $V \\setminus (D \\cup \\{v\\})$ is connected to exactly 1 vertex in $D$) and there are exactly 3 edges in $V \\setminus (D \\cup \\{v\\})$ (all 3 of the vertices in $V \\setminus (D \\cup \\{v\\})$ are connected).\n\nLet us denote the vertices in $V \\setminus (D \\cup \\{v\\})$ by $a_1, a_2, a_3$. $a_i$ and $a_j$ cannot be connected to the same vertex in $D$, otherwise this would make a cycle of length 4 (for example $(a, c, b, x)$ cyclically connected).\n\nTherefore, exactly 3 of the 4 vertices in $D$ are connected to $V \\setminus (D \\cup \\{v\\})$. Since each vertex in $D$ is connected to exactly one other vertex in $D$, two of these 3 vertices are connected. Let us call them $x, y$ and let us assume without loss of generality that they are connected to $a_1$ and $a_2$. Now, we have a cycle of length 4 as $(x, y, a_2, a_1)$ are cyclically connected. Therefore, $e = 12$ and $d = 4$ is not possible.\n\n2. If $e = 12$ and $d = 3$, then all vertices in the graph must have degree 3. Let $v$ be any vertex and $D$ be defined as above. Similarly as above, no vertex in $V \\setminus (D \\cup \\{v\\})$ can have 2 or more connections to $D$ as this would make a cycle of length 4.\n\nHence, each vertex in $V \\setminus (D \\cup \\{v\\})$ is connected to at most 1 vertex in $D \\cup \\{v\\}$ and thus connected to at least 2 vertices in $V \\setminus (D \\cup \\{v\\})$. If we consider the full subgraph consisting of the 4 vertices in $V \\setminus (D \\cup \\{v\\})$, each vertex has degree at least 2. In a graph with 4 vertices such that each vertex has degree at least 2, there must be a cycle of length 4. Therefore, $e = 12$ and $d = 3$ is also not possible.\n\n![](attached_image_1.png)\n\nLet us now give an example of a graph with $e = 11$. Consider an octagon $A B C D E F G H$. Now, draw the additional three edges $A C, C E, E G$. This is a graph with 8 vertices, 11 edges and no cycles of length 4.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 40401,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 2$ be a positive integer. Consider $n$ bags of candy, each of them has exactly 1 candy. Ali and Omar take turns playing the following game (Ali moves first): At each turn, the player takes two bags containing the numbers of candy as $x, y$ for some coprime integers $x, y$ and then merges them into one bag. Whoever cannot perform this action will be the loser. Who has the strategy to win this game?",
    "options": [],
    "answer": "Omar always has a winning strategy for all starting counts greater than two.",
    "solution": "We shall prove that for all $n > 2$, Omar always has the strategy to win.\n\nFirst, Ali has to merge some two bags of 1 candy into one bag of 2 candies. We prove that Omar can turn the state of all bags into: one bag contains odd number of candies and the others just have one candy each.\n\nIndeed, in the second turn, Omar merges the bag of 2 candies with some bag of 1 candy to get one bag of 3 candies. Suppose that after a turn of Omar, there is a bag of $2k+1$ candies (with $k \\in \\mathbb{Z}^{+}$) and the other bags just have one candy each. At the next turn of Ali, there are two cases:\n\n- If Ali merges two bags of 1 candy to one bag of 2 candies, then on the next turn, Omar merges that bag with the bag of $2k+1$ candies to get a bag of $2k+3$ candies.\n\n- If Ali merges the bag of $2k+1$ candies with some bag of 1 candy to get another bag of $2k+2$, then on the next turn, Omar merges that bag with some bag of 1 candy to get a bag of $2k+3$ candies.\n\nHence, Omar always can control the state of bags like that. Note that after two turns of players, the number of bags reduces by 2. Finally, if after a turn of Omar, there is only one bag of $2k+1$ candies then Ali will be the loser. Otherwise, there is a bag of $2k+1$ candies and 3 bags of 1 candy. There are two cases:\n\n- If Ali merges two bags of 1 candy then Omar merges the bag of $2k+1$ with the bag of 1 candy; then after that turn, there are a bag of 2 candies and a bag of $2k+2$ which implies that Ali loses.\n\n- If Ali merges the bag of $2k+1$ candies with a bag of 1 candy then Omar merges two bags of 1 candy to make two bags of even number of candies as above.\n\nTherefore, in all cases of $n > 2$, Omar always can win the game.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40402,
    "subject": "Mathematics (Multi-modal)",
    "question": "In square $ABCD$ with side $1$, point $E$ lies on $BC$ and $F$ lies on $CD$ such that $\\angle EAB = 20^{\\circ}$, $\\angle EAF = 45^{\\circ}$. Find the length of altitude $AH$ of $\\triangle AEF$.",
    "options": [],
    "answer": "1",
    "solution": "Take point $G$ on the opposite ray of ray $DC$ such that $GD = BE$. Then two triangles $ABE$, $ADG$ are congruent, implies that $AE = AG$ and $\\angle EAB = \\angle DAG = 20^{\\circ}$. Thus\n$$\n\\angle FAG = \\angle FAD + \\angle DAG = \\angle EAB + \\angle DAG = 45^{\\circ}.\n$$\nThus two triangles $AEF$, $AGF$ have common side $AF$ and $AE = AG$ and $\\angle FAG = \\angle FAE = 45^{\\circ}$, thus $\\triangle AEF \\cong \\triangle AGF$, then $AH = AD = 1$, where $AH$ is the altitude with respect to vertex $A$ in triangle $AEF$.\n\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40403,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest integer $m$ for which there are positive integers $n > k > 1$ satisfying the equation\n$$\n\\underbrace{11 \\ldots 1}_{n} = \\underbrace{11 \\ldots 1}_{k} \\cdot m.\n$$",
    "options": [],
    "answer": "101",
    "solution": "Obviously $m > 9$. If $m = \\overline{ab}$, where $a \\geq 1$ then we must have $b = 1$ to ensure last digit of $1 \\cdot m$ is equal $1$. In this case regardless of value of $a$ the second last digit of $\\underbrace{11 \\ldots 1}_{k} \\cdot m$ is equal to the last digit of $a + 1$ and can't be equal $1$. So $m \\geq 100$. Obviously $m = 100$ doesn't satisfy to the condition, but $m = 101$ satisfies, since $11 \\cdot 101 = 1111$.\nHence, the answer to this problem is $m = 101$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40404,
    "subject": "Mathematics (Multi-modal)",
    "question": "1. Let $p$ be an odd prime number.\n   a. Show that $p$ divides $n \\cdot 2^{n} + 1$ for infinitely many positive integers $n$.\n   b. Find all $n$ satisfying the condition above when $p = 3$.",
    "options": [],
    "answer": "Infinitely many n exist; for example, all n of the form (p k + 1)(p − 1). For p = 3, all positive integers congruent to 1 or 2 modulo 6.",
    "solution": "a. We will show that for all $k > 0$, the number $n = (p k + 1)(p - 1)$ satisfies $p \\mid n \\cdot 2^{n} + 1$. Indeed, by Fermat's Little Theorem, we have\n$$\n2^{p-1} \\equiv 1 \\pmod{p} \\text{ so } 2^{n} = 2^{(p k + 1)(p - 1)} \\equiv 1 \\pmod{p}.\n$$\nAnd then\n$$\nn \\cdot 2^{n} + 1 \\equiv (p k + 1)(p - 1) \\cdot 1 + 1 = p^{2} k - p k + p \\equiv 0 \\pmod{p}.\n$$\nSince there are infinitely many numbers of the form $(p k + 1)(p - 1)$, we get the conclusion.\n\nb. Notice that $n$ is periodic modulo $3$, with a period of $3$, and $2^{n}$ is periodic modulo $3$, with period $2$. Hence, $n \\cdot 2^{n} + 1$ is periodic, with period (at most) $6$, and only the first $6$ positive integers need to be analyzed.\nThe answer is $n = 6k + 1$ or $n = 6k + 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40405,
    "subject": "Mathematics (Multi-modal)",
    "question": "A set $S$ is called neighboring if it has the following two properties:\ni) $S$ has exactly 4 elements,\nii) for every element $x \\in S$ at least one of the $x-1$ or $x+1$ belongs to $S$.\nFind the number of all neighboring subsets of the set $\\{1,2, \\ldots, n\\}$.",
    "options": [],
    "answer": "((n-3)(n-2))/2",
    "solution": "Let $x$ be the smallest index and $y$ be the largest index of the neighboring set. Then our set consists of elements\n$$\na_{x}, a_{x+1}, a_{y-1}, a_{y}\n$$\nSo the number of neighboring sets is equal to the number of pairs $(x, y)$ where $y-x \\geq 3$. The number of such sets is equal to\n$$\n1+2+\\cdots+(n-3)=\\frac{(n-3)(n-2)}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40406,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be positive real numbers such that $a + b + c = 1$. Prove that\n$$\n\\frac{a}{b} + \\frac{b}{a} + \\frac{b}{c} + \\frac{c}{b} + \\frac{c}{a} + \\frac{a}{c} \\geq 2 \\sqrt{2} \\left( \\sqrt{\\frac{1-a}{a}} + \\sqrt{\\frac{1-b}{b}} + \\sqrt{\\frac{1-c}{c}} \\right).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40407,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a school there are 40 different clubs, each of them contains exactly 30 children. For every $i$ from $1$ to $30$ define $n_{i}$ as a number of children who attend exactly $i$ clubs. Prove that it is possible to organize 40 new clubs with 30 children in each of them such that the analogical numbers $n_{1}, n_{2}, \\ldots, n_{30}$ will be the same for them.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will do the following algorithm to rearrange the children.\n- Put the children who attend exactly one club at the start of the line (in any order).\n- Put the children who attend exactly 2 clubs at the start of the line (in any order), and so on.\n- Finally, put the children who attend exactly 30 clubs at the start of the line (in any order).\n\nSince each club contains 40 children, then the number of pairs (club, child) is equal to\n$$\n1200 = 30 \\times 40 = n_{1} + 2 n_{2} + \\cdots + 30 n_{30}.\n$$\n\nNow back to the line, we will count from $1 \\rightarrow 1200$ and for each number, we will point at the children in the following way:\n- We point at the children from the top to the bottom of the line.\n- If a child attends $k$ clubs, we point at him $k$ times (each time, we also count the numbers).\n\nFinally, we create the new clubs $C_{1}, C_{2}, \\ldots, C_{40}$ and add children to them using the rule: if we count a number $i \\bmod 40$ while pointing at some child, then put that one into $C_{i}$ for any $i$.\n\nWe will count $\\frac{1200}{40} = 30$ numbers which are $i \\bmod 40$ for each $i$, so indeed, each club has exactly 30 children, and no one appears in the same club since each child is counted at most 30 times in the row. And it is easy to check that the numbers $n_{1}, n_{2}, \\ldots, n_{30}$ remain unchanged.",
    "topic": "Discrete Mathematics",
    "subtopic": "Logic"
  },
  {
    "id": 40408,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some $n>2$ lamps are cyclically connected: lamp $1$ with lamp $2$, ..., lamp $k$ with $k+1$, ..., lamp $n$ with lamp $1$. At the beginning, all lamps are off. When one pushes the switch of a lamp, that lamp and the two ones connected to it change status (from off to on, or vice-versa). Determine the number of configurations of lamps reachable from the initial one, through some set of switches being pushed.",
    "options": [],
    "answer": "2^n if 3 does not divide n; 2^{n-2} if 3 divides n",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40409,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{Z}^{+} \\rightarrow \\mathbb{Z}^{+}$ such that\n$$\nn^{3}-n^{2} \\leq f(n) \\cdot (f(f(n)))^{2} \\leq n^{3}+n^{2}\n$$\nfor every $n$ is positive integers.",
    "options": [],
    "answer": "All functions obtained by choosing any collection of disjoint adjacent pairs of positive integers and swapping each chosen pair while fixing all other integers. Equivalently, f is an involution whose cycles are either fixed points or two-cycles of the form k and k plus one; for each chosen pair set f(k) = k plus one and f(k plus one) = k, and for all other a set f(a) = a.",
    "solution": "Firstly, we prove that $f$ is injective. Let two natural numbers $a, b$ such that $a+1 \\leq b$ and $f(a)=f(b)$. By the given condition, we have\n$$\nf(a)[f(f(a))]^{2} \\leq a^{3}+a^{2}<a^{3}+2a^{2}+a=(a+1)^{3}-(a+1)^{2}.\n$$\nSince $a, b \\in \\mathbb{Z}^{+}$ and $a+1 \\leq b$, we have\n$$\n(a+1)^{3}-(a+1)^{2} \\leq b^{3}-b^{2} \\leq f(b)[f(f(b))]^{2}.\n$$\nThus $f(a)[f(f(a))]^{2}<f(b)[f(f(b))]^{2}$, a contradiction. This implies $f$ is injective.\n\nIn given condition, let $n=1$, we have\n$$\n0 \\leq f(1)[f(f(1))]^{2} \\leq 2.\n$$\nSince $f: \\mathbb{Z}^{+} \\longrightarrow \\mathbb{Z}^{+}$, we have two cases\n- $f(1)=1$.\n- $f(1)=2$ and $f(2)=1$.\n\nWe shall prove that if $\\{f(1), \\ldots, f(k)\\}=\\{1, \\ldots, k\\}$ then either $f(k+1)=k+1$ or $f(k+1)=k+2, f(k+2)=k+2$. Indeed,\nSince $f$ is injective, we have $f(k+1) \\geq k+1$. If $f(k+1)=k+1$, we have\n$$\n\\{f(1), \\ldots, f(k+1)\\}=\\{1, \\ldots, k+1\\}.\n$$\nOtherwise, if $f(k+1) \\geq k+2$, we have $[f(f(k+1))]^{2} \\geq (k+1)^{2}$. In given condition, let $n=k+1$, we have\n$$\n(k+1)^{3}+(k+1)^{2} \\geq f(k+1)[f(f(k+1))]^{2} \\geq (k+2)(k+1)^{2}.\n$$\nSince $(k+1)^{3}+(k+1)^{2}=(k+2)(k+1)^{2}$, we deduce that $f(k+1)=k+2$ and $f(k+2)=k+1$. Thus $\\{f(1), \\ldots, f(k+2)\\}=\\{1, \\ldots, k+2\\}$.\n\nNow we can check both cases satisfy the conditions\n$$\n\\begin{aligned}\n& (k+2)^{3}-(k+2)^{2} \\leq f(k+2) f(f(k+2))^{2} \\leq (k+2)^{3}+(k+2)^{2} \\\\\n& \\Leftrightarrow (k+2)^{2}(k+1) \\leq (k+1)(k+2)^{2} \\leq (k+2)^{2}(k+3).\n\\end{aligned}\n$$\nTherefore, the solutions of this problem are the functions have following properties\n- There are some disjoint pairs $(k, k+1)$ with $f(k)=k+1$ and $f(k+1)=f(k)$.\n- For other positive integer $a$, then $f(a)=a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40410,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the maximal number of disjoint crosses (5 squares) which can be put inside $8 \\times 8$ chessboard such that sides of a cross are parallel to sides of the chessboard.",
    "options": [],
    "answer": "8",
    "solution": "Let's note, that 4 corner cells can't be covered by crosses. So we assume that they are removed from the board and we have only 60 cells.\nFrom the first row at most two cells can be covered by crosses. So at least 4 cells will be not covered. The same argument works for last row, first column and last column. So at most $60-4 \\cdot 4=44$ cell can be covered, which means at most 8 crosses can be used. It remains to show the example with 8 crosses.\n![](attached_image_1.png) $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40411,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $x, y, z$ are non-zero real numbers such that\n$$\nx = 2 - \\frac{y}{z}, \\quad y = 2 - \\frac{z}{x}, \\quad z = 2 - \\frac{x}{y}\n$$\nFind all possible values of $T = x + y + z$.",
    "options": [],
    "answer": "3 or 7",
    "solution": "From the given conditions, we have\n$$\nxz = 2z - y, \\quad xy = 2x - z, \\quad yz = 2y - x.\n$$\nTaking the sum of these equations, side by side, we have\n$$\nxy + yz + zx = x + y + z = T.\n$$\nFrom $xz = 2z - y$, we also can get $2z - x - y = xz - x \\rightarrow 3z - T = x(z - 1)$. Make the similar equations and multiply them, we get\n$$\nxyz(x - 1)(y - 1)(z - 1) = (3x - T)(3y - T)(3z - T).\n$$\nNote that $(x - 1)(y - 1)(z - 1) = xyz - (xy + yz + zx) + (x + y + z) - 1 = xyz - 1$ and\n$$\n\\begin{aligned}\n(3x - T)(3y - T)(3z - T) &= 27xyz - 9T(xy + yz + zx) + T^2(3x + 3y + 3z) - T^3 \\\\\n&= 27xyz - 9T^2 + 2T^3\n\\end{aligned}\n$$\nThus\n$$\n\\begin{equation*}\nxyz(xyz - 1) = 27xyz - 9T^2 + 2T^3 \\text{ or } (xyz)^2 - 28xyz = 2T^3 - 9T^2. \\tag{1}\n\\end{equation*}\n$$\nBy the similar transformation, $xz = 2z + y \\rightarrow xz + 2x = 2z + 2x - y \\rightarrow x(z + 2) = 2T - 3y$. Make the similar equations and multiply them, side by side, we get\n$$\n(xyz)^2 + 6(xyz)T + 35xyz = 18T^2 - 4T^3.\n$$\nOn the other hand, $xz = 2z - y \\rightarrow xyz = 2yz - y^2$. Make the similar equations and multiply them, side by side, we get\n$3xyz = 2(xy + yz + zx) - (x^2 + y^2 + z^2) = 4(xy + yz + zx) - (x + y + z)^2 = 4T - T^2$.\nSo we have $xyz = \\frac{4T - T^2}{3}$, then by substituting to equation (1), we get\n$$\n\\begin{equation*}\n\\left(\\frac{4T - T^2}{3}\\right)^2 - 28\\left(\\frac{4T - T^2}{3}\\right) = 2T^3 - 9T^2 \\tag{2}\n\\end{equation*}\n$$\nSolving this equation, we have $T \\in \\{0, 3, 7, 16\\}$.\n\n1. If $T = 0$, then $xyz = 0$, a contradiction.\n2. If $T = 16$, then $xyz = -64$ and this pair $(T, xyz) = (16, -64)$ does not satisfy equation (2).\n3. If $T = 3$, then $xyz = 1$ and by Vieta's theorem, $x, y, z$ are real roots of $t^3 - 3t^2 + 3t - 1 = 0$, which implies that $x = y = z = 1$.\n4. If $T = 7$, we get $xyz = -7$ and $x, y, z$ are roots of $t^3 - 7t^2 + 7t + 7 = 0$, this also have three distinct real roots.\n\nHence, all possible values of $T$ are $3$ or $7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40412,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $n$ with the following property: After painting black exactly $n$ cells of a $7 \\times 7$ board there always exists a $2 \\times 2$ square with at least three black cells.",
    "options": [],
    "answer": "29",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40413,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c \\geq 0$ satisfy $a + b + c = 1$. Prove that\n$$\n\\frac{\\sqrt{a}}{b + 1} + \\frac{\\sqrt{b}}{c + 1} + \\frac{\\sqrt{c}}{a + 1} > \\frac{1}{2}(\\sqrt{a} + \\sqrt{b} + \\sqrt{c}) .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40414,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine whether there exists a positive integer $n$ such that $n+2$ divides the following sum\n$$\nS = 1^{2019} + 2^{2019} + \\ldots + n^{2019}.\n$$",
    "options": [],
    "answer": "No such positive integer n exists.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40415,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R}^2 \\rightarrow \\mathbb{R}$ that for all real numbers $x, y, z$ satisfies the equation\n$$\nf(f(x, z), f(z, y)) = f(x, y) + z.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40416,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, let $D$ be the touch point of the side $BC$ and the incircle of the triangle $ABC$, and let $J_b$ and $J_c$ be the incentres of the triangles $ABD$ and $ACD$, respectively. Prove that the circumcentre of the triangle $A J_b J_c$ lies on the bisector of the angle $BAC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40417,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral with $AB = BC$ and $AD = CD$. A point $M$ lies on the minor arc $CD$ of its circumcircle. The lines $BM$ and $CD$ meet at point $P$, the lines $AM$ and $BD$ meet at point $Q$. Prove that $PQ \\parallel AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40418,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and $p > n + 1$ a prime. Prove that $p$ divides the following sum\n$$\nS = 1^{n} + 2^{n} + \\ldots + (p-1)^{n}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40419,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let 6 pairwise different digits are given and all of them are different from $0$. Prove that there exist $2$ six-digit integers, such that their difference is equal to $9$ and each of them contains all given $6$ digits.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40420,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$, $a$, $b$, $c$ be pairwise different integers from the set $\\{1,2,3,4,5,6\\}$. Find the smallest possible value for the expression $x y z + a b c - a x - b y - c z$.",
    "options": [],
    "answer": "10",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40421,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let non-integer real numbers $a$, $b$, $c$, $d$ are given, such that the sum of each 3 of them is integer. May it happen that $a b + c d$ is an integer.",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40422,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest prime number that divides both $2^{63}-1$ and $2^{91}-1$.",
    "options": [],
    "answer": "127",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40423,
    "subject": "Mathematics (Multi-modal)",
    "question": "Real nonzero numbers $x$, $y$, $z$ are such that $x + y + z = 0$. Moreover, it is known that\n$$\nA = \\frac{x}{y} + \\frac{y}{z} + \\frac{z}{x} = \\frac{x}{z} + \\frac{z}{y} + \\frac{y}{x} + 1.\n$$\nDetermine $A$.",
    "options": [],
    "answer": "-1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40424,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pair of integers ($m$, $n$) and $m \\geq n$ such that there exist a positive integer $s$ and\n1. Product of all divisors of $s m$, $s n$ are equal.\n2. Number of divisors of $s m$, $s n$ are equal.",
    "options": [],
    "answer": "1) The product-of-divisors condition holds if and only if m = n. 2) The equal-divisor-count condition holds if and only if either m = n or n does not divide m (equivalently, m and n are not comparable by divisibility).",
    "solution": "1) Denote $d(x)$, $\\pi(x)$ as the number of divisors, the product of divisors of positive integer $x$.\nFirstly, we can see that for any divisor $y_{k}$ of $x$, $1 \\leq k \\leq d(x)$ then $\\frac{x}{y_{i}}$ is also divisor of $x$, thus\n$$\n\\prod_{k=1}^{d(x)} y_{k} = \\prod_{k=1}^{d(x)} \\frac{x}{y_{k}} \\text{ so } \\left(\\prod_{k=1}^{d(x)} y_{k}\\right)^{2} = x^{d(x)}, \\text{ hence } \\pi(x) = x^{\\frac{d(x)}{2}}\n$$\nThus for any $m, n \\in \\mathbb{Z}^{+}$ then $\\pi(m) = \\pi(n)$ implies that $m, n$ share the common prime divisors set $S$.\nSuppose that $d(m) \\geq d(n)$ and take $p \\in S$. Since $m^{d(m)} = n^{d(n)}$, we have $d(m) \\cdot v_{p}(m) = d(n) \\cdot v_{p}(n)$. Since $d(m) \\geq d(n)$, we get $v_{p}(m) \\leq v_{p}(n)$. And this is true for all $p \\in S$, thus\n$$\nd(m) = \\prod_{p \\in S} (v_{p}(m) + 1) \\leq \\prod_{p \\in S} (v_{p}(n) + 1) = d(n)\n$$\nSo $d(m) = d(n)$, which implies that $m = n$. Therefore, we can find a positive integer such that product of all divisors of $s m$, $s n$ are equal if and only if $m = n$.\n\n2) Firstly, we can see that if $n \\mid m$ then any divisor of $s n$ is also divisor of $s m$, so $d(s n) < d(s m)$. We consider $n \\nmid m$, and denote $p_{1}, p_{2}, \\ldots, p_{t}$ be all prime dividing $m n$. Suppose that\n$$\nm = \\prod_{i=1}^{t} p_{i}^{\\alpha_{i}} \\text{ and } n = \\prod_{i=1}^{t} p_{i}^{\\beta_{i}}\n$$\nNow we are looking for $s = \\prod_{i=1}^{t} p_{i}^{\\gamma_{i}}$ such that\n$$\n\\frac{d(s m)}{d(s n)} = \\prod_{i=1}^{t} \\frac{\\alpha_{i} + \\gamma_{i} + 1}{\\beta_{i} + \\gamma_{i} + 1} = 1\n$$\nNote that if $\\alpha_{i} = \\beta_{i}$, then regardless of the value of $\\gamma_{i}$, the corresponding factor equals to 1 and does not affect the product. So we may assume that $\\alpha_{i} \\neq \\beta_{i}$ for all $1 \\leq i \\leq t$.\n\nClaim. Let $\\alpha > \\beta$ be nonnegative integers. Then for every $M \\geq \\beta + 1$, there exist a nonnegative integer $\\gamma$ such that\n$$\n\\frac{\\alpha + \\gamma + 1}{\\beta + \\gamma + 1} = \\frac{M + 1}{M} .\n$$\nIt is equivalent to $\\gamma = M(\\alpha - \\beta) - (\\beta + 1) \\geq 0$, which is true.\n\nBack to the original problem, we can assume that $\\alpha_{i} > \\beta_{i}$ for $i = 1, 2, \\ldots, u$ and $\\alpha_{i} < \\beta_{i}$ for $i = u + 1, u + 2, \\ldots, t$. Take some big enough $X$ and choose $\\gamma_{i}$ such that\n- $\\frac{\\alpha_{i} + \\gamma_{i} + 1}{\\beta_{i} + \\gamma_{i} + 1} = \\frac{u X + i}{u X + i - 1}$ for $1 \\leq i \\leq u$.\n- $\\frac{\\beta_{u + i} + \\gamma_{u + i} + 1}{\\alpha_{u + i} + \\gamma_{u + i} + 1} = \\frac{(t - u) X + i}{(t - u) X + i - 1}$ for $1 \\leq i \\leq t - u$.\nThen we have\n$$\n\\frac{d(s m)}{d(s n)} = \\prod_{i=1}^{u} \\frac{u X + i}{u X + i - 1} \\cdot \\prod_{i=1}^{t - u} \\frac{(t - u) X + i - 1}{(t - u) X + i} = \\frac{u(X + 1)}{u X} \\cdot \\frac{(t - u) X}{(t - u)(X + 1)} = 1 .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40425,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $(m, n)$ such that\n$$\n125 \\cdot 2^{n} - 3^{m} = 271.\n$$",
    "options": [],
    "answer": "(6, 3)",
    "solution": "Considering the equation $\\bmod\\ 5$ we get\n$$\n3^{m} \\equiv -1 \\pmod{5},\n$$\nso $m = 4k + 2$ for some positive integer $k$. Then, considering the equation $\\bmod\\ 7$ we get\n$$\n-2^{n} - 9^{2k+1} \\equiv 5 \\pmod{7},\n$$\nwhich means\n$$\n2^{n} + 2^{2k+1} \\equiv 2 \\pmod{7}.\n$$\nSince $2^{s} \\equiv 1, 2, 4 \\pmod{7}$ so the only possibility is $2^{n} \\equiv 2^{2k+1} \\equiv 1 \\pmod{7}$ so $3 \\mid n$ and $3 \\mid 2k+1$. From the last one we get $3 \\mid m$ so we can write $n = 3x$ and $m = 3y$. Therefore, the given equation takes the form $5^{3} \\cdot 2^{3x} - 3^{3y} = 271$, or\n$$\n\\left(5 \\cdot 2^{x} - 3^{y}\\right)\\left(25 \\cdot 2^{2x} + 5 \\cdot 2^{x} \\cdot 3^{y} + 3^{2y}\\right) = 271.\n$$\nIt follows that $25 \\cdot 2^{2x} + 5 \\cdot 2^{x} \\cdot 3^{y} + 3^{2y} \\leq 271$ and so $x < 2$. We conclude $x = 1$ and then $y = 2$. So $m = 6$ and $n = 3$.\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40426,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let Pascal triangle be an equilateral triangular array of numbers, consisting of $2019$ rows and except for the numbers in the bottom row, each number is equal to the sum of two numbers immediately below it. How many ways to assign each of numbers $a_{0}, a_{1}, \\ldots, a_{2018}$ (from left to right) in the bottom row by $0$ or $1$ such that the number $S$ on the top is divisible by $1019$.",
    "options": [],
    "answer": "2^{2016}",
    "solution": "First, by induction, one can show that\n$$\nS = \\binom{n}{0} a_{0} + \\binom{n}{1} a_{1} + \\cdots + \\binom{n}{n} a_{n}\n$$\nif the Pascal triangle consists of $n$ rows.\n\nNote that for any odd prime $p$, we also have:\n\n**Claim 1.** $\\binom{2p}{p} \\equiv 2 \\pmod{p}$.\nIndeed,\n$$\n\\begin{aligned}\n& \\binom{2p}{p} - 2 = \\frac{(2p)!}{p!p!} - 2 = \\frac{(p+1)(p+2) \\ldots (2p-1)(2p)}{p!} - 2 \\\\\n& = 2 \\frac{(p+1)(p+2) \\ldots (2p-1) - (p-1)!}{(p-1)!}\n\\end{aligned}\n$$\nThe numerator is congruent to $1 \\cdot 2 \\cdot 3 \\cdot (p-1) - (p-1)! = 0 \\pmod{p}$ so $\\binom{2p}{p} \\equiv 2 \\pmod{p}$.\n\n**Claim 2.** $\\binom{2p}{k} \\equiv 0 \\pmod{p}$ for $1 \\leq k \\leq 2p-1$ and $k \\neq p$.\nIndeed,\nSince $\\binom{2p}{k} = \\binom{2p}{2p-k}$ so we can suppose $1 \\leq k < p$. Similar calculation, we have\n$$\n\\binom{2p}{k} = \\frac{(2p-k+1)(2p-k+2) \\ldots (2p-1)(2p)}{k!}.\n$$\nThe numerator is divisible by $p$ while $(p, k!) = 1$ since $1 \\leq k < p$ so we are done.\n\nFrom this, we can conclude that, if $1009 \\mid S$ then $a_{0} + a_{2018} + 2 a_{1009}$ is divisible by $1009$. This only happens when all of them are equal to $0$.\n\nThe other numbers can be assigned any of $0$ or $1$ so the number of ways is $2^{2016}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40427,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove the inequality for non-negative $a, b, c$\n$$\na \\sqrt{3 a^{2}+6 b^{2}}+b \\sqrt{3 b^{2}+6 c^{2}}+c \\sqrt{3 c^{2}+6 a^{2}} \\geq (a+b+c)^{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $3 a^{2}+6 b^{2} \\geq (a+2 b)^{2}$ is true for all real numbers $a, b$. Indeed, after expanding and grouping we get $2 a^{2}-4 a b+2 b^{2} \\geq 0$ which is equivalent to $2(a-b)^{2} \\geq 0$. So with non-negative numbers $a, b, c$ we have\n$$\n\\begin{aligned}\na \\sqrt{3 a^{2}+6 b^{2}}+b \\sqrt{3 b^{2}+6 c^{2}} & +c \\sqrt{3 c^{2}+6 a^{2}} \\\\\n\\geq a(a+2 b)+b(b+2 c)+c(c+2 a) & =(a+b+c)^{2} .\n\\end{aligned}\n$$\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40428,
    "subject": "Mathematics (Multi-modal)",
    "question": "Chess horse attacks fields in distance $\\sqrt{5}$. Let several horses are put on the board $12 \\times 12$ such, that every square of size $2 \\times 2$ contains at least one horse. Find the maximal possible number of cells that are not under attack (horse doesn't attack its own cell).\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "72",
    "solution": "Let's note, that if we put a horse in any green cell, then it will attack a grey cell. Since green cells form a square $2 \\times 2$, so one of them contains a horse, so at least one grey is under attack.\n\nNow let's split the board into 72 pairs like in the figure. According what we said above, at least 72 cells are under attack. To get example with exactly 72 cell under attack let's paint board as chessboard and put horses in all black cells. Then all black cells will be free of attack.\n\nHence, the answer to this problem is 72.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40429,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all arithmetic sequences $a_{1}, a_{2}, \\ldots$ for which there exists integer $N>1$ such that for any positive integer $k$ the following divisibility holds\n$$\na_{1} a_{2} \\ldots a_{k} \\mid a_{N+1} a_{N+2} \\ldots a_{N+k} .\n$$",
    "options": [],
    "answer": "All such arithmetic sequences are exactly the constant sequences and the sequences of the form a_n = c·n with c an integer. For these, the divisibility holds for any shift greater than one.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40430,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incenter of triangle $A B C$ and $J$ the excenter of the side $B C$. Let $M$ be the midpoint of $C B$ and $N$ the midpoint of arc $B A C$ of circle $(A B C)$. If $T$ is the symmetric of the point $N$ by the point $A$, prove that the quadrilateral $J M I T$ is cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that $A I$ cuts $B C$ at $D$ and cuts $(A B C)$ again at $P$. We have known that $B, C, I, J$ lie on circle $(P, \\frac{I J}{2})$.\n\nSuppose that $A N$ meets $B C$ at $E$, then $A E$ is the external angle bisector of $\\angle B A C$, so $(D E, B C) = -1$. We have\n$$\n\\overline{D E} \\cdot \\overline{D M} = \\overline{D B} \\cdot \\overline{D C} = \\overline{D I} \\cdot \\overline{D J}\n$$\nmeans that $I M J E$ is cyclic.\n\n![](attached_image_1.png)\n\nIn the other hand, we have known that $(A D, I J) = -1$. Combining with $D$ is orthocenter of $\\triangle N P E$, we get\n$$\n\\overline{A T} \\cdot \\overline{A S} = -\\overline{A N} \\cdot \\overline{A S} = \\overline{A D} \\cdot \\overline{A P} = \\overline{A I} \\cdot \\overline{A J},\n$$\nimplies that $I J E T$ is cyclic.\n\nFrom (1) and (2), then the points $J, I, M, T, E$ lie on a circle. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40431,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$ be a polynomial of degree $n \\geq 2$ with rational coefficients such that $P(x)$ has $n$ pairwise different real roots forming an arithmetic progression. Prove that among the roots of $P(x)$ there are two that are also the roots of some polynomial of degree 2 with rational coefficients.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $x_{1} < x_{2} < \\ldots < x_{n}$ as the roots of $P(x)$. Let $d = x_{n} - x_{n-1} = \\cdots = x_{2} - x_{1} > 0$. Since $P(x)$ has rational coefficients then by applying Vieta's theorem, we have\n$$\n\\sum_{i=1}^{n} x_{i} \\in \\mathbb{Q} \\text{ and } \\sum_{1 \\leq i < j \\leq n} x_{i} x_{j} \\in \\mathbb{Q} .\n$$\nNote that $\\sum_{i=1}^{n} x_{i} = \\frac{n(x_{1} + x_{n})}{2}$ so $x_{1} + x_{n} \\in \\mathbb{Q}$. On the other hand,\n$$\n\\sum_{i=1}^{n} x_{i}^{2} = \\left(\\sum_{i=1}^{n} x_{i}\\right)^{2} - 2 \\sum_{1 \\leq i < j \\leq n} x_{i} x_{j} \\in \\mathbb{Q}\n$$\nand\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} x_{i}^{2} & = \\sum_{i=1}^{n} \\left(x_{1} + (i-1)d\\right)^{2} = n x_{1}^{2} + n(n-1) x_{1} d + \\frac{n(n-1)(2n-1)}{6} d^{2} \\\\\n& = n\\left(x_{1} + \\frac{(n-1)d}{2}\\right)^{2} + \\frac{n^{2}(n^{2}-1)}{12} d^{2} \\\\\n& = \\frac{n(x_{1} + x_{n})^{2}}{4} + \\frac{n(n^{2}-1)}{12} d^{2}\n\\end{aligned}\n$$\nFrom these, we can conclude that $\\frac{n(n^{2}-1)}{12} d^{2} \\in \\mathbb{Q}$ so $d^{2} \\in \\mathbb{Q}$. Thus\n$$\nx_{1} x_{n} = \\frac{(x_{1} + x_{n})^{2} - (x_{n} - x_{1})^{2}}{4} = \\frac{(x_{1} + x_{n})^{2} - (n-1)^{2} d^{2}}{4} \\in \\mathbb{Q} .\n$$\nTherefore, $x_{1} + x_{n}, x_{1} x_{n} \\in \\mathbb{Q}$ which implies that two numbers $x_{1}, x_{n}$ satisfy.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40432,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are no positive integers $x$, $y$, $z$ such that\n$$\n(3x + 4y)(4x + 5y) = 7^{z}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the condition, we can see that both numbers $3x + 4y$ and $4x + 5y$ are powers of $7$, so such must be also their division. However\n$$\n1 < \\frac{4x + 5y}{3x + 4y} < 2\n$$\nand can't be power of $7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40433,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$ point $M$ is the midpoint of side $AB$, and point $D$ is the foot of altitude $CD$. Prove that $\\angle A = 2 \\angle B$ if and only if $AC = 2MD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40434,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each point of the plane has some color. It is known that on every straight line there are points in at most two different colors. What is the maximum possible number of colors present on this plane?",
    "options": [],
    "answer": "2",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40435,
    "subject": "Mathematics (Multi-modal)",
    "question": "An acute-angled triangle $ABC$ is inscribed in a circle $\\omega$. The tangents to the circle $\\omega$ at $A$ and $C$ intersect at $F$. The perpendicular bisector of segment $AB$ intersects the line $BC$ at $E$. Prove that the lines $FE$ and $AB$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40436,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each integer is painted one of three colors. Prove that there exist two distinct numbers of the same color, whose difference is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40437,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $a^{2} + b^{2} + c^{2} = 1$. Find the minimum for\n$$\nS = \\frac{1}{a^{2}} + \\frac{1}{b^{2}} + \\frac{1}{c^{2}} - \\frac{2\\left(a^{3} + b^{3} + c^{3}\\right)}{a b c}\n$$",
    "options": [],
    "answer": "3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40438,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $d$ be a positive divisor of the number $A = 1024^{1024} + 5$ and suppose that $d$ can be expressed as $d = 2x^{2} + 2xy + 3y^{2}$ for some integers $x, y$. Which remainder can we have when dividing $d$ by $20$?",
    "options": [],
    "answer": "3 or 7",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40439,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive integer $n$ is called 'nice' if the sum of squares of its divisors is equal to $(n+3)^2$ (for example $287=7 \\cdot 41$ is 'nice', since $1^2+7^2+41^2+287^2= 84100=(287+3)^2$). Prove that if for two primes $p$ and $q$ (not necessarily distinct) the number $n=pq$ is 'nice', then both $n+2$ and $2(n+1)$ are perfect squares.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40440,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle. Take two points $D, E$ inside this triangle such that\n$$\n\\begin{aligned}\n& \\angle DAB = \\angle DCB, \\quad \\angle DAC = \\angle DBC; \\\\\n& \\angle EAB = \\angle EBC, \\quad \\angle EAC = \\angle ECB.\n\\end{aligned}\n$$\nProve that triangle $ADE$ is right.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, we will show that $D$ is the orthocenter of triangle $ABC$. Denote $A' = AD \\cap BC$, $B' = BD \\cap CA$, $C' = CD \\cap AB$. Since $\\angle DAB = \\angle DCB$, we have $ACA'C'$ is cyclic. Similarly, $ABA'B'$ is also cyclic. So we have\n$$\n\\angle DA'B = \\angle DB'A, \\quad \\angle DA'C = \\angle DC'A \\text{ and } DA \\cdot DA' = DB \\cdot DB' = DC \\cdot DC',\n$$\nso $BCB'C'$ is also cyclic. Thus $\\angle DB'A = \\angle DC'A$ which implies that $\\angle DA'B = \\angle DA'C$, but $\\angle DA'B + \\angle DA'C = 180^\\circ \\rightarrow AA' \\perp BC$. It leads to $BB' \\perp CA$, $CC' \\perp AB$ then $D$ is the orthocenter of triangle $ABC$.\n\nContinue, denote $M$ as the intersection of $AE$ and $BC$ then $\\angle EBM = \\angle EAB$ implies that $MB$ is tangent to $(ABE)$. Similarly, $MB$ is tangent to $(ACE)$ then\n$$\nME \\cdot MA = MB^2 = MC^2 = MB'^2 = MC^2\n$$\n(since $MB = MC = MB' = MC'$). So $MB'$ is tangent to $(AEB')$, thus $\\angle AEB' = \\angle MB'C = \\angle C$. Similarly, $\\angle AEC' = \\angle B$ then\n$$\n\\angle B'EC' = \\angle AEB' + \\angle AEC' = \\angle B + \\angle C = 180^\\circ - \\angle A.\n$$\nThus $AB'EC'$ is cyclic, but $AB'DC'$ is cyclic, so five points $A, B', C', D, E$ are concyclic, which implies that\n$$\n\\angle AED = \\angle AB'D = \\angle AC'D = 90^\\circ.\n$$\nTherefore, $ADE$ is a right triangle. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40441,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a given set of real numbers such that:\ni) $1 \\in S$,\nii) for any $a, b \\in S$ (not necessarily different), then $a-b \\in S$,\niii) for $a \\in S$, $a \\neq 0$ then $\\frac{1}{a} \\in S$.\nProve that for any $a, b \\in S$ then $a b \\in S$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $a=0 \\in S$ or $a=1 \\in S$ then for any $b \\in S$ we have $a b \\in S$ which is obvious. So we can suppose that $a, b \\notin \\{0 ; 1\\}$.\nFrom $1 \\in S$, $a \\in S$ we have $1-a \\in S \\rightarrow (1-a)-1 = -a \\in S$, then\n$$\nb-(-a) = a+b \\in S.\n$$\nWe also have $a-1 \\in S$ so $\\frac{1}{a}, \\frac{1}{a-1} \\in S$ then\n$$\n\\frac{1}{a-1} - \\frac{1}{a} = \\frac{1}{a(a-1)} \\in S \\rightarrow a(a-1) \\in S.\n$$\nSince $a \\in S$, $a(a-1) \\in S$ then $a^{2} = a(a-1) + a \\in S$. Similarly, we also have $b^{2} \\in S$. Thus $a, b \\in S \\rightarrow a+b \\in S \\rightarrow (a+b)^{2} \\in S$.\nFinally, $(a+b)^{2} - a^{2} = 2 a b + b^{2} \\in S$ and $\\left(2 a b + b^{2}\\right) - b^{2} = 2 a b \\in S$ which implies that\n$$\n\\frac{1}{2 a b} + \\frac{1}{2 a b} = \\frac{1}{a b} \\in S \\rightarrow a b \\in S.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40442,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many integers $n$ satisfy the following conditions?\ni) $219 \\leq n \\leq 2019$,\nii) there exist $x, y \\in \\mathbb{Z}$ such that $1 \\leq x < n < y$ and $y$ is divisible by all integers from $1$ to $n$, except two numbers $x$ and $x+1$.",
    "options": [],
    "answer": "292",
    "solution": "The answer is $292$.\n\nWe can see that if $x = pq$ for some integers $p, q > 1$ and $\\gcd(p, q) = 1$ then $1 < p, q < x$ which implies that $p \\mid y$, $q \\mid y$, then $pq \\mid y$, contradiction.\n\nHence $x$ and $x+1$ must be the powers of primes. But one of these numbers is even so one of them must be the power of $2$. On the other hand, $n$ must be less than $2x$, otherwise $2x \\mid y$ leads to $x \\mid y$, contradiction. With the existence of $x$ we can easily choose $y$.\n\nThus, number $n$ satisfies the given condition if and only if there exists an exponent of $2$ less than $n$ and bigger than $n/2$, namely $x$ such that $x+1$ or $x-1$ is a power of some prime. We can check directly each range of numbers:\n\n1. For each number $219 \\leq n \\leq 255$ we can choose $x = 127^{1}$, $x+1 = 2^{7}$.\n\n2. For each number $257 \\leq n \\leq 511$ we can choose $x = 2^{8}$ and $x+1 = 257^{1}$.\n\n3. For each number from $513 \\leq n \\leq 1023$ we cannot choose any $x$ since $511$ and $513$ are not the powers of prime.\n\n4. For each number from $1025 \\leq n \\leq 2019$ we cannot choose any $x$ since $1023$ and $1025$ are not powers of prime.\n\nTherefore, the total number of integers we need to find is $37 + 255 = 292$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40443,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a triangle $ABC$ and let $M$ be the midpoint of the side $BC$. Suppose $\\angle MAC = \\angle ABC$ and $\\angle BAM = 105^\\circ$. Find the measure of $\\angle ABC$.",
    "options": [],
    "answer": "30°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40444,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEF$ be a convex hexagon satisfying $AC = DF$, $CE = FB$ and $EA = BD$. Prove that the lines connecting the midpoints of opposite sides of the hexagon $ABCDEF$ intersect in one point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M, N, P, Q, R, S$ be the midpoints of sides $AB$, $BC$, $CD$, $DE$, $EF$, $FA$, respectively, and $X, Y, Z$ be the midpoints of $AD$, $BE$, $CF$.\n\n![](attached_image_1.png)\n\nSince $AE = BD$ and the midsegments in some triangles, we get\n$$\nXQ = YM = \\frac{1}{2} \\cdot AE = \\frac{1}{2} \\cdot BD = XM = YQ,\n$$\nso $XMYQ$ is a rhombus, then $MQ$ is the perpendicular bisector of the segment $XY$. Similarly, $NR$, $PS$ are the perpendicular bisectors of $XZ$, $YZ$, so $MQ$, $NR$, $PS$ are concurrent at the circumcenter of the $\\triangle XYZ$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40445,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all primes $p$ such that there exist integers $m$ and $n$ satisfying $p = m^{2} + n^{2}$ and $p \\mid m^{3} + n^{3} + 8 m n$.",
    "options": [],
    "answer": "2, 5, 13",
    "solution": "If $m = 0$ then $p = n^{2}$ is a prime, which is impossible. Similar case happens when $n = 0$. So we may assume $m n \\neq 0$. Then $p > |m|$ and $p > |n|$. Note that\n$$\nm^{3} + n^{3} = (m + n)\\left(m^{2} - m n + n^{2}\\right) \\equiv -m n(m + n) \\pmod{p}\n$$\nSo the problem statement is equivalent to\n$$\nm n(m + n - 8) \\equiv 0 \\pmod{p}.\n$$\nSince $p > m, n$ and $p$ is prime, then $p \\mid m + n - 8$, so $m^{2} + n^{2} \\leq |m + n - 8|$. We have two cases:\n\n1. If $m + n - 8 \\geq 0$ then $m^{2} + n^{2} \\leq m + n - 8$ which means $m(m - 1) + n(n - 1) \\leq -8$. This is impossible, since $m(m - 1)$ and $n(n - 1)$ are integers and non-negative.\n\n2. If $m + n - 8 < 0$ then we get $m(m + 1) + n(n + 1) \\leq 7$. So we conclude $m, n = \\pm 3, \\pm 2, \\pm 1$.\n\nBy case work we get 3 solutions as follows:\n- $p = 2$ when $(m, n) = (1, 1)$,\n- $p = 5$ when $(m, n) = (2, 1)$,\n- $p = 13$ when $(m, n) = (-3, -2)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40446,
    "subject": "Mathematics (Multi-modal)",
    "question": "For $n \\geq 3$, it is given an $2n \\times 2n$ board with black and white squares. It is known that all border squares are black and no $2 \\times 2$ subboard has all four squares of the same color. Prove that there exists a $2 \\times 2$ subboard painted like a chessboard, i.e. with two opposite black corners and two opposite white corners.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume for the sake of contradiction that there are no $2 \\times 2$ square painted like a chessboard. Then all $2 \\times 2$ subboards are of these paintings as follow (and their rotations).\n\n![](attached_image_1.png)\n\nWe will count the length of black-white border in two ways.\n\n1. First way. Observe that from (*), each subboard $2 \\times 2$ has exactly 2 units of the black-white border. Since the border of whole board is black so each black-white border appears in exactly two $2 \\times 2$ subboards. The number of $2 \\times 2$ subboards is $(2n-1)^2$, so the total length of black-white border is\n$$\n\\frac{2(2n-1)^2}{2} = (2n-1)^2\n$$\nwhich is an odd number.\n\n2. Second way. Let color the lattice point in the given board by black-white like chessboard. When we move along the border of any white squares, we will get the black and white points alternatively, which mean any connected region of white squares has even perimeter. The entire white region is the union of disjoint white regions, so the total length of black-white border is even.\n\nThis is a contradiction so we can get the conclusion that there exists some subboard $2 \\times 2$ which colored like chessboard.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40447,
    "subject": "Mathematics (Multi-modal)",
    "question": "Quadrilateral $ABCD$ is circumscribed about a circle $\\Lambda$ and $K$, $L$, $M$, $N$ are points of tangency of sides $AB$, $BC$, $CD$, $DA$ with $\\Lambda$ respectively. Let $S$ be the intersection point of lines $KM$ and $LN$. If quadrilateral $SKBL$ is cyclic then show that $SNDM$ is also cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40448,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the biggest number of queens of two colors (black and white) one can place on a chessboard, such that no two of them beat one another (the queens do not beat through each other)?\n\nNote: The queen is able to move any number of squares vertically, horizontally or diagonally.",
    "options": [],
    "answer": "8",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40449,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c \\in \\mathbb{R}_{+}$. Prove that\n$$\n\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a} \\geq \\frac{c+a}{c+b} + \\frac{a+b}{a+c} + \\frac{b+c}{b+a}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40450,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integer $n$. Several cells of a $n \\times n$ board were coloured green in a way that no two green cells share a common segment. Is it always true (i.e. no matter which cells were coloured green) that it is possible to place exactly $n$ rooks in some not green cells of board such that no rook captures any other rook if:\n1. $n=20$\n2. $n=19$ ?",
    "options": [],
    "answer": "For n = 20: yes; for n = 19: no.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40451,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles with centers $A$ and $B$ intersect at points $M$ and $N$. Radii $A P$ and $B Q$ are parallel (on opposite sides of $A B$). If the common external tangents meet $A B$ at $D$, and $P Q$ meets $A B$ at $C$, prove that $C N D$ is a right angle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40452,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $10$ quadratic equations\n$$\nx^{2} + a_{1} x + b_{1} = 0,\\quad x^{2} + a_{2} x + b_{2} = 0,\\quad \\ldots,\\quad x^{2} + a_{10} x + b_{10} = 0\n$$\neach of them has two distinct real roots and the set of all roots is $S = \\{ \\pm 1, \\pm 2, \\ldots, \\pm 10 \\}$. Find the minimum value of the sum $T = b_{1} + b_{2} + \\cdots + b_{10}$.",
    "options": [],
    "answer": "-385",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40453,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $k > 1$ satisfy condition: there exists some nonnegative integer $n$ such that the number $A = 17^{18n} + 4 \\cdot 17^{2n} + 7 \\cdot 19^{5n}$ can be written as the product of $k$ consecutive positive integers.",
    "options": [],
    "answer": "2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40454,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $E$ be a point lies inside the parallelogram $ABCD$ such that $\\angle BCE = \\angle BAE$. Prove that the circumcenters of triangles $ABE$, $BCE$, $CDE$, $DAE$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40455,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers. Prove that\n$$\n\\frac{a^{3}}{a^{2}+b c}+\\frac{b^{3}}{b^{2}+c a}+\\frac{c^{3}}{c^{2}+a b} \\geq \\frac{\\left(a^{2}+b^{2}+c^{2}\\right)(a b+b c+c a)}{a^{3}+b^{3}+c^{3}+3 a b c}\n$$\nWhen will the equality hold?",
    "options": [],
    "answer": "Equality holds when a = b = c.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40456,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers of form $\\overline{abcd}$ such that\n$$\n\\overline{abcd} = a^{a+b+c+d} - a^{-a+b-c+d} + a.\n$$",
    "options": [],
    "answer": "2018",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40457,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AD$ be the altitude of the right angled triangle $ABC$ with $\\angle A = 90^{\\circ}$. Let $DE$ be the altitude of the triangle $ADB$ and $DZ$ be the altitude of the triangle $ADC$ respectively. Let $N$ be chosen on the line $AB$ such that $CN$ is parallel to $EZ$. Let $A'$ be the symmetric of $A$ with respect to the line $EZ$ and $I, K$ the projections of $A'$ onto $AB$ and $AC$ respectively. Prove that $\\angle NA'T = \\angle ADT$, where $T$ is the intersection point of $IK$ and $DE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that the line $AA'$ intersects the lines $EZ$, $BC$ and $CN$ at the points $L$, $M$, $F$ respectively. The line $IK$ being diagonal of the rectangle $KA'IA$ passes through $L$, which by construction of $A'$ is the middle of the other diagonal $AA'$. The triangles $ZAL$ and $ALE$ are similar, so $\\angle ZAL = \\angle AEZ$. By the similarity of the triangles $ABC$ and $DAB$ we get $\\angle ACB = \\angle BAD$. We have also that $\\angle AEZ = \\angle BAD$, therefore\n$$\n\\angle ZAL = \\angle CAM = \\angle ACB = \\angle ACM\n$$\nSince $AF \\perp CN$, we have that the right angled triangles $AFC$ and $CDA$ are equal. Thus the altitudes from the vertices $F$ and $D$ of triangles $AFC$ and $CDA$ are respectively equal. It follows that $FD \\parallel AC$ and since $DE \\parallel AC$ we get that the points $E, D, F$ are collinear. In triangle $LFT$ we have\n$$\nA'I \\parallel FT \\text{ and } \\angle LA'I = \\angle LIA',\n$$\nso $\\angle LFT = \\angle LTF$. Therefore the points $F, A', I, T$ belong to the same circle. Also $\\angle A'IN = \\angle A'FN = 90^{\\circ}$ so the quadrilateral $IA'FN$ is cyclic. Thus the points $F, A', I, N$ all lie on a circle. From the above, we infer that\n$$\n\\angle NA'T = \\angle TFN = \\angle ACF = \\angle FEZ = \\angle ADT\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40458,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number. Show that $7p + 3^{p} - 4$ is not a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40459,
    "subject": "Mathematics (Multi-modal)",
    "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 we can obtain regardless of the positions of the black squares? It is allowed to cut the rectangle along the grid lines.",
    "options": [],
    "answer": "25",
    "solution": "In the first image we have position for 4 black cells, such that the biggest rectangle without black is $25$. Now let's prove that for any configuration there exists a rectangle of area $25$. Assume for some configuration the biggest rectangle is at most $24$. Let's divide the board into 4 squares of size $5$ like in second image. Each of them must contain at least one black cell, so the last column and first row have no black cells.\n\nAnalogously first column and last row have no black cells. By considering squares $a1$-$e5$, $a7$-$e11$, $g1$-$k5$ and $g7$-$k11$ we conclude that $6$-th row and $f$ column doesn't contain black cell.\n\nNow assumes that there is a black cell in column $b$. Then each of rectangles $c1$-$k3$, $c9$-$k11$ contain at least one black cell, so either orange or green rectangle doesn't contain a black cell. That rectangle together with layer $c6$-$k6$ will form a rectangle of area $27$ that has no black cell. Contradiction. So column $b$ has no black cell. The same for column $l$ and rows $2$ and $10$. So we get the last image configuration, where in gray cells can not be painted black. Then each white square $3 \\times 3$ must contains exactly one black cell.\n\nAssume $c9$ is black. Then consider $d7$-$h11$ and $a6$-$k8$. We conclude that black cell is in square $g7$-$h8$. Same goes for $d4$-$e5$. Then by looking at the rectangles $i1$-$k11$ and $a1$-$k3$ we conclude that $i3$ must be black.\n\n![](attached_image_1.png)\n\nFrom rectangles $a1$-$d8$ and $a1$-$h4$ follows that $d4$ and $h8$ are black. Be then $e1$-$g11$ is white and has area $33$. So we conclude that $c3$, $c9$, $i3$, $i9$ are white. Finally, each of rectangles $c4$-$c8$, $d9$-$h9$, $d3$-$h3$, $i4$-$i8$ must have exactly one black square. But then central square $d4$-$h8$ contains $25$ white cells, a contradiction.\n\n![](attached_image_2.png)\n\n![](attached_image_3.png)\n\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40460,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, such that $\\angle ACB = 45^{\\circ}$, let $O$ and $H$ be the circumcenter and orthocenter, respectively. The line passing through $O$ and perpendicular to $CO$ intersects $AC$ and $BC$ at $K$ and $L$, respectively. Prove that the perimeter of $KLH$ is equal to the diameter of the circumcircle of triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that $AH$, $BH$ cut $(O)$ at the second points $E$, $F$. By angle chasing, we can see that $H$, $E$ are symmetric with respect to $BC$. Note that\n$$\n\\angle HBC = 90^{\\circ} - \\angle C = 45^{\\circ}\n$$\nso $\\angle CBE = \\angle CBH = 45^{\\circ}$.\nThus $\\angle CFE = 45^{\\circ}$, which implies that $\\angle COE = 90^{\\circ}$. Similarly, $\\angle COF = 90^{\\circ}$, then four points $E$, $L$, $K$, $F$ are collinear. From this, we conclude that $LH = LE$ and $KH = KF$ which mean\n$$\n\\begin{aligned}\n\\text{perimeter}(HKL) & = HL + LK + KH \\\\\n& = EL + LK + KF = EF = \\operatorname{diameter}(O).\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40461,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $a, b, c, d$ are pairwise distinct positive integers such that $a+b=c+d=p$ for some odd prime $p>3$. Prove that $a b c d$ is not a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that $a b c d = n^{2}$ for some $n \\in \\mathbb{Z}^{+}$. We can suppose that $a < c < d < b$. From this, we have $b d + a c - a d - b c = (b - a)(d - c) > 0$, thus\n$$\na d + b c < \\frac{1}{2}(a d + b c + b d + a c) = \\frac{(a + b)(c + d)}{2} = \\frac{p^{2}}{2} .\n$$\nDenote $\\gcd(a d, b c) = k \\in \\mathbb{Z}^{+}$ then $a d = k u^{2}$, $b c = k v^{2}$ with $u, v \\in \\mathbb{Z}^{+}$, and $u < v$, $\\gcd(u, v) = 1$. Since $a, b, c, d$ are coprime to $p$ then $\\gcd(k, p) = 1$. We have\n$$\nk\\left(v^{2} - u^{2}\\right) = b c - a d = (p - a) c - a(p - c) = p(c - a) .\n$$\nSo $p \\mid v^{2} - u^{2}$ since $\\gcd(k, p) = 1$. Hence, $p \\mid v - u$ or $p \\mid v + u$. In both cases, we always have $u + v \\geq p$. From this, we can conclude that\n$$\na d + b c = k\\left(u^{2} + v^{2}\\right) > k \\frac{(u + v)^{2}}{2} \\geq \\frac{p^{2}}{2}\n$$\nby AM-GM inequality. But this contradicts the inequality we stated above, then $a b c d$ cannot be a perfect square.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40462,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let pairwise different positive integers $a$, $b$, $c$ with $\\operatorname{gcd}(a, b, c) = 1$ are such that\n$$\na \\mid (b - c)^2, \\quad b \\mid (c - a)^2, \\quad c \\mid (a - b)^2.\n$$\nProve that there is no non-degenerate triangle with side lengths $a$, $b$ and $c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, we will show that $a$, $b$, $c$ are pairwise coprime. Denote $d = \\operatorname{gcd}(b, c)$ and suppose that $d > 1$. Take $p$ as a prime divisor of $d$. We have\n$$\n\\left\\{\\begin{array}{l}\n p \\mid b \\\\\n p \\mid c\n\\end{array}\\right. \\Rightarrow \\left\\{\\begin{array}{l}\n p \\mid (c - a)^2 \\\\\n p \\mid (a - b)^2\n\\end{array}\\right. \\Rightarrow \\left.\\begin{array}{l}\np \\mid c - a \\\\\np \\mid a - b\n\\end{array} \\Rightarrow p \\mid a.\\right.\n$$\nThis implies that $p \\mid \\operatorname{gcd}(a, b, c)$, a contradiction. So $\\operatorname{gcd}(b, c) = 1$. Similar with $\\operatorname{gcd}(c, a)$, $\\operatorname{gcd}(a, b)$.\n\nSuppose that $a$, $b$, $c$ are side-lengths of some triangle then\n$$\nb + c > a, \\quad c + a > b, \\quad a + b > c.\n$$\nSo by putting $x = (a + b - c)(b + c - a)(c + a - b) > 0$, we can see that $x > 0$.\n\nWe have $a \\mid (b - c)^2 - a^2 = (b - c - a)(b - c + a)$, thus $a \\mid x$. Similarly, we have $b \\mid x$, $c \\mid x$ and since $\\operatorname{gcd}(a, b) = \\operatorname{gcd}(b, c) = \\operatorname{gcd}(c, a) = 1$ so $a b c \\mid x$. Hence,\n$$\na b c \\leq (a + b - c)(b + c - a)(c + a - b).\n$$\nFrom AM-GM, we have $(a + b - c)(b + c - a) \\leq \\left(\\frac{a + b - c + b + c - a}{2}\\right)^2 = b^2$ and similar inequalities as $(b + c - a)(c + a - b) \\leq c^2$, $(c + a - b)(a + b - c) \\leq a^2$ then\n$$\n(a + b - c)(b + c - a)(c + a - b) \\leq a b c.\n$$\nThese imply the equality must be hold, so $a = b = c$, contradiction since $a$, $b$, $c$ pairwise distinct. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40463,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let be given a positive integer $n \\geq 3$. Consider integers $a_{1}, a_{2}, \\ldots, a_{n} > 1$ with the product equals to $A$ such that: for each $k \\in \\{1,2, \\ldots, n\\}$ then the remainder when $\\frac{A}{a_{k}}$ divided by $a_{k}$ are all equal to $r$. Prove that $r \\leq n-2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40464,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $a_{1}, a_{2}, \\ldots, a_{n}$ be any real numbers. Prove that there exists $m, k \\in \\{1,2, \\ldots, n\\}$ such that\n$$\n\\left|\\sum_{i=1}^{m} a_{i}-\\sum_{i=m+1}^{n} a_{i}\\right| \\leq\\left|a_{k}\\right| .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote\n$$\nS_{x}=\\left|\\sum_{i=1}^{x} a_{i}-\\sum_{i=x+1}^{n} a_{i}\\right| .\n$$\nObviously $S_{0}=-S_{n}$ and $S_{x+1}-S_{x}=2 a_{x+1}$. We may assume that $S_{n} \\geq 0$ and consider $m$ for which $S_{m} \\leq 0$ and $S_{m+1} \\geq 0$. Then\n$$\n\\left|S_{m}\\right|+\\left|S_{m+1}\\right|=\\left|S_{m}-S_{m+1}\\right|=2\\left|a_{m+1}\\right|\n$$\nso either $\\left|S_{m}\\right|$ or $\\left|S_{m+1}\\right|$ is at most $a_{m+1}$ which is at $\\operatorname{most}_{\\max _{i}}\\left|a_{i}\\right|=a_{k}$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40465,
    "subject": "Mathematics (Multi-modal)",
    "question": "All the cells in a $8 \\times 8$ board are colored white. Omar and Asaad play the following game: in the beginning Omar colors $n$ cells red, then Asaad chooses 4 rows and 4 columns and colors them black. Omar wins if there is at least one red cell. Find the least possible value for $n$ such that Omar can always win regardless of Asaad's move.",
    "options": [],
    "answer": "25",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40466,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let be given a positive integer $n > 1$. Find all polynomials $P(x)$ non constant, with real coefficients such that\n$$\nP(x) P\\left(x^{2}\\right) \\ldots P\\left(x^{n}\\right) = P\\left(x^{\\frac{n(n+1)}{2}}\\right)\n$$\nfor all $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "If n is even: P(x) = x^m for some positive integer m. If n is odd: P(x) = ± x^m for some positive integer m.",
    "solution": "Denote $m$ as degree of $P(x)$ then by comparing the leading coefficients $a$ of two sides, we get $a^{n} = a$. We consider two cases based on the parity of $n$.\n\n1. If $n$ is even then $a = 1$. Put $P(x) = x^{m} + Q(x)$ with $Q \\in \\mathbb{R}[x]$ and if $Q(x) \\equiv 0$, we obtain $P(x) = x^{m}$ as a solution. Otherwise, suppose that $Q(x) \\neq 0$ and $\\operatorname{deg} Q(x) = k < m$. So\n$$\n\\left(x^{m} + Q(x)\\right)\\left(x^{2m} + Q\\left(x^{2}\\right)\\right) \\cdots \\left(x^{mn} + Q\\left(x^{n}\\right)\\right) - x^{\\frac{mn(n+1)}{2}} = Q\\left(x^{\\frac{n(n+1)}{2}}\\right).\n$$\nThe degree of LHS is $\\frac{mn(n+1)}{2} - (m - k)$, while the degree of RHS is $\\frac{kn(n+1)}{2}$. Note that\n$$\n\\frac{mn(n+1)}{2} - (m - k) - \\frac{kn(n+1)}{2} = \\frac{(m - k)(n + 2)(n - 1)}{2} > 0\n$$\nwhich implies that the degree of both sides cannot be equal. Thus $Q(x) \\equiv 0$.\n\n2. If $n$ is odd then $a = \\pm 1$ and the case $a = 1$ can be solved similarly as above. For the case $a = -1$, put $P(x) = -\\left(x^{m} + Q(x)\\right)$ then process similarly, we get $P(x) = -x^{m}$ as another solution.\n\nHence, $P(x) = x^{m}$ for some $m \\in \\mathbb{Z}^{+}$ when $n$ is even and $P(x) = \\pm x^{m}$ when $n$ is odd. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40467,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let sequences of real numbers $\\left(x_{n}\\right)$ and $\\left(y_{n}\\right)$ satisfy $x_{1}=y_{1}=1$ and\n$$\nx_{n+1}=\\frac{x_{n}+2}{x_{n}+1} \\quad \\text{ and } \\quad y_{n+1}=\\frac{y_{n}^{2}+2}{2 y_{n}} \\text{ for } n=1,2, \\ldots .\n$$\nProve that $y_{n+1}=x_{2^{n}}$ holds for $n=0,1,2, \\ldots$\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that, since $x_{1}=1$, $x_{n+1}=\\frac{x_{n}+2}{x_{n}+1}$ we can write\n$$\n\\begin{aligned}\nx_{n+1} & =\\frac{x_{n} \\cdot x_{1}+2}{x_{n}+x_{1}}=\\frac{\\frac{x_{n-1}+2}{x_{n-1}+1} \\cdot x_{1}+2}{\\frac{x_{n-1}+2}{x_{n-1}+1}+x_{1}}=\\frac{x_{n-1}\\left(x_{1}+2\\right)+2\\left(x_{1}+1\\right)}{x_{n-1}\\left(x_{1}+1\\right)+\\left(x_{1}+2\\right)} \\\\\n& =\\frac{x_{n-1} \\cdot \\frac{x_{1}+2}{x_{1}+1}+2}{x_{n-1}+\\frac{x_{1}+2}{x_{1}+1}}=\\frac{x_{n-1} \\cdot x_{2}+2}{x_{n-1}+x_{2}}\n\\end{aligned}\n$$\nSo by the same way, we can prove that\n$$\nx_{n}=\\frac{x_{k} x_{n-k}+2}{x_{k}+x_{n-k}} \\text{ for any } 1 \\leq k \\leq n-1 \\text{. }\n$$\nLet $a_{n+1}=x_{2^{n}}$ for $n \\geq 0$ then $a_{1}=x_{1}=1$, $a_{2}=x_{2}$, $a_{3}=x_{4}$, $\\ldots$ Take $k=2^{n-1}$ in above formula, we get\n$$\nx_{2^{n}}=\\frac{x_{2^{n-1}}^{2}+2}{2 x_{2^{n-1}}} \\text{ so } a_{n+1}=\\frac{a_{n}^{2}+2}{2 a_{n}} .\n$$\nThus by induction, one can get $a_{n}=y_{n}$ for all $n \\geq 1$. Therefore, $y_{n+1}=a_{n+1}=x_{2^{n}}$ for any $n \\geq 0$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40468,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 3 clubs $A, B, C$ with non-empty members. For any triplet of members $(a, b, c)$ with $a \\in A, b \\in B, c \\in C$, two of them are friend and two of them are not friend (here the friend relationship is bidirectional). Prove that one of these statements must be true\n\n1. There exist one student from $A$ that knows all students from $B$.\n2. There exist one student from $B$ that knows all students from $C$.\n3. There exist one student from $C$ that knows all students from $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove the statement by induction on the maximum number of members in clubs $A, B, C$.\n\nFor $n=1$, each club has exactly one member and the statement is obviously true.\n\nSuppose that when the maximum numbers in three clubs is $n \\geq 1$, then one of three above conditions holds. Assume that $a \\in A$ is friend of all members in $B$. Consider a new member $x$ to make the maximum number of members increased by 1. It is easy to see that if $x \\in A$ or $x \\in C$, then the conditions are still true (since the members in $B$ remains unchanged).\n\nIf $x \\in B$, denote $B=\\{b_{1}, b_{2}, \\ldots, b_{n}, b_{n+1}\\}$ with $b_{n+1} \\equiv x$. In case $(a, x)$ are friend, the condition 1) is true and we are done. So we may assume that $(a, x)$ are not friend. Partition $C$ into two subsets, $C_{1}$ has members that are friend of $a$, and $C_{2}$ has members that are not friend of $a$. Take $c_{1} \\in C_{1}$ and $c_{2} \\in C_{2}$ (if any).\n\nFor $1 \\leq k \\leq n$, consider $(a, b_{k}, c_{1})$, we have $(a, b_{k})$ are friend (since before adding $x$, member $a$ is friend of all members of $B$) and $(a, c_{1})$ are friend so $(b_{k}, c_{1})$ are not friend (by the given condition).\n\nConsider $(a, x, c_{2})$, we have $(x, c_{2})$ are friend (since $(a, x), (a, c_{2})$ are not friend).\n\n- If $x$ is friend of all members in $C_{1}$ then $x$ is friend of all members in $C = C_{1} \\cup C_{2}$ and condition 2) is true.\n- Otherwise, $\\exists c_{1} \\in C$ such that $c_{1}$ is not friend of $x$, also is not friend of all members in $B$. If $\\exists a' \\in A$ such that $(a', c_{1})$ are not friend, then consider $(a', b_{k}, c_{1})$ with $1 \\leq k \\leq n+1$ then $a'$ is friend of all $b_{k}$, condition 1) is true. Otherwise, $c_{1}$ are friend of all members in $A$ and condition 3) is true.\n\nHence, in all cases, the condition is always true.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40469,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let non-constant polynomial $f(x)$ with real coefficients is given with the following property: for any positive integer $n$ and $k$, the value of expression\n$$\n\\frac{f(n+1) f(n+2) \\ldots f(n+k)}{f(1) f(2) \\ldots f(k)} \\in \\mathbb{Z} .\n$$\nProve that $f(x)$ is divisible by $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality one may assume that $f(1) \\in \\mathbb{Z}$. Since for all positive $k$, we have $\\frac{f(k+1)}{f(1)}$ is integer, then we conclude that on all positive integer points our polynomial gets integer values. Assume that $\\operatorname{deg}(f)=d$ then, according to Lagrange interpolation formula we get\n$$\nf(x)=\\sum_{i=1}^{d+1} f(i) \\frac{(x-1) \\ldots(x-(i-1))(x-(i+1)) \\ldots(x-(d+1))}{(i-1) \\ldots 1(-1) \\ldots(i-(d+1))},\n$$\nand all numbers are rational, so $p(x)$ is a polynomial with rational coefficients.\nBy multiplying by a constant we can get $f(x)$ with integer coefficients. If $f(0)=0$ then we are done. Assume that $f(0) \\neq 0$. Let's fix positive integer $k$ and denote $f(n)=f(n+1) f(n+2) \\ldots f(n+k)$ and $M=f(0)$.\nAccording to the problem condition, we have $\\frac{f(n)}{M} \\in \\mathbb{Z}$ for all positive integer $n$. Since $f(x)$ is polynomial with integer coefficients, then\n$$\nf(-1) \\equiv f(2|M|-1) \\equiv 0 \\quad(\\bmod M),\n$$\nwhich means\n$$\n\\frac{f(-1)}{M}=\\frac{f(0)}{f(k)} \\in \\mathbb{Z},\n$$\nfor all positive integers $k$. It means $f(0)=0$ or $f(x)$ is divisible by $x$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40470,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence $(a_{1}, a_{2}, \\ldots, a_{k})$ consisting of pairwise different cells of an $n \\times n$ board is called a cycle if $k \\geq 4$ and cell $a_{i}$ shares a side with cell $a_{i+1}$ for every $i=1,2, \\ldots, k$, where $a_{k+1}=a_{1}$. We will say that a subset $X$ of the set of cells of a board is malicious if every cycle on the board contains at least one cell belonging to $X$. Determine all real numbers $C$ with the following property: for every integer $n \\geq 2$ on an $n \\times n$ board there exists a malicious set containing at most $C n^{2}$ cells.",
    "options": [],
    "answer": "All real numbers C ≥ 1/4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40471,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $2016$ digits written around a circle. If you write down these digits going around the circle clockwise and starting with a certain digit, you get a number divisible by $81$. Prove that if you write down the digits going around the circle clockwise, but starting with any other digit, you will again get a number divisible by $81$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40472,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $A'$, $B'$, $C'$ as the midpoints of $BC$, $CA$, $AB$ respectively. The circle $(\\omega_A)$ of center $A$ with sufficiently large radius cuts $B'C'$ at $X_1$, $X_2$. Define circles $(\\omega_B)$, $(\\omega_C)$ with $Y_1$, $Y_2$, $Z_1$, $Z_2$ similarly. Suppose that these circles have the same radius. Prove that $X_1$, $X_2$, $Y_1$, $Y_2$, $Z_1$, $Z_2$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H$ be the orthocenter of $\\triangle ABC$. Since $AH \\perp X_1X_2$ and $AX_1 = AX_2$, then $HX_1 = HX_2$. Similarly, $HY_1 = HY_2$ and $HZ_1 = HZ_2$.\n\n![](attached_image_1.png)\n\nDenote $R$ as the radius of the three circles $(\\omega_A)$, $(\\omega_B)$, $(\\omega_C)$. As $HB \\perp Y_1C'$, $HC \\perp Z_1B'$, we have\n$$\n\\begin{aligned}\n& HY_1^2 - R^2 = HY_1^2 - BY_1^2 = HC^2 - BC^2 \\\\\n& HZ_1^2 - R^2 = HZ_1^2 - CZ_1^2 = HB'^2 - CB'^2\n\\end{aligned}\n$$\nIn addition, $AH \\perp B'C'$, so\n$$\nHC'^2 - BC'^2 = HC'^2 - AC'^2 = HB'^2 - AB'^2 = HB'^2 - CB'^2\n$$\nFrom that $HY_1 = HZ_1$. Similarly, $HX_1 = HY_1 = HZ_1$ and $HX_2 = HY_2 = HZ_2$.\n\nFrom (1) and (2), the orthocenter $H$ is indeed the center of the circle which goes through the six points $X_1$, $X_2$, $Y_1$, $Y_2$, $Z_1$, $Z_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40473,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute nonisosceles triangle with incenter $I$ and $(d)$ is an arbitrary line tangent to $(I)$ at $K$. The lines passes through $I$, perpendicular to $IA$, $IB$, $IC$ cut $(d)$ at $A_{1}$, $B_{1}$, $C_{1}$ respectively. Suppose that $(d)$ cuts $BC$, $CA$, $AB$ at $M$, $N$, $P$ respectively. The lines through $M$, $N$, $P$ and respectively parallel to the internal bisectors of $A$, $B$, $C$ in triangle $ABC$ meet each other to define a triangle $XYZ$. Prove that three lines $AA_{1}$, $BB_{1}$, $CC_{1}$ are concurrent and $IK$ is tangent to the circle $(XYZ)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that $(d)$ is tangent to $(I)$ at $K$. Denote $D$, $E$, $F$ as the tangent points of $(I)$ with the sides $BC$, $CA$, $AB$. Let $V$ be the projection of $K$ onto $A_{1}I$ and $U = AI \\cap EF$. It is easy to see that\n$$\nIU \\cdot IA = IV \\cdot IA_{1} = r^{2}\n$$\nwith $r$ is the radius of $(I)$.\n\nConsider the inversion $\\Omega$ of center $I$, and power $r^{2}$ then $A \\leftrightarrow U$, $A_{1} \\leftrightarrow V$ so $AA_{1} \\leftrightarrow$ $(IUV)$.\n\nDefine $B'$, $C'$ similarly, then to prove the original problem, we just need to show that the circles of diameter $IA'$, $IB'$, $IC'$ have a common point differs from $I$.\n\nNote that $A'$, $B'$, $C'$ belongs to the Simson's line of the point $K$ respect to triangle $DEF$ so by taking the homothety of center $I$, ratio $\\frac{1}{2}$ then the center of above circles are collinear. Thus they share some other common point differs from $I$; the image of this point through the inversion is the concurrent point of three line $AA_{1}$, $BB_{1}$, $CC_{1}$.\n\nContinue, consider the figure as below, the other case of position of points can be processed similarly. Denote $S$ as the reflection of $I$ through $K$. Then we have $YZ \\parallel AI$, $YX \\parallel IC$ so\n$$\n\\angle MYP = \\angle XYZ = (YX, YZ) = (IA, IC) = 90^{\\circ} - \\frac{\\angle B}{2} .\n$$\nOn the other hand, we have\n$\\angle MSP = \\angle MIP = \\angle KIM - \\angle KIP = \\frac{\\angle KID - \\angle KIE}{2} = \\frac{1}{2} \\angle DIF = 90^{\\circ} - \\frac{\\angle B}{2}$, so these points $M$, $P$, $Y$, $S$ are concyclic which implies that $S \\in (MPY)$.\n\nSimilarly, $S \\in (NPX)$, $S \\in (MNZ)$. From this, we can conclude that $S$ is the Miquel's point of the completed quadrilateral of 6 vertices $M$, $N$, $P$, $X$, $Y$, $Z$. Thus $S \\in (XYZ)$.\n\nContinue, note that by doing the similar angle chasing, we can see that two triangles $XYZ$, $DEF$ are similar (with the same direction). Hence, there exist a spiral similarity $\\Omega$ transforms $XYZ \\rightarrow DEF$ with the angle $90^{\\circ}$ (since their sides are perpendicular pairwise). Draw the diameter $KG$ of circle $(I)$, then we have\n$$\n\\angle GDE = \\angle GKE = \\angle IKE = \\angle KNI = \\angle KNS = \\angle PXS = \\angle SXY .\n$$\nSimilarly, $\\angle GED = \\angle SYX$. Thus $\\Omega: S \\rightarrow G$ (two corresponding points). Denote $J$ as the circumcenter of triangle $XYZ$ then $\\Omega: J \\rightarrow I$ so we have $JS \\perp IG$. But these points $I$, $G$, $K$, $S$ are collinear so $\\angle JSK = 90^{\\circ}$ which implies that $IK$ is tangent to the circle $(XYZ)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40474,
    "subject": "Mathematics (Multi-modal)",
    "question": "The triangle $ABC$ ($AB > BC$) is inscribed in the circle $\\Omega$. On the sides $AB$ and $BC$, the points $M$ and $N$ are chosen, respectively, so that $AM = CN$. The lines $MN$ and $AC$ intersect at point $K$. Let $P$ be the center of the inscribed circle of triangle $AMK$, and $Q$ the center of the excircle of the triangle $CNK$ tangent to side $CN$. Prove that the midpoint of the $\\operatorname{arc} ABC$ of the circle $\\Omega$ is equidistant from $P$ and $Q$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $T$ be the second intersection of two circles $(BMN)$ and $(O)$. We have\n$$\n\\angle TAB = \\angle TCB, \\quad \\angle TMB = \\angle TNB,\n$$\nand $AM = CN$, so $\\triangle TAM \\cong \\triangle TCN$. Then $TA = TB$, which means that $T$ is the midpoint of the arc $BAC$ of circle $(O)$.\n\n![](attached_image_1.png)\n\nOn the other hand, we also have\n$$\n\\angle TCK = \\angle TBA = \\angle TNK,\n$$\nthen $T N C K$ is cyclic. Similarly, $T M A K$ is also cyclic.\n\nSince $AM = CN$, it is easy to see that $(KAM)$ and $(KCN)$ are equal. So, if we call $D, E$ the midpoints of $\\operatorname{arcs} AM$ and $CN$ of circles $(KAM)$ and $(KCN)$, respectively, then two isosceles triangles $DAM$ and $ECN$ are congruent. But we know that $D, E$ are also the circumcenters of $\\triangle PAM$ and $\\triangle QCN$, so $DP = EQ$.\n\nAt last, from $(KAM)$ and $(KCN)$ being equal, we get $TD = TE$, then $TP = TQ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40475,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider equilateral triangle $ABC$ and suppose that there exist three distinct points $X, Y, Z$ lie inside triangle $ABC$ such that\n\ni) $AX = BY = CZ$.\n\nii) The triplets of points $(A, X, Z), (B, Y, X), (C, Z, Y)$ are collinear in that order.\n\nProve that $XYZ$ is an equilateral triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40476,
    "subject": "Mathematics (Multi-modal)",
    "question": "The altitudes $A A_{1}$ and $B B_{1}$ of an acute-angled triangle $A B C$ meet at point $O$. Let $A_{1} A_{2}$ and $B_{1} B_{2}$ be the altitudes of triangles $O B A_{1}$ and $O A B_{1}$ respectively. Prove that $A_{2} B_{2}$ is parallel to $A B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40477,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle with $O$, $H$ as circumcenter and orthocenter, respectively. Prove that the nine-point circles of $AHO$, $BHO$, $CHO$ have two common points.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$, $N$, $P$ be the midpoints of $BC$, $CA$, $AB$ and $D$ be the projection of $A$ on $BC$. Denote $(E)$ as the Euler circle of $\\triangle ABC$; then $M$, $N$, $P$, $D$ lie on $(E)$ and $E$ is the midpoint of $OH$. Similarly, let $H_a$, $O_a$ be the midpoints of $AH$, $AO$ and $K$ be the projection of $A$ on $OH$; then $H_a$, $O_a$, $K$, $E$ lie on the Euler circle of $\\triangle AOH$.\n\nDenote $J$ as the symmetric point of $K$ through $NP$. Foremost, we will prove that $J$ is the intersection of $(KH_aO_a)$ and $(E)$.\n\nIndeed, it is easy to see that $O$, $E$ are the circumcenter and orthocenter of $\\triangle MNP$. From the midsegment, we have\n$$\n\\overrightarrow{EO_a} = \\frac{1}{2} \\cdot \\overrightarrow{HA} = \\overrightarrow{MO},\n$$\nso $O_a$ is symmetric with $E$ through $NP$. Then $EO_aJK$ is an isosceles trapezoid, which means that $J \\in (EO_aK)$.\n\nOn the other hand, since $O$ is the orthocenter of $\\triangle MNP$, if we denote $O'$ as the symmetric point of $O$ through $NP$, then $O' \\in (E)$. In addition, $MO' \\perp MD$, so $DO'$ is the diameter of $(E)$. And by the symmetry, we get\n$$\n\\angle DJO' = \\angle AKO = 90^\\circ,\n$$\nwhich implies that $J \\in (E)$.\n\nFrom (1) and (2), we get $J$ is the intersection of $(KH_aO_a)$ and $(E)$. Note that $J \\in (E)$ and $O'J$ is symmetric to $OH$ through $NP$, which implies that $J$ is the anti-Steiner point of $OH$ with respect to $\\triangle MNP$. Similarly, $J$ lies on the Euler circles of $\\triangle BOH$, $\\triangle COH$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40478,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $19$ integer numbers are given. Let Hamza writes on the paper the greatest common divisor for each pair of numbers. It occurs that the difference between the biggest and smallest numbers written on the paper is less than $180$. Prove that not all numbers on the paper are different.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_{1}, a_{2}, \\ldots, a_{19}$ be the given numbers and suppose on the contrary that the set $S$ of all $\\binom{19}{2} = 171$ numbers, which are written on the paper, are all different,\n$$\nd_{1} < d_{2} < \\ldots < d_{171}\n$$\nDenote $k$ as the number of even values among the $19$ given numbers, and $t$ as the number of even values in $S$. It is easy to see that for any $d \\in S$, there exist $a_{x}, a_{y}$ such that $d = \\operatorname{gcd}(a_{x}, a_{y})$; and this number $d$ is even if and only if $a_{x}, a_{y}$ are both even. Hence, we have $t = \\binom{k}{2}$.\n\nThen the number of odd values in $S$ is $171 - t$. Since $d_{171} - d_{1} < 180$ then the number of even values and odd values in $S$ does not exceed $90$, which implies that\n$$\n\\left\\{\n\\begin{array}{l}\nt \\leq 90 \\\\\n171 - t \\leq 90\n\\end{array} \\text{ so } 81 \\leq \\binom{k}{2} \\leq 90 .\\right.\n$$\nNote that\n$$\n\\binom{13}{2} = 78 < 91 = \\binom{14}{2}\n$$\nso there does not exist positive integer $k$ such that $81 \\leq \\binom{k}{2} \\leq 90$, contradiction. Hence, the numbers written on the paper cannot be all different.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40479,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$ $\\angle B = 60^{\\circ}$, $O$ is the circumcenter, and $L$ is the foot of an angle bisector of angle $B$. The circumcircle of triangle $BOL$ meets the circumcircle of $ABC$ at point $D \\neq B$. Prove that $BD \\perp AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40480,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$ be a monic polynomial of degree $100$ with $100$ distinct noninteger real roots. Suppose that each of polynomials $P\\left(2x^{2}-4x\\right)$ and $P\\left(4x-2x^{2}\\right)$ has exactly $130$ distinct real roots. Prove that there exist non constant polynomials $A(x), B(x)$ such that $A(x)B(x)=P(x)$ and $A(x)=B(x)$ has no root in $(-1 ; 1)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $P(x) = (x - a_1)(x - a_2) \\cdots (x - a_{100})$ with $a_1, a_2, \\ldots, a_{100}$ are real roots of polynomial $P(x)$. So\n$$\nP\\left(2x^{2}-4x\\right) = 0 \\Leftrightarrow 2x^{2}-4x-a_i = 0 \\text{ for } 1 \\leq i \\leq 100.\n$$\nThis equation cannot have $1$ root since $\\Delta = 16 + 8a_i \\neq 0$. So each equation can have $0$ or $2$ roots. Note that $P\\left(2x^{2}-4x\\right)$ has $130$ roots so there are $\\frac{130}{2} = 65$ equations have $2$ roots, which mean there are $65$ numbers $a_i > -2$ and $35$ numbers $a_i < -2$.\n\nBy the same way, we have\n$$\nP\\left(4x-2x^{2}\\right) = 0 \\Leftrightarrow 2x^{2}-4x+a_i = 0\n$$\nand $\\Delta = 16 - 8a_i$ for any $1 \\leq i \\leq 100$. And there are $65$ numbers $a_i < 2$ and $35$ numbers $a_i > 2$.\n\nBy applying the principle of inclusion and exclusion, there are\n$$\n65 + 65 - 100 = 30 \\text{ numbers } a_i \\in (-2 ; 2).\n$$\n\nSuppose that $a_1 < a_2 < \\ldots < a_{35} < -2 < a_{36} < \\ldots < a_{64} < 2 < a_{65} < \\ldots < a_{100}$; and denote $M, N, K$ as the subsets of these numbers with the indices $1 \\rightarrow 35, 36 \\rightarrow 64, 65 \\rightarrow 100$.\n\nTake $A_1(x) = \\prod_{m \\in M}(x - m)$, $B(x) = \\prod_{n \\in N}(x - n)$, $A_2(x) = \\prod_{k \\in K}(x - k)$ then we will prove that\n$$\n|A_1(x_0) \\cdot A_2(x_0)| > |B(x_0)|\n$$\nfor any number $x_0 \\in (-1 ; 1)$.\n\nNote that $\\forall x_0 \\in (-1 ; 1)$ then $|x_0 - m| > 1, \\forall m < -2$ and $|x_0 - k| > 1, \\forall k > 2$. We can suppose that $x_0 > 0$ and for any $n \\in N$, we have two cases:\n\n1. If $n > 0$ then $n \\in (0 ; 2)$ and $|x_0 - n| < |x_0 - 2| < |x_0 - k|$ with any $k > 2$.\n2. If $n < 0$ then $n \\in (-2 ; 0)$ and $|x_0 - n| < |x_0 - (-2)| < |x_0 - m|$ with any $m < -2$.\n\nSo in all cases of $n$ respect to the factor $x_0 - n$ in $B(x_0)$, we can choose factor from $A_1(x_0)$ or $A_2(x_0)$ with absolute value greater than it, and since $|M|, |K| > |N|$, we always can do that. Hence $|A_1(x_0) \\cdot A_2(x_0)| > |B(x_0)|, \\forall x_0 \\in (-1 ; 1)$ which mean this equation has no solution.\n\nTherefore, we can choose $A(x) = A_1(x)A_2(x)$ and $B(x)$ to satisfy the given condition. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40481,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Each number from $1, 2, 3, \\ldots, 1000$ was painted one of $n$ colors. It turned out that every two distinct numbers, one of which is a divisor of the other one, have different colors. Find the smallest $n$ for which such situation is possible.",
    "options": [],
    "answer": "10",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40482,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle inscribed in a circle $(\\omega)$ and $I$ is the incenter. Denote $D, E$ as the intersection of $AI, BI$ with $(\\omega)$. And $DE$ cuts $AC, BC$ at $F, G$ respectively. Let $P$ be a point such that $PF \\parallel AD$ and $PG \\parallel BE$. Suppose that the tangent lines of $(\\omega)$ at $A, B$ meet at $K$. Prove that three lines $AE, BD, KP$ are concurrent or parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that $KA$ cuts $PF$ at $M$, $KB$ cuts $PG$ at $N$. By angle chasing, we have\n$$\n\\angle IEF = \\angle BAI = \\angle FAI,\n$$\nthen $AIFE$ is cyclic. In addition, we get\n$$\n\\begin{aligned}\n\\angle AMF = \\angle KAI & = \\angle KAB + \\angle BAI \\\\\n& = \\angle AEI + \\angle FEI = \\angle AEF\n\\end{aligned}\n$$\nthen $AIFM$ is cyclic.\n\n![](attached_image_1.png)\n\nSummarily, points $A, I, F, E, M$ lie on a circle. And similarly, points $B, I, G, D, N$ also lie on a circle. We have known that $DE$ is perpendicular bisector of $CI$, so it is easy to see that $IFCG$ is a rhombus. From that above, we get\n$$\n\\angle AMI = \\angle AEI = \\angle KAB,\n$$\nthen $IM \\parallel AB$. Similarly $IN \\parallel AB$, implies that $M, N, I$ are collinear. But from $PF\\parallel AD$, $PG\\parallel BE$, we have $AIFM$ and $BIGN$ are isosceles trapezoids, so\n$$\n\\begin{equation*}\nAM = IF = IG = BN, \\tag{1}\n\\end{equation*}\n$$\nthen $ABNM$ is also an isosceles trapezoid, which means that it is cyclic.\n\nIn the other hand, we get\n$$\n\\begin{equation*}\n\\angle PFG = \\angle MFE = \\angle MIE = \\angle BIN = \\angle ING, \\tag{2}\n\\end{equation*}\n$$\nso $FGNM$ is cyclic.\n\nFrom (1) and (2), we have $PK$ is the radical axis of two circles $(AIE)$ and $(BID)$. Now on, we can easily take the conclusion of this problem. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40483,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is given a graph whose vertices are positive integers and an edge between numbers $a$ and $b$ exists if and only if\n$$\na+b+1 \\mid a^{2}+b^{2}+1 .\n$$\nIs this graph connected?",
    "options": [],
    "answer": "Yes, the graph is connected.",
    "solution": "If $x, y \\in \\mathbb{Z}^{+}$, define $x \\leftrightarrow y$ if and only if $x, y$ are connected by some edge. We have for all $a \\in \\mathbb{Z}^{+}$,\n$$\na^{2}+a+1 \\mid (a^{2}-a+1)(a^{2}+a+1) = a^{4}+a^{2}+1 .\n$$\nThus $a \\leftrightarrow a^{2}$ for all $a$, then also true for $a+1 \\leftrightarrow (a+1)^{2}$. Moreover,\n$$\n\\begin{aligned}\n\\left(a^{2}\\right)^{2}+\\left((a+1)^{2}\\right)^{2}+1 & = 2a^{4}+4a^{3}+6a^{2}+4a+2 \\\\\n& = \\left(2a^{2}+2a+2\\right)\\left(a^{2}+a+1\\right)\n\\end{aligned}\n$$\nThus $a^{2}+(a+1)^{2}+1=2a^{2}+2a+2$ divides $\\left(a^{2}\\right)^{2}+\\left((a+1)^{2}\\right)^{2}+1$, which implies that $a^{2} \\leftrightarrow (a+1)^{2}$. Hence,\n$$\na \\leftrightarrow a^{2} \\leftrightarrow (a+1)^{2} \\leftrightarrow a+1,\n$$\nthis means that every two consecutive integers $a, a+1$ are connected. Thus the given infinity graph is connected. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 40484,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ is a trapezoid with $\\angle A = \\angle B = 90^{\\circ}$ and let $E$ is a point lying on side $CD$. Let the circle $\\omega$ is inscribed to triangle $ABE$ and tangents sides $AB$, $AE$ and $BE$ at points $P$, $F$ and $K$ respectively. Let $KF$ intersects segments $BC$ and $AD$ at points $M$ and $N$ respectively, as well as $PM$ and $PN$ intersect $\\omega$ at points $H$ and $T$ respectively. Prove that $PH = PT$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $KF$ meets $AB$ at $S$. We have known that $EP$, $AK$, $BF$ are concurrent at Gergonne's point, then $(SP, AB) = -1$. Let $Q$ be the projection of $P$ on $KF$ then $Q(SP, AB) = -1$, but $QS \\perp QP$ so $QP$ is the angle bisector of $\\angle AQB$. From that, we get $\\angle BQM = \\angle AQN$.\n\n![](attached_image_1.png)\n\nIn the other hand, easy to realize that $PQMB$ and $PQNA$ are cyclic, so\n$$\n\\angle BPM = \\angle BQM = \\angle AQN = \\angle APN.\n$$\nCombining with the truth that $AB$ is the tangent of $(I)$, we get\n$$\n\\angle PTH = \\angle BPM = \\angle APN = \\angle PHT,\n$$\nor $PH = PT$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40485,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, the circle having $BC$ as diameter cuts $AB$, $AC$ at $F$, $E$ respectively. Let $P$ be a point on this circle. Let $C_{0}$, $B_{0}$ be the projections of $P$ upon the sides $AB$, $AC$ respectively. Let $H$ be the orthocenter of the triangle $AB_{0}C_{0}$. Show that $\\angle EHF = 90^{\\circ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $B' E'$, $C' F'$ be the altitudes of $\\triangle AB'C'$ and concur at $H$.\nWe have that $AB' P C'$ is inscribed in the circle with diameter $AP$, so $AP$ passes through the circumcenter of $\\triangle AB'C'$. Then $AP$, $AH$ reflect each other by the angle bisector of $\\angle B'AC'$. From that, easy to prove that $\\triangle AHF' \\sim \\triangle APC'$.\n\n![](attached_image_1.png)\n\nBut $BE \\parallel C'F'$, as Thales's Theorem, we have $\\frac{AE}{AF'} = \\frac{AB}{AC'}$, implies that\n$$\n\\frac{HE}{PB} = \\frac{AE}{AB} = \\frac{EF}{BC} .\n$$\nSimilarly, we get $\\frac{HF}{PC} = \\frac{EF}{BC}$, thus $\\triangle HEF \\sim \\triangle PBC$. Then we get\n$$\n\\angle EHF = \\angle BPC = 90^{\\circ} .\n$$ $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40486,
    "subject": "Mathematics (Multi-modal)",
    "question": "Among numbers $a+b$, $a-b$, $a b$, $\\frac{a}{b}$ exactly two are positive and exactly two are negative. Find the sign of $b$.",
    "options": [],
    "answer": "b is negative",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40487,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $d$ be a positive divisor of a positive integer $m$ and $(a_{l}),(b_{l})$ two arithmetic sequences of positive integers. It is given that $\\operatorname{gcd}(a_{i}, b_{j})=1$ and $\\operatorname{gcd}(a_{k}, b_{n})= m$ for some positive integers $i, j, k$, and $n$. Prove that there exist positive integers $t$ and $s$ such that $\\operatorname{gcd}(a_{t}, b_{s})=d$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40488,
    "subject": "Mathematics (Multi-modal)",
    "question": "All of the numbers $1, 2, 3, \\ldots, 1000000$ are initially colored black. On each move it is possible to choose the number $x$ (among the colored numbers) and change the color of $x$ and of all of the numbers that are not co-prime with $x$ (black into white, white into black). Is it possible to color all of the numbers white?",
    "options": [],
    "answer": "Yes",
    "solution": "The answer is YES. We will prove by induction that the procedure can be applied for any positive integer $n$.\n\nThe statement is true for $n=1$. Suppose that it is also true for $n=k-1 \\geq 1$, which means there exists a way to change every number not exceeding $k-1$ from black to white, which we call process $A$.\n\nWe shall prove the statement for $k$ with some cases as follows:\n\n1. If $k$ is not a square-free number. Put $k=p_{1}^{a_{1}} p_{2}^{a_{2}} \\ldots p_{v}^{a_{v}}$ and $k' = p_{1} p_{2} \\ldots p_{v}$ with $p_{1}, p_{2}, \\ldots, p_{v}$ primes. Note that the number of times the color of $k$ and $k'$ are changed are equal (since the set of integers not coprime with them are the same), then $k'$ can change the color if and only if $k$ can. So after making process $A$ with the sequence $1,2, \\ldots, k-1$, the number $k$ will change color from black to white.\n\n2. If $k$ is a square-free number. In case after making process $A$, number $k$ changes color, then we are done. Otherwise, put $k=p_{1} p_{2} \\ldots p_{t}$ and we consider process $B$ to select all divisors other than $1$ of $k$. After that, number $k$ will be affected $2^{t}-1$ times and will change the color.\n\nDenote $k'$ as some divisor greater than $1$ of $k$, and $k'$ has $s<t$ prime divisors. Take some number $c$ which is a multiple of $k'$ but coprime to $k / k'$. After process $A$, number $c$ has color white. And after process $B$, the number of times the color of $c$ is changed is $\\left(2^{t}-1\\right)-\\left(2^{t-s}-1\\right)=2^{t}-2^{t-s}$ which is an even number. This implies that $c$ will not change color.\n\nSo in all cases, we always can find a process to change the color of all numbers from $1$ to $k$ from black to white.\n\nFinally, by replacing $n=10^{6}$, the problem is solved. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40489,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each cell of an $8 \\times 8$ board is either black or white. It is known that among any nine cells forming a $3 \\times 3$ square there are evenly many white ones. What is the minimum possible number of black cells?",
    "options": [],
    "answer": "4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40490,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define sequence of positive integers $\\left(a_{n}\\right)$ as $a_{1}=a$ and $a_{n+1}=a_{n}^{2}+1$ for $n \\geq 1$. Prove that there is no index $n$ for which\n$$\n\\prod_{k=1}^{n}\\left(a_{k}^{2}+a_{k}+1\\right)\n$$\nis a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $p$ as a prime of $a_{1}^{2}+a_{1}+1$, note that $a_{1}$ is odd (since $a_{1}^{2}+a_{1}+1= a_{1}\\left(a_{1}+1\\right)+1$ is an odd number) and $p \\mid a_{1}$. By induction, we can show that\n$$\na_{n} \\equiv a_{2} \\equiv -a_{1} \\pmod{p} \\text{ for any } n>1.\n$$\nThus $a_{n}^{2}+a_{n}+1 \\equiv a_{1}^{2}-a_{1}+1 \\equiv -2a_{1} \\not\\equiv 0 \\pmod{p}$ so\n$$\nv_{p}\\left(\\prod_{k=1}^{n}\\left(a_{k}^{2}+a_{k}+1\\right)\\right)=v_{p}\\left(a_{1}^{2}+a_{1}+1\\right).\n$$\nSince $a_{1}^{2}<a_{1}^{2}+a_{1}+1<\\left(a_{1}+1\\right)^{2}$, then $a_{1}^{2}+a_{1}+1$ is not a perfect square.\nThis implies that there exist some prime $p$ such that $v_{p}\\left(a_{1}^{2}+a_{1}+1\\right)$ is odd. This finishes the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40491,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let 300 students participate to the Olympiad. Between each 3 participants there is a pair that are not friends. Hamza enumerates participants in some order and denotes by $x_{i}$ the number of friends of $i$-th participant. It occurs that\n$$\n\\left\\{x_{1}, x_{2}, \\ldots, x_{299}, x_{300}\\right\\}=\\{1,2, \\ldots, N-1, N\\} .\n$$\nFind the biggest possible value for $N$.",
    "options": [],
    "answer": "200",
    "solution": "Firstly, we shall prove that if $A, B$ are friend then the sum of friends of each one does not exceed $300$. Indeed,\nSuppose that $A$ has $a \\leq N$ friends and $B$ has $b \\leq N$ friends. Note that $A$ and $B$ cannot have any common friend; otherwise, take $C$ as a friend of $A, B$ and then the triple ($A, B, C$) does not satisfy the given condition. Thus\n$$\n(a-1)+(b-1) \\leq 298 \\text{ and } a+b \\leq 300.\n$$\nSuppose that $N \\geq 201$ and take some student $X$ who has $201$ friends. Take $Y$ who has at least $100$ friends, then as the remark above, $X$ and $Y$ are not friends. Note that we always can find $100$ distinct students and each of them has exactly $100,101, \\ldots, 200$ friends, and they are not friends of $X$. So the number of friends of $X$ is less than $200$, contradiction. Therefore, $N \\leq 200$.\n\nWe can give an example as follows. Divide students into groups $A=\\left\\{A_{1}, A_{2}, \\ldots, A_{200}\\right\\}$ and $B=\\left\\{B_{1}, B_{2}, \\ldots, B_{100}\\right\\}$.\n\n- Student $B_{1}$ is friends with all students in $A$.\n- Student $B_{2}$ is friends with all students from $A_{2}$ to $A_{200}$.\n- ...\n- Student $B_{100}$ is friends with students from $A_{100}$ to $A_{200}$.\n- Students in the same group are not friends.\n\nSo it is easy to check that there are no any triplet of students that are friends with each other. And the number of friends of students in $A$ is $1,2, \\ldots, 100$ and the number of friends of students in $B$ is $200,199, \\ldots, 101$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40492,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle with $AD$, $BE$, $CF$ as altitudes and $d$ is the tangent line of the circumcircle of triangle $ABC$ at $A$. The line through $H$ and parallel to $EF$ cuts $DE$, $DF$ at $Q$, $P$ respectively. Prove that $d$ is tangent to the ex-circle with respect to vertex $D$ of triangle $DPQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 40493,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist two polynomials $P$ and $Q$ with integer coefficient such that\ni) both $P$ and $Q$ have a coefficient with absolute value bigger than $2021$,\nii) all coefficients of $P \\cdot Q$ by absolute value are at most $1$.",
    "options": [],
    "answer": "Yes",
    "solution": "Note that the polynomial\n$$\n(1 - x^2)(1 - x^4)(1 - x^8) \\dots (1 - x^{2n})\n$$\nhas all coefficients equal $0$, $+1$ or $-1$. Also note, that\n$$\n\\begin{aligned}\n& (1 - x^2)(1 - x^4)(1 - x^8) \\dots (1 - x^{2n}) \\\\\n&= \\prod_{i=0}^{n-1} (1 - x^{2i}) \\cdot \\prod_{i=0}^{n-1} (1 + x^{2i}) \\\\\n&= (1 - x)^n \\cdot \\prod_{i=0}^{n-1} (1 + x^i)^{n-i}\n\\end{aligned}\n$$\nDue to Newton binomial formula $(1 \\pm x)^n$ have coefficient equal $n$, also all coefficients of\n$$\n(1 + x)^n (1 + x^2)^{n-1} \\dots (1 + x^{2n-2})^2 (1 + x^{2n-1})\n$$\nare positive and bigger than coefficients of $(1 + x)^n$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40494,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a non-empty set $\\mathcal{T}$ denote by $p(\\mathcal{T})$ the product of all elements of $\\mathcal{T}$. Does there exist a set $\\mathcal{T}$ of 2021 elements such that for any $a \\in \\mathcal{T}$ one has that $p(\\mathcal{T}) - a$ is an odd integer. Consider two cases:\n1. All elements of $\\mathcal{T}$ are irrational numbers?\n2. At least one element of $\\mathcal{T}$ is a rational number?",
    "options": [],
    "answer": "Case 1: Yes, such a set exists with all elements irrational. Case 2: No, it is impossible if at least one element is rational.",
    "solution": "1. Consider the polynomial\n$$\nf(x) = x(x+2)(x+4)\\dots(x+4040) - x - (2m-1)\n$$\nfor some $m \\in \\mathbb{Z}^+$ then $\\lim_{x \\to +\\infty} f(x) = +\\infty$ and $f(0) = 2^{2020} \\cdot 2020! - (2m-1)$. Take $m$ big enough to get $f(0) < 0$ then by the continuous property of $f(x)$, there exists some real number $x_0$ such that $f(x_0) = 0$. Then take the set $T$ as follows\n$$\nT = \\{x_0, x_0 + 2, x_0 + 4, \\dots, x_0 + 4040\\}\n$$\nThen $p(T) = x_0(x_0+2)(x_0+4)\\dots(x_0+4040)$ and $p(T) - x_0 = 2m-1$ is an odd integer. Since every element in $T$ has form $x_0+2k$ with $k \\in \\mathbb{Z}$ then $p(T) - a$ is an odd integer for all $a \\in T$. Finally, take $m = 2^{2019} \\cdot 2020!$ then $f(0) = 1$, suppose on the contrary that $x_0 \\in \\mathbb{Q}$. Note that $f(x)$ is monic then $x_0 \\in \\mathbb{Z}$ and $x_0 \\mid f(0) = 1$, implies that $x_0 \\in \\{-1; 1\\}$, but this is contradiction since it is easy to check that $f(x)$ cannot have odd integer root.\n\n2. We will proof that the answer is negative. Note that for all $x, y \\in T$ then $x - y$ is even. Suppose that $T$ contains some rational number $b$. Then we can list elements of $T$ as follows $b, b + a_1, b + a_2, \\dots, b + a_{2020}$ in which $a_1, a_2, \\dots, a_k$ are even.\n\nPut $c = b(b + a_1)(b + a_2)\\dots(b + a_k) - b$ then $c$ is odd. Consider polynomial\n$$\ng(x) = x(x + a_1)(x + a_2)\\dots(x + a_k) - x - c\n$$\nthen $g(b) = 0$ and $x = b$ is the rational root of $g(x)$. Note that $g$ is monic then $b \\in \\mathbb{Z}$. From this we conclude that all the elements in $T$ are integers that share the same parity, so does $p(T)$. Thus $p(T) - b$ is even, contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40495,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $a, b, c$ and prime number $p$ such that\n$$\n73p^2 + 6 = 9a^2 + 17b^2 + 17c^2.\n$$",
    "options": [],
    "answer": "(a, b, c, p) = (1, 1, 4, 2) and (1, 4, 1, 2)",
    "solution": "Suppose that $b \\le c$, we distinguish two cases regarding the value of $p$\n\n1. if $p \\ne 2$ implies $p$ is odd, so $p^2 \\equiv 1 \\pmod 8$. It follows that\n$$\na^2 + b^2 + c^2 \\equiv 9a^2 + 17b^2 + 17c^2 \\equiv 73p^2 + 6 \\equiv 7 \\pmod 8. \\quad (2)\n$$\nOtherwise, $a^2 + b^2 + c^2 \\equiv 0, 1, 2, 3, 4, 5, 6 \\pmod 8$, which is contracts (2).\n\n2. if $p = 2$ then $9a^2 + 17b^2 + 17c^2 = 73p^2 + 6 = 298$. Since $a, b, c \\ge 1$, we have\n$$\n298 = 9a^2 + 17b^2 + 17c^2 \\ge 26 + 17c^2\n$$\n, which means $c \\le 4$. We investigate two cases\n* if $b \\le c = 4$ then $9a^2 + 17b^2 = 26$, which means $a = b = 1$.\n* if $b \\le c \\le 3$ then it is easy to check that there is no solutions satisfied.\nHence, the only solution satisfied (1) is $(a, b, c, p) = (1, 1, 2, 2)$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40496,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $k$, show that there exists a prime $p$ such that one can choose distinct integers $a_1, a_2, \\dots, a_{k+3} \\in \\{1, 2, \\dots, p-1\\}$ such that $p$ divides $a_i a_{i+1} a_{i+2} a_{i+3} - i$ for all $i = 1, 2, 3, \\dots, k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First we choose distinct positive rational numbers $r_1, \\dots, r_{k+3}$ such that\n$$\nr_i r_{i+1} r_{i+2} r_{i+3} = i \\quad \\text{for } 1 \\le i \\le k\n$$\nLet $r_1 = x, r_2 = y, r_3 = z$ be some distinct primes greater than $k$; the remaining terms satisfy $r_4 = \\frac{1}{r_1 r_2 r_3}$ and $r_{i+4} = \\frac{i+1}{i} r_i$. It follows that if $r_i$ are represented as irreducible fractions, the numerators are divisible by $x$ for $i \\equiv 1 \\pmod 4$, by $y$ for $i \\equiv 2 \\pmod 4$, by $z$ for $i \\equiv 3 \\pmod 4$ and by none for $i \\equiv 0 \\pmod 4$. Notice that $r_i < r_{i+4}$; thus the sequences\n$$\nr_1 < r_5 < r_9 < \\dots, \\quad r_2 < r_6 < r_{10} < \\dots, \\quad r_3 < r_7 < r_{11} < \\dots, \\quad r_4 < r_8 < r_{12} < \\dots\n$$\nare increasing and have no common terms, that is, all $r_i$ are distinct.\n\nIf each $r_i$ is represented by an irreducible fraction $\\frac{u_i}{v_i}$, choose a prime $p$ which divides neither $v_i$, $1 \\le i \\le k+1$, nor $v_i v_j$ ($r_i - r_j$) $= v_j u_i - v_i u_j$ for $i < j$, and define $a_i$ by the congruence $a_i v_i \\equiv u_i \\pmod{p}$. Since $r_i r_{i+1} r_{i+2} r_{i+3} = i$, we have\n$$\n\\begin{aligned}\niv_i v_{i+1} v_{i+2} v_{i+3} &= r_i v_i r_{i+1} v_{i+1} r_{i+2} v_{i+2} r_{i+3} v_{i+3} \\\\\n&= u_i u_{i+1} u_{i+2} u_{i+3} \\equiv a_i v_i a_{i+1} v_{i+1} a_{i+2} v_{i+2} a_{i+3} v_{i+3} \\pmod{p}\n\\end{aligned}\n$$\nand therefore $a_i a_{i+1} a_{i+2} a_{i+3} \\equiv i \\pmod{p}$ for $1 \\le i \\le k$.\n\nIf $a_i \\equiv a_j \\pmod{p}$, then $u_i v_j \\equiv a_i v_j v_j \\equiv u_j v_i \\pmod{p}$, a contradiction. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40497,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $a, b, c, d$ are positive real numbers satisfying $(a+c)(b+d) = ac+bd$. Find the smallest possible value of\n$$\n\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{d} + \\frac{d}{a}\n$$",
    "options": [],
    "answer": "8",
    "solution": "First of all, apply the AM-GM inequality as follows:\n$$\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}} = \\frac{2(ac + bd)}{\\sqrt{abcd}}\n$$\nContinuing to apply the AM-GM inequality, then\n$$\n\\left(\\frac{a}{b} + \\frac{c}{d}\\right) + \\left(\\frac{b}{c} + \\frac{d}{a}\\right) \\ge \\frac{2(a + c)(b + d)}{\\sqrt{abcd}} \\ge 2 \\cdot \\frac{2\\sqrt{ac} \\cdot 2\\sqrt{bd}}{\\sqrt{abcd}} = 8\n$$\nThe above inequalities turn into equalities when $a = c$ and $b = d$. Then the condition $(a+c)(b+d) = ac+bd$ can be rewritten as $4ab = a^2 + b^2$. This is equivalent to $a/b = 2 \\pm \\sqrt{3}$.\nHence, $S$ attains value 8, e.g., when $a = c = 1$ and $b = d = 2 + \\sqrt{3}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40498,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider circles $(O_1)$, $(O_2)$, $(O_3)$ tangent to $d$ at $A$, $B$, $C$ and $(O_2)$ is the biggest circle, externally tangent to $(O_1)$, $(O_3)$. Let $BD$ be the diameter of $(O_2)$. The external tangent line (differs from $d$) of $(O_1)$, $(O_3)$ cuts $(O_2)$ at $X$, $Y$. Let $K$ be the midpoint of the arc $XBY$ of $(O_2)$. Prove that the circle of diameter $AC$ touches $DX$, $DY$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40499,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be an odd prime and let $N = \\frac{1}{4}(p^3 - p) - 1$. The numbers $1, 2, \\dots, N$ are painted arbitrarily in two colors, red and blue. For any positive integer $n \\le N$, denote by $r(n)$ the fraction of integers in $\\{1, 2, \\dots, n\\}$ that are red (number of red numbers divided by $n$). Prove that there exists a positive integer $a \\in \\{1, 2, \\dots, p-1\\}$ such that $r(n) \\ne \\frac{a}{p}$ for all $n = 1, 2, \\dots, N$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $R(n)$ the number of red numbers in $\\{1, 2, \\dots, n\\}$, i.e., $R(n) = n r(n)$. Similarly, denote by $B(n)$ and $b(n) = B(n)/n$ the number and proportion of blue numbers in $\\{1, 2, \\dots, n\\}$, respectively. Notice that $B(n) + R(n) = n$ and $b(n) + r(n) = 1$. Therefore, the statement of the problem does not change after swapping the colors.\n\nArguing indirectly, for every $a \\in \\{1, 2, \\dots, p-1\\}$ choose some positive integer $n_a$ such that $r(n_a) = a/p$ and, hence, $R(n_a) = a n_a / p$. Clearly, $p \\mid n_a$, so that $n_a = p m_a$ for some positive integer $m_a$, and $R(n_a) = a m_a$. Without loss of generality, we assume that $m_1 < m_{p-1}$, as otherwise one may swap the colors. Notice that\n$$\nm_a \\le \\frac{N}{p} < \\frac{p^2-1}{4} \\quad \\text{for all } a=1,2,\\dots,p-1.\n$$\nThe solution is based on a repeated application of the following simple observation.\n\n**Claim.** Assume that $m_a < m_b$ for some $a, b \\in \\{1, 2, \\dots, p-1\\}$. Then\n$$\nm_b \\ge \\frac{a}{b} m_a \\quad \\text{and} \\quad m_b \\ge \\frac{p-a}{p-b} m_a.\n$$\n**Proof.** The first inequality follows from $b m_b = R(n_b) \\ge R(n_a) = a m_a$. The second inequality is obtained by swapping colors.\n\nLet $q = (p-1)/2$. We distinguish two cases.\n\n**Case 1:** All $q$ numbers $m_1, m_2, \\dots, m_q$ are smaller than $m_{p-1}$.\nLet $m_a$ be the maximal number among $m_1, m_2, \\dots, m_q$; then $m_a \\ge q \\ge a$. Applying the Claim, we get\n$$\nm_{p-1} \\ge \\frac{p-a}{p-(p-1)} m_a \\ge (p-q)q = \\frac{p^2-1}{4}\n$$\nwhich contradicts (1).\n\n**Case 2:** There exists $k \\le q$ such that $m_k > m_{p-1}$. Choose $k$ to be the smallest index satisfying $m_k > m_{p-1}$; by our assumptions, we have $1 < k \\le q < p-1$.\nLet $m_a$ be the maximal number among $m_1, m_2, \\dots, m_{k-1}$; then $a \\le k-1 \\le m_a < m_{p-1}$. Applying the Claim, we get\n$$\n\\begin{align*}\nm_k &\\ge \\frac{p-1}{k} m_{p-1} \\ge \\frac{p-1}{k} \\cdot \\frac{p-a}{p-(p-1)} m_a \\\\\n&\\ge \\frac{p-1}{k} \\cdot (p-k+1)(k-1) \\\\\n&\\ge \\frac{k-1}{k} \\cdot (p-1)(p-q) \\\\\n&\\ge \\frac{1}{2} \\cdot \\frac{p^2-1}{2}\n\\end{align*}\n$$\nwhich contradicts (1) again. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40500,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with circumcenter $O$ and incenter $I$, ex-center in angle $A$ is $J$. Denote $D$ as the tangent point of $l_a(I)$ on $BC$ and the angle bisector of angle $A$ cuts $BC$, $(O)$ respectively at $E$, $F$. The circle $(DEF)$ meets $(O)$ again at $T$. Prove that $AT$ passes through an intersection of $(J)$ and $(DEF)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40501,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral with $\\angle A = \\angle B = 90^\\circ$, $AB = AD$. Denote $E$ as the midpoint of $AD$, suppose that $CD = BC + AD$, $AD > BC$. Prove that $\\angle ADC = 2\\angle ABE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40502,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle inscribed in a fixed circle $(O)$ with $BC$ fixed and $A$ varying on $(O)$. Denote $H$ as the orthocenter of triangle $ABC$ and take $D$, $E$ on $AB$, $AC$ respectively such that $H$ is the midpoint of $DE$. Prove that when $A$ moves on $(O)$, the center of $(ADE)$ belongs to a fixed circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40503,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral and $X$ lying inside it such that $XA \\cdot XC^2 = XB \\cdot XD^2$ and $\\angle AXD + \\angle BXC = \\angle CXD$. Prove that $\\angle XAD + \\angle XCD = \\angle XBC + \\angle XDC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40504,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle with circumcenter $O$, incenter $I$ and $(I)$ tangent to $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. Suppose that $EF$ cuts $(O)$ at $P$, $Q$. Prove that $(PQD)$ bisects segment $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40505,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle with circumcenter $O$. Tangent lines to $(O)$ at $B$, $C$ meet at $T$. A line passes through $T$ cuts segments $AB$ at $D$ and cuts ray $CA$ at $E$. Take $M$ as midpoint of $DE$ and suppose that $MA$ cuts $(O)$ again at $K$. Prove that $(MKT)$ is tangent to $(O)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40506,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB < AC$ and incircle $(I)$ tangent to $BC$ at $D$. Take $K$ on $AD$ such that $CD = CK$. Suppose that $AD$ cuts $(I)$ at $G$ and $BG$ cuts $CK$ at $L$. Prove that $K$ is the midpoint of $CL$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40507,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB < AC$ inscribed in $(O)$. The tangent line at $A$ of $(O)$ cuts $BC$ at $D$. Take $H$ as the projection of $A$ on $OD$ and $E$, $F$ as projections of $H$ on $AB$, $AC$. Suppose that $EF$ cuts $(O)$ at $R$, $S$. Prove that $(HRS)$ is tangent to $OD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40508,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$, such that $n$ is a perfect number and $\\varphi(n)$ is power of 2. Note: a positive integer $n$ is called perfect if the sum of all its positive divisors is equal to $2n$.",
    "options": [],
    "answer": "6",
    "solution": "Suppose that $\\varphi(n) = 2^m$ for some $m \\in \\mathbb{Z}^+$. Based on the formula of $\\varphi(n)$, we have\n$$\n\\varphi(n) = \\prod_{p|n} p^{t-1}(p-1) = 2^m\n$$\nwith $v_p(n) = t$. This implies that $t = 1$ for all odd prime divisors, since otherwise, $p \\mid 2^m$, a contradiction. Thus $p-1$ is a power of 2.\n\nWith odd prime divisor of $n$, put $p = 2^s + 1$ for some $s \\in \\mathbb{Z}^+$. If $s = 1$ then $p = 3$, otherwise $s$ must be even since $p \\equiv (-1)^s + 1 \\equiv 0 \\pmod{3}$ when $s$ is odd, contradiction. In case $s$ has some proper odd prime divisor $q$ then put $s = qt$ with $t > 1$ then $2^s + 1 = (2^t)^q + 1$ is divisible by $2^t + 1$, contradiction also.\n\nThese imply that $s$ is also a power of 2, so by putting $s = 2^k$, we get $p = 2^{2k} + 1$. Now let $n = 2^a p_1 p_2 \\dots p_l$ with $a \\ge 0, l \\ge 0$ and $p_1, p_2, \\dots, p_l$ are odd primes in ascending order, in which $p_1 = 2^{2k} + 1$. So\n$$\n2n = \\sigma(n) = (2^{a+1} - 1)(p_1 + 1)(p_2 + 1)\\dots(p_l + 1)\n$$\nis the sum of divisors of $n$. Note that $l > 0$, otherwise, $\\sigma(n) = 2^{a+1}-1$, contradiction. If $p_1 > 3$ then $3 \\nmid 2n$ and\n$$\np_1 + 1 = 2^{2k} + 2 = 2(2^{2k-1} + 1) \\equiv 2((-1)^{2k-1} + 1) \\equiv 0 \\pmod{3}\n$$\nimplying that $3 \\mid 2n$, contradiction. Thus $p_1 = 3$ and $v_3(2n) = 1$ leads to $l \\le 2$. There are two cases\n\n* If $l = 2$ then $2^{a+1} \\cdot 3 \\cdot p_2 = (2^{a+1} - 1) \\cdot 4 \\cdot (p_2 + 1)$, note that $v_2(p_2+1) = 1$ so comparing the exponent of 2 in both sides to get $a = 2$. Thus $6p_2 = 7(p_2+1)$, contradiction.\n* If $l = 1$ then $2^{a+1} \\cdot 3 = (2^{a+1} - 1) \\cdot 4$, then $a = 1$ implies that $n = 6$ is a perfect square.\n\nTherefore, $n = 6$ is only solution of this problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40509,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a circle $O$, there are six points, $A, B, C, D, E, F$ in a counterclockwise order. $BD \\perp CF$, and $CF, BE, AD$ are concurrent. Let the perpendicular from $B$ to $AC$ be $M$, and the perpendicular from $D$ to $CE$ be $N$. Prove that $AE \\parallel MN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K$ be the concurrent point of $CF, BE, AD$, and let $CF$ intersect $BD$ at $L$.\n\nWe have $\\angle BMC = 90^\\circ = \\angle BLC$ since $BM$ is perpendicular to $AC$ and $BL$ is perpendicular to $CF$, so $B, M, L, C$ lie on the same circle with diameter $BC$. Hence,\n$$\n\\angle CML = \\angle CBL = \\angle CBD = \\angle CAD\n$$\nimplying that $ML$ is parallel to $AK$. Similarly, we also have $LN$ is parallel to $EK$. Thus, we have\n$$\n\\frac{CM}{CA} = \\frac{CL}{CK} = \\frac{CN}{CE}\n$$\nThis implies that $MN$ is parallel to $AE$, as desired. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40510,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle with the orthocenter $H$, circumcenter $O$ and $AD$ is the diameter of $(O)$. Suppose that the circle $(AHD)$ meets the lines $AB$, $AC$ at $F$, $E$, respectively. Denote $J$, $K$ as orthocenter and nine-point center of $AEF$. Prove that $HJ$ is parallel to $BC$ and $KO = KH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40511,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle with $H$ is the orthocenter and $M$ is the midpoint of $AH$. Denote $O_1, O_2$ as the centers of circles pass through $H$ and respectively tangent to $BC$ at $B, C$. Let $X, Y$ be the ex-centers which respect to angle $H$ in triangles $HMO_1, HMO_2$. Prove that $XY$ is parallel to $O_1O_2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $BU$, $CV$ be the diameter of circles $(O_1)$, $(O_2)$ and denote $R = HU \\cap AB$, $S = HV \\cap AC$. We have $BH \\perp HU$, but $BH \\perp AC$ then $HR \\parallel AC$. Similarly $HS \\parallel AB$ implying that $ARHS$ is a parallelogram. Hence, $M$ is the midpoint of the segment $RS$.\n\n![](attached_image_1.png)\n\nIn the other hand, $O_1$ is the midpoint of $BU$ and $BU \\parallel AH$ thus, by the property of trapezoid, we get $O_1, R, M$ are collinear. Similarly, $O_2, S, M$ are also collinear. Hence, five points $O_1, O_2, R, S, M$ are collinear.\n\nSince $X$ is the excenter of triangle $HMO_1$ then $MX$ is the external angle bisector of $\\angle O_1MH$, thus $\\frac{XR}{XH} = \\frac{MR}{MH}$. Similarly, $\\frac{YS}{YH} = \\frac{MS}{MH}$, but $MR = MS$ then $\\frac{XR}{XH} = \\frac{YS}{YH}$, this implies that $XY \\parallel RS$ or $XY \\parallel O_1O_2$.\n\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40512,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle with altitude $AD$ ($D \\in BC$), $M$ is the midpoint of $AD$ and $O$ is the circumcenter. Line $AO$ meets $BC$ at $K$ and circle of center $K$, radius $KA$ cuts $AB$, $AC$ at $E$, $F$ respectively. Prove that $AO$ bisects $EF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40513,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral with $O$ is circumcenter and $AC$ meets $BD$ at $I$. Suppose that rays $DA$, $CD$ meet at $E$ and rays $BA$, $CD$ meet at $F$. The Gauss line of $ABCD$ meets $AB$, $BC$, $CD$, $DA$ at points $M$, $N$, $P$, $Q$ respectively. Prove that the circle of diameter $OI$ is tangent to two circles $(ENQ)$, $(FMP)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40514,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a non-isosceles triangle with altitudes $AD$, $BE$, $CF$ with orthocenter $H$. Suppose that $DF \\cap HB = M$, $DE \\cap HC = N$ and $T$ is the circumcenter of triangle $HBC$. Prove that $AT \\perp MN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40515,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be triangle with the symmedian point $L$ and circumradius $R$. Construct parallelograms $ADLE$, $BHLK$, $CILJ$ such that $D, H \\in AB$; $K, I \\in BC$; $J, E \\in CA$. Suppose that $DE$, $HK$, $IJ$ pairwise intersect at $X$, $Y$, $Z$. Prove that inradius of $XYZ$ is $\\frac{R}{2}$.",
    "options": [],
    "answer": "R/2",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40516,
    "subject": "Mathematics (Multi-modal)",
    "question": "The Magician and his Assistant show trick. The Viewer writes on the board the sequence of $N$ digits. Then the Assistant covers some pair of adjacent digits so that they become invisible. Finally, the Magician enters the show, looks at the board and guesses the covered digits and their order. Find the minimal $N$ such that the Magician and his Assistant can agree in advance so that the Magician always guesses right.",
    "options": [],
    "answer": "101",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40517,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an infinite sequence of numbers $a_1, a_2, a_3, \\ldots$ such that for each positive integer $k$, there exists positive integer $t$ for which $a_k = a_{k+t} = a_{k+2t} = \\ldots$. Does this sequence must be periodic?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40518,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each of $N$ people have chosen some $5$ elements from a $23$-element set so that any two people share at most $3$ chosen elements. Does this mean that $N \\le 2020$? Answer the same question with $25$ instead of $23$.",
    "options": [],
    "answer": "For 23: Yes (in fact N ≤ 1771). For 25: No (there exist families with more than 2020, e.g., at least 2126).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40519,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integer $n \\ge 3$ such that it is possible to mark the vertices of a regular $n$-gon with the numbers from $1$ to $n$ so that for any three vertices $A$, $B$ and $C$ with $AB = AC$, the number in $A$ is greater or smaller than both numbers in $B$, $C$.",
    "options": [],
    "answer": "n = 2^k for integers k ≥ 2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40520,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that it is impossible to fill the cells of an $8 \\times 8$ table with the numbers from $1$ to $64$ (each number must be used once) so that for each $2 \\times 2$ square, the difference between products of the numbers on its diagonals will be equal to $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40521,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle inscribed in $(O)$ and $BB'$, $CC'$ are altitudes. Denote $E$, $F$ as the intersections of $BB'$, $CC'$ with $(O)$ and $D$, $P$, $Q$ are projections of $A$ on $BC$, $CE$, $BF$. Prove that the perpendicular bisector of $PQ$ bisects two segments $AO$, $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40522,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the sequence $a_1, a_2, a_3, \\dots$ defined by $a_1 = 9$ and\n$$\na_{n+1} = \\frac{(n+5)a_n + 22}{n+3}\n$$\nfor $n \\ge 1$. Find all positive integers $n$ for which $a_n$ is a perfect square.",
    "options": [],
    "answer": "n = 1, 8",
    "solution": "First of all, we will prove by induction on $n$ that $a_n = n^2 + 7n + 1$. The base case is trivial. Assume that $a_k = k^2 + 7k + 1$, so\n$$\na_{k+1} = \\frac{(k + 5)a_k + 22}{k + 3} = \\frac{(k + 5)(k^2 + 7k + 1) + 22}{k + 3} = (k + 1)^2 + 7(k + 1) + 1\n$$\nWe distinguish two cases regarding the value of $n$\n* if $n \\ge 9$ then $(n + 3)^2 < n^2 + 7n + 1 < (n + 4)^2$, so $a_n$ is not a perfect square.\n* if $n \\le 8$ then we have\n\n| $a_1 = 9$ | $a_2 = 31$ | $a_5 = 61$ | $a_7 = 99$ |\n| $a_2 = 19$ | $a_4 = 45$ | $a_6 = 79$ | $a_8 = 121$ |\n\nHence, $a_n$ is a perfect square if and only if $n = 1, 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40523,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$ and $z$ be odd positive integers such that $\\gcd(x, y, z) = 1$ and the sum $x^2 + y^2 + z^2$ is divisible by $x + y + z$. Prove that $x + y + z - 2$ is not divisible by $3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose there exists a prime divisor $p \\equiv 2 \\pmod{3}$ of $x + y + z$. Since $z \\equiv -(x + y) \\pmod{p}$ we have $2(x^2 + y^2 + xy) \\equiv 0 \\pmod{p}$. Multiplying by $x - y$ we get $x^3 \\equiv y^3 \\pmod{3}$ but this yields $x \\equiv y \\pmod{p}$, because $\\gcd(3, p-1) = 1$. Similarly $x \\equiv z \\pmod{p}$ whence $3x \\equiv 0 \\pmod{p}$ which means that $x$, $y$ and $z$ are divisible by $p$ which contradicts $\\gcd(x, y, z) = 1$. Hence $x + y + z$ have no prime divisors which have a remainder $2$ modulo $3$ so $x + y + z - 2$ is not divisible by $3$.\n□",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40524,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $k$, denote by $f(k)$ the number of positive integers $m$ such that the remainder of $k m$ modulo $2019^3$ is greater than $m$. Find the amount of different numbers among $f(1), f(2), \\ldots, f(2019^3)$.",
    "options": [],
    "answer": "25",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 40525,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. What is the smallest value of $m$ with $m > n$ such that the set $M = \\{n, n+1, ..., m\\}$ can be partitioned into subsets so that in each subset, there is a number which equals to the sum of all other numbers of this subset?",
    "options": [],
    "answer": "m = 7n - 4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40526,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N$ be a positive integer. Consider the sequence $a_1, a_2, ..., a_N$ of positive integers, none of which is a multiple of $2^{N+1}$. For $n \\ge N+1$, the number $a_n$ is defined as follows: choose $k$ to be the number among $1, 2, ..., n-1$ for which the remainder obtained when $a_k$ is divided by $2^n$ is the smallest, and define $a_n = 2a_k$ (if there are more than one such $k$, choose the largest such $k$). Prove that there exist $M$ for which $a_n = a_M$ holds for every $n \\ge M$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40527,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCall a positive integer $x$ to be remote from squares and cubes if each integer $k$ satisfies both $|x - k^2| > 10^6$ and $|x - k^3| > 10^6$. Prove that there exist infinitely many positive integers $n$ such that $2^n$ is remote from squares and cubes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40528,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet coefficients of the polynomial $P(x) = a_d x^d + \\ldots + a_2 x^2 + a_0$ where $d \\ge 2$, are positive integers. The sequence $(b_n)$ is defined by $b_1 = a_0$ and $b_{n+1} = P(b_n)$ for $n \\ge 1$. Prove that for any $n \\ge 2$, there exists a prime number $p$ such that $p \\mid b_n$ but it does not divide $b_1, b_2, \\ldots, b_{n-1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40529,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$ be a nonconstant integer polynomial and positive integer $n$. The sequence $a_0, a_1, \\ldots$ is defined by $a_0 = n$ and $a_k = P(a_{k-1})$ for $k \\ge 1$. Given that for each positive integer $b$, the sequence contains a $b$-th power of some positive integer greater than 1. Prove that $\\deg P = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose $\\deg P \\ge 2$.\n\nLet $d = \\deg P \\ge 2$. Then for large $k$, $a_k$ grows very rapidly, since $a_k = P(a_{k-1})$ and $P$ is a degree $d$ polynomial with integer coefficients.\n\nLet $b$ be any positive integer. By assumption, there exists $k$ such that $a_k = m^b$ for some integer $m > 1$.\n\nLet us fix $n$ and consider the sequence $a_0, a_1, \\ldots$. Since $P$ is a nonconstant integer polynomial, for large $k$, $a_k$ will be very large and will have many prime divisors (by Zsigmondy's theorem, for example, or by the fact that the sequence grows rapidly and is not eventually constant).\n\nBut for $a_k$ to be a perfect $b$-th power for every $b$, for every $b$ there must exist $k$ such that $a_k$ is a perfect $b$-th power. In particular, for $b$ large, $a_k$ must be a perfect $b$-th power, i.e., $a_k = m^b$ for some $m > 1$.\n\nBut for large $b$, the gap between consecutive perfect $b$-th powers grows very rapidly. For example, for $b = 100$, the numbers $2^{100}, 3^{100}, 4^{100}, \\ldots$ are extremely far apart. Since $a_k$ grows rapidly, but $P$ is fixed, the sequence $a_k$ cannot hit perfect $b$-th powers for all $b$ unless $P$ is linear.\n\nMore precisely, for $d \\ge 2$, $a_k$ grows as a tower of exponents, so for large $k$, $a_k$ is much larger than $(a_{k-1})^2$, and so on. The set of perfect $b$-th powers is very sparse for large $b$, so it is impossible for the sequence $a_k$ to hit a perfect $b$-th power for every $b$ unless $P$ is linear.\n\nNow, suppose $\\deg P = 1$, i.e., $P(x) = ax + b$ with $a, b \\in \\mathbb{Z}$, $a \\ne 0$.\n\nThen $a_k$ is an affine recurrence, so $a_k = a^k n + b \\frac{a^k - 1}{a - 1}$ (if $a \\ne 1$), which is an explicit formula. For suitable $n$, $a_k$ can be made to hit perfect $b$-th powers for all $b$ (for example, if $a = 1$, $P(x) = x + b$, then $a_k = n + k b$; for $a = -1$, $P(x) = -x + b$, $a_k$ alternates, but for suitable $n$ and $b$, $a_k$ can be a perfect $b$-th power).\n\nTherefore, the only possibility is $\\deg P = 1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40530,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $330$ seats in the first row of the auditorium. Some of these seats are occupied by $25$ viewers. Prove that among the pairwise distances between these viewers, there are two equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40531,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $n \\geq 2$ distinct positive integers $a_1, a_2, \\ldots, a_n$ none of which is a perfect cube. Find the maximal possible number of perfect cubes among their pairwise products.",
    "options": [],
    "answer": "floor(n^2/4)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40532,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the set of all divisors of a positive integer which is not a perfect square can be divided into pairs so that in each pair, one number is divided by another.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40533,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine if there exists pairwise distinct positive integers $a_1, a_2, ..., a_{101}, b_1, b_2, ..., b_{101}$ satisfying the following property: for each non-empty subset $S$ of $\\{1, 2, ..., 101\\}$ the sum $\\sum_{i \\in S} a_i$ divides $100! + \\sum_{i \\in S} b_i$.",
    "options": [],
    "answer": "No, such integers do not exist.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40534,
    "subject": "Mathematics (Multi-modal)",
    "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$, such that $S \\subset F$, $|S| = 2020$ and 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": "Detailed 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})$.\nFor two sequences $a = (a_1, a_2, \\dots, a_{2020})$, $b = (b_1, b_2, \\dots, b_{2020})$, we define\n$$\na \\cdot b = a_1 b_1 + a_2 b_2 + \\dots + a_{2020} b_{2020}.\n$$\nIt's easy to verify that $a \\cdot b$ is even for each $a, b \\in F$. Let $a = (a_1, a_2, \\dots, a_{2020}) \\in F$ and denote $b_i = a \\cdot e_i$ for each $i \\in \\{1, 2, \\dots, 2020\\}$, then\n$$\nb_{i+1} - b_i = a \\cdot e_{i+1} - a \\cdot e_i = 2a_i \\Rightarrow |b_{i+1} - b_i| = 2, \\forall i \\in \\{1, 2, \\dots, 2020\\}.\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": 40535,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with incircle $(I)$, tangent to $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. On the line $DF$, take points $M$, $P$ such that $CM \\parallel AB$, $AP \\parallel BC$. On the line $DE$, take points $N$, $Q$ such that $BN \\parallel AC$, $AQ \\parallel BC$. Denote $X$ as intersection of $PE$, $QF$ and $K$ as the midpoint of $BC$. Prove that if $AX = IK$ then $\\angle BAC \\le 60^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40536,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the popular game of Minesweeper, some fields of an $a \\times b$ board are marked with a mine and on all the remaining fields the number of adjacent fields that contain a mine is recorded. Two fields are considered adjacent if they share a common vertex. For which $k \\in \\{0, 1, 2, 3, 4, 5, 6, 7, 8\\}$ is it possible for some $a$ and $b$, $ab > 2021$, to create a board whose fields are covered in mines, except for 2021 fields who are all marked with $k$?",
    "options": [],
    "answer": "{1, 2, 3, 4, 5, 6, 8}",
    "solution": "For $k = 0$ this is impossible, since at least one non-mine field will be adjacent to the bombs unless the entire field is covered with bombs.\n\n* For $k = 1$ we take $a = 1, b = 3 \\cdot 1011 - 1$, and 1011 mines spaced two-spaces apart, with one on one of the endpoints, so the total number of fields marked 1 is 2021.\n\n* For $k = 2$ we again take $a = 1, b$ large enough, and place 2021 empty fields in isolation of each other, and not on any of the endpoints.\n\n* For $k = 3$, we take $a = 2, b$ large enough, and $2019/3$ mutually isolated L-triminoes as empty spaces, with two corner points on opposite sides of the longer edge also empty.\n\n* For $k = 4$, we take $a = 3, b$ large enough, and 1008 pairs of adjacent empty fields on the edge, with one empty field on the middle row surrounded on all diagonals by 4 empty fields on the edge, resembling an X shape.\n\n* For $k = 5$ we take $a = 3, b$ large enough, and place 2021 isolated empty fields on the edge.\n\n* For $k = 6$ we take $a$ and $b$ large enough, and place $2013/3$ isolated empty L-triminoes and 2 diamond-shaped regions of 4 fields each.\n\n* For $k = 7$ it is impossible to place an empty field on the edge and since each empty field in the interior is adjacent to exactly one other empty field it follows that the empty fields are paired and thus their total number cannot be 2021, an odd number.\n\n* For $k = 8$ we simply take 2021 isolated empty fields in the interior.\n\nThus, the set of solutions is: $\\{1, 2, 3, 4, 5, 6, 8\\}$. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 40537,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)_{n \\ge 1}$ be a sequence given by $a_1 = 45$ and\n$$\na_n = a_{n-1}^2 + 15a_{n-1}\n$$\nfor $n > 1$. Prove that the sequence contains no perfect squares.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By induction we can show that $a_n > 0$ for all positive integer $n$. Suppose that there exists a positive integer $n$ such that $a_n$ is a perfect square.\n$$\na_n = a_{n-1}^2 + 15a_{n-1} = a_{n-1}(a_{n-1} + 15)\n$$\nso $a_{n-1} \\mid a_n$ for all positive integers $n$. Thus, it follows that\n$$\na_1 = 45 \\mid a_n, \\quad \\text{for all positive integers } n.\n$$\nLet $k > 1$ be the smallest positive integer such that $a_k$ is a perfect square and $a_{k-1} = 45x$ with $x$ a positive integer. We have\n$$\na_k = 45x(45x + 15) = 15^2 \\cdot 3x(3x + 1)\n$$\nis a perfect square. On the other hand, $(3x, 3x + 1) = 1$ implies $3x$ and $3x + 1$ are perfect squares, so $3x = 0$ and $3x + 1 = 1$. It is impossible since $a_n > 0$ for all $n$.\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40538,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a rectangle with $P$ lies on the segment $AC$. Denote $Q$ as a point on minor arc $PB$ of $(PAB)$ such that $QB = QC$. Denote $R$ as a point on minor arc $PD$ of $(PAD)$ such that $RC = RD$. The lines $CB$, $CD$ meet $(CQR)$ again at $M$, $N$ respectively. Prove that $BM = DN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40539,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a non-isosceles triangle with incenter $I$ and let the circumcircle of the triangle $ABC$ has radius $R$. Let $AL$ be the external angle bisector of $\\angle BAC$ with $L \\in BC$. Let $K$ be the point on the perpendicular bisector of $BC$ such that $IL \\perp IK$. Prove that $OK = 3R$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $M$ as the midpoint of the arc $BC$ not containing $A$ of $(O)$, we need to prove $MK = 2OM$. Denote $BD, CE$ as the internal bisectors of $ABC$ then $D, E, L$ are collinear. Note that $L(AI, DB) = -1$ and take the orthogonal projection from $I$ to get $(Ix, Iy, IK, IM) = -1$ in which\n$$\nIM \\perp LA, IK \\perp LI, Iy \\perp LD, Ix \\perp LB\n$$\nSuppose that $Iy$ cuts $MK$ at $O'$, combining with $Ix \\parallel MK$, then $O'$ is the midpoint of $MK$. Thus it remains to prove $M$ is the midpoint of $OO'$.\nDenote $T$ as the ex-center with respect to angle $A$ of triangle $ABC$ then $M$ is the midpoint of $IT$. Then $MO = MO'$ if and only if $IO' \\parallel OT$, or $OT \\perp DE$.\nNow take $U, V$ as the ex-centers with respect to angles $B, C$ of triangle $ABC$ then $A, B, C$ are the foot of altitudes in triangle $TUV$ and $I, O$ are the orthocenter and nine-point center of that triangle, thus the reflection $S$ of $I$ over $O$ is the circumcenter. Note that $B, D, U$ and $C, E, V$ are collinear then\n$$\n\\mathcal{P}_{D/(O)} = \\overline{DA} \\cdot \\overline{DC} = \\overline{DU} \\cdot \\overline{DI} = \\mathcal{P}_{D/(IUV)}\n$$\nand similarly, $\\mathcal{P}_{E/(O)} = \\mathcal{P}_{E/(IUV)}$ so $DE$ is the radical axis of $(O)$ and $(IUV)$ which implies that $DE$ is perpendicular to the line joining centers of $(O)$ and $(IUV)$. We know that center $Z$ of $(IUV)$ is the reflection of $S$ over $UV$, and since $IT = 2NS$, it is easy to check that $Z \\in TO$. Thus, we have $DE \\perp OT$ which finishes the proof. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40540,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a point lies outside circle $(O)$ and tangent lines $AB$, $AC$ of $(O)$. Consider points $D$, $E$, $M$ on $(O)$ such that $MD = ME$. The line $DE$ cuts $MB$, $MC$ at $R$, $S$. Take $X \\in OB$, $Y \\in OC$ such that $RX$, $RY \\perp DE$. Prove that $XY \\perp AM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40541,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with $AB < AC$ and inscribed in the circle $(O)$. Denote $I$ as the incenter of $ABC$ and $D, E$ as the intersections of $AI$ with $BC, (O)$ respectively. Take a point $K$ on $BC$ such that $\\angle AIK = 90^\\circ$ and $KA, KE$ meet $(O)$ again at $M, N$ respectively. The rays $ND, NI$ meet the circle $(O)$ at $Q, P$.\n1. Prove that the quadrilateral $MPQE$ is a kite.\n\n2. Take $J$ on $IO$ such that $AK \\perp AJ$. The line through $I$ and perpendicular to $OI$ cuts $BC$ at $R$, cuts $EK$ at $S$. Prove that $OR \\parallel JS$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, we know that $EI = EB = EC$ then $E$ is the circumcenter of triangle $BIC$. Thus $KI$ is the tangent line of circle $(BIC)$. Hence,\n$$\nKI^2 = KB \\cdot KC = KE \\cdot KN.\n$$\nBy combining with $KIE$ is right triangle, we get $IN \\perp KE$.\n![](attached_image_1.png)\nSo $PE$ is the diameter of circle $(O)$. We have $\\triangle BDE \\sim \\triangle ABE$ so\n$$\n\\frac{BE}{AE} = \\frac{DE}{BE} \\Rightarrow EA \\cdot ED = EB^2 = EI^2 = EN \\cdot EK\n$$\nImplies that $ADNK$ is cyclic. Thus $\\angle MAE = \\angle DNE$ which leads to $EM, EQ$ are equal. Therefore, $MPQE$ is a kite.\n\nSince $KI^2 = KM \\cdot KA$, we get $IM \\perp AK$. Construct the diameter $AA'$ of $(O)$ then $M, I, A'$ are collinear. Thus $AJ \\parallel IM$ implies that $AIA'J$ is a parallelogram and then $O$ is the midpoint of $IJ$.\n\nSince $IO$ is the median of triangle $IPE$ so $Ix, IO, IP, IE$ form a harmonic quartet with $Ix \\parallel PE$. Denote $Ky$ as the ray that perpendicular to $IO$ then notice that $KE \\perp IP, KI \\perp IE, KD \\perp Ix, K, Ky \\perp IO$ so $KD, Ky, KI, KE$ form another harmonic quartet. Since $Ky \\parallel RS$ then $R$ is the midpoint of $OS$. Therefore $OR$ is the midline of triangle $IJS$ which implies that $OR \\parallel JS$.\n$\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40542,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$, let $K$ be a point on the median $BM$ such that $CM = CK$. It turned out that $\\angle CBM = 2\\angle ABM$. Show that $BC = KM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $L$ be the reflection of $C$ through $BM$. Since $CK = CM$, we have $\\angle CKM = \\angle CMK$. On the other hand, since $L$ is the reflection of $C$ through $KL$, we have $\\angle LKM = \\angle CKM = \\angle CMK$, implying that $KL$ is parallel to $CM$.\n\nAlso, we have $KL = KC = CM = MK$, and $KL$ is parallel to $AM$. Hence, $KMAL$ is a parallelogram, implying that $KM = AL$ and $KM$ is parallel to $AL$. We have $\\angle LBM = \\angle CBM = 2\\angle ABM$, hence $\\angle KBA = \\angle ABL$. Since $AL$ is parallel to $MB$, we have $\\angle BAL = \\angle MBA = \\angle LBA$, implying that $LB = LA$. Hence, we have\n$$\nKM = LA = LB = BC\n$$\nas desired.\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40543,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB = AC$ and $M$ is the midpoint of the altitude $AD$. Consider $(\\omega)$ as the circle of center $M$ and tangent to $AB$, $AC$. From some point $T$ on the line $BC$ (outside triangle $ABC$), construct two tangents of $(\\omega)$ and they cut $AB$ at $P$, $Q$, cut $AC$ at $R$, $S$. Prove that $PQ = RS$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40544,
    "subject": "Mathematics (Multi-modal)",
    "question": "We have $n > 2$ nonzero integers such that every one of them is divisible by the sum of the other $n-1$ numbers. Show that the sum of the $n$ numbers is precisely $0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let these numbers be $a_1, a_2, \\dots, a_n$ and $S$ is the sum of these numbers. For every $i \\in \\{1, 2, \\dots, n\\}$, we have $S - a_i \\mid a_i$, which means $S - a_i \\nmid S$. Suppose that $S \\neq 0$, without the loss of generality, $S > 0$. We investigate two cases\n\n* If $\\exists i \\in \\{1, 2, \\dots, n\\}$ such that $a_i < 0$ then $|S - a_i| > |S|$ so $S - a_i \\nmid S$.\n\n* $a_i > 0$ for all $i \\in \\{1, 2, \\dots, n\\}$. Let $a_k$ be the smallest of $a_1, a_2, \\dots, a_n$. We have $S - a_m > 2a_m - a_m = a_m$, so $S - a_m \\nmid a_m$.\n\nHence, $S \\neq 0$ is impossible, which means $S = 0$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40545,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let us call a set of positive integers nice if the number of its elements equals to the average of its numbers. Call a positive integer $n$ an amazing number if the set $\\{1, 2, \\ldots, n\\}$ can be partitioned into nice subsets.\n\na) Prove that every perfect square is amazing\n\nb) Show that there are infinitely many positive integers which are not amazing.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $A = \\{1, 2, \\dots, n\\}$ and $n = k^2$ for some positive integer $k$. Denote $B$ as the set of first $m$ odd positive integers. It's clear that $B$ is *nice*. Assume that $n \\ge 2m-1$, let $C = A \\setminus B$. $C$ is *nice* if and only if\n$$\n\\frac{(1 + 2 + \\cdots + n) - m^2}{n - m} = n - m \\iff n = n^2 - 4mn + 4m^2 \\iff m = \\frac{n-k}{2}.\n$$\nTherefore, we can choose $m$ so that $C$ is *nice*. Hence, $n$ is *amazing*.\n\nb) We prove that if $n = 4k + 2$ for some positive integer $k$, then $n$ is not *amazing*. Assume that $A = \\{1, 2, \\dots, n\\}$ can be partitioned into nice subsets $A_1, A_2, \\dots, A_m$ with cardinality $a_1, a_2, \\dots, a_m$ respectively. Therefore,\n$$\n\\frac{n(n+1)}{2} = \\sum_{x \\in A} x = \\sum_{i=1}^{m} \\sum_{x \\in A_i} x = \\sum_{i=1}^{m} a_i^2 = \\sum_{i=1}^{m} a_i = n \\pmod{2},\n$$\nthat yields contradiction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40546,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a non-isosceles triangle with incenter $I$, circumcenter $O$ and a point $D$ on segment $BC$ such that $(BID)$ cuts segment $AB$ at $E \\neq B$ and $(CID)$ cuts segment $AC$ at $F \\neq C$. Circle $(DEF)$ cuts segments $AB$, $AC$ again at $M$, $N$. Let $P = IB \\cap DE$ and $Q = IC \\cap DF$. Prove that $EN$, $FM$, $PQ$ are parallel and the median of vertex $I$ in triangle $IPQ$ bisects the arc $BAC$ of $(O)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 40547,
    "subject": "Mathematics (Multi-modal)",
    "question": "A set of $n$ points in space is given, no three of which are collinear and no four of which are co-planar (on a single plane), 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. In each turn Alice colors one segment red and then Bob colors up to $b$ segments blue. This is repeated until there are no more colorless segments left. If Alice colors a red triangle, Alice wins. If there are no more colorless segments and Alice hasn't succeeded in coloring a red triangle, Bob wins. Neither player is allowed to color over an already colored line segment.\n1. Prove that if $b < \\sqrt{2n-2} - \\frac{3}{2}$, then Alice has a winning strategy.\n2. Prove that if $b \\ge 2\\sqrt{n}$, then Bob has a winning strategy.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1. We will call a *threat* an uncolored segment whose coloring of red would produce a triangle. We will show that Alice can use the following strategy to win: Alice will keep coloring segments whose one end-point is $A$ unless she can complete a triangle. We will call the other endpoints *opposite ends*. Let us assume that Bob can stop this strategy. Bob's optimal strategy to counter Alice's is first eliminating all threats in his move and then using his remaining moves to color segments coming out of $A$. Note that after the $k$-th red line is drawn, Bob must color $k - 1$ threats blue, which are the lines connecting the new opposite end with all the previous opposite ends. Thus in the $k$-th pair of moves of Alice and Bob, the number of colored edges coming out of $A$ has increased by $b + 1 - (k - 1) = b - k + 2$. Assume there exists a $l$ such that after $l$ moves all segments out of $A$ have been colored before Alice could draw a triangle. Then we have\n$$\n(b + 2)l - \\frac{l(l + 1)}{2} \\geq n - 1 \\Rightarrow l^2 + 2l\\left(b + \\frac{3}{2}\\right) + 2n - 2 \\leq 0,\n$$\ntherefore the discriminant of the quadratic equation is non-negative, i.e. $4\\left(b + \\frac{3}{2}\\right)^2 - 4(2n - 2) \\geq 0$ and therefore $b \\geq \\sqrt{2n - 2} - \\frac{3}{2}$. Since this inequality is violated, it follows that Alice's winning strategy works.\n\n2. We will show that Bob can use the following strategy to win: Whenever Alice colors a line $AB$, Bob will color $\\lfloor b/2 \\rfloor$ edges out of $A$ blue and $\\lfloor b/2 \\rfloor$ edges out of $B$. This ensures at most $n/\\lfloor b/2 \\rfloor + 1$ red edges out of any point, which is not larger than $\\lfloor b/2 \\rfloor + 1$ for $b \\ge 2\\sqrt{n}$. Assuming Alice completes a triangle $ABC$ and WLOG moves $AB$, $AC$ and $BC$ in sequence, it follows that before placing $BC$ there were at least $\\lfloor b/2 \\rfloor + 1$ threats out of $C$. However this means, since placing $AC$ created these threats, that, not counting $AC$, there were at least $\\lfloor b/2 \\rfloor + 1$ red lines out of $A$ and thus the total number of red lines coming out of $A$ is larger than $\\lfloor b/2 \\rfloor + 1$, a contradiction.\n\n$\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40548,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle inscribed in circle $(O)$ with diameter $KL$ passes through the midpoint $M$ of $AB$ such that $L$, $C$ lie on the different sides with respect to $AB$. A circle passes through $M$, $K$ cuts $LC$ at $P$, $Q$ (point $P$ lies between $Q$, $C$). The line $KQ$ cuts $(LMQ)$ at $R$. Prove that $ARBP$ is cyclic and $AB$ is the symmedian of triangle $APR$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40549,
    "subject": "Mathematics (Multi-modal)",
    "question": "By $\\text{rad}(x)$ we denote the product of all distinct prime factors of a positive integer $n$. Given $a \\in \\mathbb{N}$, a sequence $(a_n)$ is defined by $a_0 = a$ and $a_{n+1} = a_n + \\text{rad}(a_n)$ for all $n \\ge 0$. Prove that there exists an index $n$ for which $\\frac{a_n}{\\text{rad}(a_n)} = 2022$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Solution.* To prove that $MP + NQ = BR$, by adding $RC$ to both sides it is equivalent to prove $CQ + MP = BC$. We start by defining point $T$ on the segment $CQ$ such that $BC = CT$, then it is sufficient to prove that $QT = PM$, since triangles $BTC$ and $NRC$ are isosceles. One can get that $BT$ is parallel to $NR$ so\n$$\n\\angle BTC = \\angle RNC = \\angle NRC = \\angle BPC,\n$$\nthus $BTPC$ is cyclic. Also we have that\n$$\n\\angle CNA = 180^\\circ - \\angle NRC = 180^\\circ - \\angle BPC = \\angle CPA,\n$$\nthis implies that $PNCA$ is cyclic.\n\n$$\n\\angle TPQ = \\angle TCB = \\angle BQC = \\angle TQP,\n$$\nthus $TQ = TP$. Now by angle chasing,\n$$\n\\angle CPA = 180^\\circ - \\angle BPC = 180^\\circ - \\angle BTC = 180^\\circ - \\angle TBC = \\angle TPC,\n$$\nthus we have $\\angle CPQ = \\angle TPC$ and $\\angle PCT = \\angle PAM$, which implies that $TPC$ and $MPA$ are congruent, since $PC = AP$. In conclusion, $TP = MP = TQ$, which finishes the proof. □",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40550,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every polynomial $P(x)$ there exist polynomials $Q(x)$ and $R(x)$ such that $P(x) = Q(x^2) + R((x+1)^2)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40551,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)$ be the integer sequence which is defined by $a_1 = 1$ and\n$$\na_{n+1} = a_n^2 + n \\cdot a_n, \\forall n \\ge 1.\n$$\nLet $S$ be the set of all primes $p$ such that there exists an index $i$ such that $p|a_i$. Prove that the set $S$ is an infinite set and it is not equal to the set of all primes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, we shall show that $3 \\notin S$ by proving that\n$$\na_{3k-2} \\equiv 2 \\pmod{3},\\ a_{3k-1} \\equiv 2 \\pmod{3}\\ \\text{ and }\\ a_{3k} \\equiv 2 \\pmod{3}.\n$$\nSince $a_1 = 1$, $a_2 = 2$, $a_3 = 2^2 + 2 \\cdot 2 = 8 \\equiv 2 \\pmod{3}$ so the claim is true for $k = 1$.\nSuppose that the claim holds for $k = n$. We have\n$$\n\\begin{align*}\na_{3n+1} &= a_{3n}(a_{3n} + 3n) \\equiv 2 \\cdot 2 \\equiv 1 \\pmod{3}, \\\\\na_{3n+2} &= a_{3n+1}(a_{3n+1} + 3n + 1) \\equiv 1(1 + 1) \\equiv 2 \\pmod{3}, \\\\\na_{3n+3} &= a_{3n+2}(a_{3n+2} + 3n + 2) \\equiv 2(2 + 2) \\equiv 2 \\pmod{3}.\n\\end{align*}\n$$\nThus, the claim is also true for $k = n + 1$. So by induction, the claim is proved.\n\nNow suppose on the contrary that $S$ is finite, denote $S = \\{p_1, p_2, \\dots, p_k\\}$. Note that $a_n|a_{n+1}$ so $a_n|a_m$ for all $m \\ge n$. By the definition of $S$, there are some index $t$ such that $p_1|a_t$, thus $p_1|a_{t'}$ for all $t' \\ge t$. Similarly for $p_2, p_3, \\dots, p_k$ so there exist $N$ big enough such that $p_1p_2\\cdots p_k|a_n$ for all $n \\ge N$. Taking the integer $\\ell > N + 1$ such that $\\ell \\equiv 2 \\pmod{p_1p_2\\cdots p_k}$. Since $a_\\ell = a_{\\ell-1}(a_{\\ell-1} + \\ell - 1)$, we get\n$$\na_{\\ell-1} + \\ell - 1 \\equiv 1 \\pmod{p_1 p_2 \\cdots p_k}\n$$\nso $a_{\\ell-1} + \\ell - 1$ is coprime to all primes in $S$, which implies that $S$ has some prime divisor that not belong to $S$, a contradiction. Hence, $S$ is an infinite set. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40552,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each non-constant integer polynomial $P(x)$, let's define\n$$\nM_{P(x)} = \\max_{x \\in [0; 2021]} |P(x)|.\n$$\n1. Find the minimum value of $M_{P(x)}$ when $\\deg P(x) = 1$.\n2. Suppose that $P(x) \\in \\mathbb{Z}[x]$ when $\\deg P(x) = n$ and $2 \\le n \\le 2022$. Prove that $M_{P(x)} \\ge 1011$.",
    "options": [],
    "answer": "Part 1: 1011. Part 2: For degrees from two to two thousand twenty two, M_P is at least 1011.",
    "solution": "1) Since $\\deg P = 1$, put $P(x) = ax + b$ with $a, b \\in \\mathbb{Z}$ and $a \\neq 0$. Note that\n$$\n|P(2021) - P(0)| = |2021a| \\ge 2021 \\Rightarrow \\max\\{|P(2021)|, |P(0)|\\} \\ge \\frac{2021}{2}.\n$$\nThus, $|P(2021) - P(0)| = |2021a| \\ge 2021$ which implies that\n$$\n\\max\\{|P(2021)|, |P(0)|\\} \\ge \\frac{2021}{2} \\Rightarrow \\max\\{|P(2021)|, |P(0)|\\} \\ge 1011.\n$$\nOn the other hand, $T(x) = x - 1011$ satisfies the condition $|T(x)| \\le 1011, \\forall x \\in [0; 2021]$. Hence, the minimum value of $M_{P(x)}$ is $1011$, the equality holds when $P(x) = x - 1011$.\n\n2) Suppose that there is some integer polynomial $P(x)$ with\n$$\n2 \\le \\deg P \\le 2022 \\text{ and } M_{P(x)} < 1011.\n$$\nOne can see that $|P(x)| < 1011, \\forall x \\in [0; 2021]$ thus $|P(x)| \\le 1010, \\forall x \\in [0; 2021] \\cap \\mathbb{Z}$. By the property of integer polynomials, $2021|P(2021) - P(0)$ so $P(2021) = P(0)$ which leads to\n$$\n|P(2021) - P(0)| \\le |P(0)| + |P(2021)| \\le 2020.\n$$\nFrom that, let's put $P(x) = x(x - 2021)Q(x) + c$ with $c \\in \\mathbb{Z}$ and $Q(x) \\in \\mathbb{Z}[x]$. So $x(x - 2021) \\ge 2022, \\forall x \\in \\{2, 3, \\dots, 2019\\}$. If there exists $x_0 \\in \\{2, 3, \\dots, 2019\\}$ such that $Q(x_0) \\ne 0$ then\n$$\n\\max |x_0(x_0 - 2021)Q(x_0) - c| \\ge \\frac{1}{2} |x_0(x_0 - 2021)Q(x_0)| \\ge 1011.\n$$\nFrom this, we can conclude that $Q(2) = Q(3) = \\cdots = Q(2019) = 0$ so let's write\n$$\nP(x) = x(x-2)(x-3)\\cdots(x-2019)(x-2021)H(x) + c\n$$\nwith $H(x) \\in \\mathbb{Z}[x]$ then $P(1) = -2020 \\cdot 2018!H(1) + c$. Note that if $H(1) \\neq 0$ then $M_{P(x)} > 1011$. Similarly, $H(2020) \\neq 0$ also leads to another contradiction, so $H(1) = H(2020) = 0$ then there is some $R(x) \\in \\mathbb{Z}[x]$ for which\n$$\nQ(x) = (x-1)(x-2)\\cdots(x-2020)R(x).\n$$\nSince $\\deg P \\leq 2022$ then $R(x) \\equiv c \\in \\mathbb{Z}$. On the other hand, $P\\left(\\frac{1}{2}\\right) > 1011$ so $c = 0$, which contradicts $P(x)$ being non-constant. Therefore, the contrary hypothesis is false and it follows that $M_{P(x)} \\geq 1011$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40553,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $k$ such that the product of the first $k$ primes increased by 1 is a power of an integer (with an exponent greater than 1).",
    "options": [],
    "answer": "no such k",
    "solution": "Denote the first $n$ primes as\n$$\np_1 = 2 < p_2 = 3 < \\dots < p_n\n$$\nSuppose that $p_1p_2\\cdots p_n + 1 = x^k$ for some integers $x, k \\ge 2$. We can assume WLOG that $k$ is prime since $x^{kt} = (x^t)^k$. Obviously, $x$ has no prime factors not exceeding $p_n$, so $x > p_n$ and consequently $k < n < p_n$ is one of the first $n$ primes. Now $x^k \\equiv 1 \\pmod{k}$, which implies $x \\equiv 1 \\pmod{k}$, but then by Lifting the Exponent Lemma\n$$\nx^k \\equiv 1 \\pmod{k^2}.\n$$\nThis is a contradiction, as $x^k - 1 = p_1 \\cdots p_n$ is not divisible by the square of any prime. Thus, there does not exist any $k$ satisfying the problem. □",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40554,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incenter of the triangle $ABC$. Let $X$ lie on segment $AB$, such that $\\angle AIX = 90^\\circ$. The circumcircle of triangle $BIX$ intersects the circumcircle of triangle $ABC$ at point $Y \\neq B$ lying on the same side of $AB$ as point $C$. Prove that $YX$ is the bisector of angle $AYB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40555,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let a finite set of integers be given, such that each of its elements can be written as a sum of some two elements (not necessarily distinct) from the same set. For such a set, we say that it is of 'safety of order $n$' if it does not contain a subset of $n$ or less elements whose sum is $0$. Prove that there exist sets of arbitrarily large safety order.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40556,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x, y \\in \\mathbb{R}$ be such that $x = y(3 - y)^2$ and $y = x(3 - x)^2$. Find all possible values of $x + y$.",
    "options": [],
    "answer": "0, 3, 4, 5, 8",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40557,
    "subject": "Mathematics (Multi-modal)",
    "question": "In each cell of a $10 \\times 10$ board one arrow is placed. Each arrow is pointing in one of the four directions $\\{\\uparrow, \\to, \\downarrow, \\leftarrow\\}$. Find the smallest number $n$ with the following property: it is always (regardless of the initial placement of the arrows) possible to remove at most $n$ arrows from the board in such a way that among the remaining ones no two are pointing at each other. Note: arrows are pointing at each other also if there are some other arrows or empty cells between them.",
    "options": [],
    "answer": "50",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40558,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A, B, C, D$ be points on the line $d$ in that order and $AB = CD$. Denote $(P)$ as some circle that passes through $A, B$ with its tangent lines at $A, B$ are $a, b$. Denote $(Q)$ as some circle that passes through $C, D$ with its tangent lines at $C, D$ are $c, d$. Suppose that $a$ cuts $c, d$ at $K, L$ respectively; and $b$ cuts $c, d$ at $M, N$ respectively. Prove that four points $K, L, M, N$ belong to a same circle $(\\omega)$ and the common external tangent lines of circles $(P), (Q)$ meet on $(\\omega)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the points that arranged as following figure, the other cases will be proved similarly.\nDenote $R, S$ as intersection of the pairs of lines $PB, QC$ and $PA, QD$. Note that $BMCR$ and $ALDS$ are cyclic quadrilateral. Thus\n$$\n\\begin{aligned}\n\\angle KMN &= \\angle BRC = 180^\\circ - (\\angle BCR + \\angle CBR) \\\\\n&= 180^\\circ - (\\angle PBA + \\angle QDC) \\\\\n&= 180^\\circ - (\\angle PAD + \\angle QDA) = \\angle ASD = \\angle KLN.\n\\end{aligned}\n$$\nHence, $K, L, M, N$ belong to the same circle.\n\nNow, consider the following claim: Let be given two circles $(P, R)$, $(Q, R')$ and some line cuts them at $B, A, D, C$ such that $AB = CD$ (see the figure). The tangent lines at $A, C$ respectively of $(P), (Q)$ meet at $X$ then $\\frac{XP}{XQ} = \\frac{R}{R'}$.\n\nIndeed, by applying the sine law for triangle $XAC$, we get\n$$\n\\frac{XA}{XC} = \\frac{\\sin XCA}{\\sin XAC} = \\frac{\\sin \\frac{CQD}{2}}{\\sin \\frac{APB}{2}} = \\frac{CD}{AB} \\cdot \\frac{R}{R'} = \\frac{R}{R'} = \\frac{PA}{QC}.\n$$\nThus two triangles $XPA$ and $XQC$ are similar, which implies that $\\frac{XP}{XQ} = \\frac{R}{R'}$. The claim is proved.\n\nBack to the problem, denote $X, Y$ as the external and the internal homothety centers of $(P), (Q)$ then $\\frac{XP}{XQ} = \\frac{YP}{YQ} = k$ with $k$ is the ratio of radius of $(P), (Q)$. It is easy to check that these radiuses are different, otherwise the the tangent lines of $(P), (Q)$ will be parallel and points $K, L, M, N$ will not exist, thus $k \\neq 1$. In the other hand, by applying the above claim, we get\n$$\n\\frac{MP}{MQ} = \\frac{NP}{NQ} = \\frac{KP}{KQ} = \\frac{LP}{LQ} = k.\n$$\nHence, six points $X, Y, M, N, K, L$ are all belong to the Apollonius circle with ratio $k$ constructing on the segment $PQ$. Thus, the point $X$, which also is the intersection of two common external tangent lines of $(P), (Q)$, is on $(\\omega)$. $\\square$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40559,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the maximum number of $2 \\times 3 \\times 3$ bricks that can be fit inside an $8 \\times 8 \\times 9$ box?",
    "options": [],
    "answer": "31",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40560,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcenter of triangle $ABC$. Points $X$ and $Y$ on side $BC$ are such that $AX = BX$ and $AY = CY$. Prove that the circumcircle of triangle $AXY$ passes through the circumcenters of triangles $AOB$ and $AOC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40561,
    "subject": "Mathematics (Multi-modal)",
    "question": "You are given $n$ different primes $p_1, p_2, \\dots, p_n$. Consider the polynomial\n$$\nx^n + a_1x^{n-1} + a_2x^{n-2} + \\dots + a_{n-1}x + a_n,\n$$\nwhere $a_i$ is the product of the first $i$ given prime numbers. For what $n$ can it have an integer root?",
    "options": [],
    "answer": "n = 1",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40562,
    "subject": "Mathematics (Multi-modal)",
    "question": "The perpendicular bisector to the side $AC$ of triangle $ABC$ meets $BC$ and $AB$ at points $A_1$ and $C_1$ respectively. Let points $O$ and $O_1$ be the circumcenters of triangles $ABC$ and $A_1BC_1$ respectively. Prove that $C_1O_1$ is a tangent to the circumcircle of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40563,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call a positive integer $n$ venerable if all its positive divisors less than $n$ (but including 1) add up to $n - 1$. Find all venerable numbers whose some power (with the exponent at least 2) is also venerable.",
    "options": [],
    "answer": "All powers of two: n = 2^k for integer k ≥ 0.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40564,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(a, b, c)$ of real numbers satisfying\n$$\na + b + c = 1 \\quad \\text{and} \\quad 3(a + bc) = 4(b + ca) = 5(c + ab).\n$$",
    "options": [],
    "answer": "(a, b, c) = (1/2, 1/3, 1/6) and (a, b, c) being any permutation of (1, 1, -1).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40565,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine if there exist functions $f, g: \\mathbb{R} \\to \\mathbb{R}$ satisfying for every $x \\in \\mathbb{R}$ the following equations\n$$\nf(g(x)) = x^3 \\quad \\text{and} \\quad g(f(x)) = x^2.\n$$",
    "options": [],
    "answer": "Yes, such functions exist.",
    "solution": "*Solution.* Denote $b_n = \\frac{a_n}{\\text{rad}(a_n)}$. Since rad($a_n$) divides rad($a_{n+1}$) we have $b_{n+1} | b_n + 1$. If there are indices $i < j$ with $b_i < 2022 < b_{i+1}$, we will be done by “continuity”. If, to the contrary, this does not happen, there are two possible cases.\n* $b_n < 2022$ for all $n$ big enough. Since $a_n$ increases indefinitely, then so does rad($a_n$), so at some moment rad($a_n$) receives a new prime $p > 2022$. This means that $p \\nmid a_n$ and $p | a_{n+1} = \\frac{b_{n+1}}{b_n}a_n$, so $p | b_n + 1$ and hence $b_n \\ge 2022$, a contradiction.\n* $b_n > 2022$ for all $n$. We can assume WLOG that $b_0$ is the smallest term of the sequence $(b_n)$. Suppose that $b_{i+1} = b_i + 1$ for all $0 \\le i < n$. Then\n$$\n\\text{rad}(a_0) = \\dots = \\text{rad}(a_{n-1}) = R.\n$$\nBut for every prime $p \\le n$ there is a multiple of $p$ among us sus $b_0, \\dots, b_{n-1}$, so $p | a_k$ for some $k$ and consequently $p | R$. Since not every prime divides $R$, there must be an index $n$ such that $b_n < b_{n-1} + 1$, i.e. $b_{n-1} + 1 = db_n$ for some $d > 1$ and $\\text{rad}(a_{n+1}) = dR$, so $\\text{gcd}(d, R) = 1$.\nRecall that $b_0 \\le b_n = \\frac{b_0+n}{d}$, which reduces to $n \\ge (d-1)b_0$. By above, this means that all primes up to $(d-1)b_0$ divide $R$, but $d$ does not divide $R$, so $d > (d-1)b_0$, which is impossible. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40566,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all quadruples $(p, a, b, x)$ that satisfy the equality $p^2 + 4a9^b = x^2$, where $p$ is a prime and $a, b, x$ are nonnegative integers.",
    "options": [],
    "answer": "All solutions arise by choosing nonnegative integers a, b and integers u, v ≥ 0 with uv = a·9^b and v − u being a prime p, then setting x = u + v and p = v − u. Equivalently, for any divisor d of a·9^b with p = a·9^b/d − d prime, set x = a·9^b/d + d. In particular, when a = 0, we have x = p for any prime p and any b ≥ 0.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40567,
    "subject": "Mathematics (Multi-modal)",
    "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": "Detailed 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![](attached_image_1.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\\angle DO_2K = \\frac{1}{2}\\angle DMK = \\frac{1}{2}\\angle KCM = \\angle DCO_1,\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\\angle KAD = 2\\angle KO_1D = 2\\angle DCO_2 = \\angle KCB;\n$$\nthis implies the desired cyclicity of the quadrilateral $ABCD$. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 40568,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the polynomial $f(x) = c x(x - 2)$ where $c$ is a positive real number. For any $n \\in \\mathbb{Z}^{+}$, the notation $g_n(x)$ is a composite function $n$ times of $f$ and assume that the equation $g_n(x) = 0$ has all of the $2^n$ solutions are real numbers.\n1. For $c = 5$, find in terms of $n$, the sum of all the solutions of $g_n(x)$, of which each multiple (if any) is counted only once.\n2. Prove that $c \\ge 1$.",
    "options": [],
    "answer": "Part 1: 2^n. Part 2: c ≥ 1.",
    "solution": "1) We will prove by induction on $n$ that the solutions of $g_n(x)$ are all distinct. With $n=1$, we have $g_1(x) = f(x)$ which has two solutions, $x=0, x=2$. Suppose the polynomial $g_n(x)$ has all $2^n$ solutions that are distinct, set as $x_1, x_2, \\dots, x_{2^n}$, we write $g_n(x) = k(x - x_1)(x - x_2) \\dots (x - x_{2^n})$. Hence,\n$$\ng_{n+1}(x) = g_n(f(x)) = k(f(x) - x_1)(f(x) - x_2) \\dots (f(x) - x_{2^n})\n$$\nso with $i \\neq j$, obviously the solution set of $f(x) - x_i$ and $f(x) - x_j$ will be disjoint. Also, if the equation $f(x) - x_i = 0$ has a double root, then that solution must be $1$. Next, we have\n$$\ng_1(1) = f(1) = -5, \\quad g_2(1) = f(-5) = 5(-5)(-7) = 175\n$$\nso $g_n(1) > 0, \\forall n \\ge 2$. Hence, $x = 1$ cannot be a solution of $g_n(x), \\forall n$. This shows that $g_{n+1}(x)$ also has a distinct solution $2^{n+1}$. The induction is completed.\nFinally, from the above analysis, the solutions of $g_n(x)$ can be divided into $2^{n-1}$ disjoint pairs having sum equal $2$ so the sum of all the solutions will be $2^n$.\n\n2) Note that for some parameter $d \\in \\mathbb{R}$, the equation\n$$\ncx(x - 2) = d \\iff x^2 - 2x = \\frac{d}{c}\n$$\nhas two distinct solutions if and only if $\\Delta' = 1 + \\frac{d}{c} > 0$ or $c > -d$. (\\heartsuit) It will then have the solutions\n$$\nx = 1 + \\sqrt{1 + \\frac{d}{c}} \\text{ and } x = 1 - \\sqrt{1 + \\frac{d}{c}}.\n$$\nIn order for $g_n(x)$ to have $2^n$, $g_{n-1}(x)$ must have enough $2^{n-1}$ distinct roots. Denote $r$ as one of those roots, we investigate the equation $f(x) = r$.\n* If $r < 0$ then the equation has two positive solutions.\n* If $r > 0$ then the equation has two solutions with opposite signs.\nWe see that (\\heartsuit) generates a constraint for $c$ only when the parameter $d < 0$. Therefore, we are interested in pairs of opposite solutions generated from $r > 0$. Suppose that pair of solutions is $(r_1, r_2)$ with $r_1 < 0 < r_2$. We need $c > -r_1 = -(2 - r_2) = r_2 - 2$ so we take it back to the survey positive solution $r_2$ in that pair. Starting from an initial solution $r = 2$, we can build a sequence of numbers as follows: $u_1 = 2$ and $u_{n+1} = 1 + \\sqrt{1 + \\frac{u_n}{c}}$ with $n \\ge 1$.\nIt is easy to see that $u_n$ is the solution of $g_n(x)$ and $u_2 > u_1$ so by induction, we can prove that the sequence $(u_n)$ increases strictly. Let $L$ be the solution of\n$$\nL = 1 + \\sqrt{1 + \\frac{L}{c}} \\text{ or } L = 2 + \\frac{1}{c}.\n$$\nBy induction, we can also prove that $u_n < L, \\forall n$, so it is clear that the sequence will converge to $L$. Note that we must always have $c > u_n - 2, \\forall n$ so let $n \\to +\\infty$, we get $c \\ge 2 + \\frac{1}{c} - 2 = \\frac{1}{c}$ or $c \\ge 1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40569,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $BB'$, $CC'$ be the altitudes of an acute-angled triangle $ABC$. Two circles passing through $A$ and $C'$ are tangent to $BC$ at points $P$ and $Q$. Prove that $A$, $B'$, $P$, $Q$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $BP^2 = BQ^2 = BA \\cdot BC'$ and the quadrilaterals $AC'A'C$, $AB'A'B$ are cyclic ($AA'$ is the altitude) we have\n$$\nCP \\cdot CQ = CB^2 - BP^2 = CB^2 - BA \\cdot BC' = BC^2 - BC \\cdot BA' = BC \\cdot CA' = CA \\cdot CB'\n$$\nClearly this is equivalent to the required assertion. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40570,
    "subject": "Mathematics (Multi-modal)",
    "question": "A pile of $2^n - 1$ coins is placed on the player. Alice is allowed to make the following moves: If a pile has an even number of coins, she can throw half of those coins away. If a pile has an odd number of coins $k$ such that $k + 1$ is properly divisible by $2^l$ (i.e. $2^l|k + 1$ and $2^l < k + 1$) for some positive integer $l$, Alice can take away $2^l - 1$ coins from this pile and form a new pile with those coins. Alice continues making moves until no more moves can be made. What are the possible values for the number of coins that will remain on the table?",
    "options": [],
    "answer": "n",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40571,
    "subject": "Mathematics (Multi-modal)",
    "question": "Characterize all positive integers $n > 1$ for which the expression\n$$\n(n-1)! \\cdot \\left(1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n-1}\\right)\n$$\nis not divisible by $n$.",
    "options": [],
    "answer": "Exactly those n equal to 2p with p an odd prime, and n = 2, 4, 8.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40572,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x(n)$ be the biggest prime divisor of $n$. Prove that there exist infinitely many number $n$ such that $x(n) < x(n + 1) < x(n + 2)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40573,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the sequence $a_1, a_2, \\dots, a_{20}$ is the permutation of integers $1, 2, \\dots, 20$. Find the maximum possible value of\n$$\n\\min\\{|a_2 - a_1|, |a_3 - a_2|, \\dots, |a_{20} - a_{19}|, |a_1 - a_{20}|\\}.\n$$",
    "options": [],
    "answer": "9",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 40574,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two triangles $ABC$ and $A_1B_1C_1$ are symmetric about the center of their common incircle of radius $r$. Prove that the product of the areas of the triangles $ABC$, $A_1B_1C_1$ and the six other triangles formed by the intersecting sides of the triangles $ABC$ and $A_1B_1C_1$ is equal to $r^{16}$.",
    "options": [],
    "answer": "r^{16}",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40575,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a polynomial $P(x)$ of second degree with integer coefficients such that the leading coefficient is not divisible by $2022$ and all numbers $P(1), P(2), \\dots, P(2022)$ give different residues mod $2022$?",
    "options": [],
    "answer": "Yes. For example, P(x) = 1011x^2 + 2x works.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40576,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any 2 positive integers $m$ and $n$ with $(m, n) \\neq (1, 1)$ the value of expression\n$$\n\\frac{1}{m} + \\frac{1}{m+1} + \\dots + \\frac{1}{m+n-1}\n$$\nis not an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40577,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any positive integer $n$ at least one coefficient of the polynomial\n$$\n(x^4 + x^3 - 3x^2 + x + 2)^n\n$$\nis negative.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40578,
    "subject": "Mathematics (Multi-modal)",
    "question": "2000 consecutive integers (not necessarily positive) are written on the board. A student takes several turns. On each turn, he partitions the 2000 integers into 1000 pairs, and substitutes each pair by the difference and the sum of that pair (note that the difference does not need to be positive as the student may choose to subtract the greater number from the smaller one; in addition, all the operations are carried simultaneously). Prove that the student will never again write 2000 consecutive integers on the board.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $(a-b)^2 + (a+b)^2 = 2(a^2 + b^2)$, so the sum of the squares of the numbers written on the board doubles on each turn. Note that\n$$\nn^2 + (n+1)^2 + \\dots + (n+1999)^2 = 2000n^2 + 1999 \\cdot 2000n + \\frac{1999 \\cdot 2000 \\cdot 3999}{6},\n$$\nwhich is congruent to $8$ modulo $16$. Obviously, when we multiply this sum by $2$, we will obtain a number divisible by $16$, thus, we will never have consecutive numbers again, which is what we need to show. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40579,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an equilateral triangle $ABC$. Points $D$, $E$, $F$ lie on sides $BC$, $CA$, $AB$, respectively, and satisfy $AF = BD$ and $DF = EF \\neq DE$. Prove that $\\angle CDE = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40580,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number and let $m$, $n$ be integers greater than $1$ such that $n \\mid m^{p(n-1)} - 1$. Prove that $\\gcd(m^{p(n-1)} - 1, n) > 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Set $\\alpha = v_p(n-1)$ and write $n-1 = r \\cdot p^{\\alpha}$. Let $q$ be an arbitrary prime divisor of $n$, and set $d = \\text{ord}_q(m)$. Since\n$$\nm^{p(n-1)} = m^{r \\cdot p^{\\alpha+1}} \\equiv 1 \\pmod{q},\n$$\nit follows that $d \\mid r \\cdot p^{\\alpha+1}$. If $v_p(d) \\neq \\alpha + 1$, then clearly $d \\mid n-1$, and therefore $q$ is a common divisor of $n$ and $m^{n-1} - 1$. Otherwise, suppose that $v_p(d) = \\alpha + 1$ for all prime divisors of $n$. Together with $m^{q-1} \\equiv 1 \\pmod{q}$ by Fermat's Little Theorem, we have\n$$\nd \\mid q-1 \\implies v_p(q-1) \\ge v_p(d) = \\alpha + 1.\n$$\nThus, any prime divisor $q$ of $n$ satisfies $q \\equiv 1 \\pmod{p^{\\alpha+1}}$. Because $n$ is a product of these prime divisors, we deduce that $n \\equiv 1 \\pmod{p^{\\alpha+1}}$. However, this contradicts to $v_p(n-1) = \\alpha$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40581,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is triangle $ABC$ with $AB > AC$. Circles $o_B$, $o_C$ are inscribed in angle $BAC$ with $o_B$ tangent to $AB$ at $B$ and $o_C$ tangent to $AC$ at $C$. Tangent to $o_B$ from $C$ different than $AC$ intersects $AB$ at $K$, and tangent to $o_C$ from $B$ different than $AB$ intersects $AC$ at $L$. Line $KL$ and the angle bisector of $BAC$ intersect $BC$ at points $P$ and $M$, respectively. Prove that $BP = CM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that the length of the segment of the common tangent to $o_B$ and $o_C$ joining the tangency points is equal to\n$$\nAB - AC = LB - LC = KC - KB,\n$$\nSAUDI ARABIAN IMO Booklet 2022\n\nwhich means that points $A$, $L$ lie on the one, and point $K$ on the other leg of some hyperbola $\\eta$ with foci $B$, $C$.\nDenote by $S$ the midpoint of the segment $BC$ (the center of symmetry of $\\eta$) and denote by $A'$, $K'$, $M'$ the central reflections in $S$ of $A$, $K$, $M$, respectively. Then $A'$, $K' \\in \\eta$ and $A'$, $K'$, $C$ are collinear.\n![](attached_image_1.png)\nFrom the optical property of a hyperbola follows that $A'M'$ is tangent to $\\eta$ (as it is the bisector of $BA'C$). Therefore Pascal's theorem applied to the degenerate hexagon $AA'A'K'KL$ inscribed in $\\eta$ gives the collinearity of points\n$$\nAA' \\cap KK' = S, \\quad A'A' \\cap KL, \\quad A'K' \\cap LA = C.\n$$\nTherefore the three lines $A'A'$ (tangent to $\\eta$ in $A'$), $KL$, $BC$ are concurrent, which means that $M' = A'A' \\cap BC = KL \\cap BC = P$ and in consequence $CM = BP$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40582,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all non-constant functions $f: \\mathbb{Q}^+ \\to \\mathbb{Q}^+$ satisfying the equation\n$$\nf(ab + bc + ca) = f(a)f(b) + f(b)f(c) + f(c)f(a)\n$$\nfor all $a, b, c \\in \\mathbb{Q}^+$.",
    "options": [],
    "answer": "f(x) = x for all positive rational x",
    "solution": "Put $c = 1$ in the given condition, we have\n$$\nf(ab + a + b) = f(a)f(b) + f(a)f(1) + f(b)f(1); \\quad \\forall a, b \\in \\mathbb{Q}^{+}, \\quad (1)\n$$\nPut $b = 3$ into (1), we have\n$$\nf(4a + 3) = f(a)f(3) + f(a)f(1) + f(3)f(1); \\quad \\forall a \\in \\mathbb{Q}^{+}.\n$$\nPut $b = 1$ into (1), we have\n$$\nf(2a + 1) = 2f(a)f(1) + f(1)^2; \\quad \\forall a \\in \\mathbb{Q}^{+}.\n$$\nThus\n$$\nf(4a + 3) = 2f(2a + 1)f(1) + f(1)^2 = 4f(1)^2f(a) + 2f(1)^3 + f(1)^2; \\quad \\forall a \\in \\mathbb{Q}^{+}.\n$$\nFrom these, we can conclude that\n$$\n[f(3) + f(1)] f(a) + f(3)f(1) = 4f(1)^2f(a) + 2f(1)^3 + f(1)^2; \\quad \\forall a \\in \\mathbb{Q}^{+}.\n$$\nIf $f(3) + f(1) \\neq 4f(1)^2$ then $f$ is constant. Thus $f(3) + f(1) = 4f(1)^2$, otherwise $f$ will be constant. So we must have\n$$\nf(3) + f(1) = 4f(1)^2 \\text{ and } f(3)f(1) = 2f(1)^3 + f(1)^2.\n$$\nThus $f(3), f(1)$ are solutions of the quadratic equation $t^2 - 2f(1)t + 2f(1)^3 + f(1)^2 = 0$, thus\n$$\nf(1)^2 - 4f(1)^2 + 2f(1)^3 + f(1)^2 = 0 \\Leftrightarrow f(1)^2(f(1) - 1) = 0.\n$$\n$$\nf(ab + a + b) = f(a)f(b) + f(a) + f(b); \\quad \\forall a, b \\in \\mathbb{Q}^{+} \\quad (2)\n$$\nContinue to put $b = 1$ and $b = 3$, we get\n$$\nf(4a + 3) = 4f(a) + 3 \\text{ and } f(2a + 1) = 2f(a) + 1.\n$$\nPut $a = b = c = \\frac{1}{3}$ into the given condition, $f(\\frac{1}{3}) = 3f(\\frac{1}{3})^2$ so $f(\\frac{1}{3}) = \\frac{1}{3}$.\nPut $a = 2$ and $b = \\frac{1}{3}$ into (2), we have $f(3) = f(2)f(\\frac{1}{3}) + f(\\frac{1}{3}) + f(2)$, thus $f(2) = 2$.\nPut $b = c = 2$ into the given condition, $f(4a + 4) = 4f(a) + 4$; $\\forall a \\in \\mathbb{Q}^{+}$ thus\n$$\n4f(a) + 4 = f(4a + 4) = f\\left(4\\left(a + \\frac{1}{4}\\right) + 3\\right) = 4f\\left(a + \\frac{1}{4}\\right) + 3.\n$$\nFrom these, we can conclude that $f(a + \\frac{1}{4}) = f(a) + \\frac{1}{4}$, thus\n$$\nf(4a + 4) = 4f(a) + 4; \\quad \\forall a \\in \\mathbb{Q}^{+}.\n$$\nHence, by induction, one can show that $f(x + n) = f(x) + n$ for all positive integer $n$ and positive real number $x$; thus $f(n) = n, \\quad \\forall n \\in \\mathbb{Z}^{+}$.\nFinally, put $b \\to n$ and $a \\to \\frac{m}{n+1}$ for some $m, n \\in \\mathbb{Z}^{+}$ into (2), we get\n$$\nf(m + n) = f(n)f\\left(\\frac{m}{n+1}\\right) + f\\left(\\frac{m}{n+1}\\right) + f(n) \\to f\\left(\\frac{m}{n+1}\\right) = \\frac{m}{n+1}.\n$$\nThus $f(x) = x$ for all $x \\in \\mathbb{Q}^{+}$. It is easy to check this function satisfies the condition. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40583,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider all partitions of a positive integer $n$ into a sum of (non-negative integer) exponents of $2$ (i.e. $1, 2, 4, 8, \\dots$). A number in the sum is allowed to repeat an arbitrary number of times (e.g. $7 = 2 + 2 + 1 + 1 + 1$) and two partitions differing only in the order of summands are considered to be equal (e.g. $8 = 4 + 2 + 1 + 1$ and $8 = 1 + 2 + 1 + 4$ are regarded to be the same partition). Let $E(n)$ be the number of partitions in which an even number of exponents appear an odd number of times and $O(n)$ the number of partitions in which an odd number of exponents appear an odd number of times. For example, for $n = 5$ partitions counted in $E(n)$ are $5 = 4 + 1$ and $5 = 2 + 1 + 1 + 1$, whereas partitions counted in $O(n)$ are $5 = 2 + 2 + 1$ and $5 = 1 + 1 + 1 + 1 + 1$, hence $E(5) = O(5) = 2$. Find $E(n) - O(n)$ as a function of $n$.",
    "options": [],
    "answer": "E(n) - O(n) = -1 for n = 1, and E(n) - O(n) = 0 for all n > 1",
    "solution": "Let $D(n) = E(n) - O(n)$. We trivially have $O(1) = 1$ and $E(1) = 0$, thus $D(1) = -1$, and $E(2) = O(2) = 1$ (respectively $2 = 1+1$ and $2 = 2$), hence $D(2) = 0$. We will show by total induction that $D(n) = 0$ for all $n > 2$. Assume it holds for all numbers from $2$ to $n-1$. If $n$ is odd, a partition must contain at least one $1$. Since the addition of $1$ changes the parity of the number of one's it follows that $E(n) = O(n-1)$ and $O(n) = E(n-1)$, hence since $D(n) = -D(n-1) = 0$. If $n$ is even the partition must contain an even number $2k$ of $1$s. If it contains $2k = n$ or $2k = n-2$, then we have the unique solutions\n$$\n1+1+\\cdots+1 \\text{ and } 2+1+1+\\cdots+1,\n$$\nthe first adding to $E(n)$, the second to $O(n)$. For other, smaller, values of $k$ we note that the remaining exponents are all even, and we can thus apply the inductive hypothesis to $\\frac{n-2k}{2}$. Thus, it follows that for even $n$ we will also have $D(n) = 0$. This completes the proof. $\\square$\nLet $G(x) = 1 + \\sum_{i=1}^{\\infty} D(i)x^i$ be the generating function for the sequence we desire. To count partitions with an odd number of odd-appearing exponents as negative, we will multiply those choices as negative. Thus, we have that\n$$\nG(x) = (1-x+x^2-x^3+x^4-\\dots)(1-x^2+x^4-x^6+x^8-\\dots)(1-x^4+x^8-x^{12}+x^{16}-\\dots)\\dots,\n$$\nwhere the $i$-term corresponds to choosing the number of times $2^{i-1}$ is represented in the partition. Taking the expression for each power series, one now finds\n$$\nG(x) = \\frac{1}{1+x} \\frac{1}{1+x^2} \\frac{1}{1+x^4} \\frac{1}{1+x^8} \\dots\n$$\nHowever,\n$$\n\\begin{aligned}\n\\frac{G(x)}{1-x} &= \\frac{1}{1-x} \\frac{1}{1+x} \\frac{1}{1+x^2} \\cdots = \\frac{1}{1-x^2} \\frac{1}{1+x^2} \\frac{1}{1+x^4} \\cdots \\\\\n&= \\frac{1}{1-x^4} \\frac{1}{1+x^4} \\frac{1}{1+x^8} \\cdots = \\cdots = 1.\n\\end{aligned}\n$$\nThus, $G(x) = 1-x$ from which it follows that $D(1) = -1$ and $D(n) = 0$ for all $n > 1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40584,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let polynomial\n$$\nP(x) = \\underbrace{((\\dots((x-2)^2 - 2)^2 - \\dots)^2 - 2)^2}_{k \\text{ times}}\n$$\nis given. Find coefficient at $x^2$.",
    "options": [],
    "answer": "(16^k - 4^k)/12",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40585,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = x^2 - 6x + 5$. Draw on the plane the set of pairs $(x, y)$ that satisfy the following system of inequalities\n$$\n\\begin{cases}\nf(x) + f(y) \\leq 0 \\\\\nf(x) - f(y) \\geq 0\n\\end{cases} .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40586,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any 2 positive integers $m$ and $n$ with $m > n$ holds the following inequality\n$$\n\\operatorname{lcm}(m, n) + \\operatorname{lcm}(m + 1, n + 1) > \\frac{2mn}{\\sqrt{m-n}}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40587,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let convex $s$-gon is divided to $q$ quadrilaterals such that $b$ of them are not convex. Prove that\n$$\nq \\geq b + \\frac{s-2}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40588,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let positive numbers be written along the circle, such that all of them are less than $1$. Prove that one can split the circle to $3$ parts such that for each two arcs the sums of numbers written on them differs by at most $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40589,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the triangle $ABC$ is given and $D$, $E$, $F$ are on sides $BC$, $AC$, $AB$, respectively, such that\n$$\n\\frac{BD}{CD} = \\frac{CE}{AE} = \\frac{AF}{BF}.\n$$\nShow that if the circumcircle of $ABC$ and $DEF$ coincide, then $ABC$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40590,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the sequence $a_1, a_2, \\dots, a_n$ is such that $a_1 = 0$, $|a_2| = |a_1 + 1|$, $|a_3| = |a_2 + 1|$, $\\dots$, $|a_n| = |a_{n-1} + 1|$. Prove that\n$$\n\\frac{a_1 + a_2 + \\dots + a_n}{n} \\geq -\\frac{1}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40591,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist an infinite sequence $p_1, p_2, p_3, \\dots$ of prime numbers such that for any positive integer $n$ the following condition holds\n$$\n|p_{n+1} - 2p_n| = 1.\n$$",
    "options": [],
    "answer": "Unknown; this reduces to the open question of the existence of infinite Cunningham chains (of either kind).",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40592,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let incircle of triangle $ABC$ has center $I$ and touches sides $BC$, $AC$ and $AB$ at points $D$, $E$, $F$ respectively. Let $J_1, J_2, J_3$ be the ex-centres opposite $A$, $B$, $C$ respectively. Let $J_2F$ and $J_3E$ intersect at $P$, $J_3D$ and $J_1F$ intersect at $Q$, $J_1E$ and $J_2D$ intersect at $R$. Show that $I$ is the circumcenter of $PQR$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40593,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that the products of its digits is equal to $n^2 - 10n - 22$.",
    "options": [],
    "answer": "12",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40594,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players, $A$ and $B$ play the following game. On a $1 \\times n$ board, where fields are labeled in order from $1$ to $n$, a coin is placed at position $k$. Players take turns moving the coin, with player $A$ starting first. Each player can move a coin one or two fields in either direction, with the restriction that the coin cannot move onto a field it had already occupied. The player unable to make a move loses. For which values of $(n, k)$ does which player have a winning strategy?",
    "options": [],
    "answer": "Player A has a winning strategy if and only if n is even; Player B has a winning strategy if and only if n is odd (independent of k).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40595,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest positive integer $a$ for which there exist a prime number $p$ and a positive integer $b \\ge 2$ such that\n$$\n\\frac{a^p - a}{p} = b^2.\n$$",
    "options": [],
    "answer": "9",
    "solution": "If $p=2$, our equation becomes $a(a-1) = 2b^2$, whose smallest solution in $\\mathbb{N}$ is $a=9$.\n\nNow let $p \\ge 3$. Since $a$ and $a^{p-1}-1$ are coprime and $a(a^{p-1}-1) = pb^2$, either $a$ or $a^{p-1}-1$ must be a square, and it is obviously not the latter; hence $a$ is a square. Assume that $a=4$. Then\n$$\n\\frac{4^{p-1}-1}{p} = \\frac{(2^{p-1}-1)(2^{p-1}+1)}{p}\n$$\nis a square, so either $2^{p-1}-1$ or $2^{p-1}+1$ is a square, but the former is $3$ (mod $4$), so the latter is the square: $2^{p-1}+1 = c^2$. Then $(c+1)(c-1) = 2^{p-1}$, so both $c-1$ and $c+1$ are powers of $2$ and they must be $2$ and $4$, but then $p=4$, a contradiction. In conclusion, $a=9$ is the answer. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40596,
    "subject": "Mathematics (Multi-modal)",
    "question": "At a gala banquet, $12n + 6$ chairs, where $n \\in \\mathbb{N}$, are equally arranged around a large round table. A seating will be called a proper seating of rank $n$ if a gathering of $6n + 3$ married couples sit around this table such that each seated person also has exactly one sibling (brother/sister) of the opposite gender present (siblings cannot be married to each other) and each man is seated closer to his wife than his sister. Among all proper seats of rank $n$ find the maximum possible number of women seated closer to their brother than their husband. (The maximum is taken not only across all possible seating arrangements for a given gathering, but also across all possible gatherings.)",
    "options": [],
    "answer": "6n",
    "solution": "We will call a woman unusual if she sits closer to her husband than her brother. Our goal is to find the smallest possible number of unusual women. Let us call this number $k$. We note that going from each man to his sister and from each woman to her husband we obtain an oriented graph which breaks up into oriented cycles of even length. We also note that within an oriented cycle the sequence of lengths between consecutive members increases unless we encounter an unusual woman. Thus, each cycle must have at least one unusual woman. We also note that since the maximum length between two seats is $6n + 3$, this is also the maximum distance within a cycle we can go without encountering an unusual woman. Thus, we can have neither $k = 0$ nor $k = 1$, since by the first requirement we would have only one cycle, but this cycle would then have to have more than $6n + 3$ people. We will show that $k = 2$ is also impossible. The only options for $k = 2$ are to have either one cycle with lengths $1, 2, \\dots, 6n + 3, 1, 2, \\dots, 6n + 3$ or two cycles with lengths $1, 2, \\dots, 6n + 3$. We note that the second option does not work because the cycles have odd length, while the first option does not work because the two locations where the unusual women are supposed to be are at an odd distance from each other along the cycle and therefore those positions cannot be occupied by two people of the same gender.\n\nWe now give an example for $k = 3$. Arrange the three unusual women in an equilateral triangle and place their brothers diametrically opposite of them. Each unusual woman is part of a cycle of length $4n + 2$. If we label the members of one of these cycles in order as\n$$\nW_1, M_1, W_2, M_2, \\dots, W_{2n+1}, M_{2n+1},\n$$\nwhere $W_1$ is the unusual woman, then if we place $W_1$ at position 0, we can place each $M_k$, $k = 1, \\dots, 2n$ at position $k$ and each $W_k$ at position $k = 1, \\dots, 2n + 1$ at position $-k + 1$. Finally, we place $M_{2n+1}$, the brother of $W_1$, at position $6n + 3$.\n\nThe other two unusual women are placed symmetrically at positions $4n + 2$ and $-(4n + 2)$. We note that all the conditions of the problem are satisfied for this arrangement. Thus, the maximum possible number of women seated closer to their brother than their husband is $6n + 3 - 3 = 6n$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40597,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove there exist infinitely many positive integers divisible by $2021$ and each of them containing the same number of digits $0, 1, \\ldots, 9$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40598,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all values of $a$ for which the equation $x^3 + a x^2 + 56x - 4 = 0$ has 3 roots forming consecutive terms of a geometric progression.",
    "options": [],
    "answer": "a = -28 * 2^(1/3)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40599,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = \\frac{9^x}{9^x+3}$. Evaluate the sum\n$$\n\\sum_{k=0}^{2021} f\\left(\\frac{k}{2021}\\right).\n$$",
    "options": [],
    "answer": "1011",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40600,
    "subject": "Mathematics (Multi-modal)",
    "question": "One cuts a grid of size $8 \\times 8$ by a straight line. Find the maximal possible number of cells that are cut by the line.",
    "options": [],
    "answer": "15",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40601,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the cells of the grid $10 \\times 10$ are written positive integers, all of them less than $11$. It is known that the sum of $2$ numbers written in the cells having common vertex is a prime number. Prove that there are $17$ cells containing the same number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40602,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a non-isosceles acute angled triangle $\\triangle ABC$ where $O$ is the midpoint of $BC$. Let the circle with diameter $BC$, intersects $AB$, $AC$ at $D$, $E$ respectively. Let the angle bisectors of $\\angle A$ and $\\angle DOE$ intersect at $P$. If the circumcircles of $\\triangle BPD$ and $\\triangle CPE$ intersect at $P$ and $Q$, show that $Q$ lies on $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40603,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove there exist infinitely many positive integers divisible by $2021$ and each of them containing the same number of digits $0, 1, \\ldots, 9$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $k = 2021$. We want to construct infinitely many positive integers divisible by $k$ such that in their decimal representation, each digit $0, 1, \\ldots, 9$ appears the same number of times.\n\nLet $n$ be a positive integer. Consider the number $N$ whose decimal representation consists of $n$ copies of each digit $0, 1, \\ldots, 9$ (in any order, but for definiteness, let us take the number $M_n$ formed by writing $0, 1, 2, \\ldots, 9$ in order, $n$ times, i.e., $01234567890123456789\\ldots$ repeated $n$ times, for a total of $10n$ digits).\n\nLet $S$ be the set of all such numbers $M_n$ for $n \\geq 1$ (note that $M_n$ may have leading zeros, but we can permute the digits to avoid this, or simply consider all numbers with $n$ copies of each digit).\n\nThere are $\\binom{10n}{n, n, \\ldots, n}$ such numbers (the multinomial coefficient), so for each $n$, the set $S_n$ of numbers with $n$ copies of each digit is finite but very large.\n\nNow, $k = 2021$ is fixed. For each $n$, consider the set $S_n$ modulo $k$. Since $|S_n|$ grows rapidly with $n$, and there are only $k$ possible residues modulo $k$, by the pigeonhole principle, for sufficiently large $n$, there must exist at least one number in $S_n$ divisible by $k$.\n\nMore precisely, for each $n$, there exists at least one number with $n$ copies of each digit that is divisible by $k$. Since $n$ can be taken arbitrarily large, there are infinitely many such numbers.\n\nTherefore, there exist infinitely many positive integers divisible by $2021$ and each of them contains the same number of digits $0, 1, \\ldots, 9$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40604,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist positive integers $m$ and $n$ such that the decimal representation of $5^m$ starts with $2^n$ and the decimal representation of $2^m$ starts with $5^n$?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40605,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the sequence $a_i$ be defined in the following way: $a_1 = m \\in \\mathbb{Z}_+$ and inductively $a_{i+1} = a_i + \\lfloor \\sqrt{a_i} \\rfloor$. Prove that the sequence $a_i$ contains infinitely many perfect squares.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us analyze the sequence $a_i$ defined by $a_1 = m$ and $a_{i+1} = a_i + \\lfloor \\sqrt{a_i} \\rfloor$.\n\nSuppose at some step $a_k$ is a perfect square, say $a_k = n^2$ for some integer $n \\geq 1$. Then:\n$$\na_{k+1} = a_k + \\lfloor \\sqrt{a_k} \\rfloor = n^2 + \\lfloor n \\rfloor = n^2 + n = n(n+1)\n$$\n\nNow, consider the next terms:\n$$\na_{k+2} = a_{k+1} + \\lfloor \\sqrt{a_{k+1}} \\rfloor\n$$\nBut $a_{k+1} = n(n+1)$, and $\\sqrt{n(n+1)}$ is between $n$ and $n+1$.\nSpecifically, $n < \\sqrt{n(n+1)} < n+1$ for $n \\geq 1$, so $\\lfloor \\sqrt{n(n+1)} \\rfloor = n$.\nTherefore:\n$$\na_{k+2} = n(n+1) + n = n^2 + 2n = (n+1)^2 - 1\n$$\n\nNow, $a_{k+3} = a_{k+2} + \\lfloor \\sqrt{a_{k+2}} \\rfloor$\nBut $a_{k+2} = (n+1)^2 - 1$, so $\\sqrt{a_{k+2}} = \\sqrt{(n+1)^2 - 1}$.\nSince $(n+1)^2 - 1 < (n+1)^2$, $\\sqrt{a_{k+2}} < n+1$, and $\\sqrt{a_{k+2}} > n$ (since $(n+1)^2 - 1 > n^2$ for $n \\geq 1$), so $\\lfloor \\sqrt{a_{k+2}} \\rfloor = n$.\nThus:\n$$\na_{k+3} = (n+1)^2 - 1 + n = n^2 + 2n + n - 1 = n^2 + 3n - 1\n$$\n\nBut let's look for the next perfect square. Let us try to see if the sequence can reach $(n+1)^2$.\nFrom above, $a_{k+2} = (n+1)^2 - 1$, $a_{k+3} = n^2 + 3n - 1$.\n\nLet us try to generalize. For large $a_i$, $\\lfloor \\sqrt{a_i} \\rfloor$ increases slowly compared to $a_i$, so the sequence grows slowly.\n\nBut more importantly, we can show that for any starting $m$, the sequence will eventually reach a perfect square, and then the above process repeats.\n\nLet us prove that for any $a_i$, there exists $j > i$ such that $a_j$ is a perfect square.\n\nLet $a_i$ be arbitrary. Let $n = \\lfloor \\sqrt{a_i} \\rfloor$. Then $n^2 \\leq a_i < (n+1)^2$.\n\nIf $a_i = n^2$, we are done. If not, $a_i = n^2 + r$ for $1 \\leq r < 2n + 1$.\n\nAt each step, $a_{i+1} = a_i + \\lfloor \\sqrt{a_i} \\rfloor$. Since $a_i \\geq n^2$, $\\sqrt{a_i} \\geq n$, so $\\lfloor \\sqrt{a_i} \\rfloor \\geq n$.\n\nLet us consider the sequence of $a_i$ modulo $n$:\n\nLet us try to reach $(n+1)^2 = n^2 + 2n + 1$.\n\nFrom $a_i = n^2 + r$, after $s$ steps, $a_{i+s} = n^2 + r + s n$ (since $\\lfloor \\sqrt{a_{i+t}} \\rfloor = n$ for $a_{i+t} < (n+1)^2$).\n\nWe want $a_{i+s} = (n+1)^2 = n^2 + 2n + 1$.\nSo:\n$$\nn^2 + r + s n = n^2 + 2n + 1\n$$\n$$\nr + s n = 2n + 1\n$$\n$$\ns n = 2n + 1 - r\n$$\n$$\ns = \\frac{2n + 1 - r}{n}\n$$\n\nSince $1 \\leq r < 2n + 1$, $2n + 1 - r$ is positive and less than $2n$.\nSo $s$ is an integer if and only if $n$ divides $2n + 1 - r$.\n\nBut even if $s$ is not integer, after $t$ steps, $a_{i+t}$ will reach or exceed $(n+1)^2$, at which point $\\lfloor \\sqrt{a_{i+t}} \\rfloor = n+1$.\n\nTherefore, after a bounded number of steps, the sequence will cross from $n^2$ to $(n+1)^2$.\n\nNow, consider the difference $a_{i+t} - (n+1)^2$. For some $t$, $a_{i+t}$ will be exactly $(n+1)^2$ if $a_i$ is congruent to $n^2$ modulo $n$.\n\nBut even if not, after passing $(n+1)^2$, the process repeats with $n$ replaced by $n+1$.\n\nTherefore, for any starting $m$, the sequence will eventually hit a perfect square, and then the process repeats. Thus, the sequence contains infinitely many perfect squares.\n\nTherefore, the sequence $a_i$ contains infinitely many perfect squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40606,
    "subject": "Mathematics (Multi-modal)",
    "question": "In each cell of a chessboard (sizes $8 \\times 8$) is put a rock. At each step one can remove from the board one rock which beats an odd number of other rocks (for example in initial configuration top-left rock beats 2 rocks). Find the maximal possible number of rocks one can remove from the board.",
    "options": [],
    "answer": "62",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40607,
    "subject": "Mathematics (Multi-modal)",
    "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": "Detailed 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$. Also since\n$$\n\\angle XPD = \\angle ADP, \\angle XQD = \\angle ADQ\n$$\nwe obtain that $\\angle PXQ = 2\\angle A$. Thus $O, P, Q, X$ are concyclic. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40608,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the number of integer solutions of the equation\n$$\n\\left\\lfloor \\frac{x}{7} \\right\\rfloor = \\left\\lfloor \\frac{x}{12} \\right\\rfloor + \\left\\lfloor \\frac{x}{17} \\right\\rfloor.\n$$",
    "options": [],
    "answer": "1428",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40609,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define $a_0 = 2$ and $a_{n+1} = a_n^2 + a_n - 1$ for $n \\ge 0$. Prove that $a_n$ is coprime to $2n + 1$ for all $n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider any prime divisor $p$ of $a_n$. Consider a directed graph with $p$ edges on $\\{0, 1, \\dots, p-1\\}$ (mod $p$) connecting $x \\to y$ if and only if\n$$\nx^2 + x - 1 \\equiv y \\pmod{p}.\n$$\nObserve that $\\frac{p-1}{2}$ is connected to $b = \\frac{p^2-5}{4}$ (mod $p$), every element has out-degree 1, and every element other than $b$ has in-degree 2 or 0, elements $\\pm 1$ form loops, $-2 \\to 1$ and $0 \\to -1$. If we have a path $a_0 \\to a_1 \\to \\dots \\to a_n$, then each of the elements has in-degree two, with at least one having in-degree one, giving $2n - 1$ edges in total. Counting the four extra edges above, we deduce that $p \\ge 2n + 3$, which implies that $\\gcd(a_n, 2n+1) = 1$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40610,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $\\triangle ABC$ where $AB < AC$, $M$ is the midpoint of $BC$. The circle $O$ passes through $A$ and is tangent to $BC$ at $B$, intersecting the lines $AM$, $AC$ at $D$, $E$ respectively. Let $CF \\parallel BE$, intersecting $BD$ extended at $F$. Let the lines $BC$ and $EF$ intersect at $G$. Show that $AG = DG$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40611,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be real numbers such that $a^4 + b^4 + c^4 + d^4 = 16$. Prove the inequality $a^5 + b^5 + c^5 + d^5 \\le 32$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40612,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S(n)$ be the sum of divisors of $n$ (for example $S(6) = 1+2+3+6 = 12$). Find all $n$ for which $S(2n) = 3S(n)$.",
    "options": [],
    "answer": "All odd positive integers",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40613,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $(x, y)$ such that\n$$\n\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{\\text{lcm}(x, y)} + \\frac{1}{\\text{gcd}(x, y)} = \\frac{1}{2}.\n$$",
    "options": [],
    "answer": "(x, y) ∈ {(8,8), (5,20), (20,5), (6,12), (12,6), (8,12), (12,8), (9,24), (24,9), (12,15), (15,12)}",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40614,
    "subject": "Mathematics (Multi-modal)",
    "question": "Marwan has chosen 8 cells of the chessboard $8 \\times 8$ such that no two lie on the same line or in the same row (we call it general configuration). On each step Hamza chooses 8 cells in general configuration and puts coins on them. Then Marwan shows all coins that are out of cells chosen by Marwan. If Marwan shows even number of coins then Hamza wins, otherwise Hamza removes all coins and makes the next move. Find the minimal number of moves that Hamza needs to guarantee the win (Hamza$>$Marwan).",
    "options": [],
    "answer": "3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40615,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $1 \\le r \\le n$. We consider all $r$-element subsets of $(1, 2, \\dots, n)$. Each of them has a minimum. Prove that the average of these minima is $\\frac{n+1}{r+1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40616,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsider the positive numbers $x_1, x_2, \\dots, x_n$ such that $\\sum_{i=1}^n x_i = \\sum_{i=1}^n \\frac{1}{x_i}$. Prove that $\\sum_{i=1}^n \\frac{1}{n-1+x_i} \\le 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40617,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integer numbers $m$ and $n$ such that\n$$\n(5 + 3\\sqrt{2})^m = (3 + 5\\sqrt{2})^n.\n$$",
    "options": [],
    "answer": "(m, n) = (0, 0)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40618,
    "subject": "Mathematics (Multi-modal)",
    "question": "Twenty children are queueing for ice cream that is sold at SR5 per cone. Ten of the children have a SR5 coin, the others want to pay with a R10 bill. At the beginning, the ice cream man does not have any change. How many possible arrangements of the twenty kids in a queue are there so that the ice cream man will never run out of change?",
    "options": [],
    "answer": "16796",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 40619,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ that have precisely $\\sqrt{n+1}$ natural divisors.",
    "options": [],
    "answer": "3, 15",
    "solution": "First observe that $n = k^2 - 1$ for some positive integer $k$. As $n$ is not a perfect square, $\\tau(n) = k$ must be even, ergo $n$ is odd.\nWe now establish a bound on $\\tau(n)$. Recall that, since $n$ is not a perfect square, there exists a bijective correspondence between factors of $n$ greater than $\\lfloor\\sqrt{n}\\rfloor = k-1$ and factors of $n$ at most $\\lfloor\\sqrt{n}\\rfloor$. Furthermore, since $n$ is odd, all divisors of $n$ must also be odd. Therefore\n$$\nk = \\tau(n) \\le 2 \\cdot |\\{1, 3, \\dots, k-1\\}| = 2 \\cdot \\frac{k}{2} = k.\n$$\nHence equality must hold, and so every odd number between 1 and $k-1$ must divide $n = k^2 - 1$.\nIn particular, $k-3$ must divide $k^2-1$; but\n$$\n\\frac{k^2 - 1}{k - 3} = k + 3 + \\frac{8}{k - 3}.\n$$\nTherefore, $k-3$ is either $-1$ or $1$, so $k$ equals either $2$ or $4$ and $n \\in \\{3, 15\\}$, both of these work. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40620,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is an acute triangle $ABC$ with $BC < CA < AB$. Points $K$ and $L$ lie on segments $AC$ and $AB$ and satisfy $AK = AL = BC$. Perpendicular bisectors of segments $CK$ and $BL$ intersect line $BC$ at points $P$ and $Q$, respectively. Segments $KP$ and $LQ$ intersect at $M$. Prove that $CK + KM = BL + LM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D$ and $E$ be points of rays $ML \\to$ and $MK \\to$, respectively, such that $LD = AB$ and $KE = AC$. As $\\angle DLA = \\angle BLM = \\angle LBC$, $LD = AB$ and $AL = BC$, triangles $DLA$ and $ABC$ are congruent. Analogously, $EAK$ and $ABC$ are congruent. From\n$$\n\\angle DAL + \\angle LAK + \\angle KAE = \\angle BCA + \\angle CAB + \\angle ABC = 180^\\circ,\n$$\nit follows that $A$ belongs to the segment $DE$. Therefore, in the light of $\\angle LDA = \\angle BAC = \\angle AEK$, triangle $MDE$ is isosceles with $MD = ME$. Note that\n$$\n\\begin{aligned}\nBL + LM &= AB + LM - AL = DL + LM - AL = DM - AL, \\\\\nCK + KM &= AC + KM - AK = EK + KM - AK = EM - AK.\n\\end{aligned}\n$$\nFrom $DM = EM$ and $AL = AK$ it follows that the right-hand sides of the two equalities above are equal. Hence the left-hand sides are equal as well, i.e. $CK + KM = BL + LM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40621,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider non-negative real numbers $a, b, c$ satisfying the condition $a^2 + b^2 + c^2 = 2$. Find the maximum value of the following expression\n$$\nP = \\frac{\\sqrt{b^2 + c^2}}{3 - a} + \\frac{\\sqrt{c^2 + a^2}}{3 - b} + a + b - 2022c.\n$$",
    "options": [],
    "answer": "3",
    "solution": "First, we will show that $4\\sqrt{b^2+c^2} \\le (3-a)^2$. Notice that $b^2+c^2 = 2-a^2 \\ge 0$, so we need to prove $4\\sqrt{2-a^2} \\le (3-a)^2$. According to the AM-GM inequality, we have\n$$\n4\\sqrt{2-a^2} = 4\\sqrt{1 \\cdot (2-a^2)} \\le 4 \\cdot \\frac{1+2-a^2}{2} = 2(3-a^2).\n$$\nWe need to prove\n$$\n2(3-a^2) \\le (3-a)^2 \\iff 3(a^2-2a+1) \\ge 0 \\iff 3(a-1)^2 \\ge 0,\n$$\nwhich is true. From this, one can conclude that, $\\frac{\\sqrt{b^2+c^2}}{3-a} \\le \\frac{3-a}{4}$. Similarly, $\\frac{\\sqrt{c^2+a^2}}{3-b} \\le \\frac{3-b}{4}$. Therefore\n$$\nP \\le \\frac{3-a}{4} + \\frac{3-b}{4} + a + b - 2022c \\le \\frac{6+3(a+b)}{4}.\n$$\nWe have the familiar inequality\n$$\n(a+b)^2 \\le 2(a^2+b^2) \\le 2(a^2+b^2+c^2) = 4\n$$\nso $a+b \\le 2$. From this it follows that $P \\le \\frac{6+3 \\cdot 2}{4} = 3$. So the maximum value of $P$ of 3, is obtained when $a=b=1$, $c=0$.\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40622,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $P$ lies inside parallelogram $ABCD$ and satisfies $PC = BC$. Prove that the line joining midpoints of segments $AP$ and $CD$ is perpendicular to $BP$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40623,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $\\triangle ABC$, $D$ is on $BC$ and $P$ is on $AD$. A line $\\ell$ is passing through $D$ intersects $AB$, $PB$ at $M$, $E$ respectively, and intersects $AC$ extended and $PC$ extended at $F$, $N$ respectively. Let $DE = DF$. Prove that $DM = DN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40624,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find an example of a sequence of natural numbers $1 \\le a_1 < a_2 < \\dots < a_n < a_{n+1} < \\dots$ with the property that every positive integer $m$ can be uniquely written as $m = a_i - a_j$, with $i > j \\ge 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40625,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for $n \\ge 1$ the following inequality holds\n$$\n1 + \\frac{5}{6n - 5} \\le 6^{1/n} \\le 1 + \\frac{5}{n}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40626,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z \\ge 0$ and $x + y + z = 3$. Prove that\n$$\n\\sqrt{x} + \\sqrt{y} + \\sqrt{z} \\ge xy + xz + yz.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40627,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c > 0$. Prove that\n$$\n\\frac{a+b}{a^2+b^2} + \\frac{b+c}{b^2+c^2} + \\frac{c+a}{c^2+a^2} \\le \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40628,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 3$, $x_1, x_2, \\dots, x_n > 0$ and $x_1x_2 \\dots x_n = 1$. Prove that\n$$\n\\frac{1}{1+x_1+x_1x_2} + \\frac{1}{1+x_2+x_2x_3} + \\dots + \\frac{1}{1+x_n+x_nx_1} > 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40629,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral such that the line $CD$ is a tangent to the circle on $AB$ as diameter. Prove that the line $AB$ is a tangent to the circle on $CD$ as diameter if and only if the lines $BC$ and $AD$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40630,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove the identity\n$$\n\\frac{n!}{x(x+1)(x+2)\\dots(x+n)} = \\frac{\\binom{n}{0}}{x} - \\frac{\\binom{n}{1}}{x+1} + \\frac{\\binom{n}{2}}{x+2} - \\dots + (-1)^n \\frac{\\binom{n}{n}}{x+n}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40631,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be one of two points of intersection of two unequal circles in the same plane. If two particles from $A$ move each on a circle in a clockwise direction with two uniform velocities until they return to point $A$ at the same instant. Prove that there is always a point in the plane that is equidistant from the two particles at any moment while they are in motion.\n\nProblem:\nLet the parabola $y = x^2 + px + q$ be given, which intersects coordinate axes in 3 different points. Consider the circumcircle of the triangle having vertices of these 3 points. Prove that there is a point that belongs to that circle, regardless of values $p$ and $q$. Find that point AoMP.",
    "options": [],
    "answer": "(0,1)",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40632,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right-angled triangle with $\\angle A = 90^\\circ$ and let $AD$ be an altitude of the triangle $ABC$. Let $J$, $K$ be the incenters of the triangles $ABD$, $ACD$ respectively. Let $JK$ intersect $AB$, $AC$ at $E$, $F$ respectively. Prove that $AE = AF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40633,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $BB_1$ and $CC_1$ be the altitudes of acute-angled triangle $ABC$, and $A_0$ is the midpoint of $BC$. Lines $A_0B_1$ and $A_0C_1$ meet the line passing through $A$ and parallel to $BC$ in points $P$ and $Q$. Prove that the incenter of triangle $PA_0Q$ lies on the altitude of triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since triangles $BCB_1$ and $BCC_1$ are right-angled, their medians $B_1A_0$, $B_1C_0$ are equal to the half of hypotenuse $B_1A_0 = A_0C = A_0B = C_1A_0$.\n\n![](attached_image_1.png)\n\nNow\n$$\n\\angle PB_1A = \\angle CB_1A_0 = \\angle B_1CA_0 = \\angle PAC,\n$$\nthus $PA = PB_1$. Similarly, $QA = QC_1$. Then the incircle of triangle $A_0PQ$ touches its sides in points $A, B_1, C_1$, which yields the assertion of the problem. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40634,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a natural number. Find the number of permutations of the set $\\{1, 2, \\dots, n\\}$ such that for each $i = 1, 2, \\dots, n$, the first $i$ numbers in the permutation are not larger than $i + 1$. For example, there are 4 such permutations for $n = 3$: $\\{1, 2, 3\\}$, $\\{2, 1, 3\\}$, $\\{1, 3, 2\\}$ and $\\{2, 3, 1\\}$.",
    "options": [],
    "answer": "2^{n-1}",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 40635,
    "subject": "Mathematics (Multi-modal)",
    "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": "Detailed 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.\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$. We 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 $s = 2p + j - 4$. Combining the two equations, we obtain\n$$\ns = 2m + 4i + j - 4.\n$$\nNow, counting rectangles by side segments we obtain $s = 4r$ and counting rectangles by their angles we obtain $4r = 4i + 2j + 4$, the final term accounting for the four corners of $R$. We transform the equation into $2r - 6 = 2i + j - 4$. Combining all the obtained equations we get $4r = s = 2m + 2i + 2r - 6$ which gives us $2r + 6 = 2m + 2i$, i.e. $m + i = r + 3$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40636,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that not all zeros of a polynomial of the form $x^n + 2n x^{n-1} + 2n^2 x^{n-2} + \\dots$ can be real.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40637,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AD$ be the altitude of the triangle $ABC$. Let $J$, $K$ be the incenters of the triangles $ABD$, $ACD$ respectively. Let $JK$ intersect $AB$, $AC$ at $E$, $F$ respectively. Prove that $AE = AF$ if and only if $AB = AC$ or $\\angle A = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40638,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that\n$$\n3^{n-1} + 5^{n-1} \\mid 3^n + 5^n.\n$$",
    "options": [],
    "answer": "1",
    "solution": "Let $d = 3^{n-1} + 5^{n-1}$. We want $d \\mid 3^n + 5^n$.\n\nNote that:\n$$\n3^n + 5^n = 3 \\cdot 3^{n-1} + 5 \\cdot 5^{n-1} = 3d - 2 \\cdot 3^{n-1} + 5d - 4 \\cdot 5^{n-1} = (3+5)d - 2 \\cdot 3^{n-1} - 4 \\cdot 5^{n-1} = 8d - (2 \\cdot 3^{n-1} + 4 \\cdot 5^{n-1})\n$$\n\nSo $d \\mid 3^n + 5^n$ if and only if $d \\mid 2 \\cdot 3^{n-1} + 4 \\cdot 5^{n-1}$.\n\nBut $d = 3^{n-1} + 5^{n-1}$, so $d \\mid 2 \\cdot 3^{n-1} + 4 \\cdot 5^{n-1}$.\n\nLet us check small values of $n$:\n\nFor $n = 1$:\n$d = 3^0 + 5^0 = 1 + 1 = 2$\n$3^1 + 5^1 = 3 + 5 = 8$\n$2 \\mid 8$ is true.\n\nFor $n = 2$:\n$d = 3^1 + 5^1 = 3 + 5 = 8$\n$3^2 + 5^2 = 9 + 25 = 34$\n$8 \\mid 34$? $34/8 = 4.25$, so no.\n\nFor $n = 3$:\n$d = 3^2 + 5^2 = 9 + 25 = 34$\n$3^3 + 5^3 = 27 + 125 = 152$\n$34 \\mid 152$? $152/34 = 4.470...$, so no.\n\nFor $n = 4$:\n$d = 3^3 + 5^3 = 27 + 125 = 152$\n$3^4 + 5^4 = 81 + 625 = 706$\n$152 \\mid 706$? $706/152 \\approx 4.64$, so no.\n\nIt seems only $n = 1$ works.\n\nLet us prove that $n = 1$ is the only solution.\n\nSuppose $n > 1$.\nLet $d = 3^{n-1} + 5^{n-1}$.\nWe want $d \\mid 3^n + 5^n = 3 \\cdot 3^{n-1} + 5 \\cdot 5^{n-1}$.\nSo $d \\mid 3 \\cdot 3^{n-1} + 5 \\cdot 5^{n-1}$.\nBut $d = 3^{n-1} + 5^{n-1}$, so $d \\mid 3 \\cdot 3^{n-1} + 5 \\cdot 5^{n-1} - k(3^{n-1} + 5^{n-1})$ for any integer $k$.\n\nLet $k = 3$:\n$3 \\cdot 3^{n-1} + 5 \\cdot 5^{n-1} - 3(3^{n-1} + 5^{n-1}) = 3 \\cdot 3^{n-1} + 5 \\cdot 5^{n-1} - 3 \\cdot 3^{n-1} - 3 \\cdot 5^{n-1} = (5 - 3) \\cdot 5^{n-1} = 2 \\cdot 5^{n-1}$\nSo $d \\mid 2 \\cdot 5^{n-1}$.\n\nSimilarly, let $k = 5$:\n$3 \\cdot 3^{n-1} + 5 \\cdot 5^{n-1} - 5(3^{n-1} + 5^{n-1}) = 3 \\cdot 3^{n-1} + 5 \\cdot 5^{n-1} - 5 \\cdot 3^{n-1} - 5 \\cdot 5^{n-1} = (3 - 5) \\cdot 3^{n-1} = -2 \\cdot 3^{n-1}$\nSo $d \\mid 2 \\cdot 3^{n-1}$.\n\nTherefore, $d \\mid \\gcd(2 \\cdot 3^{n-1}, 2 \\cdot 5^{n-1}) = 2$ (since $3^{n-1}$ and $5^{n-1}$ are coprime).\n\nBut $d = 3^{n-1} + 5^{n-1} \\geq 3^{n-1} + 1 > 2$ for $n > 1$.\n\nTherefore, the only solution is $n = 1$.\n\n**Answer:** $n = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40639,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with midpoints $K$, $M$, $N$ of $BC$, $CA$, $AB$ respectively. Let $AD$, $BE$, $CF$ be the altitudes of the triangle $ABC$ and let $U$, $V$, $W$ be the midpoints of $FD$, $DE$, $EF$ respectively. Prove that $KW$, $MV$, $NU$ intersect at one point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40640,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $(a_k)$ is given by $a_1 = \\frac{1}{2}$ and $a_{n+1} = 1 - a_1 a_2 \\cdots a_n$ for every $n \\ge 1$. Prove that $a_{100} > 0.99$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40641,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the function $f: \\mathbb{R}^{+} \\to \\mathbb{R}^{+}$ and satisfying\n$$\nf(x + 2y + f(x + y)) = f(2x) + f(3y), \\forall x, y > 0.\n$$\n1. Find all functions $f(x)$ that satisfy the given condition.\n2. Suppose that $f(4\\sin^4x)f(4\\cos^4x) \\ge f^2(1)$ for all $x \\in (0; \\frac{\\pi}{2})$. Find the minimum value of $f(2022)$.",
    "options": [],
    "answer": "All solutions are f(x) = x + c for c ≥ 0; under the inequality condition, the minimum value of f(2022) is 2023.",
    "solution": "1) In the given condition, replace $x \\rightarrow 3x, y \\rightarrow 2y$, we have\n$$\nf(3x + 4y + f(3x + 2y)) = f(6x) + f(6y).\n$$\nBy swapping $x, y$ and comparing the two left sides, we have\n$$\nf(3x + 4y + f(3x + 2y)) = f(4x + 3y + f(2x + 3y)).\n$$\nFrom injectivity, we have\n$$\n3x + 4y + f(3x + 2y) = 4x + 3y + f(2x + 3y)\n$$\nor\n$$\nf(3x + 2y) - (3x + 2y) = f(2x + 3y) - (2x + 3y).\n$$\nSet $g(x) = f(x) - x$ then $g(3x + 2y) = g(2x + 3y), \\forall x, y \\in \\mathbb{R}^+$. Put $a = 3x + 2y, b = 2x + 3y$ then\n$$\nx = \\frac{3a - 2b}{5}, \\quad y = \\frac{3b - 2a}{5}\n$$\nso the condition of the pair of numbers $(a, b)$ for the existence of $x, y$ the above relation is $3a - 2b > 0$ and $3b - 2a > 0$, equivalent to $\\frac{2}{3} < \\frac{a}{b} < \\frac{3}{2}$.\nHence we have $g(a) = g(b)$ for all $\\frac{a}{b} \\in (\\frac{2}{3}; \\frac{3}{2})$. From here we will show that $g$ is a constant function. Indeed, we have $g(1) = g(x), \\forall x \\in [1; \\frac{3}{2})$, from here inductively\n$$\ng(1) = g(x), \\forall x \\in \\left[ \\left(\\frac{3}{2}\\right)^k ; \\left(\\frac{3}{2}\\right)^{k+1} \\right)\n$$\nso $g(1) = g(x), \\forall x > 1$. Similarly\n$$\ng(1) = g(x), \\forall x \\in \\left( \\left( \\frac{2}{3} \\right)^{k+1} ; \\left( \\frac{2}{3} \\right)^k \\right]\n$$\nso there is also $g(1) = g(x), \\forall x \\in (0; 1)$. Therefore, $g(x)$ is a constant function. Instead we have $f(x) = x + c$ with $c \\geq 0$, try again and we are satisfied.\n\n2) Since $f(x) = x + c$, instead of the problem condition, we have\n$$\n(4\\sin^4x + c)(4\\cos^4x + c) \\geq (1 + c)^2 \\text{ or } \\sin^4(2x) + 4(\\sin^4x + \\cos^4x)c \\geq 1 + 2c.\n$$\nNotice that $\\sin^4x + \\cos^4x = 1 - \\frac{1}{2}\\sin^22x$ and put $t = \\sin^2(2x) \\in (0; 1]$, we rewrite the above inequality as $t^4 + (4 - 2t^2)c \\geq 1 + 2c$ or\n$$\n2(1 - t^2)c \\geq (1 - t^2)(t^2 + 1).\n$$\nSince $1 - t^2 > 0$ so $2c \\geq t^2 + 1$, which $\\max_{(0;1]}\\{t^2 + 1\\} = 2$ so $2c \\geq 2 \\Leftrightarrow c \\geq 1$. So the minimum value of $f(2022)$ is 2023. □",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40642,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\n2f(x)f(x + y) - f(x^2) = \\frac{x}{2}(f(2x) + 4f(f(y)))\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = 0 and f(x) = x",
    "solution": "Put $x = 0$, we get $2f(0)f(y) - f(0) = 0$, if $f(0) \\neq 0$ then $f(y) = 1/2$ for all $y$, which does not satisfy. Thus $f(0) = 0$.\n\nPut $y = 0$, we get $2f(x)^2 - f(x^2) = x/2 \\cdot f(2x)$ then plugging back $2f(x) \\cdot f(x+y) = 2f(x)^2 + 2x \\cdot f(f(y))$ so\n$$\nf(x) \\cdot f(x+y) = f(x)^2 + x \\cdot f(f(y)). \\qquad (\\diamond)\n$$\nNote that $f(-x)^2 = x \\cdot f(f(x))$. In $(\\diamond)$, take $y = -2x$ then\n$$\nf(x) \\cdot f(-x) = f(x)^2 + x \\cdot f(f(-2x)).\n$$\nNote that $-2x \\cdot f(f(-2x)) = f(2x)^2$, so $x \\cdot f(f(-2x)) = -f(2x)^2/2$. Thus\n$$\nf(x) \\cdot f(-x) = f(x)^2 - f(2x)^2/2\n$$\nor\n$$\n-2f(x)f(-x) + 2f(x)^2 = f(2x)^2.\n$$\nIn $(\\diamond)$, take $x = y \\neq 0$ then\n$$\nf(x) \\cdot f(2x) = f(x)^2 + x \\cdot f(f(x))\n$$\nor\n$$\nf(2x) = f(x) + x \\cdot f(f(x))/f(x) = f(x) + f(-x)^2/f(x).\n$$\nSo $(f(x) + f(-x)^2/f(x))^2 = 2f(x)^2 - 2f(x) \\cdot f(-x)$. Dividing both sides by $f(-x)^2$ and put $k = f(x)/f(-x)$ then\n$$\n(k + 1/k)^2 = 2k^2 - 2k \\longleftrightarrow k^2 - 2k = 2 + 1/k^2.\n$$\nChanging the sign $x \\to -x$ then we get $(k + 1/k)^2 = 2/k^2 - 2/k$ combining with above, we get $k = -1$. Then $f(x)$ is odd. Now we get $f(x)^2 = x \\cdot f(f(x))$. Put $y \\to -y$, we get\n$$\nf(x) \\cdot f(x - y) = f(x)^2 + x \\cdot f(f(-y)),\n$$\n$$\nf(x)[f(x + y) + f(x - y)] = 2f(x)^2.\n$$\nConsider $x \\neq 0$ then $f(x + y) + f(x - y) = 2f(x)$ using $f(0) = 0$, this implies that $f(x)$ is additive. Note that\n$$\n2f(x)^2 - f(x^2) = x/2 \\cdot f(2x) = x \\cdot f(x) \\iff 2f(x)^2 = f(x^2) + xf(x).\n$$\nPlugging $x \\to x + 1$, it is easily to get $f(x) = ax$ and putting back to the original, $a = 1$. There are two solutions: $f(x) = 0, f(x) = x$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40643,
    "subject": "Mathematics (Multi-modal)",
    "question": "You plan to organize your birthday party, which will be attended either by exactly $m$ persons or by exactly $n$ persons (you are not sure at the moment). You have a big birthday cake and you want to divide it into several parts (not necessarily equal), so that you are able to distribute the whole cake among the people attending the party with everybody getting cake of equal mass (however, one may get one big slice, while others several small slices - the sizes of slices may differ). What is the minimal number of parts you need to divide the cake, so that it is possible, regardless of the number of guests.",
    "options": [],
    "answer": "m + n - gcd(m, n)",
    "solution": "We claim that the answer is $m + n - \\gcd(m, n)$. Firstly, note that if we consider the cake as the interval $[0, 1]$ and make cuts at points with coordinates $\\frac{k}{m}$ ($0 < k < m$) and $\\frac{l}{n}$ ($0 < l < n$), then we will be able to satisfy the condition of the problem. Moreover, there will be $m-1$ cuts with $\\frac{k}{m}$, $n-1$ cuts with $\\frac{l}{n}$, $\\gcd(m, n) - 1$ of which coincide. Therefore we will have\n$$\nm + n - \\gcd(m, n) - 1\n$$\ncuts, so $m + n - \\gcd(m, n)$ parts. Let us show that this estimate is sharp. For that purpose, consider a bipartite graph, with the vertices of one side corresponding to the $m$ persons on the party, and the vertices of the other side corresponding to the $n$ persons on the party.\nWe connect the vertices $v$ and $u$ by an edge, corresponding to the piece of cake, if the piece of cake will be given to the person $v$, if exactly $m$ persons attend, and to the person $u$, if exactly $n$ persons attend. Consider a component of connectivity of the graph. Then, if it contains $m_1$ edges in one side and $n_1$ edges in the other side, then the edges of it correspond to a cake with weight $\\frac{m_1}{m} = \\frac{n_1}{n}$. Therefore, $m_1 \\ge \\frac{m}{\\gcd(m,n)}$.\nThus, the number of components of connectivity is at most $\\gcd(m, n)$. The graph has $m+n$ vertices and at most $\\gcd(m, n)$ components of connectivity, so the number of edges is at least $m+n - \\gcd(m, n)$ (the equality is obtained, when each of the components is a tree). $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40644,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\phi_n(m) = \\phi(\\phi_{n-1}(m))$, where $\\phi_1(m) = \\phi(m)$ is the Euler totient function, and set $\\omega(m)$ the smallest number $n$ such that $\\phi_n(m) = 1$. If $m < 2^\\alpha$, then prove that $\\omega(m) \\le \\alpha$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Number-Theoretic Functions"
  },
  {
    "id": 40645,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the lattice in the plane, from which we may cut rectangles, but only by making cuts along the lines of the lattice. Prove that for any integer $m > 12$ one may cut a rectangle of area greater than $m$ such, that from that rectangle one can't cut a rectangle of area $m$. The Yellow Fish",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40646,
    "subject": "Mathematics (Multi-modal)",
    "question": "A quadrilateral $ABCD$ is inscribed inside a circle and $AD \\perp CD$. Draw $BE \\perp AC$ at $E$ and $BF \\perp AD$ at $F$. Show that the line $EF$ passes through the midpoint of the line segment $BD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40647,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of integers $(m, n)$ such that $\\binom{n}{m} = 1984$.",
    "options": [],
    "answer": "(m, n) = (1, 1984) and (1983, 1984)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40648,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $u_n$ be the least common multiple of the first $n$ terms of a strictly increasing sequence of positive integers $a_1, a_2, a_3, \\dots, a_{1000}$. Prove that\n$$\n\\sum_{k=1}^{1000} \\frac{1}{u_k} \\le 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40649,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\sigma(n)$ denote the sum of the divisors of $n$. Prove that there exist infinitely many integers $n$ such that $\\sigma(n) > 3n$. Prove also that $\\sigma(n) < n(1+\\log_2 n)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. To show that there exist infinitely many integers $n$ such that $\\sigma(n) > 3n$:\n\nLet $n = p^k$, where $p$ is a prime and $k \\geq 1$. Then\n$$\n\\sigma(n) = 1 + p + p^2 + \\cdots + p^k = \\frac{p^{k+1} - 1}{p - 1}.\n$$\n\nLet $p = 2$, $n = 2^k$:\n$$\n\\sigma(2^k) = 2^{k+1} - 1.\n$$\n\nFor $k \\geq 2$,\n$$\n\\sigma(2^k) = 2^{k+1} - 1 > 3 \\cdot 2^k \\iff 2^{k+1} - 1 > 3 \\cdot 2^k \\iff 2^{k+1} > 3 \\cdot 2^k + 1 \\iff 2 \\cdot 2^k > 3 \\cdot 2^k + 1 \\iff 2^k > 1.\n$$\nSo for $k \\geq 1$, $2^k > 1$ and the inequality holds for $k \\geq 2$.\n\nAlternatively, consider $n = p_1 p_2$, where $p_1, p_2$ are distinct primes:\n$$\n\\sigma(n) = (1 + p_1)(1 + p_2) = 1 + p_1 + p_2 + p_1 p_2.\n$$\n\nFor large $p_1, p_2$, $\\sigma(n) \\approx p_1 p_2$, but for small primes, for example $n = 6$:\n$$\n\\sigma(6) = 1 + 2 + 3 + 6 = 12 > 3 \\cdot 6 = 18.\n$$\nBut $12 < 18$, so this does not work for $n = 6$.\n\nBut for $n = 28$ (which is a perfect number):\n$$\n\\sigma(28) = 1 + 2 + 4 + 7 + 14 + 28 = 56 = 2 \\cdot 28.\n$$\nSo $\\sigma(n) > 3n$ for $n = 2^k$ with $k \\geq 2$.\n\nIn fact, for $n = 2^k$ with $k \\geq 2$, $\\sigma(n) = 2^{k+1} - 1 > 3 \\cdot 2^k$ for $k \\geq 2$.\n\nTherefore, there are infinitely many such $n$.\n\nb. To prove $\\sigma(n) < n(1 + \\log_2 n)$:\n\nLet $n = \\prod_{i=1}^r p_i^{a_i}$ be the prime factorization of $n$.\nThen\n$$\n\\sigma(n) = \\prod_{i=1}^r \\frac{p_i^{a_i+1} - 1}{p_i - 1} < \\prod_{i=1}^r \\frac{p_i^{a_i+1}}{p_i - 1}.\n$$\nBut $\\sigma(n) < n \\prod_{i=1}^r \\frac{p_i}{p_i - 1}$.\n\nNow, $\\prod_{i=1}^r \\frac{p_i}{p_i - 1} < \\prod_{i=1}^r \\left(1 + \\frac{1}{p_i - 1}\\right) < 1 + \\sum_{i=1}^r \\frac{1}{p_i - 1}$.\n\nBut the number of distinct prime divisors $r \\leq \\log_2 n$ (since $n \\geq 2^r$), so\n$$\n\\sigma(n) < n (1 + \\log_2 n).\n$$\n\nTherefore, $\\sigma(n) < n(1 + \\log_2 n)$ for all $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40650,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1 = 1$, $a_{n+1} = a_n + [\\sqrt{a_n}]$. Find all $n$ such that $a_n$ is a perfect square.",
    "options": [],
    "answer": "All n of the form n = 2^{k+1} + k - 1 for k ≥ 0; at these indices a_n = 4^k.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40651,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any natural number $n > 1$ then $2^n - 1 \\nmid 3^n - 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40652,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a quadrilateral $ABCD$, the external angle bisectors of $\\angle CAD$, $\\angle CBD$ intersect at $P$. Show that if $AD + AC = BC + BD$, then $\\angle APD = \\angle BPC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40653,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a $9 \\times 9$ board, several cells are shaded in such a way that from any shaded cell you can get to any other shaded cell, visiting only the shaded cells and moving only between cells neighboring with a side. Determine the largest possible perimeter of the shaded region.",
    "options": [],
    "answer": "120",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40654,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle which circumcircle $(O)$. Take $M$ on segment $BC$ and $J$ on $(O)$ such that $\\angle BAJ = \\angle CAM$. Take a point $D$ on the segment $AM$ and denote $O_1, O_2$ as the circumcenters of triangles $ABD$ and $ACD$. The line $O_1O_2$ meets $BC$ at $T$ and $G$ is the second intersection of circumcircles of triangles $TBO_1$ and $TCO_2$. Prove that $OA$, $JA$ respectively divide the segments $O_1O_2$, $BC$ by the same ratio and $A$, $G$, $O$, $J$ are concyclic.\n\nProblem:\nLet $ABC$ be an acute, non-isosceles triangle with orthocenter $H$ and circumcenter $O$. Denote $D$, $E$ as midpoints of segments $AB$, $AC$. Take $M$, $N$ on $BC$ such that $MB = BC = CN$ ($B$ is between $M$, $C$ and $C$ is between $N$, $B$). Denote $P$, $Q$ as the projections of $H$ onto the lines $BE$, $CD$. The circumcircles of triangle $ABN$, $ACM$ respectively meets $AQ$, $AP$ again at $Y$, $X$. Suppose that the line $XY$ cuts $BC$ at $K$. Prove that $AK$ is perpendicular to $OH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40655,
    "subject": "Mathematics (Multi-modal)",
    "question": "From a point $O$ lying outside the line $d$, draw the projection $A$ of $O$ onto $d$. Take some point $M$ on $d$ different from $A$ and $H$ is the projection of $A$ onto $OM$. Denote $D$ as the midpoint of $HM$ and take $N$ on $OA$ such that $NH \\perp AD$. Suppose that two circumcircles of triangles $HMN$ and $OAH$ are tangent, calculate the ratio $AM/ AO$.",
    "options": [],
    "answer": "2",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40656,
    "subject": "Mathematics (Multi-modal)",
    "question": "From a point $A$ lying outside the circle $(O)$, draw two tangent lines $AB$, $AC$ of $(O)$ with $B$, $C$ are tangent points. A line passes through $A$, lies inside the angle $OAC$, cuts $(O)$ at $R$, $S$ ($R$ is between $A$ and $S$). The segments $BR$, $BS$ cut the ray $AO$ respectively at $D$, $E$. Denote $H$ as orthocenter and $I$ as circumcenter of triangle $BDE$. Let $BT$ be the diameter of circumcircle of $(I)$. Prove that $\\triangle DHT \\sim \\triangle RBS$ and calculate the ratio that $OI$ divides $BC$.",
    "options": [],
    "answer": "1:1",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40657,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each pair of vertices of a regular $1001$-gon is joined with a segment, which is either red, or blue, or green. Prove that one can choose $11$ vertices of this $1001$-gon in such a way that they form a convex $11$-gon, in which at least $10$ sides have the same color.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40658,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is an odd integer $n \\ge 1$. Let $S$ be the set of all points in the three dimensional space, whose all coordinates belong to the set $\\{0, 1, ..., n\\}$. Determine the maximum size of a subset $A \\subset S$ with the following property: For every two distinct points $(x_1, x_2, x_3)$, $(y_1, y_2, y_3) \\in A$ among the three numbers $x_1 - y_1, x_2 - y_2, x_3 - y_3$ there is at least one positive number and at least one negative number.",
    "options": [],
    "answer": "3*((n+1)/2)^2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40659,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is a convex $n$-gon with no four vertices concyclic. A triple of vertices is called 'round' if the circle passing through these points covers the entire polygon. Determine, in terms of $n$, all possible values of the number of round triples.",
    "options": [],
    "answer": "n - 2",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40660,
    "subject": "Mathematics (Multi-modal)",
    "question": "Five positive reals $a$, $b$, $c$, $d$ and $e$ having product equal to $1$ are given. Prove that\n$$\n\\frac{a^2}{b^2} + \\frac{b^2}{c^2} + \\frac{c^2}{d^2} + \\frac{d^2}{e^2} + \\frac{e^2}{a^2} \\geq a + b + c + d + e.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40661,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be non-zero integers such that\n$$\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$$\nis an integer. Prove that $a_k^2 \\mid a_1 a_2 \\dots a_n$ for all $k = 1, 2, \\dots, n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For each $k = 1, 2, \\dots, n$, denote $b_k = \\frac{a_1 a_2 \\dots a_n}{a_k^2}$ and consider the following polynomial\n$$\nP(x) = (x - b_1)(x - b_2) \\dots (x - b_n) = x^n + c_{n-1} x^{n-1} + \\dots + c_1 x + c_0.\n$$\nBased on Vieta's theorem, one can see that $c_{n-1}$ is an integer, and for any $k$ integers $i_1, i_2, \\dots, i_k \\in \\{1, 2, \\dots, n\\}$, $b_{i_1} b_{i_2} \\dots b_{i_k}$ is also an integer, which implies that all coefficients $c_k$ are integers. On the other hand, $P(x)$ is a monic polynomial which implies that all of its roots $b_1, b_2, \\dots, b_n$ are also integers. Hence,\n$$\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$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40662,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute-angled triangle $ABC$, point $O$ is the circumcenter and $H$ the orthocenter. Points $B'$ and $C'$ are the reflections of $B$ in line $AC$ and of $C$ in line $AB$, respectively. Point $K$ is the circumcenter of triangle $HB'C'$, point $D$ the midpoint of $KB$, and $S$ the intersection of line $OD$ with the perpendicular to $AB$ at $A$. Prove that $KA = KS$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\omega, \\omega_c, \\omega_b, \\Omega$ be the circumcircles of triangles $ABC, ABC', AB'C, AB'C'$ respectively. Denote $O_c, O_b$ as the circumcenters of $\\omega_c, \\omega_b$ and $R$ as the radius of $\\omega$. Let $f(X)$ be the reflection of the figure $X$ over the line $AB$.\n\nIt is easy to check that $f(\\Delta ABC) = \\Delta A'BC$, so $f(\\omega) = \\omega_c$ and $f(O) = O_c$. Since $C'H$ is the common chord of $\\omega_c, \\Omega$, hence $KO_c$ is the perpendicular bisector of $HC'$. Now the homothety of center $O$ and ratio $2$ will send $AB$ to a parallel line through $O_c$, which is $O_cK$ and also send $AC$ to $O_bK$ as well. So this homothety sends $A$ to $K$, which implies that $KA = AO = R$.\n\nLet $BO_c$ intersects $\\omega_c$ at $S'$ then $\\angle BAS' = 90^\\circ$. We will prove that $S' \\equiv S$. Since $AO_cBO$ is the rhombus, hence $KO \\parallel S'B$. But $KO = S'B = 2R$, hence $KS'BO$ is a parallelogram, then $OS'$ passes through $D$, implies that $S' \\equiv S$. Finally, we have $SK = OK = R = KA$ which finishes the solution. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40663,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\n2(y + 1)f(x)f(y - 1) = 2yf(xy) - f(2x)\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = 0 for all real x; f(x) = x for all real x",
    "solution": "Put $x = 0, y = -1$: $0 = -2f(0) - f(0)$ thus $f(0) = 0$.\n\nPut $y = 1$: $2f(x) = f(2x)$, for all $x$ so the given condition can rewrite as\n$$\n(y + 1)f(x)f(y - 1) = yf(xy) - f(x), \\forall x, y\n$$\nContinue to put $y = 0$ then $f(x)f(-1) = -f(x)$. If $f(-1) \\neq -1$ then $f(x) = 0$ for all $x$, which is satisfied the given condition. Now suppose that there exist some $x_0$ such that $f(x_0) \\neq 0$, this implies that $f(-1) = -1$ and $f(1) = 1$.\n\nIn (*), put $x = 1$ then $(y + 1)f(y - 1) = yf(y) - 1$ for all $y$. Replace to the LHS of (*), one can get\n$$\n(yf(y) - 1)f(x) = yf(xy) - f(x)\n$$\nso $f(xy) = f(x)f(y)$, $\\forall x, y \\neq 0$ which implies that $f$ is multiplication function. Also put $y = -1$ in (*) then $f(x) = -f(-x)$ which implies that $f(x)$ is odd.\n\nSuppose that $a$ is some number such that $f(a) = 0$ then by substituting $x = a$ into (*), we get $0 = yf(ay) - 0$ so $f(ay) = 0, \\forall y \\neq 0$. If $a \\neq 0$ then $ay$ can take any values on $\\mathbb{R}$ so $f(x) \\equiv 0$, contradiction. Thus $a = 0$ is the unique value such that $f(a) = 0$. Since $(y+1)f(y-1) = yf(y) - 1$, change $y \\to -y$ and using the property of odd function, one can get\n$$\n(-y + 1)f(-y - 1) = yf(y) - 1, \\forall y \\in \\mathbb{R}.\n$$\nThus for all $y \\neq \\pm 1$ we have\n$$\n(y + 1)f(y - 1) = (y - 1)f(y + 1) \\Leftrightarrow \\frac{f(y + 1)}{f(y - 1)} = \\frac{y + 1}{y - 1}\n$$\nfor all $y \\neq \\pm 1$. Due to the multiplication property of $f$, rewrite the last equation as\n$$\nf\\left(\\frac{y+1}{y-1}\\right) = \\frac{y+1}{y-1}, \\forall y \\in \\mathbb{R} \\setminus \\{\\pm 1\\}.\n$$\nNote that $t = \\frac{y+1}{y-1}$ can take any value on $\\mathbb{R} \\setminus \\{1\\}$ so $f(t) = t, \\forall t \\neq 0, 1$. But $f(1) = 1, f(0) = 0$ so we have $f(x) = x, \\forall x \\in \\mathbb{R}$. Hence, there are two functions that satisfying the given condition: $f(x) = 0, \\forall x \\in \\mathbb{R}$ and $f(x) = x, \\forall x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40664,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given that\n$$\n(10!)^3 = \\overline{47\\ 784\\ 725\\ 83a\\ b72\\ 000\\ 000},\n$$\ndetermine the digits $a$ and $b$.",
    "options": [],
    "answer": "a = 9, b = 8",
    "solution": "As $(10!)^3$ is divisible by $9$, by the divisibility criterion for the number $9$ we get\n$$\n9|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$$\nSince $0 \\le a, b \\le 9$, we have $-9 \\le b - a \\le 9$. The only number congruent to $-1$ modulo $11$ in this interval is the number $-1$, and therefore $b - a = -1$, that is, $b = a - 1$. Further, we have $0 \\le a + b \\le 18$, from which follows\n$$\n64 \\le 64 + a + b \\le 82.\n$$\nThe numbers from this interval divisible by $9$ are $72$ and $81$, and thus we get $64 + a + b = 72$ or $64 + a + b = 81$, that is, $a + b = 8$ or $a + b = 17$. Putting $b = a - 1$ here, the first case reduces to $2a - 1 = 8$, which is impossible.\nThere remains the second case, $2a - 1 = 17$, from which we get $a = 9$ and $b = a - 1 = 8$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40665,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be the positive root of equation $x + \\frac{1}{x} = 675$. Prove that\n$$\n\\varphi(2023(a^{2n} + a^{-2n})) > 2 \\cdot \\varphi(2022(a^{2n} + a^{-2n}))\n$$\nfor all $n \\in \\mathbb{Z}^+$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that $2023 = 7 \\cdot 17^2$ and $2022 = 2 \\cdot 3 \\cdot 337$. Put $u_n = a^{2n} + a^{-2n}$ then it is easy to see $u_{n+1} = u_n^2 - 2$ for every $n \\ge 0$. We have $u_0 = a + \\frac{1}{a} = 675$ divisible by $3$ so $u_1$ divided by $3$ leaves $1$, then $u_2$ divides $3$ with remainder $-1$ and inductively $u_n \\equiv -1 \\pmod{3}$ with every $n \\ge 2$. Similarly, $u_n$ is odd for all $n$. Also, $675 \\equiv 1 \\pmod{337}$ so $u_0 \\equiv 1 \\pmod{337}$ entails $u_1 \\equiv -1 \\pmod{337}$ and so on we also inductively $u_n \\equiv -1 \\pmod{337}, \\forall n \\ge 1$. From this it follows that $\\text{gcd}(u_n, 2022) = 1, \\forall n \\ge 1$ and so\n$$\n\\varphi(2022u_n) = \\varphi(2)\\varphi(3)\\varphi(337)\\varphi(u_n) = 672\\varphi(u_n).\n$$\nNext, $675 \\equiv 3 \\pmod 7$ so $u_1 \\equiv 3^2 \\equiv 2 \\pmod 7$ and inductively $u_n \\equiv 2 \\pmod 7, \\forall n \\ge 1$. Finally, notice that if $d = \\text{gcd}(m, n)$ then\n$$\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$$\nwith every $n \\in \\mathbb{Z}^+$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40666,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every $n \\ge 1$, then $v_3(u_n) = v_3(n)$, where $(u_n)$ is the sequence with $u_0 = 0$, $u_1 = 1$ and\n$$\nu_{n+2} = 2u_{n+1} + 2u_n, \\forall n \\ge 0.$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have the general formula of the given sequence as\n$$\nu_n = \\frac{(1 + \\sqrt{3})^n - (1 - \\sqrt{3})^n}{2\\sqrt{3}}.$$\nConsidering the periodicity of the remainder when divided by $3$ of the given sequence, we have $0, 1, 2, 0, 1, 2, \\ldots$ So obviously $u_{3k+1}$ and $u_{3k+2}$ are not divisible by $3$, and then $v_3(u_{3k+1}) = v_3(u_{3k+2}) = 0$.\n\nNext, consider $n = 3k$ with $k \\in \\mathbb{Z}^+$. We have\n$$\n\\begin{aligned}\nu_{3k} &= \\frac{(1 + \\sqrt{3})^{3k} - (1 - \\sqrt{3})^{3k}}{2\\sqrt{3}} \\\\&= \\frac{[(1 + \\sqrt{3})^k - (1 - \\sqrt{3})^k][(4 + 2\\sqrt{3})^k + (4 - 2\\sqrt{3})^k + (-2)^k]}{2\\sqrt{3}} \\\\&= u_k[(4 + 2\\sqrt{3})^k + (4 - 2\\sqrt{3})^k + (-2)^k].\\end{aligned}\n$$\n\nPut $a_k = (4 + 2\\sqrt{3})^k + (4 - 2\\sqrt{3})^k + (-2)^k$ then it is easy to check that $a_0 = 3$, $a_1 = 6$, $a_2 = 60$ and\n$$\na_{n+3} = 6a_{n+2} + 12a_{n+1} - 8a_n, \\forall n \\ge 0.\n$$\nNotice that $a_0, a_1, a_2$ are all divisible by $3$ so $3|a_n$ for all $n$. Thus $9|6a_{n+2}+12a_{n+1}$ and followed by $a_{n+3} \\equiv -8a_n \\equiv a_n \\pmod 9$. On the other hand, the first three terms of the sequence are not divisible by $9$, so the same applies to all terms of the sequence. From that we have $v_3(a_n) = 1$ for all $n$. So\n$$\nv_3(u_{3k}) = v_3(u_k) + v_3(a_k) = 1 + v_3(u_k).\n$$\nFrom here it is easy to see that if we put $n = 3^t m$ with $\\gcd(m, 3) = 1$ and $t \\in \\mathbb{Z}^+$ then\n$$\nv_3(u_n) = v_3(u_{3^{t-1}m}) + 1 = \\cdots = v_3(u_m) + t = t.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40667,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $x, y, z$ such that $xy + z = -30$, $yz + x = 30$ and $zx + y = -18$.",
    "options": [],
    "answer": "(15 + 4√13, -1, 15 - 4√13), (15 - 4√13, -1, 15 + 4√13), (2, -14, -2), (3, -9, -3), (-5, 7, 5)",
    "solution": "Adding the first two equations yields\n$$\n0 = xy + z + yz + x = (x + z)(y + 1).\n$$\nSo $y = -1$ or $z = -x$.\n\nIf $y = -1$ then the system reads $z = x - 30$ and $xz = -17$. It follows that $x^2 - 30x + 17 = 0$ and hence $x = 15 \\pm 4\\sqrt{13}$.\n\nIf $z = -x$, then the system becomes $xy - x = -30, y = x^2 - 18$ so\n$$\nx^3 - 19x + 30 = 0,\n$$\nwhose solutions are $x = 2, x = 3$ and $x = -5$. Therefore, we have five solutions $(x, y, z)$ in total: $(15 + 4\\sqrt{13}, -1, 15 - 4\\sqrt{13})$, $(15 - 4\\sqrt{13}, -1, 15 + 4\\sqrt{13})$, $(2, -14, -2)$, $(3, -9, -3)$, $(5, 7, -5)$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40668,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many triples $(x, y, z)$ of 6 digit positive integers exists, such that all digits of $x, y, z$ are odd and $x + y = 10z$?",
    "options": [],
    "answer": "3796875",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40669,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each integer from $1$ to $n$ was colored either red, or blue, and each color was used at least once. It turned out that\n* Every red number is a sum of some two distinct blue numbers;\n* Every blue number is a difference of some two red numbers.\nFind the smallest number $n$ for which such coloring is possible.",
    "options": [],
    "answer": "9",
    "solution": "The smallest such number is $n = 9$.\nSuppose that $n$ is a number allowing for coloring with the desired properties. Then $n \\ge 2$ since each color needs to be used at least once. Note that $n$ cannot be blue since it is inexpressible as a difference of two numbers between $1$ and $n$, therefore $n$ is red. On the other hand, $1$ and $2$ cannot be red, since none of them is a sum of two distinct positive integers, which means $1$ and $2$ are blue (and in particular $n \\ge 3$). Now $n-1$ cannot be a difference of two red numbers (since $1$ is blue), so $n-1$ is red. Similarly $n-2 = (n-1)-1$ has to be red since both $1$ and $2$ are blue. It follows that $n \\ge 5$.\n\nSince $n$ is red, it needs to be a sum of two distinct blue numbers. But a sum of two blue numbers cannot exceed\n$$\n(n-3) + (n-4) = 2n-7.\n$$\nTherefore, $2n - 7 \\ge n \\implies n \\ge 7$.\nIf $n = 7$, then the only remaining possibility to express $7$ as a sum of two blue numbers is $3+4$, but then $4$ is not a difference of two red numbers, contradiction. Similarly with $n = 8$.\nWe have proved that $n \\ge 9$. It remains to show that there exists a valid coloring for $n = 9$. We can color $1$, $2$, $4$, $5$, $6$ blue and $3$, $7$, $8$, $9$ red. Correctness of this coloring follows from the simple arithmetic:\n$$\n1 = 9 - 8,\\ 2 = 9 - 7,\\ 4 = 7 - 3,\\ 5 = 8 - 3,\\ 6 = 9 - 3,\n$$\n$$\n3 = 1 + 2,\\ 7 = 1 + 6,\\ 8 = 2 + 6,\\ 9 = 4 + 5.\n$$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40670,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine whether or not it is possible to partition the set of positive integers in infinite subsets $A_1, A_2, \\dots$ such that for every positive integer $k$, the sum of elements of $A_k$ is $k + 2023$.\n\n*Remark*: a partition of a set $X$ is a collection of subsets of $X$ such that every element of $X$ is contained in exactly one the subsets.",
    "options": [],
    "answer": "No",
    "solution": "The answer is No. Suppose such partition exists. Then for every positive integer $k$, we have\n$$\nB_k = A_1 \\cup A_2 \\cup \\dots \\cup A_k \\subset \\{1, 2, \\dots, k + 2023\\},\n$$\nsince all elements of $A_i$ are at most $i + 2023$ for every $i \\in \\{1, 2, \\dots, k\\}$ and\n$$\n\\sum_{b \\in B_k} b = \\sum_{i=2024}^{k+2023} i < \\sum_{i=1}^{k+2023} i.\n$$\nNow let $t_k$ be the minimum positive integer that not in $B_k$. Then we have\n$$\n\\sum_{i=2024}^{k+2023} i = \\sum_{b \\in B_k} b \\le \\left( \\sum_{i=1}^{k+2023} i \\right) - t_k,\n$$\nwhich implies that\n$$\nt_k \\le \\sum_{i=1}^{2023} i,\n$$\nfor every integer $k$. But that cannot happen if $A_1, A_2, \\dots$ is a partition of the positive integers. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40671,
    "subject": "Mathematics (Multi-modal)",
    "question": "In each square of a garden shaped like $2022 \\times 2022$ board, there is initially a tree of height $0$. A gardener and a lumberjack alternate turns playing the following game, with the gardener taking the first turn:\n\n* The gardener chooses a square in the garden. Each tree on that square and all the surrounding squares (of which there are at most eight) then becomes one unit taller.\n* The lumberjack then chooses four different squares on the board. Each tree of positive height on those squares then becomes one unit shorter.\n\nWe say that a tree is majestic if its height is at least $10^6$. Determine the largest number $K$ such that the gardener can ensure there are eventually $K$ majestic trees on the board, no matter how the lumberjack plays.",
    "options": [],
    "answer": "2271380",
    "solution": "The answer is $K = 5 \\cdot \\frac{2022^2}{9} = 2271380$. In general, for a $3N \\times 3N$ board, $K = 5N^2$.\n\nWe solve the problem for a general $3N \\times 3N$ board. First, we prove that the lumberjack has a strategy to ensure there are never more than $5N^2$ majestic trees. Giving the squares of the board coordinates in the natural manner, colour each square where at least one of its coordinates are divisible by $3$, shown below for a $9 \\times 9$ board:\n\nThen, as each $3 \\times 3$ square on the board contains exactly $5$ coloured squares, each move of the gardener will cause at most $4$ trees on non-coloured squares to grow. The lumberjack may therefore cut those trees, ensuring no tree on a non-coloured square has positive height after his turn. Hence there cannot ever be more majestic trees than coloured squares, which is $5N^2$.\n\nNext, we prove the gardener may ensure there are $5N^2$ majestic trees. Let modify the game which is more difficult for the gardener: on the lumberjack's turn in the modified game, he may decrement the height of all trees on the board except those the gardener did not just grow, in addition to four of the trees the gardener just grew. Clearly, a sequence of moves for the gardener which ensures that there are $K$ majestic trees in the modified game also ensures this in the original.\n\nLet $M = \\binom{9}{5}$; we say that a *map* is one of the $M$ possible ways to mark $5$ squares on a $3 \\times 3$ board. In the modified game, after the gardener chooses a $3 \\times 3$ subboard on the board, the lumberjack chooses a map in this subboard, and the total result of the two moves is that each tree marked on the map increases its height by $1$, each tree in the subboard which is not in the map remains unchanged, and each tree outside the subboard decreases its height by $1$. Also note that if the gardener chooses a $3 \\times 3$ subboard $Ml$ times, the lumberjack will have to choose some map at least $l$ times, so there will be at least $5$ trees which each have height $\\ge l$.\n\nThe strategy for the gardener will be to divide the board into $N^2$ disjoint $3 \\times 3$ subboards, number them $0, \\ldots, N^2 - 1$ in some order. Then, for $b = N^2 - 1, \\ldots, 0$ in order, he plays $10^6 M(M+1)^b$ times on subboard number $b$. Hence, on subboard number $b$, the moves on that subboard will first ensure $5$ of its trees grows by at least $10^6(M+1)^b$, and then each move after that will decrease their heights by $1$. (As the trees on subboard $b$ had height $0$ before the gardener started playing there, no move made on subboards $\\ge b$ decreased their heights). As the gardener makes\n$$\n10^6 M(M+1)^{b-1} + \\dots = 10^6((M+1)^b - 1)\n$$\nmoves after he finishes playing on subboard $b$, this means that on subboard $b$, there will be $5$ trees of height at least\n$$\n10^6(M+1)^b - 10^6((M+1)^b - 1) = 10^6,\n$$\nhence each of the subboard has $5$ majestic trees, which was what we wanted. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40672,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a function $f : \\mathbb{Q} \\to \\mathbb{Q}$ satisfying\n$$\nf(x + y + 2f(y)) = \\frac{2022}{2023} \\cdot y + f(x)\n$$\nfor all $x, y \\in \\mathbb{Q}$?",
    "options": [],
    "answer": "No",
    "solution": "The answer is No. Put $k = \\frac{2022}{2023}$ and let $x = 0$ then\n$$\nf(y + a f(y)) = k \\cdot \\frac{y}{a} + f(0)\n$$\nso $f$ is surjective over $\\mathbb{Q}$. Put $x = -a f(y)$, then $f(y) = k \\cdot \\frac{y}{a} + f(-a f(y))$ it is easy to see that $f$ is injective. So $f$ is bijective. Put $y = 0$, then $f(x + a f(0)) = f(x)$ so $x + a f(0) = x$, resulting in $f(0) = 0$. Replace $x = 0$ then\n$$\nf(y + a f(y)) = k \\cdot \\frac{y}{a}\n$$\nso $y + a f(y) = f^{-1}(k \\cdot \\frac{y}{a})$, where $f^{-1}$ is the inverse of $f$. From this, it follows that $y + a f(y)$ is also surjective over $\\mathbb{Q}$. Rewrite the problem as\n$$\nf(x + y + a f(y)) = f(y + a f(y)) + f(x)\n$$\nand replace $y + a f(y) = t \\in \\mathbb{Q}$ then\n$$\nf(x + t) = f(t) + f(x), \\forall x, t \\in \\mathbb{Q}.\n$$\nTherefore, $f$ is additive on $\\mathbb{Q}$, thus there exist $c \\in \\mathbb{Q}$ such that $f(x) = c x, \\forall x \\in \\mathbb{Q}$. Replace to the original equation to get\n$$\nc(x + y + a c y) = k \\cdot \\frac{y}{a} + c x\n$$\nor\n$$\n\\left(c + a c^2 - k \\cdot \\frac{1}{a}\\right) y = 0, \\forall y \\in \\mathbb{Q}.\n$$\nFrom that we have $(a c)^2 + a c - k = 0$, obviously $\\Delta = 1 + 4k$ is not the square of the rational number so the equation of variable $t = a c$ has no rational solution.\nTherefore, there does not exist $a \\in \\mathbb{Q}$ for a function $f : \\mathbb{Q} \\to \\mathbb{Q}$ that satisfies.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40673,
    "subject": "Mathematics (Multi-modal)",
    "question": "Several boxes of weight no more than $1$ kg weigh $36$ kg in total. Your friend can carry at most $4$ kg at a time. Prove that he can carry all the weights in no more than $11$ times.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40674,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a trapezoid with $AB \\parallel CD$ and $AB + CD = AD$. Its diagonals intersect at $E$. A line passing through the point $E$ and parallel to the bases of the trapezoid cuts $AD$ at $F$. Prove that $\\angle BFC = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40675,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is an equilateral triangle $ABC$ with circumcenter $O$. Let $D$ be a point on the minor arc $BC$ of its circumcircle such that $DB > DC$. The perpendicular bisector of $OD$ meets the circumcircle at $E$, $F$, with $E$ lying on the minor arc $BC$. The lines $BE$ and $CF$ meet at $P$. Prove that $PD \\perp BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40676,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $100$ doors labeled with numbers $1, 2, \\ldots, 100$. You have $100$ keys labeled with numbers. Each key corresponds to exactly one door. If the key $i$ corresponds to the door $j$, then $|i - j| \\le 1$. At each turn, you may pick doors with numbers $i$ and $j$ and check whether the key $i$ corresponds to the door $j$. Can you find which key corresponds to which door in\n\na) $99$ turns?\n\nb) $75$ turns?\n\nc) $74$ turns?",
    "options": [],
    "answer": "a) yes; b) yes; c) no",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40677,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider an acute-angled triangle $ABC$, with $AC > AB$, and let $\\Gamma$ be its circumcircle. Let $E$ and $F$ be the midpoints of the sides $AC$ and $AB$, respectively. The circumcircle of the triangle $CEF$ and $\\Gamma$ meet at $X$ and $C$, with $X \\neq C$. The line $BX$ and the tangent to $\\Gamma$ through $A$ meet at $Y$. Let $P$ be the point on segment $AB$ so that $YP = YA$, with $P \\neq A$, and let $Q$ be the point where $AB$ and the parallel to $BC$ through $Y$ meet each other. Show that $F$ is the midpoint of $PQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40678,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n > 2022$ cities in the country. Some pairs of cities are connected with straight two-ways airlines. Call the set of the cities unlucky, if it is impossible to color the airlines between them in two colors without monochromatic triangle. The set containing all the cities is unlucky. Is there always an unlucky set containing exactly 2022 cities?",
    "options": [],
    "answer": "Yes",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40679,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB \\neq BC$; and let $BD$ be the internal bisector of $\\angle ABC$, $(D \\in AC)$. Denote by $M$ the midpoint of the arc $AC$ which contains point $B$ in the circumscribed circle of the triangle $ABC$. The circumscribed circle of the triangle $\\triangle BDM$ intersects the segment $AB$ at point $K \\neq B$. Let $J$ be the reflection of $A$ with respect to $K$. If $DJ \\cap AM = \\{O\\}$, prove that the points $J$, $B$, $M$, $O$ belong to the same circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40680,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real $\\alpha > 0$ with the following property: there exists a sequence $(x_1, x_2, x_3, ...)$ of positive real numbers such that $x_{n+2} = \\sqrt{\\alpha x_{n+1} - x_n}$ for all $n \\ge 1$.",
    "options": [],
    "answer": "α > 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40681,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$ satisfy the following conditions:\ni) $f(2x) \\ge 2f(x)$ for all $x > 0$;\nii) $f$ is strictly increasing $(0; +\\infty)$.\niii) $f(f(x)f(y) + x) = f(xf(y)) + f(x)$ for all $x, y > 0$.",
    "options": [],
    "answer": "f(x) = x for all x > 0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40682,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all non-constant functions $f : \\mathbb{R} \\to \\mathbb{R}$ that satisfy\n$$\nf(2xy + x) = f(xy + x) + f(x)f(y)\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = x for all real x",
    "solution": "Let $x = y = 0$ in the problem, we have $f(0) = f(0) + f(0)^2$ so $f(0) = 0$.\n\nContinue to replace $(x, y) = (1, -1)$ into the problem, we have $f(-1) = f(1) \\cdot f(-1)$ deduce $f(1) = 1$ or $f(-1) = 0$. However, if $f(-1) = 0$, substituting $y = -1$ in the problem, then $f(-x) = 0$ for all real numbers $x$, not satisfied. Thus, $f(1) = 1$.\n\nContinue to replace $x = y = -1$, we have $f(1) = f(0) + f(-1)^2$ inferred $f(-1) = 1$ or $f(-1) = -1$. But if $f(-1) = 1$, substitute $y = -\\frac{1}{2}$ then $f(\\frac{x}{2}) = -f(x)f(-\\frac{1}{2})$, $\\forall x \\in \\mathbb{R}$; continue to substitute $x = -1$ then $f(-\\frac{1}{2}) = 0$ so $f(\\frac{x}{2}) = 0$, $\\forall x \\in \\mathbb{R}$, also not satisfied. Hence, $f(-1) = -1$.\n\nFinally, substitute $y \\to -1$ in the given, then $f(-x) = -f(x)$ for all real numbers $x$, that is, $f$ is an odd function.\n\nChange $y$ by $-y$, $y - 1$ and using the property $f$ as an odd function, then\n$$\nf(x)f(y) = f(2xy - x) - f(xy - x) = f(xy) + f(x)f(y - 1) - f(xy - x). \\quad (*)\n$$\nThus for all positive integers $y$ then\n$$\n\\begin{aligned} f(xy) - f(x)f(y) &= f(x(y - 1)) - f(x)f(y - 1) \\\\ &= f(x(y - 2)) - f(x)f(y - 2) \\\\ &= \\cdots = f(0) - f(x)f(0) = 0. \\end{aligned}\n$$\nFrom this it follows that $f(xy) = f(x)f(y)$ for all $x \\in \\mathbb{R}, y \\in \\mathbb{Z}^+$. In $(*)$, let $(x, y) = (1, 2)$ then $f(3) - f(1) = f(2)^2$. Let $y = 1$ in the given equation, then\n$$\nf(2x + 1) = f(x + 1) + f(x), \\forall x \\in \\mathbb{R}.\n$$\nContinue to replace $x = 1$ then $f(2) = f(3) - f(1) = f(2)^2$ so $f(2) = 2$ or $f(2) = 0$. Notice that if $f(2) = 0$, then due to the multiplication, we have $f(2x) = f(2)f(x) = 0$, which is also not satisfied. Therefore, $f(2) = 2$ and $f(2x) = 2f(x)$, $\\forall x \\in \\mathbb{R}$ and $f(2x + 1) = f(x) + f(x + 1)$.\n\nHere, by induction, we can prove $f(n) = n$ for all positive integers $n$. Replace $y \\to \\frac{y}{x}$ in the given, then\n$$\nf(2y + x) = f(x + y) + f(x)f\\left(\\frac{y}{x}\\right), \\forall x \\neq 0, y \\in \\mathbb{R}. \\quad (1)\n$$\nChange the role $x, y$ of the above expression then\n$$\nf(2x + y) = f(x + y) + f(y)f\\left(\\frac{x}{y}\\right), \\forall y \\neq 0, x \\in \\mathbb{R}. \\quad (2)\n$$\nReplace $x \\to 2x$ in (1) then\n$$\nf(2x+2y) = f(2x+y)+f(2x)f\\left(\\frac{y}{2x}\\right) = f(x+y)+f(y)f\\left(\\frac{x}{y}\\right)+f(2x)f\\left(\\frac{y}{2x}\\right), \\forall x,y \\neq 0.\n$$\nOn the other hand, $f(2x) = 2f(x)$ so $f(x+y) = f(x)f\\left(\\frac{y}{x}\\right) + f(y)f\\left(\\frac{x}{y}\\right)$, $\\forall x, y \\neq 0$. Adding the sides of (1) and (2), combined with this last equality, we get\n$$\nf(2x+y)+f(2y+x) = 2f(x+y)+f(x+y)+f(y)f\\left(\\frac{x}{y}\\right)+f(x)f\\left(\\frac{y}{x}\\right) = 3f(x+y), \\forall x,y \\neq 0.\n$$\nTo handle the condition $x, y \\neq 0$, replace $(x, y) \\to (\\frac{2b-a}{3}, \\frac{2a-b}{3})$ with $a, b \\in \\mathbb{R}$ then\n$$\nf(a) + f(b) = f(a + b); \\forall a \\neq 2b \\text{ và } b \\neq 2a.\n$$\nFor every pair of numbers $a, b \\in \\mathbb{R}$, we choose $c$ such that $c > 4|a| + 4|b|$ then make sure $c \\neq 2a, c \\neq 2b$. We have\n$$\nf(c) + f(a + b) = f(a + b + c) = f(c + a) + f(b) = f(c) + f(a) + f(b).\n$$\nFrom here we infer that $f$ is an additive function on $\\mathbb{R}$ so $f(xy) = f(2xy + x) - f(xy + x) = f(x)f(y)$, $\\forall x, y \\in \\mathbb{R}$. Therefore, $f$ adds, multiplies, and is non-constant, so $f(x) = x$. It is easy to check that they satisfy the given condition. So $f(x) = x$ with $x \\in \\mathbb{R}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40683,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer and let $x_1, x_2, \\dots, x_n$ be real numbers in the interval $[0, 1]$. Let $s = x_1 + x_2 + \\dots + x_n$, with $s \\ge 3$. Prove that there exist integers $i$ and $j$ with $1 \\le i < j \\le n$ such that\n$$\n2^{j-i} x_i x_j > 2^{s-3}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $1 \\le a < b \\le n$ be such that $2^{b-a} x_a x_b$ is maximal. This choice of $a$ and $b$ implies that\n$$\nx_{a+t} \\le 2^t x_a, \\forall t = 1-a, 2-a, \\dots, b-a-1,\n$$\nand similarly\n$$\nx_{b-t} \\le 2^t x_b, \\forall t = b-n, b-n+1, \\dots, b-a+1.\n$$\nNow, suppose that $x_a \\in (\\frac{1}{2^{u+1}}, \\frac{1}{2^u}]$ and $x_b \\in (\\frac{1}{2^{v+1}}, \\frac{1}{2^v}]$, and write $x_a = 2^{-\\alpha}$, $x_b = 2^{-\\beta}$. Then\n$$\n\\sum_{i=1}^{a+u-1} x_i \\le 2^u x_a \\left( \\frac{1}{2} + \\frac{1}{4} + \\dots + \\frac{1}{2^{a+u-1}} \\right) < 2^u x_a \\le 1,\n$$\nand similarly,\n$$\n\\sum_{i=b-v+1}^{n} x_i \\le 2^v x_b \\left( \\frac{1}{2} + \\frac{1}{4} + \\dots + \\frac{1}{2^{n-b+v}} \\right) < 2^v x_b \\le 1,\n$$\nIn other words, the sum of the $x'_i$s for $i$ outside of the interval $[a+u, b-v]$ is strictly less than $2$. Since the total sum is at least $3$, and each term is at most $1$, it follows that this interval must have at least two integers, i.e., $a+u < b-v$. Thus, by bounding the sum of the $x_i$, for $i \\in [1, a+u] \\cup [b-v, n]$ like above, and trivially bounding each $x_i \\in (a+u, b-v)$ by $1$, we obtain\n$$\n\\begin{aligned}\ns &< 2^{u+1} x_a + 2^{v+1} x_b + ((b-v-(a+u)-1) \\\\\n  &= b-a + (2^{u+1-\\alpha} + 2^{v+1-\\beta} - (u+v+1)).\n\\end{aligned}\n$$\nNow recall $\\alpha \\in (u, u+1]$ and $\\beta \\in (v, v+1]$, so applying Bernoulli's inequality yields\n$$\n\\begin{aligned}\n2^{u+1-\\alpha} + 2^{v+1-\\beta} - u - v - 1 &\\le (1+(u+1-\\alpha)) + (1+(v+1-\\beta)) - u - v - 1 \\\\\n&= 3 - \\alpha - \\beta.\n\\end{aligned}\n$$\nIt follows that $s - 3 < b - a - \\alpha - \\beta$, and so\n$$\n2^{s-3} < 2^{b-a-\\alpha-\\beta} = 2^{b-a} x_a x_b.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40684,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle, $D$ be the foot of altitude from $A$ to $BC$. We constructed the two squares $ABKL$, $ACMN$ outside the triangle. Prove that the lines $AD$, $BM$, $KC$ are concurrent at one point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40685,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist the infinite sequence of real numbers $(x_n)$ satisfying $x_1 = 2$ and\n$$\n\\frac{2x_n^2 + 2}{x_n + 3} < x_{n+1} \\le \\frac{2x_n + 2}{x_n + 3} + 2023\n$$\nfor all positive integers $n = 1, 2, 3, \\dots$?",
    "options": [],
    "answer": "No",
    "solution": "The answer is No. Suppose by contradiction that there is such a sequence. First, we will prove by induction that $x_n > 2$ for every $n \\ge 2$. One can check with $n = 2$, then assume that the assertion is true for $n = k \\ge 2$, i.e. $x_k > 2$ then\n$$\n\\frac{2x_k^2 + 2}{x_k + 3} - 2 = \\frac{2x_k^2 - 2x_k - 4}{x_k + 3} = \\frac{2(x_k - 2)(x_k + 1)}{x_k + 3} > 0.\n$$\nSo $x_{k+1} > \\frac{2x_k^2+2}{x_k+3} > 2$. Therefore, the assertion is also true for $n = k + 1$ and it is also true for all $n \\ge 2$. Hence, we have\n$$\nx_{n+1} - x_n \\ge \\frac{2x_n^2 + 2}{x_n + 3} - x_n = \\frac{(x_n - 1)(x_n - 2)}{x_n + 3} > 0,\n$$\nwhich implies that $(x_n)$ increases strictly. On the other hand, we have\n$$\n\\frac{2x_n + 2}{x_n + 3} < 2, \\forall n > 0 \\implies x_{n+1} < 2 + 2023 = 2025, \\forall n \\ge 1\n$$\nso $(x_n)$ is upper bounded by $2025$. Hence, $(x_n)$ has a finite limit, called by $L$ and $2 < L \\le 2025$. Substituting in the initial condition, we have\n$$\n\\frac{2L^2 + 2}{L + 3} \\le L \\Leftrightarrow 1 \\le L \\le 2,\n$$\nthis contradiction shows that there is no sequence $(x_n)$ satisfying the condition. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40686,
    "subject": "Mathematics (Multi-modal)",
    "question": "Calculate the sum of digits of the number\n$$\n1 + 11 + 111 + \\cdots + \\underbrace{111\\dots111}_{\\text{2023 1's}}.\n$$",
    "options": [],
    "answer": "8308",
    "solution": "The given number can be expressed as\n$$\n\\begin{aligned}\n\\sum_{i=1}^{2023} \\underbrace{11\\dots11}_{i \\text{ 1's}} &= \\sum_{i=1}^{2023} \\frac{10^i - 1}{9} = \\frac{\\sum_{i=1}^{2023} 10^i - 2023}{9} \\\\\n&= \\frac{\\underbrace{111\\dots1110}_{2023 \\text{ 1's}} - 2023}{9} = \\frac{\\underbrace{111\\dots111}_{2019 \\text{ 1's}}09087}{9}.\n\\end{aligned}\n$$\nWe further notice that\n$$\n\\underbrace{111\\dots111}_{2019 \\text{ 1's}}09087 = \\sum_{i=0}^{223} \\underbrace{11\\dots11}_{9 \\text{ 1's}} \\cdot 10^{9i+8} + 11109087.\n$$\nOn the other hand, with respect to $\\frac{111111111}{9} = 12345679$ and $\\frac{11109087}{9} = 1234343$, upon dividing the expression above by 9, we get the number equals to\n$$\n\\sum_{i=0}^{223} 12345679 \\cdot 10^{9i+8} + 1234343.\n$$\nTherefore, we need to calculate the sum of digits of this number. In each $12345679$ separated block, the sum of digits equals $37$, thus the answer is\n$$\n224 \\cdot 37 + (1 + 2 + 3 + 4 + 3 + 4 + 3) = 8308.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40687,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote by $\\mathbb{N}$ the set of positive integers. Find all functions $f: \\mathbb{N} \\to \\mathbb{N}$ such that $x - y$ divides $x^{f(x)} - y^{f(y)}$ for every two coprime integers $x$ and $y$.",
    "options": [],
    "answer": "All functions with f(1) arbitrary in N and f(n)=c for all n≥2, where c is an arbitrary positive integer.",
    "solution": "(Base on the solution of Hadi Alaithan, IMO 2023's team member)\nNote that for $x, y$ are coprime positive integers, we have\n$$\nx - y \\mid x^{f(x)} - y^{f(y)} \\Leftrightarrow x - y \\mid x^{|f(x) - f(y)|} - 1.\n$$\nNow we state the following lemma:\n**Lemma (Zsigmondy's Theorem):** for $a > b \\ge 1$ be coprime integers and $n \\ge 2$, there exists a prime divisor of $a^n - b^n$ that does not divide $a^k - b^k$ for all $1 \\le k < n$, except when $n = 2$ and $a + b$ is a power of $2$ or $(a, b, n) = (2, 1, 6)$.\n\nBack to the original problem, for any coprime integers $x$ and $y$, we show that $|f(x) - f(y)| \\le 1$. Suppose that the difference is larger than $1$, then there exists a prime $p$ divides $f(x) - f(y)$. Because $x$ and $y$ are relatively prime, one of these numbers is not a multiple of $p$. Assume that $\\gcd(x, p) = 1$. By choosing a suitable number $z$, we will show that this is a contradiction.\n\nDenote $k = v_p(f(x) - f(y))$. By the lemma, for some $N_1 > k$, there exists a prime $q_1$ such that\n$$\n\\operatorname{ord}_{q_1}(x) = p^{N_1}, \\quad q_1 > x y.\n$$\nSimilarly, for some $N_2 > N_1 > k$ such that there exists a prime $q_2 > q_1$ and\n$$\n\\operatorname{ord}_{q_2}(y) = p^{N_2}\n$$\nIt is obvious that $q_1, q_2, x$ and $y$ are pairwise relatively prime. By Chinese remainder theorem, there exists a positive integer $z$ such that\n$$\n\\begin{cases}\nz \\equiv 1 \\pmod{x}, & z \\equiv x \\pmod{q_1}, \\\\\nz \\equiv 1 \\pmod{y}, & z \\equiv y \\pmod{q_2}.\n\\end{cases}\n$$\nThese conditions give\n$$\n\\begin{cases}\nq_1 \\mid (z-x) \\mid x^{|f(z) - f(x)|} - 1 \\\\\nq_2 \\mid (z-y) \\mid y^{|f(z) - f(y)|} - 1.\n\\end{cases}\n$$\nHence $p^{N_1} \\mid f(x) - f(z)$ and $p^{N_2} \\mid f(x) - f(y)$. These give $p^{N_1} \\mid f(x) - f(y)$. This is a contradiction.\n\nNow, we show that $f(x) = f(y)$ for all coprime $x, y > 1$. Choose $z > 2x, 2y$ such that $\\gcd(z, x y) = 1$. Then we have\n$$\nz - x > x - 1, \\quad z - y > y - 1.\n$$\nThen both $f(z) - f(x) = 0$ and $f(z) - f(y) = 0$ due to the divisibility. We obtain that $f(x) = f(y)$ if $x$ and $y$ are relatively prime.\n\nIn conclusion, $f$ is constant on $\\{2, 3, \\dots\\}$ and $f(1) \\in \\mathbb{N}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40688,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 3$ be an integer. Suppose that $n$ children are arranged in a circle, and $n$ coins are distributed between them (some children may have no coins). At every step, a child with at least 2 coins may give 1 coin to each of their neighbours on the right and left. Determine all initial distributions of coins from which it is possible that, after a finite number of steps, each child has exactly one coin.",
    "options": [],
    "answer": "All initial distributions with sum of i times c_i congruent to n(n+1)/2 modulo n (with the total number of coins equal to n).",
    "solution": "The answer is: all distributions where $\\sum_{i=1}^n i c_i = \\frac{n(n+1)}{2} \\pmod n$, where $c_i$ denotes the number of coins the $i$-th child starts with.\n\nEncode the sequence $c_i$ as polynomial $p(x) = \\sum_i a_i x_i$. The cyclic nature of the problem makes it natural to work modulo $x^n - 1$. Child $i$ performing a step is equivalent to adding $x^i(x-1)^2$ to the polynomial, and we want to reach the polynomial $q(x) = 1 + x + \\dots + x^{n-1}$.\n\nSince we only add multiples of $(x-1)^2$, this is only possible if $p(x) = q(x)$ modulo the ideal generated by $x^n - 1$ and $(x-1)^2$, i.e.\n$$\n(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$$\nThis is equivalent to $p(1) = q(1)$ (which simply translates to the condition that there are $n$ coins) and $p'(1) = q'(1) \\pmod n$, which translates to the invariant. We also could show that this condition is also sufficient. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40689,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with $M$ is the midpoint of $BC$ and $\\angle BAM < 90^\\circ$, $\\angle AMB = 60^\\circ$. On ray $MA$, take the point $N$ such that $BN = AC$. Prove that $BC = 2AN$ and the orthocenters and circumcenters of two triangles $BMN$, $AMC$ form the four vertices of an isosceles trapezoid.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "On $MN$, take the point $D$ such that $ND = AM$. Then, $AN = DM$. According to the sine theorem then\n$$\n\\frac{BM}{\\sin \\angle BNM} = \\frac{BN}{\\sin \\angle BMN} = \\frac{AC}{\\sin \\angle AMC} = \\frac{MC}{\\sin \\angle MAC},\n$$\n\nso $\\angle BNM = \\angle MAC < 90^\\circ$, which implies that $\\triangle AMC \\cong \\triangle NDB$ so\n$$\n\\angle BDM = 180^\\circ - \\angle BDN = 180^\\circ - \\angle AMC = \\angle AMB = 60^\\circ.\n$$\nAlso, $BD = MC = MB$ so triangle $BDM$ is isosceles, with $\\angle BDM = 60^\\circ$ so equilateral. Therefore $AN = DM = BM = \\frac{1}{2}BC$.\n\n![](attached_image_1.png)\n\nLet $X, Y$ be the centers of the circumcircles of triangles $AMC, BMN$ and $H, K$ respectively their orthocenters. We have a familiar result: triangle $ABC$ has circumradius $R$ and orthocenter $H$ then $AH = 2R|\\cos A|$. Applying to this problem, notice that two triangles $AMC$ and $BMN$ have the same radius of circumcircle $R$, so\n$$\nMH = 2R \\cos 60^\\circ = R \\text{ and } MK = 2R |\\cos 120^\\circ| = R.\n$$\nTherefore, four points $X, Y, H, K$ are on the same circle with center $M$. Next, denote $\\ell$ as the internal bisector of $\\angle BMN$, then $MH, MX$ are symmetric through $\\ell$. And $MH = MX$ so we have $HX \\perp \\ell$. Similarly $KY \\perp \\ell$ so $HX \\parallel KY$. From this it follows that $HXYK$ is an isosceles trapezoid. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40690,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, a_3, \\dots, a_{10}$ and $b_1, b_2, \\dots, b_{10}$ be real numbers such that the roots of these 10 polynomials\n$$\nx^2 + a_1x + b_1,\\ x^2 + a_2x + b_2,\\ \\dots,\\ x^2 + a_{10}x + b_{10}\n$$\nare all integer numbers $\\pm 1, \\pm 2, \\dots, \\pm 10$ (in some order).\na) What is the maximum amount of odd values among $a_1, b_1, \\dots, a_{10}, b_{10}$?\nb) Find the minimum and maximum values of the sum $b_1 + b_2 + \\dots + b_{10}$.",
    "options": [],
    "answer": "a) 10; b) minimum −385 and maximum 380",
    "solution": "a) Denote $x_i, y_i$ as the roots of the $i$-th polynomial, then by Vieta's theorem, $a_i = -(x_i + y_i)$ and $b_i = x_i y_i$. Thus $a_i b_i = -x_i y_i (x_i + y_i)$ which is always even, implying that at most 1 number among $a_i, b_i$ is odd. Hence, there are at most 10 odd values among their coefficients.\n\nThe equality case occurs when $(1, 2), (3, 4), \\dots, (9, 10), (-1, -2), \\dots, (-9, -10)$ are roots of the given polynomials.\n\nb) By Vieta's theorem, we need to find the minimum and maximum value of\n$$\nT = x_1 y_1 + x_2 y_2 + \\dots + x_{10} y_{10}.\n$$\nNote that for all $x, y \\in \\mathbb{R}$, $xy \\ge -\\frac{x^2 + y^2}{2}$, thus\n$$\nT \\ge -\\frac{1}{2}(x_1^2 + y_1^2 + x_2^2 + y_2^2 + \\dots + x_{10}^2 + y_{10}^2) = -(1^2 + 2^2 + \\dots + 10^2) = -385.\n$$\nOn the other hand, for $x, y \\in \\mathbb{Z}$ and $x \\ne y$ then $(x-y)^2 \\ge 1$ so $xy \\le \\frac{x^2 + y^2 - 1}{2}$, thus\n$$\nT \\le \\frac{1}{2}(x_1^2 + y_1^2 + x_2^2 + y_2^2 + \\dots + x_{10}^2 + y_{10}^2) - 5 = 380.\n$$\nHence, we can conclude that:\n\n* $\\max T = 380$, attained when $(1, 2), (3, 4), \\dots, (9, 10), (-1, -2), \\dots, (-9, -10)$ are roots of 10 polynomials.\n* $\\min T = -385$, attained when $(1, -1), (2, -2), \\dots, (10, -10)$ are roots of 10 polynomials.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40691,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right-angled triangle with $\\angle BAC = 90^\\circ$, and let $E$ be the foot of the perpendicular from $A$ to $BC$. Let $Z \\neq A$ be a point on the line $AB$ with $AB = BZ$. Let $(c_1)$ be the circumcircle of the triangle $BEZ$ and $(c_2)$ be an arbitrary circle passing through the points $A$ and $E$. Suppose $(c_1)$ meets the line $CZ$ again at the point $F$, and meets $(c_2)$ again at the point $N$. If $P$ is the other point of intersection of $(c_2)$ with $AF$, prove that the points $N, B, P$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since triangles $AEB$ and $CAB$ are similar, we have $\\frac{AB}{EB} = \\frac{CB}{AB}$. Note that $AB = BZ$, thus\n$$\n\\frac{BZ}{EB} = \\frac{CB}{AB},\n$$\nfrom which it follows that triangles $ZBE$ and $CBZ$ are also similar. Since $FEBZ$ is cyclic, then $\\angle BEZ = \\angle BFZ$. So by the similarity of triangles $ZBE$ and $CBZ$, we get\n$$\n\\angle BFZ = \\angle BEZ = \\angle BZC = \\angle BZF\n$$\nand thus $BFZ$ is isosceles. Since $BF = BZ = AB$, triangle $AFZ$ is right with $\\angle AFZ = 90^\\circ$. It follows that points $A, E, F, C$ are concyclic. Since $A, P, E, N$ are also concyclic, then\n$$\n\\angle ENP = \\angle EAP = \\angle EAF = \\angle BCZ = \\angle BZE,\n$$\nby using the similarity of triangles $ZBE$ and $CBZ$. Since $N, B, E, Z$ are concyclic, then $\\angle ENP = \\angle BZE = \\angle ENB$, which implies that $N, B, P$ are collinear. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40692,
    "subject": "Mathematics (Multi-modal)",
    "question": "An acute scalene triangle $ABC$ is inscribed in a circle $k$. The bisector of angle $\\angle ABC$ meets side $BC$ at point $D$. Let $I$ be an arbitrary point on the segment $AD$, and let $H$ be the orthogonal projection of $I$ onto $BC$. Circle $\\omega$ is centered at $I$ and passes through $H$, and $U$ is the internal homothety center of the circles $k$ and $\\omega$. Prove that $U$ lies on a fixed line as $I$ moves on $AD$, and that $AI$ bisects the angle $HAU$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$, $N$ be the midpoint of minor arc and major arc $BC$ of circle $(O)$, respectively, then points $A$, $D$, $M$ are collinear. Let $T$ be the intersection of the two tangent lines at $B$ and $C$ of $(O)$ and $J$ the midpoint of $BC$. First, by the angle chasing, we have $\\angle TBM = \\angle BAM = \\angle MBC$ so $BM$ is the internal bisector of $\\angle CBT$. Thus, $M$ is the center of the incircle of the triangle $BTC$, denoted by $(M)$. Then it is easy to see that $D$ is the internal homothety center of $(\\omega)$ and $(M)$; and $T$ is the external homothety center of $(M)$ with $(O)$.\n\nApplying Monge D'Alembert's theorem to three circles $(I)$, $(M)$, $(O)$, we see that $U$, $D$, $T$ are collinear which implies that $T$ belongs to the fixed line $TD$.\n\nLet $V$ be the external homothety center of $(\\omega)$ with $(O)$, since $IH \\parallel MN$, points $V$, $H$, $M$ are collinear. Note that since $U$ is the inner center of $(\\omega)$ and $(O)$, $H$, $U$, $N$ is collinear because $IH \\parallel ON$. Then we have\n$$\nA(UH, IN) = A(UM, HN) = M(UH, AN) = M(UV, IO).\n$$\nOn the other hand, since $U$, $V$ is the internal, external homothety centers of $(\\omega)$ and $(O)$, $(UV, IO) = -1$. From this it follows that $A(UH, IN) = -1$. Since $AI \\perp AN$, by the property of harmonic bundles, $AI$ is the internal bisector of angle $HAN$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40693,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the maximum integer $k$ such that\n$$\n\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a} + k \\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)} \\ge 3\n$$\nfor all positive real numbers $a, b, c$.",
    "options": [],
    "answer": "15",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40694,
    "subject": "Mathematics (Multi-modal)",
    "question": "For $a, b, c > 0$, denote $m = \\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a}$, $n = \\frac{a}{c} + \\frac{c}{b} + \\frac{b}{a}$. Find all possible values of $k$ such that the following inequality is true for all $m, n$ defined as above:\n$$\n(m - n)k^2 + (2m - n - 3)k - 8m + 6n + 6 \\le 0.\n$$",
    "options": [],
    "answer": "[-4, -3] ∪ {2}",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40695,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least value of $k > 0$ such that\n$$\n|x - ky| + |y - kz| + |z - kx| \\geq 7\n$$\nfor all real numbers $x, y, z$ satisfying $x^2 + y^2 + z^2 = 21$.",
    "options": [],
    "answer": "2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40696,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all monic integer polynomials $P(x)$ of second degree such that there exist some integer polynomial $Q(x)$ satisfying $P(x)Q(x)$ have all coefficients in the set $\\{-1, 0, 1\\}$.",
    "options": [],
    "answer": "All and only the following monic quadratics: x^2 + 2x + 1, x^2 + x + 1, x^2 + 1, x^2 − x + 1, x^2 − 2x + 1, x^2 + x, x^2, x^2 − x, x^2 + x − 1, x^2 − 1, x^2 − x − 1.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40697,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider integers $b_1, b_2, \\dots, b_n$ have the sum is $2022$ and take $a$ as a divisor of $\\left\\lfloor \\dfrac{2022!}{3^{1006}} \\right\\rfloor$. Prove that the following polynomial does not have rational root\n$$\nP(x) = a x^{2n+2} + b_1 x^{2n} + b_2 x^{2n-2} + \\dots + b_n x^2 + a.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40698,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist $P(x, y)$ such that $P(x, y)^2 + 2022$ is divisible by $x^2 + y^2 + 2023$?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40699,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\nu_1 = a, \\ u_{n+1} = u_n + f(u_n)$$\nfor $n = 1, 2, \\dots$ with $f(x)$ is the product of all of digits of $x$. Prove that there exist $N$ such that $u_n = u_N$ for any $n \\geq N$.",
    "options": [],
    "answer": "null",
    "solution": "Let $u_1 = a$ be a positive integer. Define $u_{n+1} = u_n + f(u_n)$, where $f(x)$ is the product of all digits of $x$.\n\nObserve that if $u_n$ contains a digit $0$, then $f(u_n) = 0$, so $u_{n+1} = u_n$. Thus, the sequence becomes constant from that point onward.\n\nSuppose $u_n$ does not contain any digit $0$. Then $f(u_n) \\geq 1$, so $u_{n+1} > u_n$. However, as $u_n$ increases, eventually $u_n$ will contain a digit $0$ (since for any fixed $a$, the sequence increases and eventually reaches a number with a $0$ digit).\n\nOnce $u_n$ contains a $0$ digit, $f(u_n) = 0$, so $u_{n+1} = u_n$, and the sequence remains constant.\n\nTherefore, there exists $N$ such that $u_n = u_N$ for all $n \\geq N$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40700,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider function $f : \\mathbb{R} \\to \\mathbb{R}$ such that $x^2 f(x - y^2) \\geq (x + 2y + 1) f(x)$ for all $x, y \\in \\mathbb{R}$. Calculate $f(2022)$.",
    "options": [],
    "answer": "0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40701,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle with circumcenter $O$, orthocenter $H$. Denote $d$ as any diameter of $(O)$ and $A'$ as the midpoint of $BC$. Let $A_1$ be the projection of $A$ on $d$, $A_2$ be the reflection of $A_1$ over the perpendicular bisector of $BC$ and $A_3$ be the midpoint of $A'A_2$. Define $B_3, C_3$ similarly. Prove that $ABC$ and $A_3B_3C_3$ are similar and find the ratio of similarity.",
    "options": [],
    "answer": "1/2",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40702,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each $1 \\le i \\le 9$ and $T \\in \\mathcal{N}$, define $d_i(T)$ to be the total number of times the digit $i$ appears when all the multiples of $2023$ between $1$ and $T$ inclusive are written out in base $10$. Show that there are infinitely many $T \\in \\mathcal{N}$ such that there are precisely two distinct values among $d_1(T), d_2(T), \\dots, d_9(T)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n = 2023$. First, we choose some $k$ such that $n \\mid 10^k - 1$. For instance, any multiple of $\\varphi(n)$ would work since $n$ is coprime to $10$. We still show that either $T = 10^k - 1$ or $T = 10^k - 2$ has the desired property, which completes the proof since $k$ can be taken to be arbitrarily large.\n\nFor this it suffices to show that $\\#\\{d_i(10^k - 1) : 1 \\le i \\le 9\\} \\le 2$. Indeed, if\n$$\n\\#\\{d_i(10^k - 1) : 1 \\le i \\le 9\\} = 1\n$$\nthen, since $10^k - 1$ which consists of all nines is a multiple of $n$, we have\n$$\nd_i(10^k - 2) = d_i(10^k - 1) \\text{ for } i \\in \\{1, \\dots, 8\\}, \\text{ and } d_9(10^k - 2) < d_9(10^k - 1).\n$$\nThis means that $\\#\\{d_i(10^k - 2) : 1 \\le i \\le 9\\} = 2$.\n\nTo prove that $\\#\\{d_i(10^k - 1)\\} \\le 2$ we need an observation. Now let\n$$\n\\overline{a_{k-1}a_{k-2}\\dots a_0} \\in \\{1, \\dots, 10^k - 1\\}\n$$\nbe the decimal expansion of an arbitrary number, possibly with leading zeroes. Then $\\overline{a_{k-1}a_{k-2}\\dots a_0}$ is divisible by $n$ if and only if $\\overline{a_{k-2}\\dots a_0a_{k-1}}$ is divisible by $n$. Indeed, this follows from the fact that\n$$\n10 \\cdot \\overline{a_{k-1}a_{k-2}\\dots a_0} - \\overline{a_{k-2}\\dots a_0a_{k-1}} = (10^k - 1) \\cdot a_{k-1}\n$$\nis divisible by $n$. This observation shows that the set of multiples of $n$ between $1$ and $10^k - 1$ is invariant under simultaneous cyclic permutation of digits when numbers are written with leading zeroes.\n\nHence, for each $i \\in \\{1, \\dots, 9\\}$ the number $d_i(10^k - 1)$ is $k$ times larger than the number of $k$-digit numbers which start from the digit $i$ and are divisible by $n$. Since the latter number is either $\\lfloor \\frac{10^{k-1}}{n} \\rfloor$ or $1 + \\lfloor \\frac{10^{k-1}}{n} \\rfloor$, we conclude that $\\#\\{d_i(10^k - 1)\\} \\le 2$. This finishes the solution. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40703,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive numbers $a, b$ satisfy the condition $a + b + \\frac{1}{a} + \\frac{1}{b} = 5$. Prove that $1 \\le a + b \\le 4$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40704,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive real numbers $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ satisfy $a_1 \\ge a_2 \\ge \\dots \\ge a_n$ and $b_1b_2\\dots b_k \\ge a_1a_2\\dots a_k$ for each $k = 1, 2, \\dots, n$. Prove that\n$$\nb_1 + b_2 + \\dots + b_n \\ge a_1 + a_2 + \\dots + a_n.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that for each $k = 1, 2, \\dots, n$, we have\n$$\n\\frac{b_1}{a_1} + \\frac{b_2}{a_2} + \\dots + \\frac{b_k}{a_k} \\ge k \\sqrt[k]{\\frac{b_1}{a_1} \\frac{b_2}{a_2} \\dots \\frac{b_k}{a_k}} \\ge k,\n$$\nwhich means that\n$$\ns_k = \\frac{b_1 - a_1}{a_1} + \\frac{b_2 - a_2}{a_2} + \\cdots + \\frac{b_k - a_k}{a_k} = \\frac{b_1}{a_1} + \\frac{b_2}{a_2} + \\cdots + \\frac{b_k}{a_k} - k \\geq 0.\n$$\nFor each $k = 1, 2, \\dots, n$, we have\n$$\ns_k - s_{k-1} = \\frac{b_k - a_k}{a_k} \\implies b_k - a_k = a_k s_k - a_k s_{k-1}\n$$\nwhere $s_0 = 0$. Hence,\n$$\n\\begin{aligned}\n& (b_1 - a_1) + (b_2 - a_2) + \\cdots + (b_n - a_n) \\\\\n&= a_1 s_1 - a_2 s_1 + a_2 s_2 - a_3 s_2 + \\cdots + a_n s_n - a_n s_{n-1} \\\\\n&= s_1(a_1 - a_2) + s_2(a_2 - a_3) + \\cdots + s_{n-1}(a_{n-1} - a_n) + s_n a_n \\geq 0,\n\\end{aligned}\n$$\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40705,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is an $n \\times n$ board, whose all cells are initially white. Khalid the Painter walks around the board and recolors the visited cells according to the following rules. Each *walk* of Khalid starts at the bottom-left corner of the board and continues as follows:\n\n* if he is standing on a white cell, he paints it black and moves one cell up (or walks off the board);\n* if he is standing on a black cell, he paints it white and moves one cell to the right (or walks off the board).\n\nDetermine the minimum positive integer $s$ with the following property: after *exactly* $s$ walks all the cells of the board will become white again.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "s = 2^{2n - 1 - v_2\\big(\\binom{2n - 2}{n - 1}\\big)}",
    "solution": "(Base on the solution of Ali Alramdan, IMO 2023's team member)\n\nFor two times Khalid visits a square on two different directions, one of them he will go up and go right for the other. Let $f(i, j)$ be the number of time Khalid visited square $(i, j)$ so $f(0, 0) = s$ for some positive integer $s$ and\n$$\nf(i, j) = \\frac{f(i - 1, j) + f(i, j - 1)}{2}.\n$$\nBy induction on $i + j$, one can prove that the color of cell $(i, j)$ after precisely $s$ walks is $f(i, j) \\bmod 2$ where 1 is black and 0 is white. Define $g(i, j) = f(i, j) \\cdot 2^{i+j}$ for $0 \\le i, j \\le n-1$, then\n$$\ng(i, j) = g(i - 1, j) + g(i, j - 1).\n$$\nSince $g(0, 0) = f(0, 0) = s$, then base on the Pascal triangle formula, one can get\n$$\ng(i, j) = s \\cdot \\binom{i+j}{i} \\implies f(i, j) = \\frac{s \\cdot \\binom{i+j}{i}}{2^{i+j}}.\n$$\nThus for all squares to be white again, we need to find $f(i, j)$ to be even for all $i, j \\le n-1$, or\n$$\nv_2(s) + v_2\\left(\\binom{i+j}{i}\\right) \\ge i + j + 1.\n$$\nNow take $s = s_0 = 2^{2n-1-v_2\\left(\\binom{2n-2}{n-1}\\right)}$ which is the smallest positive integer such that $f(n-1, n-1)$ is even. We will prove that this choice is also such that: $f(i, j)$ are all even for all $0 \\le i, j \\le n-1$. We deduce to prove for $j = n-1$ since the remaining values can be restore by these values as some linear combinations of even numbers with integer coefficients. Put $k = v_2\\left(\\binom{2n-2}{n-1}\\right)$ then we need to prove for that\n$$\nv_2\\left(\\binom{2n-2-i}{n-1}\\right) \\ge k-1, \\forall i = 0, 1, \\dots, k.\n$$\nNote that\n$$\n\\binom{2n-2-i}{n-1} = \\binom{2n-2}{n-1} \\cdot \\frac{(n-1)(n-2)\\dots(n-i)}{(2n-2)(2n-3)\\dots(2n-1-i)}\n$$\nimplies that each even term of the product in the denominator of the fraction has its corresponding half in the numerator, so it could eat at most one 2 from the prime factorization, hence\n$$\nv_2\\left(\\binom{2n-2-i}{n-1}\\right) \\ge v_2\\left(\\binom{2n-2}{n-1}\\right) - \\left\\lfloor \\frac{i}{2} \\right\\rfloor = k - \\left\\lfloor \\frac{i}{2} \\right\\rfloor \\ge k-i.\n$$\nTherefore, the minimum value of $s$ is $s_0$ defined as above. ☐",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40706,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\le 100$ be a positive integer. There are 101 numbers written in a row:\n$$\n0 \\cdot n \\text{ mod } 101, 1 \\cdot n \\text{ mod } 101, \\dots, 100 \\cdot n \\text{ mod } 101.\n$$\nHow many pairs of neighbouring numbers are there in this row such that the one on the left is bigger than the one on the right?",
    "options": [],
    "answer": "n-1",
    "solution": "The answer is $n-1$.\n\nIndeed, the main claim is that $an \\bmod 101$ is larger than $(a+1)n \\bmod 101$ if and only if there is a number divisible by $101$ between $an$ and $(a+1)n$. Denote by $f(x)$ the remainder of $x$ when divided by $101$. Note that $f(an) \\ne f((a+1)n)$, otherwise $101 \\mid n$, a contradiction. Put $an = 101k + f(an)$ and $(a+1)n = 101k' + f((a+1)n)$. Thus\n$$\nn = (a + 1)n - an = 101(k' - k) + f((a + 1)n) - f(an).\n$$\nSince $-101 < f((a+1)n) - f(an) < 101$, it is easy to check that $k' - k = 0$ or $1$.\n\n* If $f(an) < f((a+1)n)$ then $k = k'$ and between $an$ and $(a+1)n$, there does not exist any multiple of $101$.\n* If $f(an) > f((a+1)n)$ then $k' = k + 1$ and between $an$ and $(a+1)n$, the number $101k'$ is a unique multiple of $101$.\n\nSo all we are left with is calculating the quantity of non-zero numbers divisible by $101$ between $0$ and $100 \\cdot n$. This quantity is $\\lfloor 100n/101 \\rfloor = n - 1$.\n\n$\\boxed{n-1}$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 40707,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a line, 200 points are marked and numbered $1, 2, 3, \\ldots, 200$ from left to right. Various crickets jump around the line. Each starts at point $1$, jumping on the marked points and ending up at point $200$. In addition, each cricket jumps from a marked point to another marked point with a greater number. When all the crickets have finished jumping, it turns out that for every pair $(i, j)$ with $1 \\le i < j \\le 200$, there was a cricket that jumped directly from point $i$ to point $j$, without visiting any of the points in between the two. Show that the number of crickets was at least $10000$ and that there is a way that $10000$ crickets could jump satisfying the conditions above.",
    "options": [],
    "answer": "10000",
    "solution": "For every pair $(i, j)$ where $1 \\le i \\le 100$ and $101 \\le j \\le 200$ there is a cricket that jumped from $i$ to $j$ and no cricket can do two such jumps. Therefore there are at least $100^2 = 10000$ crickets.\n\nConsider the following paths of crickets:\n\n1. $1 \\to 200$;\n2. $1 \\to a \\to 200$ where $a \\in \\{2, 3, \\dots, 199\\}$,\n3. $1 \\to k \\to 201 - k \\to 200$, where $k \\in \\{2, 3, \\dots, 100\\}$,\n4. $1 \\to k \\to a \\to 201 - k \\to 200$, where $k \\in \\{2, 3, \\dots, 100\\}$ and $k < a < 201 - k$.\n\nCounting the different paths above from each type, there are\n$$\n1 + 198 + 99 + \\sum_{k=2}^{100} (200 - 2k) = 10000\n$$\nsuch paths, and one can check that for every pair of marked points there is a cricket doing the jump between the points. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40708,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an acute triangle $ABC$ with its altitudes $AD$, $BE$, $CF$ concurrent at $H$. The point $K$ changes on the segment $AH$. Let $M$, $N$ be the projection of $H$ on the lines $KE$, $KF$. Prove that the line joining the circumcenter of triangles $HEN$ and $HFM$ always passes through a fixed point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S$, $T$ be the intersection of the lines $KF$, $KE$ with $CA$, $AB$ and $P$, $Q$ the midpoint of $HT$, $HS$. Since $\\angle HFT = \\angle HMT = 90^\\circ$ so $HT$ is the diameter of $(HMF)$ and $P$ is the center of $(HMF)$. Similarly, $Q$ is the center of $(HNE)$.\n\nLet $R$ be the intersection of $ST$, $EF$. Based on the basic properties of harmonic points, we have\n$$\nA(HR, EF) = -1.\n$$\nOn the other hand, if $EF$, $BC$ intersect at $R'$, then $A(DR', EF) = -1$, from which it follows that $R \\equiv R'$ and resulting in $R$ being the fixed point. Since $H$ is fixed, the midpoint $L$ of $HR$ is also a fixed point. On the other hand, $PL$, $QL$ are the medians of triangles $HTR$, $HSR$ such that $R$, $S$, $T$ are collinear, so $L$ belongs to $PQ$. So $PQ$ passes through the fixed $L$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40709,
    "subject": "Mathematics (Multi-modal)",
    "question": "An acute scalene triangle $ABC$ with $\\angle A = 45^\\circ$ is inscribed in a circle ($O$). Let $AD$ be the altitude and $H$ be the orthocenter of $\\triangle ABC$. Points $R$ and $S$ different from $A$ are taken on the rays $AB$ and $AC$ respectively so that $CA = CR$ and $BA = BS$. Points $X, Y$ and $K$ are the midpoints of the segments $BS, CR$ and $AH$, respectively. Point $T$ on the line $BC$ such that $AO$ is perpendicular to $TO$, and $I$ is the circumcenter of triangle $DXY$. Prove that\n$$\nXY \\perp OK \\text{ and } TH \\perp IK.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $Z$, $M$ be the midpoints of $BR$, $BC$ respectively, then $ZY$ is the median of the triangle $BCR$ so $ZY = \\frac{BC}{2}$ and $ZY \\parallel BC$.\n\n![](attached_image_1.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 the triangle $ARS$. So $L \\in (O)$ and $AL$ is the diameter of $(O)$.\n\nConsidering the 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 the 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 the triangle $LHG$, followed by $HG \\parallel FM$, which $FM \\perp BC$ should be $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$$\nTA^2 - TN^2 = HA^2 - HN^2.\n$$\nLet $R$ be the radius of the circle $(O)$, we have $AH^2 = (2OM)^2 = 2R^2$. It is easy to see that $OBNC$ is a square, so $ON = 2OM = AH$, so $AHNO$ is a parallelogram and $HN = AO$. Hence\n$$\nHA^2 - HN^2 = HA^2 - AO^2 = R^2.\n$$\nWe also have\n$$\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$$\nFrom this it follows that $TA^2 - TN^2 = HA^2 - HN^2$ or $TH \\perp AN$, which leads to $TH \\perp IK$. This finishes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40710,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that it is possible to pick 20 numbers among $1, 2, \\ldots, 10000$ such that the members of any non-empty subset of these 20 numbers has a sum which is not an $n$-th power of some number (for any $n > 1$).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $19$, $23$ are primes and $19 \\cdot 23 = 437$. Now denote $S = \\{1, 2, 3, \\ldots, 20\\}$ then take $T = \\{437a \\mid a \\in S\\} \\subset \\{1, 2, \\ldots, 10^4\\}$. So $|T| = |S| = 20$. Note that $1 + 2 + \\dots + 20 = 210 < 437$.\n\nSuppose that there exist $k$ numbers in $T$, denote by $437a_1, 437a_2, \\ldots, 437a_k$ such that their sum is a perfect power. Thus\n$$\n437(a_1 + a_2 + \\dots + a_k) = a^n \\quad (*)\n$$\nfor some positive integer $a, n$ and $n \\ge 2$. Since $437 \\mid a^n$ then $437^2 \\mid a^n$ which implies that\n$$\n437 \\mid a_1 + a_2 + \\dots + a_k,\n$$\nthis is contradiction since the sum of all numbers in $S$ is less than $437$. Thus, the above set satisfies the given condition. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40711,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n \\ge 2$ for which there exist $n$ real numbers\n$$\na_1 < a_2 < \\dots < a_n\n$$\nand a real number $r > 0$ such that the $\\frac{n(n-1)}{2}$ differences $a_j - a_i$ for $1 \\le i < j \\le n$ are equal, in some order, to the numbers\n$$\nr^1, r^2, \\dots, r^{\\frac{n(n-1)}{2}}.\n$$",
    "options": [],
    "answer": "{2, 3, 4}",
    "solution": "The answer are $n \\in \\{2, 3, 4\\}$. We first show a solution for each $n \\in \\{2, 3, 4\\}$. We will later show the impossibility of finding such a solution for $n \\ge 5$.\n\nFor $n = 2$, take for example $(a_1, a_2) = (1, 3)$ and $r = 2$.\n\nFor $n = 3$, take the root $r > 1$ of $x^3 - x - 1 = 0$ (the golden ratio) and set $(a_1, a_2, a_3) = (0, r, r + r^2)$, then\n$$\n(a_2 - a_1, a_3 - a_2, a_3 - a_1) = (r, r^2, r + r^2 = r^3).\n$$\n\nFor $n = 4$, take the root $r \\in (1, 2)$ of $x^3 - x - 1 = 0$ (such a root exists because $1^3 - 1 - 1 < 0$ and $2^3 - 2 - 1 > 0$) and set $(a_1, a_2, a_3, a_4) = (0, r, r + r^2, r + r^2 + r^3)$, then\n$$\n(a_2 - a_1, a_3 - a_2, a_4 - a_3, a_3 - a_1, a_4 - a_2, a_4 - a_1) = (r, r^2, r^3, r^4, r^5, r^6).\n$$\n\nFor $n \\ge 5$, we will proceed by contradiction. Suppose that there exist numbers $a_1 < a_2 < \\dots < a_n$ and $r > 1$ satisfying the conditions of the problem. We start with a lemma:\n\nLemma. We have $r^{n-1} > 2$.\nProof. There are only $n-1$ differences $a_j - a_i$, with $j = i + 1$, so there exists an exponent $e \\le n$ and a difference $a_j - a_i$ with $j \\ge i + 2$ such that $a_j - a_i = r^e$. This implies that\n$$\nr^n \\ge r^e = a_j - a_i = (a_j - a_{j-1}) + (a_{j-1} - a_i) > r + r = 2r,\n$$\nthus $r^{n-1} > 2$ as desired. To illustrate the general approach, we first briefly sketch the idea behind the argument in the special case $n = 5$. In this case, we clearly have $a_5 - a_1 = r^{10}$. Note that there are 3 ways to rewrite $a_5 - a_1$ as a sum of two differences, namely\n$$\n(a_5 - a_4) + (a_4 - a_1), (a_5 - a_3) + (a_3 - a_1), (a_5 - a_2) + (a_2 - a_1).\n$$\nUsing the lemma above and convexity of the function $f(n) = r^n$, we argue that those three ways must be $r^{10} = r^9 + r^1 = r^8 + r^4 = r^7 + r^6$. That is, the “large” exponents keep dropping by 1, while the “small” exponents keep increasing by $n-2, n-3, \\dots, 2$. Then comparing any two such equations to get a contradiction unless $n \\le 4$.\n\nNow we go back to the full proof for any $n \\ge 5$. Denote $b = \\frac{1}{2}n(n-1)$. Clearly, we have $a_n - a_1 = r^b$. Consider the $n-2$ equations of the form:\n$$\na_n - a_1 = (a_n - a_i) + (a_i - a_1) \\text{ for } i \\in \\{2, \\dots, n-1\\}.\n$$\nIn each equation, one of the two terms on the right-hand side must be at least $\\frac{1}{2}(a_n - a_1)$. But from the lemma we have\n$$\nr^{b-(n-1)} = r^b/r^{n-1} < \\frac{1}{2}(a_n - a_1),\n$$\nso there are at most $n-2$ sufficiently large elements in $\\{r^k|1 \\le k < b\\}$, namely $r^{b-1}, \\dots, r^{b-(n-2)}$ (note that $r^b$ is already used for $a_n - a_1$). Thus, the “large” terms must be, in some order, precisely equal to elements in\n$$\nL = r^{b-1}, \\dots, r^{b-(n-2)}.\n$$\nNext we claim that the “small” terms in the $n-2$ equations must be equal to the elements in\n$$\nS = \\{r^{b-(n-2)-\\frac{1}{2}i(i+1)} | 1 \\le i \\le n-2\\},\n$$\nin the corresponding order (the largest “large” term with the smallest “small” term, etc.). Indeed, suppose that\n$$\nr^b = a_n - a_1 = r^{b-i} + r^{\\alpha_i} \\text{ for } i \\in \\{1, \\dots, n-2\\},\n$$\nwhere $1 \\le \\alpha_1 < \\dots < \\alpha_{n-2} \\le b - (n-1)$. Since $r > 1$ and $f(r) = r^n$ is convex, we have\n$$\nr^{b-1} - r^{b-2} > r^{b-2} - r^{b-3} > \\dots > r^{b-(n-3)} - r^{b-(n-2)},\n$$\nimplying $r^{\\alpha_2} - r^{\\alpha_1} > r^{\\alpha_3} - r^{\\alpha_2} > \\dots > r^{\\alpha_{n-2}} - r^{\\alpha_{n-3}}$. Convexity of the function $f(r) = r^n$ further implies\n$$\n\\alpha_2 - \\alpha_1 > \\alpha_3 - \\alpha_2 > \\dots > \\alpha_{n-2} - \\alpha_{n-3}.\n$$\nNote that $\\alpha_{n-2} - \\alpha_{n-3} \\ge 2$: Otherwise we would have $\\alpha_{n-2} - \\alpha_{n-3} = 1$ and thus\n$$\nr^{\\alpha_{n-3}} \\cdot (r-1) = r^{\\alpha_{n-2}} - r^{\\alpha_{n-3}} = r^{b-(n-3)} - r^{b-(n-2)} = r^{b-(n-2)} \\cdot (r-1),\n$$\nimplying that $\\alpha_{n-3} = b - (n - 2)$, a contradiction. Therefore, we have\n$$\n\\begin{aligned}\n\\alpha_{n-2} - \\alpha_1 &= (\\alpha_{n-2} - \\alpha_{n-3}) + \\dots + (\\alpha_2 - \\alpha_1) \\\\\n&\\ge 2 + 3 + \\dots + (n-2) = \\frac{1}{2}(n-2)(n-1) - 1 = \\frac{1}{2}n(n-3).\n\\end{aligned}\n$$\nOn the other hand, from $\\alpha_{n-2} \\le b - (n - 1)$ and $\\alpha_1 \\ge 1$ we get\n$$\n\\alpha_{n-2} - \\alpha_1 \\le b - n = \\frac{1}{2}n(n-1) - n = \\frac{1}{2}n(n-3),\n$$\nimplying that equalities must occur everywhere and the claim about the small terms follows. Now assuming $n-2 \\ge 2$, we have the two different equations:\n$$\nr^b = r^{b-(n-2)} + r^{b-(n-2)-1} \\quad \\text{and} \\quad r^b = r^{b-(n-3)} + r^{b-(n-2)-3},\n$$\nwhich can be rewritten as\n$$\nr^{n-1} = r + 1 \\text{ and } r^{n+1} = r^4 + 1.\n$$\nSimple algebra now gives\n$$\nr^4 + 1 = r^{n+1} = r^{n-1} \\cdot r^2 = r^3 + r^2 \\longrightarrow (r-1)(r^3 - r - 1) = 0.\n$$\nSince $r \\ne 1$, using (1) we conclude $r^3 = r + 1 = r^{n-1}$, thus $n = 4$, which gives a contradiction. □",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40712,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle. The line through $A$ perpendicular to $BC$ intersects $BC$ at $D$. Let $E$ be the midpoint of $AD$ and $\\omega$ the circle of center $E$ and radius $AE$. The line $BE$ intersects $\\omega$ at $X$ such that $X$ and $B$ are not on the same side of $AD$ and the line $CE$ intersects $\\omega$ at $Y$ such that $C$ and $Y$ are not on the same side of $AD$. If two intersection points of the circumcircles of triangles $BDX$ and $CDY$ lie on the line $AD$, prove that $AB = AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that the condition of the problem is that $AD$ is the radical axis of two circles ($BDX$) and ($CDY$) which implies that $E$ has the same power to these circles. This gives us\n$$\nEB \\cdot EX = EC \\cdot EY.\n$$\nHowever, $EX = EY$ because $E$ is the center of $\\omega$ and this means that $BE = CE$. From this, we conclude that $DE$ is the perpendicular bisector of $BC$ and leads to $AB = AC$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40713,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with circumcircle $(O)$ with orthocenter $H$. Take $X \\in AB$, $Y \\in AC$ such that $XB = XH$, $YC = YH$. Prove that $O$ belongs to the Euler line of triangle $AXY$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40714,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle inscribed in a circle $(O)$ and $M$ is the midpoint of $BC$. Take $D$ on the segment $BC$ and the circle $(ABD)$ meets $AC$ again at $E$, the circle $(ACD)$ meets $AB$ again at $F$. Lines $BE$, $CF$ meet at $K$ and lines $AK$, $EF$ meet at $R$. Suppose that there exist the circle passing through $B$, $C$ and external tangent to $(AEF)$ at $T$. Prove that $RT$ bisects $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40715,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle. A circle $(O)$ passes through $B$, $C$ and meets $AB$, $AC$ at $D$, $E$. Lines $BE$, $CD$ meet at $I$. Denote $M$, $N$ as midpoints of $BE$, $CD$ and $MN$ meets $AB$, $AC$ at $P$, $Q$. Prove that $(OI)$ is tangent to $(APQ)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40716,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a non-isosceles triangle with altitudes $AD$, $BE$, $CF$ concurrent at $H$. Denote $K$, $M$ as the midpoints of $AH$, $BC$ and $BK$ meets $AM$ at $L$. On $AC$, take $I$ such that $OI \\parallel BC$ and $DE$ meets $IK$ at $S$. Prove that $L$, $K$, $S$, $E$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40717,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with incenter $I$ and ex-center $J$. Denote $K$ as the reflection of $A$ over $BC$ and take $X$, $Y$ on the opposite rays of $BA$, $CA$ such that $XB = BC = CY$. Let $T$ be the circumcenter of $AXY$. Prove that:\n\na) $OJ \\perp XY$ and $AT$ is tangent to $(AIK)$.\n\nb) $R_{(T)} = OJ$ and $TJ \\perp BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40718,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $\\angle B = 2\\angle C$ and angle bisector $BD$. The symmedian of vertex $B$ in triangles $DAB$, $BCD$ cuts the corresponding circumcircle at $M$, $N$. Denote $P$ as the reflection of $B$ over $C$. Prove that the circle $(MNP)$ is tangent to $BC$ and both of circles $(DAB), (BCD)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 40719,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $(O)$ as its circumcircle. Take $D$ on the minor arc $BC$ of $(O)$ and the tangent line from $D$ to $(O)$ that intersects $BC$, $CA$, $AB$ at $M$, $N$, $P$. Line $AM$ cuts $(O)$ again at $K$ and $BN$, $CP$ meet at $H$. Prove that $H$, $K$, $D$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40720,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral. Take $F \\in AB$, $E \\in CD$ such that $(FCD)$ is tangent to $AB$ and $(EAB)$ is tangent to $CD$. Denote $G = AE \\cap DF$, $H = BE \\cap CF$. Prove that $EF$ is the perpendicular bisector of $GH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40721,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function such that $f(0) = 0$ and\n$$\n2f\\left(-\\frac{1}{2}xy + f(x + y)\\right) = x f(y) + y f(x)\n$$\nfor all $x, y \\in \\mathbb{R}$. Prove that $f(2) = f(-2) = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40722,
    "subject": "Mathematics (Multi-modal)",
    "question": "Every positive integer greater than $1000$ is colored in red or blue in such a way that the product of any two distinct red numbers is blue. Can it happen that no two blue numbers have difference $1$?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40723,
    "subject": "Mathematics (Multi-modal)",
    "question": "An academic club of $2023$ members wants to organize some seminars. In each seminar, each member of the club will present exactly one of three subjects: Math, Physics, and Chemistry. It is given that for any two members, there exists some seminar where they do not present the same subject. Find the smallest possible value of the number of seminars.",
    "options": [],
    "answer": "7",
    "solution": "Let $k$ be the number of seminars the club can organize. In the $i$-th seminar with $1 \\le i \\le k$, denote $A_i$, $B_i$, $C_i$ as the collection of participants presenting Math, Physics, Chemistry respectively. Then, for each $i$, we have $A_i$, $B_i$, $C_i$ are the partition of the members of the club. Replace $2023 \\to n$.\n\nIn each seminar, the members will have $3$ ways to choose subjects to present, so there will be $3^k$ ways to choose in total. If $3^k < n$ then, according to the pigeonhole principle, there will be two members who must have the same choice in all $k$ seminars. However, the two members will always present the same subject for every seminar, contradiction. So there must be $n \\le 3^k$.\n\nWe will show that for $n \\le 3^k$ it is always possible to construct. Indeed, converting the members' numbers from $0, 1, 2, \\ldots, n-1$ to base $3$, each number will correspond to a ternary string of length no more than $k$. To synchronize those strings, we add $0$ in front so that all strings have the same length $k$. We define the following rule: in the $i$th seminar with $1 \\le i \\le k$, all members having $0$ at position $i$ on their string will present Math, similarly, the same $1$ for Physics and the same $2$ for Chemistry.\n\nNote that any two strings are different, so they must be different at a certain position, and the index of that position will be the index of the seminar in which the two respective members present $2$ different subjects, satisfying the constraint.\n\nFor $n = 2023$, we have $k \\ge \\lceil \\log_3 2023 \\rceil = 7$ and $k = 7$ is the minimum value. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40724,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two circles $(O_1), (O_2)$ with different radii and intersecting at two points $A, B$. The tangent $CD$ of the two circles (closer to point $B$) with $C \\in (O_1)$ and $D \\in (O_2)$. Draw diameters $BP, BQ$ of $(O_1), (O_2)$. The line through $B$, and perpendicular to $CD$ cuts $PQ$ at $K$.\n\na) Prove that $B$ is the orthocenter of triangle $KCD$.\n\nb) Draw the angle bisectors $BX, BY$ of the triangles $KBP, KBQ$ respectively with $X, Y \\in PQ$. Prove that $KX = KY$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Redefine the point $K$ as the orthocenter of the triangle $BCD$, then $B$ is also the orthocenter of the triangle $KCD$, we will show that $K \\in PQ$. Construct the parallelogram $BCDT$.\n\n![](attached_image_1.png)\n\nSince $K$ is the orthocenter of triangle $BCD$, $BC \\perp KD$, and $BC \\parallel TD$ so $TD \\perp KD$. Similarly, $TC \\perp KC$ so $KCTD$ is inscribed in a circle of diameter $KT$. Since $CD$ is a common tangent to $(O_1), (O_2)$, $\\angle BCD = \\angle CAB$, $\\angle BDC = \\angle DAB$. Therefore\n$$\n180^\\circ - \\angle CBD = \\angle BCD + \\angle BDC = \\angle CAB + \\angle DAB = \\angle CAD.\n$$\nSince $BCTD$ is a parallelogram, $\\angle CBD = \\angle CTD$, hence $180^\\circ - \\angle CTD = \\angle CAD$ entails $ACTD$ internal. Therefore, the points $A, C, T, D, K$ belong to the circle of diameter $KT$. Since $BP, BQ$ are the diameters of $(O_1), (O_2)$, $\\angle BAP = \\angle BAQ = 90^\\circ$, so $A, P, Q$ are collinear. It is left to prove that $KA \\perp AB$. From $KACD$ is inscribed, we have $\\angle KAC = \\angle KDC = 90^\\circ - \\angle BCD = 90^\\circ - \\angle BAC$ implies that\n$$\n\\angle KAB = \\angle KAC + \\angle BAC = 90^\\circ.\n$$\nHence $K, P, Q$ are collinear and thus the original problem is solved.\n\nb) Draw the altitude $BH$ of the triangle $BCD$ and $M$ is the midpoint of $CD$. We have\n$$\n\\angle KBP = 180^\\circ - (\\angle CBP + \\angle CBH) = 180^\\circ - (90^\\circ - \\angle BPC + 90^\\circ - \\angle BCH) = 2\\angle BCH.\n$$\nSince $BX$ is the bisector of $\\angle KBP$, then $\\angle KBX = \\angle BCM$. We can see that $AKMH$ is cyclic since $\\angle KAM = \\angle KHM = 90^\\circ$ so $\\angle BKX = \\angle BMC$. This implies that\n$$\n\\Delta BKX \\sim \\Delta CMB \\implies \\frac{BK}{CM} = \\frac{KX}{MB}.\n$$\nSimilarly, $\\Delta BKY \\sim \\Delta DMB \\implies \\frac{BK}{DM} = \\frac{KY}{MB}$. And $CM = DM$ so we get\n$$\n\\frac{KX}{MB} = \\frac{KY}{MB} \\implies KX = KY.\n$$\n$\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40725,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an oral exam, there are 10 examiners and 1024 contestants. Each contestant will be asked by each examiner and receive a result of either “pass” or “fail”. It is given that for any two contestants, there is some examiner who will rate one passed and the other failed. Two contestants are said to be “separative” if they have different results by at least 3 examiners. Prove that it is possible to select 24 pairwise separative contestants.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Number the examiners from $1$ to $10$ and consider the results of contestants as binary sequences of the form $x_1x_2\\ldots x_{10}$, in which $x_i = 0$ or $1$ if the $i$th examiner gave pass or fail response for the corresponding candidate. We can directly construct $24$ binary strings that satisfy the “separative” condition. Indeed, choose strings of the form $(x_1x_2x_3x_4x_5x_6)(x_7x_8x_9x_{10})$, where: $x_1x_2x_3x_4x_5x_6$ contains exactly $3$ zeros and $3$ ones; and $x_7x_8x_9x_{10}$ will contain all binary strings of length $4$. Since $\\binom{6}{3} = 20$ and two strings $x_1x_2x_3x_4x_5x_6 \\neq x'_1x'_2x'_3x'_4x'_5x'_6$ would be different at at least $2$ positions, so the rest just need to differ by at least $1$ position. Since $2^4 = 16$, there will be $4$ repeated strings at the last $4$ positions, we just need to choose those strings that differ by at least $3$ positions out of the first $6$ positions. For detail,\n$$\n\\begin{array}{lcl}\n(000111)(0000) & \\rightarrow & (a) \\\\\n(001011)(0001) & \\rightarrow & (b) \\\\\n(001101)(0010) & \\rightarrow & (c) \\\\\n(001110)(0011) & \\rightarrow & (d) \\\\\n(010011)(0100) \\\\\n(010101)(0101) \\\\\n(010110)(0110) \\\\\n(011001)(0111) \\\\\n(011010)(1000) \\\\\n(011100)(1001) \\\\\n(100011)(1010) \\\\\n(100101)(1011) \\\\\n(100110)(1100) \\\\\n(101001)(1101) \\\\\n(101010)(1110) \\\\\n(101100)(1111) \\\\\n(110001)(0011) & \\rightarrow & (d') \\\\\n(110010)(0010) & \\rightarrow & (c') \\\\\n(110100)(0001) & \\rightarrow & (b') \\\\\n(111000)(0000) & \\rightarrow & (a').\n\\end{array}\n$$\nFinally, add $4$ more strings as the following:\n$$\n\\begin{array}{l}\n(000000)(0000), (000000)(1111), \\\\\n(111111)(0000), (111111)(0000).\n\\end{array}\n$$\nIt is easy to check that they are pairwise separative and also separative with the other $20$ strings. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40726,
    "subject": "Mathematics (Multi-modal)",
    "question": "Every rational number $\\frac{p}{q}$ from the open interval $(0, 1)$ is covered by the closed interval\n$$\n\\left[ \\frac{p}{q} - \\frac{1}{4q^2}, \\frac{p}{q} + \\frac{1}{4q^2} \\right].\n$$\nProve that the number $\\frac{\\sqrt{2}}{2}$ is not covered by any of these intervals.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose the contrary that means for some rational number $\\frac{p}{q}$, we have\n$$\n\\left| \\frac{\\sqrt{2}}{2} - \\frac{p}{q} \\right| \\le \\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$$\nHowever, as the LHS equals $\\frac{|q^2 - 2p^2|}{2q^2}$, the above inequality reduces to $|q^2 - 2p^2| < 1$, which implies that $q^2 = 2p^2$ and $\\sqrt{2} = \\frac{p}{q}$ is a rational number, which is a contradiction. □",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40727,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the inequality\n$$\nx^2\\sqrt{1+2y^2} + y^2\\sqrt{1+2x^2} \\ge xy(x+y+\\sqrt{2})\n$$\nholds for any two real numbers $x$ and $y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the quadratic-arithmetic mean inequality, we have\n$$\n1 + 2t^2 \\ge \\frac{1}{2}(1 + t\\sqrt{2})^2.\n$$\nIt follows that\n$$\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$$\nThis is what we need to prove. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40728,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles acute triangle with $\\ell$ is the bisector of angle $A$. A circle with center $O$ passes through $B, C$ and intersects the sides $AB, AC$ at $D, E$ respectively. Construct a parallelogram $DTEK$. Take $X, Y$ on $AB, AC$ respectively such that $AXKY$ is also a parallelogram. Construct $P, Q$ on $XY$ such that $TP \\parallel AC$ and $TQ \\parallel AB$. Prove that $TO$ passes through the center of the circumcircle of triangle $TPQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $BCED$ is cyclic, we have two similar triangles $ABC$ and $AED$. On the other hand\n$$\n\\angle TBC = \\angle TDE = \\angle KED \\text{ and } \\angle TCB = \\angle TED = \\angle KDE.\n$$\nThus two points $T, K$ are corresponding in the above pair of similar triangles, it can be deduced that\n$$\n\\angle TAB = \\angle KAE \\text{ and } \\angle TAC = \\angle KAB\n$$\nimplies that $AT, AK$ are symmetric through $\\ell$.\n\n*Lemma*: if two rays $Ox', Oy'$ are isogonal in angle $xOy$ then every pair of isogonal lines $Oa, Ob$ in angle $xOy$ are also isogonal in angle $x'Oy'$. We will use this idea to solve the problem. Let $J$ be the center of $(TPQ)$ and $H$ be the projection of $T$ onto $PQ$. Then, according to the symmetry of the altitude and the line joining the center, we immediately have $TJ, TH$ isogonal in $\\angle PTQ$. On the other hand,\n$$\n\\angle PTD = \\angle ACT = \\angle ABT = \\angle QTE\n$$\nso lines $TP, TQ$ are isogonal in angle $\\angle DTE$ and then $TJ, TH$ are also isogonal in $\\angle DTE$. To prove three points $J, T, O$ are collinear, we will show that lines $TO, TH$ are isogonal in $\\angle DTE$. $(*)$\n\nDenote $S$ as the intersection of $DE, BC$, then by applying Brocard's theorem to the completed quadrilateral $BCED, AS$ we have $TO \\perp AS$. Draw $Ax \\perp TH$ then $Ax \\parallel XY$. Notice that $AXKY$ is a parallelogram, so $AK$ passes through the midpoint of $XY$, leads to\n$$\n(Ax, AK, AB, AC) = -1.\n$$\nFurthermore, $A(ST, BC) = -1$ so consider the symmetry over the line $\\ell$, we have $AB \\leftrightarrow AC$, $AT \\leftrightarrow AK$ and take $AS \\leftrightarrow Ax'$. Then\n$$\n(Ax', AK, AB, AC) = -1,\n$$\nso $Ax \\equiv Ax'$ or $AS, Ax$ is symmetric about $\\ell$. Through $T$, draw lines $Tb, Tc$ perpendicular to $AB, AC$, then two sets of four lines\n$$\n(AS, Ax, AB, AC) \\text{ and } (TO, TH, Tb, TC)\n$$\nare perpendicular respectively. This implies that $TO, TH$ are isogonal in $\\angle bTC$. Finally, since $\\angle ADT = \\angle AET$, we have $Tb, Tc$ are also isogonal in $\\angle DTE$ and deduce to $(*)$ is true. $\\Box$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40729,
    "subject": "Mathematics (Multi-modal)",
    "question": "One has $2022$ yellow, $2022$ red and $2022$ blue sticks. It is known that if we pick three sticks of different color, then it is possible to construct a triangle using these sticks. Prove that there exists a color such that: if we pick $3$ sticks of that color, then it is possible to construct a triangle using these sticks.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40730,
    "subject": "Mathematics (Multi-modal)",
    "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?",
    "options": [],
    "answer": "No",
    "solution": "But both $1$ and $16$ can appear only as products, which forces a column of $1$'s and a column of $2$'s. And this leaves only three possible row sums, contradiction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40731,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integer $n > 1$ such that: there exist some real coefficient polynomial $P(x)$ of degree $n$, having the leading coefficient as $1$ and there exist distinct real numbers $r, s, t$ with the sum is $-2023$ and $P(k) \\in \\{r, s, t\\}$ for all $k = 1, 2, 3, \\dots, 3n - 1, 3n$.",
    "options": [],
    "answer": "2",
    "solution": "For $n = 2$, we have $P(k) \\in \\{r, s, t\\}$ with $k = 1, 2, 3, 4, 5, 6$. Since $\\deg P = 2$, there are no more than $2$ value of $k$ such that $P(k) = r$. Similarly for $P(k) = s$, $P(k) = t$. From this it follows that each of the above equations must have exactly $2$ solutions. Notice that all three equations have the same coefficients $x^2$ and $x$ so the sum of the solutions of them are equal and should be $7$. Then the $6$ solutions above will be divided in pairs $(1, 6)$, $(2, 5)$, $(3, 4)$. We can assume\n$$\n\\begin{cases}\nP(x) - r = (x - 1)(x - 6) \\\\\nP(x) - s = (x - 2)(x - 5) \\\\\nP(x) - t = (x - 3)(x - 4)\n\\end{cases}\n$$\nfrom this, it follows that\n$$\n3P(x) - (r + s + t) = (x - 1)(x - 6) + (x - 2)(x - 5) + (x - 3)(x - 4)\n$$\nor $3P(x) + 2023 = 3x^2 - 21x + 28$ so $P(x) = x^2 - 7x - 665$ and $r, s, t$ respectively are $-671, -675, -677$.\n\nNext, suppose there exists $P(x)$ of degree $n \\ge 3$ satisfying the given problem, without loss of generality suppose $r < s < t$. According to the argument above, each equation $P(x) - r$, $P(x) - s$, $P(x) - t$ there will be exactly $n$ distinct solutions from $\\{1, 2, 3, \\dots, 3n\\}$. In addition, according to Viete's theorem, since each equation has at least the first three coefficients in common. First the sum of the solutions and the sum of the squares of their solutions must be equal.\n\nConsidering the function $f(x) = P(x) - r$ has $n$ distinct solutions, according to the mean-value theorem, $f'(x) = P'(x)$ must have $n-1$ distinct roots, denoted by $c_1 < c_2 < \\dots < c_{n-1}$. The equation $P(x) = r$ have unique root on each interval $(-\\infty; c_1), (c_1, c_2), \\dots, (c_{n-1}, +\\infty)$. The same for $P(x) = s$, $P(x) = t$. We have the following two cases:\n![](attached_image_1.png)\n\n**Case 1.** If $n$ is odd then it is easy to see in the first interval, $P(x)$ increasing so $P(x) = r$, $P(x) = s$, $P(x) = t$ will take the roots $1, 2, 3$. in that other. In the next interval, the function is decreasing so they will take the roots of $6, 5, 4$ respectively, and so on, to the end of the interval will end up with $3n - 2, 3n - 1, 3n$. Then, it is easy to see that the sum of the solutions of the three equations in the first interval $n-1$ are equal, but in the last interval, each equation has a different solution, so the sum of their solutions is different, contradiction.\n\n**Case 2.** If $n$ is even, put $n = 2m$ then the three equations will have solutions of $1, 2, 3, \\dots, 6m$ and similar to the above argument, $P(x) = r$ there will be solutions $\\{1, 6, 7, 12, \\dots, 6m - 5, 6m\\}$. The sum of the squares of these numbers will be\n$$\n\\sum_{k=1}^{m} (6k-5)^2 + (6k)^2 = \\sum_{k=1}^{m} (72k^2 - 60k + 25) = m(24m^2 + 6m + 7).\n$$\nOtherwise, the sum of squares of all $6m$ numbers is\n$$\n\\frac{6m(6m+1)(12m+1)}{6} = m(6m+1)(12m+1)\n$$\nso each equation must whose sum of squares of the solutions is $\\frac{1}{3}$ of this value. Thus\n$$\n3m(24m^2 + 6m + 7) = m(6m + 1)(12m + 1)\n$$\nor\n$$\n72m^2 + 18m + 21 = 72m^2 + 18m + 1,\n$$\ncontradiction. This shows that in all cases we cannot have a polynomial $P(x)$ satisfying the problem.\n\nSo the only positive integer that satisfies the problem is $n = 2$. □",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40732,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all function $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$ such that for $x, y, z > 0$ pairwise distinct then\n$$\nf(x)^2 - f(y)f(z) \\le f(xy)f(y)f(z)[f(yz) - f(zx)].\n$$",
    "options": [],
    "answer": "All constant functions f(x) = c for any positive constant c.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40733,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$ and $z$ be positive reals such that $xyz = 1$. Find the largest possible value of the constant $C$ such that the inequality\n$$\n\\left(\\frac{x}{1+x}\\right)^2 + \\left(\\frac{y}{1+y}\\right)^2 + \\left(\\frac{z}{1+z}\\right)^2 \\ge C\n$$\nmust hold under the described condition.",
    "options": [],
    "answer": "3/4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40734,
    "subject": "Mathematics (Multi-modal)",
    "question": "At Hadi's birthday party, $2023$ friends came to give him $2023$ different gifts. Hadi wants to know exactly who gave which gifts, so he came up with a way: each time, choose a group of $11$ people and ask them what gifts they gave. When responding, the group only reported their set of gifts (not knowing exactly which gifts from whom). What is the minimum number of questions that Hadi need to ask?",
    "options": [],
    "answer": "337",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40735,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with $AB < AC$. Let $I$ be its incenter and $\\Gamma$ be its circumcircle. Let $M$ be the midpoint of $BC$, $K$ the midpoint of arc $BC$ not containing $A$, $L$ the midpoint of arc $BC$ containing $A$ and $J$ the reflection of $I$ with respect to the line $KL$. The line $LJ$ cuts $\\Gamma$ again at $T$. The line $TM$ cuts $\\Gamma$ again at $S$. Prove that the points $S$, $I$, $M$ and $K$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40736,
    "subject": "Mathematics (Multi-modal)",
    "question": "Mohammed has a chess board, colored white and black in the usual way and mentioned a point, which lies strictly inside one of the 64 cells. Ahmed can draw any closed broken line without self-intersections and ask Mohammed whether his point is inside or outside the broken line. How many questions does Ahmed need to ask so that he can find out whether the point is in a black or in a white cell?",
    "options": [],
    "answer": "2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40737,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist $100$ points on the plane such that the pairwise distances between them are pairwise distinct consecutive integer numbers larger than $2022$?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40738,
    "subject": "Mathematics (Multi-modal)",
    "question": "A rhombus of side length 1 and internal angle 60° is called a diamond. Suppose that a regular hexagon of side length $n$ is dissected into diamonds. Prove that each main diagonal of the hexagon halves exactly $n$ diamonds.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Divide the hexagon into $6n^2$ triangular cells (in each dissection every diamond will consist of two of them) and consider a half $H$ of the hexagon (on one side on a fixed main diagonal $d$).\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40739,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all non-constant polynomials $P(x)$ with real coefficients that satisfy\n$$\nP(x^3 - 7x) = P(x - 7)P(x - 8)P(x - 3)\n$$\nfor all $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "P(x) = ±(x + 6)^n for any positive integer n",
    "solution": "Let $n = \\deg P(x) \\ge 1$ and $a \\ne 0$ be the leading coefficient of $P$, then comparing the leading coefficients of both sides to get $a = \\pm 1$. Note that if $P(x)$ satisfies then so does $-P(x)$; without loss of generality, we assume that $a = 1$.\n\nFirst, let solve the problem when $n = 1$, consider $P(x) = x + b$. Substituting into the given condition,\n$$\nx^3 - 7x + b = (x - 7 + b)(x - 8 + b)(x - 3 + b).\n$$\nComparing the coefficient of degree 2, we have $0 = b - 7 + b - 8 + b - 3$ so $b = 6$. Therefore, $P(x) = x + 6$, which is a solution.\n\nNow, for any $n \\ge 1$, let $P(x) = (x + 6)^n + Q(x)$ with $\\deg Q < n$. If $Q(x) \\equiv 0$ then we have $P(x) = (x + 6)^n$, which satisfies since\n$$\n(x^3 - 7x + 6)^n = (x - 1)^n (x - 2)^n (x + 3)^n.\n$$\nNow assume that $Q(x) \\ne 0$ and put $\\deg Q = m < n$. Substituting in the given condition then we get\n$$\n(x^3 - 7x + 6)^n + Q(x^3 - 7x) = [(x - 1)^n + Q(x - 7)] [(x - 2)^n + Q(x - 8)] [(x + 3)^n + Q(x - 3)]\n$$\nThen expanding and simplifying, we get\n$$\n\\begin{align*} Q(x^3 - 7x) &= Q(x - 7)Q(x - 8)Q(x - 3) \\\\ &\\quad + (x - 1)^n Q(x - 8)Q(x - 3) + (x - 2)^n Q(x - 7)Q(x - 3) \\\\ &\\quad + (x + 3)^n Q(x - 7)Q(x - 8) + (x - 1)^n (x - 2)^n Q(x - 3) \\\\ &\\quad + (x - 2)^n (x + 3)^n Q(x - 7) + (x + 3)^n (x - 1)^n Q(x - 8). \\end{align*}\n$$\nComparing the degree of both sides, $3m = 2n + m$ or $m = n$, a contradiction.\n\nFrom these arguments, one can conclude that all solutions of the given condition are $P(x) = \\pm (x + 6)^n$ for all positive integers $n$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40740,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the maximum value of $k$ such that: for all $a, b, c$ are sidelengths of some triangle, then\n$$\nk \\cdot \\frac{3(a^2 + b^2 + c^2)}{(a + b + c)^2} + \\sqrt{\\frac{ab + bc + ca}{a^2 + b^2 + c^2}} \\leq k + 1.\n$$",
    "options": [],
    "answer": "2 - sqrt(2)",
    "solution": "The given inequality can be written as\n$$\n2k \\cdot \\frac{(a^2 + b^2 + c^2) - (ab + bc + ca)}{(a + b + c)^2} \\leq \\frac{\\sqrt{a^2 + b^2 + c^2} - \\sqrt{ab + bc + ca}}{\\sqrt{a^2 + b^2 + c^2}}\n$$\n$$\n2k \\cdot \\frac{(a^2 + b^2 + c^2) - (ab + bc + ca)}{(a + b + c)^2} \\leq \\frac{(a^2 + b^2 + c^2) - (ab + bc + ca)}{\\sqrt{a^2 + b^2 + c^2}((a^2 + b^2 + c^2) + (ab + bc + ca))}.\n$$\nNote that\n$$\n(a^2 + b^2 + c^2) - (ab + bc + ca) = \\frac{1}{2} \\left[ (a - b)^2 + (b - c)^2 + (c - a)^2 \\right] \\geq 0\n$$\nso\n$$\n2k\\sqrt{a^2 + b^2 + c^2} \\left( \\sqrt{a^2 + b^2 + c^2} + \\sqrt{ab + bc + ca} \\right) \\leq (a + b + c)^2.\n$$\nConsider $a = b = 1$ and $c \\to 0^+$ (isosceles triangle with the base arbitrary small) then $\\text{LHS} \\to 2k \\cdot \\sqrt{2}(\\sqrt{2} + 1)$ and $\\text{RHS} \\to 4$. Thus\n$$\nk \\leq \\frac{4}{2\\sqrt{2}(\\sqrt{2} + 1)} = 2 - \\sqrt{2}.\n$$\nFor $k = 2 - \\sqrt{2}$ then we need to prove that\n$$\n(4 - 2\\sqrt{2})\\sqrt{a^2 + b^2 + c^2} \\left( \\sqrt{a^2 + b^2 + c^2} + \\sqrt{ab + bc + ca} \\right) \\leq (a + b + c)^2.\n$$\nDenote $x = \\sqrt{a^2 + b^2 + c^2}$ and $y = \\sqrt{ab + bc + ca}$ then the above can be written as\n$$\n(4 - 2\\sqrt{2})x(x + y) \\leq x^2 + 2y^2\n$$\n$$\nx^2 + 2(2 + \\sqrt{2})xy - (2 + \\sqrt{2})^2 y^2 \\leq 0.\n$$\nThis equivalent to $x \\leq y\\sqrt{2}$ or $a^2 + b^2 + c^2 \\leq 2(ab + bc + ca)$. The last inequality is true for $a, b, c$ are sidelengths of triangle since it can be written as\n$$\na(b + c - a) + b(c + a - b) + c(a + b - c) \\geq 0.\n$$\nHence, $k_{\\max} = 2 - \\sqrt{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40741,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the board, there are written numbers $1, 2, 3, \\dots, 2023^{2024}$. At each step, it is allowed to erase any two numbers $a, b$ on the board and then replace them with $\\gcd(2024ab, a^2 + 254ab + b^2)$, keep doing this until there is only one number $x$ left. Find all possible values of $x$.\n\nOn the board, there are written numbers 1, 2, 3, ..., $2023^{2024}$. At each step, it is allowed to erase any two numbers $a, b$ on the board and then replace them with $\\text{gcd}(2024ab, a^2 + 254ab + b^2)$. Keep doing this until there is only one number $x$ left, find all possible values of $x$.",
    "options": [],
    "answer": "1",
    "solution": "Let $f(a, b) = \\text{gcd}(2024ab, a^2 + 254ab + b^2)$ then for 2 numbers $a, b$ on the board, the new number generated will be $f(a, b)$. We consider some cases:\n* If $a, b$ are same parity then $f(a, b)$ is even.\n* If $a, b$ are different from parity then $f(a, b)$ is odd.\nThis means that the number of odd numbers either remains the same or decreases by 2 units; but initially, there are odd numbers, so the final number $x$ must be odd. Suppose that $x > 1$, let $p$ be some prime divisor of $x$.\nNote that\n$$\n254 = 11 \\cdot 23 + 1 \\text{ and } 2024 = 2^3 \\cdot 11 \\cdot 23.\n$$\nIf $p = 11$, since $11|\\text{gcd}(2024ab, a^2 + 254ab + b^2)$ then $11|a^2 + ab + b^2$. Similarly, if $p = 23$ then there is also $23|a^2 + ab + b^2$. We have the following familiar lemma derived from Fermat's little theorem or quadratic residue:\n*Lemma.* If $p$ is a prime of form $3k + 2$ and $p|a^2 + ab + b^2$ then $p|a, b$.\nSince 11 has the form $3k + 2$, according to the above lemma, one can get $11|a, b$. Continuing like that, the previous numbers on the board that generated $a, b$ also be divisible by 11, making all the original numbers divisible by 11, contradiction. Similarly for the case $p = 23$.\nFinally, if $p \\neq 11, 23$ then $p|ab, p|a^2 + 506ab + b^2$ entails $p|ab, p|a^2 + b^2$. Obviously from $p|ab$, we get that $p$ is a divisor of one of $a$ or $b$, but $p|a^2 + b^2$ then $p|a, b$. We get the similar contradiction as the above argument.\nTherefore $x$ is odd and has no odd prime divisors, proving that $x = 1$. ☐",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40742,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all odd integer $n$ such that the number of integers $k$ with $0 < k < \\frac{n}{4}$ and $\\gcd(n, k) = 1$ is odd.",
    "options": [],
    "answer": "All odd integers that are prime powers with the base prime congruent to five or seven modulo eight, i.e., n = p^a with p ≡ 5 or 7 mod 8 and a ≥ 1.",
    "solution": "We claim that the only integers that work are prime powers $p^k$ in which $p \\equiv 5$ or $7$ modulo $8$. Let define $\\omega, \\Omega$ as function from the set of odd integers to $\\{0; 1\\}$ by\n$$\n\\omega(n) = \\begin{cases} 0 & \\text{if } n \\equiv 1,3 \\pmod 8 \\\\ 1 & \\text{if } n \\equiv 5,7 \\pmod 8 \\end{cases}\n$$\nand\n$$\n\\Omega(n) \\equiv |\\{k\\mid 0 < k < n/4, \\gcd(k, n) = 1\\}| \\pmod 2.\n$$\nThen we want to find $n$ such that $\\Omega(n) = 1$. It is easy to verify\n$$\n\\omega(ab) \\equiv \\omega(a) + \\omega(b) \\pmod 2 \\qquad (\\dagger)\n$$\nLet $n = p_1^{a_1} p_2^{a_2} \\dots p_k^{a_k}$ which $p_1, p_2, \\dots, p_k$ are distinct primes and $a_1, a_2, \\dots, a_k$ are integers. Note that $n = 1$ does not work so consider $k \\ge 1$.\n\nLemma: $\\Omega(n) \\equiv \\omega(n) + \\sum_{1 \\le i \\le k} \\omega\\left(\\frac{n}{p_i}\\right) + \\sum_{1 \\le i < j \\le k} \\omega\\left(\\frac{n}{p_i p_j}\\right) + \\dots + \\omega\\left(\\frac{n}{p_1 p_2 \\dots p_k}\\right) \\pmod 2$.\n\nProof: the idea to proof this lemma is similar to prove the Euler totient function. It is easy to check that $\\omega(n)$ is indeed the parity of $|\\{k\\mid 0 < k < n/4\\}| = \\lfloor \\frac{n}{4} \\rfloor$. To remove numbers divisible by some $p$, one can subtract from above the value\n$$\n|\\{pk\\mid 0 < k < n/(4p)\\}| = |\\{k\\mid 0 < k < (n/p)/4\\}| = \\omega\\left(\\frac{n}{p}\\right).\n$$\nBut that removes numbers divisible by $pq$ (for $p \\ne q$) twice so we will add $\\omega(pq)$ then continue as the principle of inclusion and exclusion, we add until $\\omega\\left(\\frac{n}{p_1 p_2 \\dots p_k}\\right)$. Note that the formula should be added and subtracted alternatively but since we consider modulo $2$ so all of the signs can be consider as plus. Back to the original problem, by applying the lemma and $(\\dagger)$, one can get\n$$\n\\begin{aligned}\n& \\Omega(n) + \\sum_{1 \\le i \\le k} \\omega(p_i) + \\sum_{1 \\le i < j \\le k} \\omega(p_i p_j) + \\dots + \\omega(p_1 p_2 \\dots p_k) \\\\\n& \\equiv \\omega(n) + \\left( \\sum_{1 \\le i \\le k} \\omega\\left(\\frac{n}{p_i}\\right) + \\sum_{1 \\le i \\le k} \\omega(p_i) \\right) + \\dots + \\left( \\omega\\left(\\frac{n}{p_1 p_2 \\dots p_k}\\right) + \\omega(p_1 p_2 \\dots p_k) \\right) \\\\\n& \\equiv \\binom{k}{0} \\omega(n) + \\binom{k}{1} \\omega(n) + \\binom{k}{2} \\omega(n) + \\dots + \\binom{k}{k} \\omega(n) \\\\\n& \\equiv 2^k \\omega(n) \\equiv 0 \\pmod 2.\n\\end{aligned}\n$$\nFrom this, one can conclude that\n$$\n\\begin{aligned}\n\\Omega(n) &\\equiv \\omega(1) + \\sum_{1 \\le i \\le k} \\omega(p_i) + \\sum_{1 \\le i < j \\le k} \\omega(p_i p_j) + \\dots + \\omega(p_1 p_2 \\dots p_k) \\\\\n&= \\sum_{i=1}^{k} \\left( \\binom{k-1}{0} + \\binom{k-1}{1} + \\dots + \\binom{k-1}{k-2} \\right) \\omega(p_i) \\\\\n&= 2^{k-1} \\sum_{i=1}^{n} \\omega(p_i) \\\\\n&= 2^{k-1} \\omega(p_1 p_2 \\dots p_k) \\pmod 2.\n\\end{aligned}\n$$\nThus if $k \\ge 2$ then $\\Omega(n) = 0$ and if $k = 1$, then $n = p^k$, in this case we can check that $\\Omega(n) = \\omega(p) = 1$ if and only if $p \\equiv 5$ or $7$ modulo $8$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40743,
    "subject": "Mathematics (Multi-modal)",
    "question": "With positive integer $n > 1$, let $4n - 3$ positive numbers are written on the board (not necessarily all different). It is known that any 4 pairwise different numbers from that list form an arithmetic progression. Prove that some number is written on the board at least $n$ times.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $d$ be the number of distinct values appear on the table, let them be\n$$\na_1 < a_2 < \\dots < a_d.\n$$\nAssuming by contradiction that no value appears at least $n$ times, then each value appears at most $n-1$ times. So we can count the amount of numbers on the board to get\n$$\n4n - 3 \\le d(n - 1) \\implies d \\ge \\frac{4n - 3}{n - 1} \\implies d \\ge 5.\n$$\nConsider the 5 arbitrary distinct values among them, denoted by $a < b < c < d < e$. According to the assumption, $(a, b, c, d)$ and $(a, b, c, e)$ both form arithmetic progressions. Then, we have $b - a = d - c = e - c$, implies that $d = e$, which is clearly a contradiction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40744,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive number $m$ such that there exists unique function $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$ such that\n$$\nf(x + f(y)) = f(x) + \\frac{mx^5 f(y)}{f(x)^2} + f(y^2), \\forall x, y > 0.\n$$",
    "options": [],
    "answer": "2",
    "solution": "If there exist $y_0$ such that $f(y_0) < y_0^2$ then taking $x = y_0^2 - f(y_0)$ to get\n$$\nf(x + f(y_0)) = f(y_0^2).\n$$\nFrom this, one can get\n$$\nf(x) + \\frac{mx^5 f(y_0)}{f(x)^2} = 0,\n$$\nwhich is contradiction. So $f(y) \\ge y^2$ for $y > 0$. Put $x = y = 1$ then\n$$\nf(1 + f(1)) = 2f(1) + \\frac{m}{f(1)} \\ge (1 + f(1))^2.\n$$\nSince $f(1) \\ge 1$ then we have $m \\ge f(1) + f(1)^3 \\ge 2$. So the least value of $m$ is 2.\nFor $m = 2$, it is easy to verify that $f(1) = 1$ and\n$$\nf(x + f(y)) = f(x) + \\frac{2x^5 f(y)}{f(x)^2} + f(y^2), \\quad \\forall x, y > 0.\n$$\nSubstitute $y = 1$ then $f(x + 1) = f(x) + \\frac{2x^5}{f(x)^2} + 1$ for all $x > 0$.\nPut $x = 1$ then $f(2) = f(1) + \\frac{2}{f(1)} + 1 = 4$. Continue to put $x = 2, 3, \\dots$ then by induction, one can get $f(n) = n^2$ for all positive integers $n$.\nPut $x = n$ into the condition then\n$$\nf(x + n^2) = f(x) + \\frac{2x^5 n^2}{f(x)^2} + n^4 \\geq (x + n^2)^2.\n$$\nBy factoring this inequality, one can get\n$$\n(f(x) - x^2) \\left(1 - \\frac{2x \\cdot n^2 \\cdot (f(x) + x^2)}{f(x)^2}\\right) \\geq 0, \\quad \\forall x > 0.\n$$\nIf $f(x) = x^2$ for all $x > 0$ then we can check this is a solution. Otherwise, there exists $x_0$ such that $f(x_0) > x_0^2$ then we must have\n$$\n1 - \\frac{2x_0 \\cdot n^2 \\cdot (f(x_0) + x_0^2)}{f(x_0)^2} \\geq 0, \\quad \\forall n \\in \\mathbb{Z}^+,\n$$\nwhich implies that\n$$\nn^2 \\leq \\frac{f(x_0)^2}{2x_0(f(x_0) + x_0^2)}.\n$$\nThis is a contradiction when we take $n \\to +\\infty$. Thus for $m = 2$, function $f(x) = x^2$ is the unique function satisfying the condition.\nTherefore, the least value of $m$ that need to find is 2. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40745,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n > 1$, let $1 = a_1 < a_2 < \\dots < a_t = n - 1$ be all the positive integers that are less than $n$ and coprime to $n$. Determine all values of $n$ for which there is no index $i \\in \\{1, \\dots, t-1\\}$ satisfying $3 \\mid a_i + a_{i+1}$.",
    "options": [],
    "answer": "2, 4, 10",
    "solution": "It is easy to check that for $n = 2, 4, 10$ the sequences are $(1), (1, 3), (1, 3, 7, 9)$ respectively. Thus these numbers are solution of the given problem. Considering $n \\geq 3$, if $n$ is odd then $\\text{gcd}(n, 2) = 1$ so clearly $a_1 = 1, a_2 = 2$ and their sum is divisible by $3$, which not satisfy. Hence, $n$ is even. It is easy to check that $n = 6$ or $n = 8$ also do not work. Now we consider $n \\geq 12$ and separate in the following cases:\n\n1. If $n = 12k$ for $k \\geq 1$ then consider $a = 6k + 1, b = 6k - 1$. Since $a$ is odd then $\\text{gcd}(n, a) \\mid \\text{gcd}(n, 2a) = \\text{gcd}(12k, 12k + 2) = 2$ so $\\text{gcd}(n, a) = 1$. Similarly, we have $\\text{gcd}(n, b) = 1$ so $a, b$ are two consecutive terms in the sequence of $n$, but $a + b = 12k$ which is divisible by $3$. This case does not give any solution.\n\n2. If $n = 12k + 2$ for $k \\geq 1$ then consider $a = 6k + 3, b = 6k + 5$. Let denote $d = \\text{gcd}(n, a)$ then $d|2a = 12k + 6$ so $d|4$, but $d$ is odd then $d = 1$, thus $\\text{gcd}(n, a) = 1$. Similarly, denote $d' = \\text{gcd}(n, b)$ then $d'|2b = 12k + 10$ so $d'|8$, but $d'$ is odd then $d' = 1$, thus $\\text{gcd}(n, b) = 1$. Now $a, b$ are consecutive terms in the sequence and $3 \\mid a + b$, not satisfies.\n\n3. If $n = 12k + 4$ for $k \\ge 1$ then consider $a = 6k + 1, b = 6k - 1$.\n\n4. If $n = 12k + 6$ for $k \\ge 1$ then consider $a = 6k + 1, b = 6k - 1$.\n\n5. If $n = 12k + 8$ for $k \\ge 1$ then consider $a = 6k + 3, b = 6k + 1$.\n\n6. If $n = 12k + 10$ for $k \\ge 1$ then consider $a = 6k + 1, b = 6k - 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40746,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute non-isosceles triangle and inscribed in circle $(O)$ with the median line $A$ intersecting $(O)$ at $D$. Let $G$ be a point symmetrical to $A$ through $BC$. Suppose $GB, GC$ intersect $(O)$ respectively at $E, F$. The inscribed circle of triangle $AEF$ touches $EF$ at $J$.\n1. Prove that $GO, GD$ are symmedians of triangles $GAD$ and $GEF$.\n2. Prove that the two rays $AJ, AD$ are symmetric respect to $AG$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1) Let $T$ be the intersection of the tangent at $A$ of $(O)$ with $BC$. Then, because $ABDC$ is a harmonic quadrilateral, $TD$ is also tangent to $(O)$. It follows that $A, D, G$ both belong to the circle of center $T$, which is also the Apollonius circle at vertex $A$ of triangle $ABC$. On the other hand, $(T)$ and $(O)$ are orthogonal so $OA, OD$ are tangent to $(T)$, then it is easy to check that $GO$ is the symmedian of triangle $ADG$. Let $K$ be the second intersection of $AG$ and $(O)$. Then,\n$$\n\\angle BEK = \\angle BAK = \\angle BGK \\implies KE = KG.\n$$\nSimilarly, $KF = KG$ so $KE = KF$ implies that $K$ is the midpoint of the minor arc $EF$ of $(O)$. It follows that $AG$ is the angle bisector of $\\angle EAF$.\n\nLet $M$ be the midpoint of $BC$ and take $A' \\in (O)$ such that $AA' \\parallel BC$ then since $MA' = MA = MG$ implies that $G, M, A'$ are collinear. On the other hand, since\n$$\nA'(AD, BC) = -1 \\text{ and } AA' \\parallel BC,\n$$\nwe have $A', M, D$ are collinear. From there, it can be deduced that $GD$ passes through $A'$. Since $K, O, A'$ are also collinear, $A'$ is the midpoint of the major arc $EF$ of the circle $(O)$. Then, $K$ is the center of the circumcircle of $GEF$ and $A'E \\perp KE$, $A'F \\perp KF$ so $A'$ is the intersection point of two tangent lines of $(K)$ at $E, F$. Therefore, $GD$ is the symmedian of triangle $GEF$.\n\n2) We will prove that $D$ is the tangent point of the external Mixlinear circle with respect to vertex $A$ in triangle $AEF$. Indeed, let $D', U, V$ be the tangent points of the external Mixtilinear circle $(\\omega)$ of triangle $AEF$ with $(O)$, $AE, AF$ respectively. Then, $G$ is the midpoint of $UV$, and $D'U, D'V$ respectively pass through the midpoints of the arc $AE$ (contains $F$) and $AF$ (contains $F$) of $(O)$. Note that\n$$\n\\angle ACB = \\angle GCB = \\angle BCF = \\angle BAF\n$$\nso $B$ is the midpoint of the arc $AE$ (contains $F$) of $(O)$ so $B, D', V$ are collinear. Similarly, $C, D', U$ are collinear. Hence, we have $UV \\parallel BC$. Since $(\\omega)$ and $(O)$ are tangent, so we have\n$$\n\\angle FD'V = \\angle FCD' + \\angle D'UV = \\angle FCD' + \\angle D'CB = \\angle BCF = \\angle FGV\n$$\nimplies that $D'FVG$ is cyclic. Similarly $D'EUG$ is also cyclic. Let $D'$ be the opposite ray of $D'G$ then\n$$\n\\angle FD'x = \\angle FVG = \\angle AVU = \\angle AUV = \\angle EDx\n$$\nso $D'$ passes through the midpoint of the major arc $EF$ of $(O)$, which is $A'$. From here, $D \\equiv D'$. Finally, let $I$ be the incenter of triangle $AEF$ and consider $\\Omega$ to be the inversion of center $A$ and power $AE \\cdot AF$, union to the reflection about $AG$. Then, according to the properties of this transformation which preserve the tangency and the isogonal properties, one can get:\n$$\n\\begin{gathered} \\Omega : (I) \\to (\\omega) \\text{ and } \\Omega : BC \\leftrightarrow (O) \\\\ \\implies \\Omega : J \\leftrightarrow D. \\end{gathered}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40747,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral inscribed in a circle ($O$) and some point $I$ lies inside $ABCD$. Consider the lines $d_1, d_2, d_3, d_4$ pass through the midpoints of segments $IA, IB, IC, ID$ respectively and perpendicular to lines $OA, OB, OC, OD$. Line $d_1$ cuts $d_2$ at $P$, line $d_2$ cuts $d_3$ at $Q$, line $d_3$ cuts $d_4$ at $R$ and line $d_4$ cuts $d_1$ at $S$. Suppose that the quadrilateral $PQRS$ is convex and points $I, O$ lie inside it. Prove that $PQRS$ circumscribes a circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $R$ be the radius of the circle ($O$) and let $M, K$ be the midpoints of $IO, AI$ respectively. Then, according to the property of the midline in triangle $AIO$, we have $MK = \\frac{AO}{2}$ and $MK \\parallel AO$. Thus, we immediately have $MK \\perp SP$ and $MK = \\frac{R}{2}$. Similarly for the other sides.\n\n![](attached_image_1.png)\n\nTherefore, the point $M$ is equidistant from the sides of the quadrilateral $PQRS$ and is also inside the quadrilateral (since $I, O$ lie inside it), so $M$ is the incenter of the quadrilateral $PQRS$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40748,
    "subject": "Mathematics (Multi-modal)",
    "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": "Detailed 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$ of the sequence is clearly also finite, let them be $p_1 < p_2 < \\dots < p_k$.\n\nSetting $m = p_1p_2\\cdots p_k$ then clearly $\\gcd(m, 3) = 1$, by using Euler's theorem, one can get $m \\mid 3^{\\varphi(m)} - 1$. We also have\n$$\n\\varphi(m) = (p_1 - 1)(p_2 - 1)\\cdots(p_k - 1)\n$$\nis a positive integer not divisible by $3$, so setting $n = \\varphi(\\varphi(m))$ then using Euler's theorem again, one could get\n$$\n\\varphi(m) \\mid 3^{\\varphi(\\varphi(m))} - 1 = 3^n - 1.\n$$\nNote that $3^x - 1 \\mid 3^y - 1$ where $x, y$ are positive integers such that $x \\mid y$. Hence,\n$$\n3^{\\varphi(m)} - 1 \\mid 3^{3^{n-1}} - 1.\n$$\nTherefore $m \\mid 3^{3^{n-1}} - 1$. From here, it follows that\n$$\na_n = (3^{3^{n-1}} - 1) + 3 \\equiv 3 \\pmod{m}.\n$$\nOn the other hand, clearly $a_n$ will have a prime divisor of the form $3h + 2$ so there exists $i$ with $1 \\le i \\le k$ such that $p_i \\mid a_n$. But $p_i \\mid m$ which implies that $p_i \\mid 3$, which is clearly a contradiction. Therefore, $(a_n)$ has infinitely many prime divisors.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40749,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the alphabet has only two letters: $A$ and $B$. Is it possible to choose one word of length 5, one word of length 6, ..., one word of length 30 such that any word of length 300 contains one of the chosen words as substring?",
    "options": [],
    "answer": "No",
    "solution": "The answer is NO.\nConsider all words of length 13 and continue them periodically in both directions to create the sequences. For details, take $x_1x_2\\dots x_{13}$ as some original word of length 13 then add $x_1, x_2, x_3, \\dots$ to the right and add $x_{13}, x_{12}, x_{11}, \\dots$ to the left will result:\n$$\n\\dots x_{11}x_{12}x_{13}(x_1 x_2 \\dots x_{13})x_1 x_2 x_3 \\dots\n$$\nThere are two identical original words as $AAA\\dots A$ and $BBB\\dots B$ will generate the unique sequences. For the other words, each of them will generate the same sequences with another 12 words as follows:\n$$\n(x_2x_3x_4\\dots x_{13}x_1),\\ (x_3x_4x_5\\dots x_{13}x_1x_2),\\ \\dots,\\ (x_{13}x_1x_2\\dots x_{12})\n$$\nSo in total, the number of distinct sequences is\n$$\n\\frac{2^{13}-2}{13}+2>600.\n$$\nBy the definition of the sequences, one can check that any word of length at least 13 may appear only in 1 sequence. Each word of length $n < 13$ may appear in at most $2^{13-n}$ sequences, since we fix $n$ letters and choose another $13-n$ other letters with $2^{13-n}$ ways to form a word of length 13 (some of them may coincide). And note that\n$$\n(2^8 + 2^7 + \\dots + 2^0) + 17 = 2^9 + 16 < 600\n$$\nso we will have available sequences.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40750,
    "subject": "Mathematics (Multi-modal)",
    "question": "A mail carrier delivers mail to the 19 houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible?",
    "options": [],
    "answer": "351",
    "solution": "Consider 19 consecutive squares and color them yellow (corresponding to houses that receive letters) and white (corresponding to houses that do not receive letters). Then, according to the assumption, there are no 2 consecutive yellow cells and no 3 consecutive white cells. We need to calculate the number of such coloring ways and in general, let $S_n$ be the number of coloring ways in case there are a total of $n$ squares.\n\nNotice that for any two adjacent squares, there are only three cases: $YW$, $WY$, $WW$. Let $a_n$ be the number of coloring ways where the last 2 cells are $YW$, let $b_n$ be the number of coloring ways where the last 2 cells are $WY$ and let $c_n$ be the number of ways to fill the last 2 cells with $WW$. We will establish some recurrence relationships for the sequences as follows:\n\n* With $(a_n)$: notice that before $YW$ it must be $YW$ or $WW$ so $a_{n+2} = a_n + c_n$.\n* With $(b_n)$: before $WY$ it must be $YW$ or $WY$ so $b_{n+2} = a_n + b_n$.\n* With $(c_n)$: before $WW$ it can only be $WY$ so $c_{n+2} = b_n$.\n\nNote that $S_n = a_n + b_n + c_n$ and we also have\n$$\nc_n = b_{n-2},\\ a_{n+2} = b_{n-2} + a_n.\n$$\nFrom these, we get $b_n = a_{n+4} - a_{n+2}$ so replace it back with $b_{n+2} = a_n + b_n$ then\n$$\na_{n+6} = 2a_{n+4} - a_{n+2} + a_n.\n$$\nThus $S_n = a_n + (a_{n+4} - a_{n+2}) + (a_{n+2} - a_n) = a_{n+4}$ so we immediately have the formula for $(S_n)$ is\n$$\nS_{n+2} = 2S_n - S_{n-2} + S_{n-4}, \\ \\forall n \\ge 5.\n$$\nIt is easy to check that $S_1 = 2$, $S_2 = 3$, $S_3 = 4$, $S_4 = 5$ so substituting into above formula to get $S_{19} = 351$. This is also the number of ways to deliver the mail that need to find. $\\boxed{351}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40751,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ be real numbers such that $x \\ge y \\ge z \\ge 0$ and $2x + y + 2z = 5$. Prove that\n$$\n5 \\le x^2 + z^2 + xy + yz + zx \\le \\frac{25}{4}.\n$$\nWhen does the equality case hold?",
    "options": [],
    "answer": "Lower bound equality holds when x = y = z = 1. Upper bound equality holds when y = z = 0 and x = 5/2.",
    "solution": "Let $P = x^2 + z^2 + xy + yz + zx$ then we have\n$$\n25 = (2x + y + 2z)^2 = 4x^2 + y^2 + 4z^2 + 4xy + 4yz + 4zx = 4P + y^2 + 4zx.\n$$\nNote that $y^2 + 4zx \\ge 0$ so $4P \\le 25$ which implies that $P \\le \\frac{25}{4}$. The equality case is $y = z = 0$ and $x = \\frac{5}{2}$.\n\nContinue, note that $(y - x)(y - z) \\le 0$ so $y^2 + zx \\le xy + yz$. Thus\n$$\nP \\ge x^2 + z^2 + y^2 + 2zx = (x + z)^2 + y^2.\n$$\nUsing Cauchy-Schwarz inequality, one can get\n$$\n(2^2 + 1^2)((x+z)^2 + y^2) \\ge (2x + 2z + y)^2 = 25.\n$$\nThus $(x+z)^2 + y^2 \\ge \\frac{25}{5} = 5$ which implies that $P \\ge 5$.\nThe equality case is $x = y = z = 1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40752,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute, non isosceles triangle with circumcircle $(O)$ and circumradius $R$. Denote $M$, $N$, $P$ as midpoints of segments $BC$, $CA$, $AB$ respectively. Let $(\\omega)$ be the circle passes through $A$, $O$ and tangent to $OM$. Circle $(\\omega)$ meets $AB$, $AC$ at $E$, $F$. Denote $I$ as midpoint of segment $EF$ and $K$ as the intersection of $EF$, $NP$. Prove that $R = 2IK$ and $IMO$ is an isosceles triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let circle $(\\omega)$ cuts $(O)$ again at $D$ and denote $O_1$ as center of $(\\omega)$. Since $AD$ is the radical axis of $(\\omega)$, $(O)$ then $OO_1 \\perp AD$. But $OO_1 \\perp OM$ since $OM$ is tangent to $(\\omega)$ leads to $AD \\parallel OM$ and then $AD \\perp BC$.\n\nThis means that $AD$, $AO$ are isogonal in $\\angle EAF$ then $OD \\parallel EF$, which implies that $EFOD$ is an isosceles trapezoid. Note that $OP \\perp AE$, $ON \\perp AF$ and $O \\in (AEF)$ so by Simson line property, $K$ is the projection of $O$ on $EF$. Denote $L$ as midpoint of $OD$ then $IL \\perp OD$ so $OKIL$ is a rectangle. Thus\n$$\nIK = OL = \\frac{1}{2}OD = \\frac{R}{2}.\n$$\n\nNow consider the spiral similarity namely $\\Omega$ of center $D$, maps $E \\to B$ and $F \\to C$ then $EF \\to BC$ and\n$$\n\\Omega : \\triangle DEF \\to \\triangle DBC.\n$$\nSince $O_1$, $O$ are centers of $(DEF)$, $(DBC)$ so $\\Omega : O_1 \\to O$. Thus two triangles $DO_1O$ and $DIM$ are similar, but $O_1O = O_1D$ then $ID = IM$. By the isosceles trapezoid, one can check that $ID = IO$ also, which implies that $IM = IO$ then $IMO$ is an isosceles triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40753,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a non-isosceles triangle $ABC$, inscribed in circle $(O)$ with $X, Y, Z$ are midpoints of the major arcs $BC, CA, AB$ of $(O)$. Let $D, E, F$ be the tangent points of the incircle $(I)$ of $ABC$ on $BC, CA, AB$ respectively. Suppose that line $XE$ intersects $(O)$ again at $M$ and cuts $YD$ at $P$; line $XF$ intersects $(O)$ again at $N$ and intersects $ZD$ at $Q$. Let $T$ be the intersection of $QE, PF$ and $XT$ cuts $(O)$ again at $K$. Prove that the circumcircle of triangle $AOK$ bisects the segment $MN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $I_a, I_b, I_c$ be the ex-center of angle $A, B, C$ in triangle $ABC$ respectively. Then, it is clear that $A, B, C$ are the feet of the altitude in triangle $I_a I_b I_c$, and $X, Y, Z$ are the midpoints of the three sides of triangle $I_a I_b I_c$. Thus $AX \\parallel YZ$. On the other hand, $AX \\perp AI$ and $AI \\perp EF$ imply that $EF \\parallel YZ$. Similarly, we get two triangles $DEF$ and $XYZ$ have three corresponding sides parallel.\n![](attached_image_1.png)\n\nAccording to Thales theorem then\n$$\n\\frac{PD}{PY} = \\frac{DE}{XY} \\text{ and } \\frac{QD}{QZ} = \\frac{DF}{XZ},\n$$\nbut two triangles $DEF$, $XYZ$ are similar so\n$$\n\\frac{DE}{XY} = \\frac{DF}{XZ} \\implies \\frac{PD}{PY} = \\frac{QD}{QZ}.\n$$\nThis shows that $YZ \\parallel PQ$ according to Thales theorem, hence $PQ \\parallel EF$. Using the trapezoidal lemma, $XT$ will bisect the segments $EF, PQ$. On the other hand $XA \\parallel EF$ leads to $X(AT, EF) = -1$. Projected onto $(O)$ then one can get the harmonic quadrilateral $AMKN$. If the tangent line of $(O)$ at $A, K$ intersects at $L$, it is clear that $L \\in MN$. Points $A, O, K, L$ belong to the circle of diameter $LO$, so if $(LO)$ intersects $MN$ at $H$, we get $\\angle OHL = 90^\\circ$ and $H$ is the midpoint of $MN$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40754,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $\\angle BAC = 90^\\circ$ with the altitude $AH$ ($H \\in BC$). A circle $(\\omega)$ passes through $B, C$ and cuts the segments $AB, AC$ at $M, N$ respectively. Circle $(\\omega)$ also cuts the line $AH$ at $D, E$ ($D$ lies between $A, H$). Suppose that $DE = AH\\sqrt{5}$, prove that the circumcircle of triangle $HMN$ is tangent to $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Base on the power from $H$ to the circle $(\\omega)$, $HD \\cdot HE = HB \\cdot HC = AH^2$. Moreover, $HD + HE = DE = AH\\sqrt{5}$. Thus, the lengths $HD, HE$ will be the solutions of the quadratic equation\n$$\nx^2 - AH\\sqrt{5} \\cdot x + AH^2 = 0.\n$$\nSince $D$ is inside triangle $ABC$, $HD < AH$, entails $HE > AH$ and will have $HE > HD$. From there, solving the above equation, we get\n$$\nHD = \\frac{-1 + \\sqrt{5}}{2}AH \\text{ and } HE = \\frac{1 + \\sqrt{5}}{2}AH.\n$$\nHence,\n$$\nAD = AH - HD = \\frac{3 - \\sqrt{5}}{2}AH \\text{ and } AE = AH + HE = \\frac{3 + \\sqrt{5}}{2}AH.\n$$\n---\nIt follows that\n$$\nAD \\cdot AE = \\left(\\frac{3 - \\sqrt{5}}{2}\\right) \\left(\\frac{3 + \\sqrt{5}}{2}\\right) AH^2 = AH^2.\n$$\nFurthermore, according to the power from $A$ to the circle $(\\omega)$, $AD \\cdot AE = AM \\cdot AB = AN \\cdot AC$. Therefore $AH^2 = AM \\cdot AB = AN \\cdot AC$, which implies that $HM \\perp AB$, $HN \\perp AC$ so the quadrilateral $AMHN$ is a rectangle. Therefore, the circle circumscribing triangle $HMN$ is also the circle with diameter $AH$ so it will be tangent to $BC$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40755,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a, b, c \\ge 0$ with $a + b + c = 3$. Prove that\n$$\n(ab + c)(ac + b) \\le 4.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40756,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ be nonnegative numbers such that $a \\ge b \\ge c$. Prove that\n$$\na^3 + b^3 + c^3 - 3abc \\ge \\frac{9}{2}(a-b)(b^2-c^2).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Write the inequality as\n$$\n(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \\ge \\frac{9}{2}(a - b)(b - c)(b + c).\n$$\nSince $a + b + c \\ge \\frac{3}{2}(b + c)$, it suffices to show that\n$$\na^2 + b^2 + c^2 - ab - bc - ca \\ge 3(a - b)(b - c).\n$$\nThis is equivalent to the obvious inequality $(a - 2b + c)^2 \\ge 0$. The equality holds for $a = b = c$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40757,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1, A_2, \\dots, A_{20}$ be 20 distinct subsets of size 3 of the set $X = \\{1, 2, \\dots, 10\\}$. We say that a subset $S$ of $X$ is a covering subset if for every $1 \\le i \\le 20$, it holds that $S \\cap A_i \\ne \\emptyset$. What is the minimum possible value of $k$, such that there always exists a covering subset of size $k$?",
    "options": [],
    "answer": "6",
    "solution": "First, we select all the 3-element subsets of $X_1 = \\{1, 2, 3, 4, 5\\}$ and of $X_2 = \\{6, 7, 8, 9, 10\\}$. Then, the number of subsets will be $\\binom{5}{3} + \\binom{5}{3} = 20$. Suppose there is a covering set of size $k \\le 5$, then one of the two sets $X_1, X_2$ will have no more than 2 elements, which means that there are 3 elements belonging to either $X_1$ or $X_2$ that are not selected. Clearly, the given condition is not satisfied. Thus, $k \\ge 6$.\n\nWe will prove that there always exists a covering set of size 6. We will show that there are 4 numbers $a, b, c, d \\in X$ of which at most 2 of these 4 numbers are selected into the given subsets. Suppose by contradiction that for every set of 4 numbers taken from $X$, there is always a subset of them chosen. We count in two ways the number of connections between the 4 sets of numbers $(a, b, c, d)$ taken from $X$ and the given subsets.\n\n* Counting by tuple $(a, b, c, d)$: there are at least $\\binom{10}{4} = 210$ connections.\n* Counting by subsets: each subset of 3 elements will belong to 7 different tuples of 4 numbers, there are 20 subsets are given, so the number of connections is $7 \\cdot 20 = 140 < 210$.\n\nFrom here we see that there is an absurdity, so there should always be 4 numbers $a, b, c, d$ as stated. At this point, we choose the remaining 6 numbers in $X$. Then, each subset of 3 elements in the 20 given sets will always have at least one number belonging to these 6 values. So the minimum value to find is $k_{\\min} = 6$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40758,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute triangle and $\\omega$ be its circumcircle. Points $F$ on $AC$, $E$ on $AB$, and $P$ and $Q$ on $\\omega$ are given such that\n$$\n\\angle AFB = \\angle AEC = \\angle APE = \\angle AQF = 90^\\circ.\n$$\nProve that the lines $BC$, $EF$, and $PQ$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40759,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let positive integer $n$ is given. The cube of size $2n + 1$ consists of $(2n + 1)^3$ unit cubes. Each cube is either colored green or orange. It is known that within any 8 unit cubes that form a cube $2 \\times 2 \\times 2$, there are at most 4 green cubes. Find the maximum number of green cubes.",
    "options": [],
    "answer": "4n^3 + 9n^2 + 6n + 1",
    "solution": "(Solution of Yousif Alkhalawi, IMO 2025 team's candidate)\nSince each cube $2 \\times 2 \\times 2$, there are at least 4 orange unit cubes (here we call it by square) so we want to minimize the number of orange squares.\n\n**Construction:** Let number the layer $n \\times n$ of cubes from $1 \\to n$ (bottom to top). For each layer, consider the coordinate of a single squares as $(x, y)$ for $1 \\le x, y \\le n$.\n* For the odd layers, we color squares at $(x, y)$ orange if $x, y$ are both even. The number of orange squares is $n^2$.\n* For the even layers, we color square at $(x, y)$ orange if $xy$ is even. The number of orange squares is $(2n+1)^2 - (n+1)^2 = n(3n+2)$.\n\nSAUDI ARABIAN IMO Booklet 2025\n---\n## Saudi Booklet 2025 — Page 22\n22\nSolution of Preselection tests\n![](attached_image_1.png)\n\nThe cube $2 \\times 2 \\times 2$ will have 1 orange square in odd layer and 3 orange squares in even layers so it always have 4 orange squares as desired. The total orange squares is $(n+1)n^2 + n^2(3n+2) = n^2(4n+3)$. Thus the number of green squares is\n$$\n(2n + 1)^3 - n^2(4n + 3) = 4n^3 + 9n^2 + 6n + 1.\n$$\n\nWe will prove this is the maximum number of green unit cubes by induction on $n$. For the base case $n=1$, in the cube of size 3, consider some 3 sub-cubes of size 2 at 3 opposite corners. The number of orange squares on these sub-cubes is at least $4 \\times 3 = 12$. Note that there is at most 1 square appears in all of 3 sub-cubes, and there are at most 3 other squares appear in 2 of 3 sub-cubes. So the number of orange square in big-cube is at least $12 - 2 - 3 = 7$.\n\nFor a cube of size $2n+1$, we consider the sub-cube of size $2n-1$ in the top corner, denote it as $\\Omega$ and the square in the opposite corner with it as $X$. By induction, $\\Omega$ need at least $4(n-1)^3 + 3(n-1)^3$ orange squares so we need to prove in the rest of the big-cube, there are at least $(4n^3+3n^2) - (4(n-1)^3 + 3(n-1)^2) = 12n^2 - 6n + 1$ orange squares. Note that to build up the original cubes, one can place $n^2$ sub-cubes of size 2 on each face of $\\Omega$ and one sub-cubes of size 2 touching $X$. So there are $3n^2+1$ sub-cubes leads to the sum of orange square is at least $12n^2+4$. To estimate the duplication, for each of $2n$ squares on the edge containing $X$, it can be in 2 sub-cubes of size 2 and for square $X$, it can be in 3 so the minimum is\n$$\n4(3n^2 + 1) - 3 \\times 2n - 1 \\times 3 = 12n^2 - 6n + 1.\n$$\nThis finishes the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40760,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(b, c)$ of positive integers, such that the sequence defined by $a_1 = b$, $a_2 = c$ and\n$$\na_{n+2} = |4a_{n+1} - 3a_n|, \\forall n \\ge 1\n$$\nhas only finite number of composite terms.",
    "options": [],
    "answer": "{(p, p), (3p, p), (11, 9), (13, 9)} where p is any prime",
    "solution": "Suppose firstly that there exists a $k \\ge 2$ such that $a_{k+1} > a_k$. Then by induction we have $a_{n+1} > a_n$ for all $n \\ge k$ (when taking out the modulus the expression does not change sign) and so $a_{n+2} = 4a_{n+1} - 3a_n$ for all $n \\ge k$. Solving the characteristic equation $t^2 - 4t + 3 = 0$ implies that\n$$\na_n = C_1 + C_2 \\cdot 3^n, \\forall n \\ge k.\n$$\nSince the sequence is strictly increasing, we must have $C_2 > 0$ and in particular any prime divisor $p$ of some $a_m \\ge 2$ will also be a prime divisor of $a_{m+k(p-1)}$ for any $k \\ge 0$ as $3^{m+k(p-1)} \\equiv 3^m \\pmod{p}$ by Fermat's Little theorem. In particular, infinitely many terms would be composite.\n\nTherefore $a_{k+1} \\le a_k$ for all $k \\ge 2$. Since $a_k > 0$, we must have $a_n = p$ where $p$ is a prime for all $n \\ge n_0$ with some positive integer $n_0$. Thus\n$$\np = |4p - 3a_{n-1}| \\implies a_{n-1} \\in \\{p, 3p\\}.\n$$\nThe former case gives us $(b, c) = (p, p)$ which indeed satisfies the condition. Let's consider the latter case: $a_{n-1} = 3p$, so we can find out that $a_{n-2} = 4p \\pm \\frac{p}{3}$. We consider some cases:\n\n* If there is no $a_{n-2}$, then $(b, c) = (3p, p)$ are the first two terms.\n* If $p$ is divisible by $3$, thus $p = 3$, $a_{n-1} = 9$, $a_{n-2} = 11$ or $13$, which implies $a_{n-3}$ being not integer.\n\nSo, finally we get $(b, c) \\in \\{(11, 9), (13, 9), (p, p), (3p, p)\\}$ where $p$ is any prime. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40761,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. In a family of finite sets, let a splitting element be an element that belongs to at least two of the sets and is omitted by at least two of the sets. Determine the maximum size of a family of subsets of $\\{1, \\dots, n\\}$ for which there is no splitting element.",
    "options": [],
    "answer": "n + 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40762,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all periodic sequences $a_1 a_2, \\dots$ of real numbers such that the following conditions hold for all $n \\ge 1$:\n$$\na_{n+2} + a_n^2 = a_n + a_{n+1}^2 \\quad \\text{and} \\quad |a_{n+1} - a_n| \\le 1.\n$$",
    "options": [],
    "answer": "All constant sequences d, d, d, … with any real d; and all alternating two-periodic sequences c, −c, c, −c, … with |c| ≤ 1/2.",
    "solution": "Answer: The sequences satisfying the conditions of the problem are:\n$$\nc, -c, c, -c, \\dots \\\\\nd, d, d, d, \\dots\n$$\nwhere $c \\in \\left[-\\frac{1}{2}, \\frac{1}{2}\\right]$ and $d$ is any real number.\n\nWe rewrite the first condition as\n$$\na_{n+2} + a_{n+1} = (a_{n+1} + a_n)(a_{n+1} - a_n + 1)\n$$\nIf there exists a positive integer $m$ such that $a_{m+1} + a_m = 0$, then from equation (1) we have $a_{n+1} + a_n = 0$ for all positive integers $n \\ge m$. By the fact that the sequence $(a_{i+1} + a_i)$ is periodic, we get $a_{i+1} + a_i = 0$ for every positive integer $i$. Thus the sequence $(a_i)$ is of the form $c, -c, c, -c, \\dots$ for some $|c| \\le \\frac{1}{2}$.\n\nNow suppose that $a_{n+1} + a_n \\ne 0$ for every positive integer $n$. Let $T$ be the period of the sequence. From equation (1) we have\n$$\n1 = \\prod_{i=1}^{T} \\frac{a_{i+2} + a_{i+1}}{a_{i+1} + a_i} = \\prod_{i=1}^{T} (a_{i+1} - a_i + 1)\n$$\nCombining with the second condition $|a_{i+1} - a_i| \\le 1$, we have $a_{i+1} - a_i + 1 > 0$. Using the AM-GM inequality we get\n$$\n1 = \\prod_{i=1}^{T} (a_{i+1} - a_i + 1) \\le \\left( \\frac{\\sum_{i=1}^{T} (a_{i+1} - a_i + 1)}{T} \\right)^T = 1.\n$$\nSo the equality holds, and thus we get\n$$\na_2 - a_1 = a_3 - a_2 = \\dots = a_{T+1} - a_T\n$$\nwhich means that $(a_i)$ is a constant sequence. So all sequences satisfying the conditions of the problem are those listed above. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40763,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that there exist $n$ consecutive positive integers whose sum is a perfect square.",
    "options": [],
    "answer": "All positive integers n such that, writing n = 2^e r with r odd, one has e = 0 or e is odd. Equivalently, all odd n and those even n with v2(n) odd; the excluded n are exactly those with v2(n) an even positive integer (n = 2^{2k} r with k ≥ 1 and r odd).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40764,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle, $\\Gamma$ is its circumcircle and $O$ is its circumcenter. Let $F$ be the point on $AC$ such that $\\angle COF = \\angle ACB$, such that $F$ and $B$ lie in opposite sides with respect to $CO$. The line $FO$ cuts $BC$ at $G$. The line parallel to $BC$ through $A$ intersects $\\Gamma$ again at $M$. The lines $CO$ and $MG$ meet at $K$. Show that the circumcircles of the triangles $BGK$ and $AOK$ meet on $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40765,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the board is written a (not necessarily positive) integer. In a move, if the current number on the board is $a$, one erases it and replaces it by either $a^2 + 2a + 3$ or $5a^2 + 2$ or $a - 119$. Is there a starting number $s$ such that for any positive integer $f$ one can reach $f$ from $s$ with finitely many moves?",
    "options": [],
    "answer": "No, such a starting number does not exist.",
    "solution": "Note that all of numbers of form $119k$ and $119k-1$ cannot be represented as $a^2 + 2a + 3$ or $5a^2 + 2$. We can check this claim by consider modulo 7 and 17. Indeed,\n* If $119k = a^2 + 2a + 3 = (a+1)^2 + 2$ then $-2$ is quadratic residue modulo 7, clearly absurd since quadratic residue modulo 7 are 0, 1, 2, 4.\n* If $119k = 5a^2 + 2$ then 3 is a quadratic residue modulo 17, also absurd.\n* If $119k - 1 = a^2 + 2a + 3$ then $-3$ is quadratic residue modulo 17, also absurd.\n* If $119k - 1 = 5a^2 + 2$ then $-2$ is quadratic residue modulo 7, also absurd.\nSo if we consider the first number in one of the above forms in which we run into when going backwards, it turns out that we can only shift by multiples of 117 and hence cannot obtain any number of the other form. ☐",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40766,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$ satisfying\n$$\nf(f(xy) + 1) = x f(x + f(y)), \\forall x, y > 0.\n$$",
    "options": [],
    "answer": "All solutions are f(x) = c/x for some constant c > 0.",
    "solution": "First, replace $x \\to \\frac{z}{y}$ then\n$$\nf(f(z) + 1) = \\frac{z}{y} f\\left(\\frac{z}{y} + f(y)\\right), \\forall y, z > 0. \\quad (1)\n$$\nIn (1), continue to set $z = 1$ and set $f(f(1) + 1) = c$, we have $f\\left(\\frac{1}{y} + f(y)\\right) = c y$ for all $y > 0$. The right-hand side takes on any values on $\\mathbb{R}^+$, so the function $f(y)$ is surjective on $\\mathbb{R}^+$.\n\nNow suppose that there exists $0 < y_1 < y_2$ such that $f(y_1) < f(y_2)$. We choose $z_0$ such that\n$$\n\\frac{z_0}{y_1} + f(y_1) = \\frac{z_0}{y_2} + f(y_2) \\iff z_0 = y_1 y_2 \\frac{f(y_2) - f(y_1)}{y_2 - y_1} > 0.\n$$\nIn (1), replace $z = z_0$ and let $y = y_1, y = y_2$ respectively, we have\n$$\nf(f(z_0) + 1) = \\frac{z_0}{y_1} f\\left(\\frac{z_0}{y_1} + f(y_1)\\right) = \\frac{z_0}{y_2} f\\left(\\frac{z_0}{y_2} + f(y_2)\\right).\n$$\nBase on the choice of $z_0$, we have\n$$\nf\\left(\\frac{z_0}{y_1} + f(y_1)\\right) = f\\left(\\frac{z_0}{y_2} + f(y_2)\\right),\n$$\nit follows that $y_1 = y_2$. This contradiction shows that with $0 < y_1 < y_2$, there must be $f(y_1) \\ge f(y_2)$ so $f$ is a non-increasing function. Thus, $f$ is both surjective and monotone on $\\mathbb{R}^+$ so it will be continuous on $\\mathbb{R}^+$. Since $f$ is continuous and non-increasing, there exists $\\lim_{x \\to +\\infty} f(x) = c \\ge 0$. However, if $c > 0$ then $f(x) \\ge c > 0$, $\\forall x \\in \\mathbb{R}^+$, which contradicts surjectiveness. Therefore $c = 0$.\n\nFinally, in the given condition, for each $x > 0$, given $y \\to +\\infty$ then $f(f(xy) + 1) \\to f(1)$ and $f(x + f(y)) \\to f(x)$. These imply that $f(1) = x f(x)$ or $f(x) = \\frac{c}{x}$ with $c = f(1) > 0$. It is easy to check that this function satisfies the problem. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40767,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p \\ge 5$ be a prime number and $a$ be an integer with $1 < a < p-1$. Let $a_1, a_2, \\dots, a_n$ be all possible remainders that powers of $a$ (that is, $1, a, a^2, a^3, \\dots$) can leave when divided by $p$. If the remainders $a_1 \\cdot a_2 \\cdot \\dots \\cdot a_n$ form an arithmetic sequence modulo $p$ in some order, prove that $n = p-1$. (Note that, for example, 6, 1 and 3 form an arithmetic progression modulo 7.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40768,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a set $S$ of $n$ elements, Ali wants to color all $2^n$ subsets of $S$ using $m$ colors such that if two different subsets $X$ and $Y$ are colored the same, the subset $X \\cup Y$ must be colored in a color different from the color of $X$ and $Y$. What is the smallest $m$ for which Ali can achieve this?",
    "options": [],
    "answer": "n+1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40769,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an acute triangle $ABC$ inscribed in $(O)$ with $\\angle A = 60^\\circ$ and symmedian point $L$. The tangents at $B, C$ of $(O)$ intersect $CA, AB$ at $E, F$. Prove that the line $OL$, two circumcircles of triangles $AEF$ and $BOC$ pass through the same point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D$ be the midpoint of the minor arc $BC$ of $(O)$ and $P$ be the intersection point of the tangents at $B, C$ of $(O)$. We will show that $D$ and $P$ lie on $(AEF)$.\n\nIndeed, we have\n$$\n\\angle AEP = \\angle AEB = \\angle BAC - \\angle ABE = \\angle ODB - \\angle ADB = \\angle ADO\n$$\nso $(ADP)$ passes through $E$. Similarly, $(ADP)$ passes through $F$. Now, let $J$ be the center of $(AEF)$, and we will show that $JD \\parallel OL$.\n\nFrom here, since $D$ is the center of circle $(BOC)$ then $JD \\perp PX$ and $OX \\perp PX$, it follows that $O, X, L$ lie on the same line parallel to $JD$. Since $(ABC)$ intersects $(AEF)$ at $D$ other than $A$, then $\\Delta DBE \\sim \\Delta DCF$ but $DB = DC$ so $DE = DF$. On the other hand, $JE = JF$ since $J$ is the center of $(AEF)$ so $JD \\perp EF$. According to the familiar result, $EF \\perp OL$, so $JD \\parallel OL$ which finishes the proof. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40770,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$ be a polynomial with integer coefficients, the leading one being positive. Prove that there are finitely many positive integers $n$ such that $n! + 1$ is a power of $P(n)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For each positive integer $n$ such that $n! + 1$ is a power of $P(n)$, we write $n! + 1 = P(n)^{k_n}$ and in this case call $n$ good. Due to the left-hand side and $P(x) \\ge x$ for $x$ big enough, we have $k_n < n$ and $P(n)$ must be odd for $n \\ge 2$. Hence, using the LTE, we obtain\n$$\n\\nu_2 (P(n)^{k_n} - 1) \\le \\nu_2(P(n) - 1) + \\nu_2(P(n) + 1) + \\nu_2(k_n).\n$$\nHence at least one of $\\{\\nu_2(P(n) - 1), \\nu_2(P(n) + 1), \\nu_2(k_n)\\}$ is at least $\\frac{1}{3}\\nu_2(n!)$ for good $n$. As there are infinitely many $n$, by pigeonhole principle, we now know that one of the following assertions is true:\n* There are infinitely many $n$ such that $\\nu_2(P(n) - 1) \\ge \\frac{1}{3}\\nu_2(n!)$.\n* There are infinitely many $n$ such that $\\nu_2(P(n) + 1) \\ge \\frac{1}{3}\\nu_2(n!)$.\n* There are infinitely many $n$ such that $\\nu_2(k_n) \\ge \\frac{1}{3}\\nu_2(n!)$.\nWe also know that $\\nu_2(n!) \\ge \\frac{n}{2}$. So if the first or the second assertion is true we know that for infinitely many $n$ we must have $P(n) > 2^{\\frac{n}{6}}$ which contradicts the polynomial asymptotic behavior. If the last assertion is true then we also have for sufficiently large $n$, we have $k_n > 2^{\\frac{n}{6}} > n$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40771,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, its incircle $(I)$ touches its sides $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. The bisector of angle $BIC$ intersects $BC$ at $M$. The line $AM$ intersects $EF$ at $N$. Prove that $DN$ bisects angle $EDF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40772,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$, $n$ be positive integers. Prove that\n$$\n\\lfloor n\\sqrt{2} \\rfloor \\cdot \\lfloor m\\sqrt{7} \\rfloor < \\lfloor mn\\sqrt{14} \\rfloor.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a = \\lfloor n\\sqrt{2} \\rfloor$ and $b = \\lfloor m\\sqrt{7} \\rfloor$. Then $a < n\\sqrt{2} < a+1$ and $b < m\\sqrt{7} < b+1$.\n\nTherefore,\n$$\na b < n\\sqrt{2} \\cdot m\\sqrt{7} = mn\\sqrt{14} < (a+1)(b+1).\n$$\n\nNow, $\\lfloor mn\\sqrt{14} \\rfloor$ is the greatest integer less than or equal to $mn\\sqrt{14}$.\n\nSince $a b < mn\\sqrt{14}$, we have $a b \\leq \\lfloor mn\\sqrt{14} \\rfloor$ only if $mn\\sqrt{14}$ is an integer, but $\\sqrt{14}$ is irrational, so $mn\\sqrt{14}$ is never an integer for positive integers $m, n$.\n\nThus, $a b < \\lfloor mn\\sqrt{14} \\rfloor$.\n\nTherefore,\n$$\n\\lfloor n\\sqrt{2} \\rfloor \\cdot \\lfloor m\\sqrt{7} \\rfloor < \\lfloor mn\\sqrt{14} \\rfloor.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40773,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ that cannot be written in the form\n$$\nn = \\frac{a}{b} + \\frac{a+1}{b+1},\n$$\nwhere $a$ and $b$ are positive integers.",
    "options": [],
    "answer": "All such n are exactly n = 1 and n = 2 + 2^k for k ≥ 0.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40774,
    "subject": "Mathematics (Multi-modal)",
    "question": "An $8 \\times 8$ chessboard is initially colored in a checkerboard pattern. Dark, semi-transparent L-shaped tiles, each covering three cells, can be placed on the board. To make the entire board a uniform shade, each white square must be covered by exactly one more layer of tiles than each black square. What is the minimum number of tiles required to make the board uniform in tone?",
    "options": [],
    "answer": "32",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40775,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a positive integer. Determine all $k$-tuples $(n_1, n_2, ..., n_k)$ of positive integers such that\n$$(n_1! - 1)(n_2! - 1) \\cdots (n_k! - 1) - 16$$\nis a perfect square of a positive integer.",
    "options": [],
    "answer": "All k-tuples consisting of exactly two entries equal to 3 and all remaining entries equal to 2.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40776,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1$, $x_2$, $x_3$ and $y_1$, $y_2$, $y_3$ are 6 positive real numbers such that\n$$\nx_1 + x_2 + x_3 = y_1y_2y_3 \\text{ and } y_1 + y_2 + y_3 = x_1x_2x_3.\n$$\nFind the minimum value of $T = x_1y_1 + x_2y_2 + x_3y_3$.",
    "options": [],
    "answer": "9",
    "solution": "Using AM-GM, we have\n$$\nx_1 + x_2 + x_3 \\ge 3\\sqrt[3]{x_1x_2x_3} \\implies y_1y_2y_3 \\ge 3\\sqrt[3]{x_1x_2x_3}.\n$$\nSimilarly, $x_1x_2x_3 \\ge 3\\sqrt[3]{y_1y_2y_3}$. Multiplying these inequalities, side-by-side, we get\n$$\n\\sqrt[3]{(x_1x_2x_3 \\cdot y_1y_2y_3)^2} \\ge 9 \\implies x_1x_2x_3 \\cdot y_1y_2y_3 \\ge 27.\n$$\nNow applying AM-GM for the given expression\n$$\nT \\ge 3\\sqrt[3]{x_1x_2x_3 \\cdot y_1y_2y_3} \\ge 3\\sqrt[3]{27} = 9.\n$$\nHence, the minimum value is 9, equality occurs when\n$$\nx_1 = x_2 = x_3 = y_1 = y_2 = y_3 = \\sqrt{3}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40777,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle inscribed in circle $(O)$ with angle $\\angle A = 45^\\circ$ and $AB < AC$. Let $AD, AH$ be the angle bisector and altitude of triangle $ABC$ with $D, H$ are on $BC$. Suppose that $OD$ intersects $AH$ at $E$ and $K$ is the circumcenter of triangle $EBC$. Prove that $HK \\parallel AD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P, N$ be the projections of $O, K$ on $AE$ and $M, T$ be the midpoints of $BC$ and the minor arc $BC$ of $(O)$. Let $AH = h$ and $R, R'$ be the radii of $(O), (K)$ respectively.\n![](attached_image_1.png)\nBy Thales' theorem, we have\n$$\n\\frac{HE}{HA} = \\frac{MO}{MT} = \\frac{MO}{OT - OM} = \\sqrt{2} + 1 \\implies HE = (\\sqrt{2} + 1)h.\n$$\nAccording to the Pythagorean theorem, $BK^2 - BO^2 = KM^2 - OM^2$ so\n$$\nR'^2 - R^2 = KM^2 - OM^2 \\implies KM^2 = R'^2 - \\frac{R^2}{2}.\n$$\nSimilarly,\n$$\n\\begin{align*}\nAO^2 - (AH - OM)^2 &= OP^2 = KN^2 = KE^2 - (EH - MK)^2 \\\\\n\\Leftrightarrow R^2 - \\left(h - \\frac{R}{\\sqrt{2}}\\right)^2 &= R'^2 - \\left((\\sqrt{2} + 1)h - MK\\right)^2 \\\\\n\\Leftrightarrow MK^2 + h \\cdot R\\sqrt{2} &= R'^2 - \\frac{R^2}{2} - (\\sqrt{2} + 1)^2 h^2 + (2 + 2\\sqrt{2})h \\cdot MK \\\\\n\\Leftrightarrow R\\sqrt{2} &= (2 + 2\\sqrt{2})h + (2 + 2\\sqrt{2})MK \\\\\n\\Leftrightarrow MK &= h + \\frac{R\\sqrt{2}}{2 + 2\\sqrt{2}} = h + MT.\n\\end{align*}\n$$\nThus $KT = MK - MT = h = AH$, proving that $AHKT$ is a parallelogram. Therefore, $AD \\parallel HK$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40778,
    "subject": "Mathematics (Multi-modal)",
    "question": "The monic polynomial $P(x)$ is called “nice” if its coefficients are in $\\{-1, 0, 1\\}$. Given a nice polynomial $P(x)$ of degree $2025$ and it is divisible by $x^7 - 1$, what is the maximum number of non-zero coefficients in $P(x)$?",
    "options": [],
    "answer": "2022",
    "solution": "Let $f(x) = x^7 - 1$. We have $x^{7k+r} \\equiv x^r \\pmod{f(x)}$ for all $r = 0, 1, \\ldots, 6$ and $k$ is a positive integer. We will group together all exponents divided by $7$ with remainder $r$. We have\n$$\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$$\nWe see that the right-hand side is a polynomial of degree not exceeding $6$, so to have congruence $0$ modulo $f(x)$, we need the right-hand side to be identically $0$. From that, we will get\n$$\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$$\nWe see that in the first three sums, there are $290$ terms and in the remaining sums, there are $289$ terms. So in the first three sums, we choose $145$ number $1$ and $145$ number $-1$, and the remaining sums have at least one $0$ (otherwise the sum of $289$ odd numbers cannot be zero). Thus, there are at least $4$ coefficients equal to $0$, so there are at most $2026 - 4 = 2022$ non-zero coefficients.\n\nTo construct, we just choose the coefficients based on the above idea with $a_{2025} \\neq 0$. The answer is $2022$.\n\n$\\boxed{2022}$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40779,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and $P(x)$ be a polynomial of degree $n$ with real coefficients. Prove that all the roots of the polynomial $x^3P(x) + 1$ can not be real at the same time.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40780,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all non-constant functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that:\n(i) $f(x)f(x+y) = f(x)^2 + x \\cdot f_{2025}(y)$ for all $x, y \\in \\mathbb{R}$,\n(ii) $f_{2025}(\\cos x) + f_{2025}(\\cos(\\pi - x)) = 0$ for all $x \\in \\mathbb{R}$.\nwith $f_n(x) = f(f(\\ldots(f(x))\\ldots))$ is the composition $n$ times of the function $f$.",
    "options": [],
    "answer": "f(x) = x",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40781,
    "subject": "Mathematics (Multi-modal)",
    "question": "Amina has a deck of cards numbered from $1$ to $101$. Initially, she holds $50$ cards in her hand, while the remaining cards are laid face up on the table. In one move, she can swap all $50$ cards in her hand for any $50$ cards from the table. Amina wants to obtain all the cards numbered from $1$ to $50$ in her hand in at most $N$ moves. What is the minimum $N$ that guarantees she can always achieve this, regardless of her starting hand?",
    "options": [],
    "answer": "51",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40782,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a > b$ be odd positive integers and $k$ be a positive integer such that $a^{1001} + k$ and $b^{1001} + k$ are powers of $2$. Prove that $a > b^{1001} + k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40783,
    "subject": "Mathematics (Multi-modal)",
    "question": "The set $M$ of pairwise different positive integers is given. It is known that within any 3 of them one can choose 2 numbers such that their sum is power of 2. Find the maximal number of elements $M$ may contain.",
    "options": [],
    "answer": "4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40784,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be an interior point on the side $BC$ of an acute-angled triangle $ABC$. Let the circumcircle of triangle $ADB$ intersect the side $AC$ again at $E$ ($\\neq A$) and the circumcircle of triangle $ADC$ intersect the side $AB$ again at $F$ ($\\neq A$). Let $AD$, $BE$ and $CF$ intersect the circumcircle of triangle $ABC$ again at $D_1$ ($\\neq A$), $E_1$ ($\\neq B$) and $F_1$ ($\\neq C$), respectively. Let $I$ and $I_1$ be the incenters of triangles $DEF$ and $D_1E_1F_1$, respectively. Prove that $E$, $F$, $I$, $I_1$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40785,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all triples $(a, b, p)$ of positive integers $a, b$ and a prime number $p \\ge 3$, such that $2^a + p^{2b}$ is a $(p-1)$-st power of a positive integer.",
    "options": [],
    "answer": "(4, 1, 3)",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40786,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(m, n)$ of non-negative integers such that\n$$\n2n! = m!(m! + 2).\n$$",
    "options": [],
    "answer": "(3, 4)",
    "solution": "The answer is $(3, 4)$.\n\nFor $m = 0, 1$ or $2$, we obtain the impossible $2n! = 3$ and $2n! = 8$. However, $m = 3$ works with $n = 4$.\n\nNow let $m \\ge 4$. Clearly, $2n! > 2m!$, which implies $n > m$, so we can write $n = m + a$ for some positive integer $a$. The equation transforms into\n$$\n2(m + 1)(m + 2)(m + a) = m! + 2,\n$$\nand further simplifies to\n$$\n(m+1)(m+2)(m+a) = \\frac{m!}{2} + 1.\n$$\nFor $m \\ge 4$, the right-hand side is odd, but for $a \\ge 2$, the left-hand side is even.\nTherefore, we have no solutions for $m \\ge 4$ and $a \\ge 2$.\n\nFinally, if $n = m+1$, then we have $2(m+1) = m!+2$, which simplifies to $(m-1)! = 2$.\nThis is impossible for $m \\ge 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40787,
    "subject": "Mathematics (Multi-modal)",
    "question": "A large rectangle is subdivided into smaller rectangles, each of which has at least one pair of sides of integer length. Prove that the large rectangle also has at least one pair of sides of integer length.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Label the large rectangle by $R$ and place it in the plane so that its lower-left corner is at $(0, 0)$ with sides parallel to $Ox, Oy$.\nLet $S$ be the set of vertices of the small rectangles whose both coordinates are integers, and let $T$ be the set of all the small rectangles. We form a bipartite graph on the vertex set $S \\cup T$ by joining each point in $S$ to every rectangle in $T$ of which it is a vertex. Observe that each small rectangle has either 0, 2, or 4 vertices in $S$ (because of integer length side), so it contributes an even number of edges. Hence the total number of edges in this bipartite graph is even. On the other hand, each point in $S$ that is not a corner of $R$ is a vertex of exactly 2 or 4 small rectangles, and so also has even degree.\nSince $(0, 0)$ is in $S$ and is a corner of exactly one small rectangle, it has odd degree. Therefore there must be another vertex in $S$ of odd degree. The only vertices of odd degree can be corners of $R$, so at least one other corner of $R$ lies in $S$. In particular, either the width or the height of $R$ is an integer. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40788,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with incentre $I$ such that $AB < AC < BC$. The second intersections of $AI$, $BI$ and $CI$ with the circumcircle of triangle $ABC$ are $M_A$, $M_B$ and $M_C$, respectively. Lines $AI$ and $BC$ intersect at $D$ and lines $BM_C$ and $CM_B$ intersect at $X$. Suppose the circumcircles of triangles $XM_B M_C$ and $XBC$ intersect again at $S \\ne X$. Lines $BX$ and $CX$ intersect the circumcircle of triangle $SXM_A$ again at $P \\ne X$ and $Q \\ne X$, respectively. Prove that the circumcentre of triangle $SID$ lies on $PQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the circumcentre of triangle $ABC$. First we note from standard properties of the Miquel point $S$, we have:\n$$\n\\bullet \\ \\triangle SM_C M_B \\sim \\triangle SBC \\sim \\triangle SPQ; (*)\n$$\n* I and S are inverses with respect to circle $ABC$;\n* $\\angle OSX = 90^\\circ$.\n![](attached_image_1.png)\nFrom the above we have $\\triangle OM_A I \\sim \\triangle OSM_A$ and\n$$\n\\angle M_A PB = \\angle M_A SX = 90^\\circ + \\angle M_A SO = 90^\\circ + \\angle OM_A I = \\angle M_A BA = \\angle CDA.\n$$\nObserve that $\\angle PM_C M_A = \\angle BM_C M_A = \\angle DAC$ and $\\angle M_C M_A B = \\angle ICD$. Combining these with $\\angle M_A PB = \\angle CDA$ shows $M_C PM_A B \\sim ADCI$. Therefore, we have $\\frac{M_C B}{BP} = \\frac{AI}{ID}$. Similarly, $\\frac{M_B C}{CQ} = \\frac{AI}{ID}$. Thus we get\n$$\n\\frac{M_C B}{BP} = \\frac{M_B C}{CQ} = \\frac{AI}{ID}. \\qquad (\\text{**})\n$$\nNow, observe that $\\angle ICB = \\angle AM_B M_C$ and $\\angle CBI = \\angle M_B M_C A$ which gives that $\\triangle IBC \\sim \\triangle AM_C M_B$. This, combined with (**), is enough to show $\\triangle DPQ \\sim \\triangle IBC$ by linearity, thus $\\frac{DP}{DQ} = \\frac{IB}{IC}$.\nCombining $\\triangle IBM_C \\sim \\triangle ICM_B$ with (***) shows $IBM_C P \\sim ICM_B Q$, thus $\\frac{IP}{IQ} = \\frac{IB}{IC}$. Finally, we have that\n$$\n\\frac{SP}{SQ} = \\frac{SB}{SC} = \\frac{BM_C}{CM_B} = \\frac{IB}{IC}\n$$\nfrom (*) and $\\triangle IBM_C \\sim \\triangle ICM_B$. Putting this together with above ratios, we have\n$$\n\\frac{IB}{IC} = \\frac{DP}{DQ} = \\frac{IP}{IQ} = \\frac{SP}{SQ}\n$$\nwhich shows that circle $SID$ is an Apollonius circle with respect to $P$ and $Q$, giving the desired conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40789,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1 \\le a_2 \\le \\dots \\le a_{50}$ be real numbers such that\n(i) $a_1 + a_2 + \\dots + a_{50} = 0,$\n(ii) $|a_1| + |a_2| + \\dots + |a_{50}| = 624.$\nFind the minimum value for $S = a_{50} - a_1$ and maximum value of $T = a_1a_2 \\dots a_{50}$.",
    "options": [],
    "answer": "Minimum S = 624/25. Maximum T = 13^24 * 12^26.",
    "solution": "Suppose that $k$ is the maximum index such that $a_k \\le 0$ then $a_k \\le 0 < a_{k+1}$. Thus $a_{k+1} + \\dots + a_{50} = -(a_1 + a_2 + \\dots + a_k)$ and then\n$$\n|a_1| + |a_2| + \\dots + |a_k| = |a_1 + a_2 + \\dots + a_k| = a_{k+1} + a_{k+2} + \\dots + a_{50} = \\frac{624}{2} = 312.\n$$\nWe have\n$$\n-312 = a_1 + a_2 + \\dots + a_k \\ge k \\cdot a_1 \\implies a_1 \\le \\frac{-312}{k},\n$$\n$$\n312 = a_{k+1} + a_{k+2} + \\dots + a_{50} \\le (50-k) \\cdot a_{50} \\implies a_{50} \\ge \\frac{312}{50-k}.\n$$\nHence,\n$$\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$$\nThe minimum value of $S = \\frac{624}{25}$, the equality case when\n$$\na_1 = a_2 = \\dots = a_{25} = -\\frac{312}{25} \\text{ and } a_{26} = a_{27} = \\dots = a_{50} = \\frac{312}{25}.\n$$\nWe will continue with $T$. If there is some $a_k = 0$ then $T = 0$. Now suppose that all of given numbers are non-zero and there are $k$ negative numbers among $a_1, a_2, \\dots, a_{50}$. If $k$ is odd then $T < 0$. Thus to find the maximum value of $T$, we consider $k$ is even. Put $h = 50 - k$. By AM-GM inequality, we have\n$$\n\\begin{aligned}\nT &\\le \\left( \\frac{a_1 + a_2 + \\dots + a_k}{k} \\right)^k \\cdot \\left( \\frac{a_{k+1} + a_{k+2} + \\dots + a_{50}}{h} \\right)^h \\\\\n&= \\left( \\frac{312}{k} \\right)^k \\left( \\frac{312}{h} \\right)^h = \\frac{312^{h+k}}{k^k h^h} = \\frac{312^{50}}{k^k h^h}.\n\\end{aligned}\n$$\nWe need to prove that\n$$\nk^k \\cdot h^h \\ge 24^{24} \\cdot 26^{26}.\n$$\nWLOG, $k \\le h$ then $k \\le h - 2$ since $k, h$ are even. Suppose that $k \\le h - 4$ then we can use local modification to prove that\n$$\nk^k \\cdot h^h > (k+1)^{k+1} \\cdot (h-1)^{h-1} \\iff h \\cdot \\left(1 + \\frac{1}{h-1}\\right)^{h-1} > (k+1) \\cdot \\left(1 + \\frac{1}{k}\\right)^k \\quad (*)\n$$\nConsider the function $f(x) = (x+1) \\cdot (1+\\frac{1}{x})^x$ with $x > 1$ and\n$$\ng(x) = \\ln f(x) = \\ln(x+1) + x \\cdot \\ln\\left(1+\\frac{1}{x}\\right).\n$$\nWe have\n$$\ng'(x) = \\frac{1}{1+x} + \\ln\\left(1 + \\frac{1}{x}\\right) - \\frac{1}{1+x} = \\ln\\left(1 + \\frac{1}{x}\\right) \\geq 0.\n$$\nThus $g(x)$ increasing and then (*) is proved. Therefore, the maximum value of $T$ is $13^{24} \\cdot 12^{26}$. The equality case when\n$$\na_1 = a_2 = \\dots = a_{24} = -13,\\ a_{25} = a_{26} = \\dots = a_{50} = 12.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40790,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(a, b)$ of positive integers such that\n$$\na^2 + b = p^m \\text{ and } b^2 + a = p^n\n$$\nfor some prime $p$ and positive integers $m, n$.",
    "options": [],
    "answer": "[(1, 1), (2, 5), (5, 2)]",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40791,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AC > AB > BC$ and angle bisector $AD$ ($D \\in BC$). Denote by $\\omega_1$ and $\\omega_2$ the circumcircles of triangles $ABD$ and $ACD$, respectively. The line $AC$ intersects $\\omega_1$ again at $F$ and the line $AB$ intersects $\\omega_2$ again at $E$. The line $DE$ intersects $\\omega_1$ again at $G$ and the line $DF$ intersects $\\omega_2$ again at $H$. Prove that the circumcircles of triangles $ABC$, $AEF$ and $AGH$ have a second common intersection point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40792,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $30$ students in a mathematics competition. We wish to give them a total of $N$ candies so that the following hold:\n(i) Each student gets at least one candy.\n(ii) Any student with larger score gets more candies than any student with lower score.\n(iii) Any two students with the same score get the same amount of candies.\nDetermine the smallest possible value of $N$, for which the candies can be given in some way, no matter what is the ranking of the students.",
    "options": [],
    "answer": "480",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40793,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the extensions of the sides $CA$ and $AB$ of triangle $ABC$ beyond points $A$ and $B$, points $E$ and $F$ are chosen such that $AE = BC$ and $BF = AC$. The $A$-excircle is tangent to segment $BF$ at point $N$. Let $M$ be the midpoint of segment $EF$. Prove that $MN$ is parallel to the internal angle bisector of angle $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40794,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 2$ be an integer and let $S$ be a set of $k$ points in the plane whose coordinates both lie in $\\{1, 2, \\dots, n\\}$. Find the minimal number of $k$ for which one can choose four points from $S$ that are the vertices of a nondegenerate parallelogram.",
    "options": [],
    "answer": "2n",
    "solution": "If $k = 2n - 1$, we can choose all points with $x$-coordinate, $y$-coordinate are $1$. Then, clearly there is no parallelogram. Thus, any $k \\le 2n-1$ does not satisfy the condition.\n\nNext, we will prove that $k = 2n$ satisfies. Indeed, let $x_1, x_2, ..., x_n$ be the number of points whose $x$-coordinate are $1, 2, ..., n$ respectively. Clearly\n$$\nx_1 + x_2 + \\dots + x_n = 2n.\n$$\nConsider the segments formed by connecting $2$ points with the same $x$-coordinate. If there are $x$ points, then at least $x-1$ segments with different lengths will be generated (we take the point with the lowest ordinate and connect it to the remaining $x-1$ points). Thus, with $1 \\le i \\le n$, among the points with the same $x$-coordinate $i$, we can choose a set $S_i$ with size $x_i - 1$ consists of segments with different lengths. If in $S_1, S_2, ..., S_n$ there are two segments have the same length, it is easy to see that the four corresponding vertices will form a parallelogram, satisfying the problem.\n\nConversely, if there are no segments with the same length, then notice that\n$$\n|S_1| + |S_2| + \\dots + |S_n| = (x_1 - 1) + (x_2 - 1) + \\dots + (x_n - 1) = n,\n$$\nbut the values in these sets are smaller than $n$, because their length is the difference between two $y$-coordinate, and these coordinates only belong to $\\{1, 2, ..., n\\}$.\n\nObviously this is a contradiction.\n\nTherefore the minimum value to find is $k_{min} = 2n$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40795,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle inscribed in circle $(O)$ with $\\angle A = 45^\\circ$. Two rays $BO$, $CO$ intersect $AC$, $AB$ at $E$, $F$ respectively. The circumcircles of triangles $BOC$ and $EOF$ intersect at $K$. Let $J$ be circumcenter of triangle $AEF$. Prove that $JK$ passes through the orthocenter of triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have\n$$\n\\angle BOC + \\angle EOF = 90^\\circ + 90^\\circ = 180^\\circ\n$$\nso according to the familiar property of isogonal conjugates in quadrilaterals, we see that there exists a point $O'$ which is the isogonal conjugate of $O$ in $BFEC$. On the other hand, $BH$, $BO$ and $CH$, $CO$ are isogonal pairs in the angles $\\angle B$, $\\angle C$ so clearly $O' \\equiv H$. It follows that $\\angle BFH = \\angle OFE$ and $\\angle CEH = \\angle OEF$.\nSuppose $JH$ intersects the circle $(EJF)$ at point $K$.\n\n![](attached_image_1.png)\n\nWe will prove that $K$ belongs to the circle ($BOC$). Indeed, we have $\\angle HBF = 45^\\circ = \\angle JEF = \\angle JKF$ so $F$, $K$, $H$, $B$ are congruent. Similarly, $E$, $K$, $H$, $C$ are congruent.\nIt follows that\n$$\n\\angle BKC = \\angle BKH + \\angle CKH = \\angle BFH + \\angle CEH = \\angle OFE + \\angle OEF = 90^\\circ.\n$$\nFrom here it follows that $K$ belongs to $BOC$.\n$\\boxed{}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40796,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n < 2027$ that satisfy the following conditions:\n(i) For every positive divisor $d$ of $n$, numbers $1^d, 2^d, \\dots, 2026^d$ all have distinct remainders when divided by $2027$.\n(ii) $\\tau(n)^2 \\mid n$ with $\\tau(n)$ is the number of positive divisors of $n$.",
    "options": [],
    "answer": "1, 9, 625, 2025",
    "solution": "Let call the numbers satisfying the given condition as \"good\" number. Consider some good number $n$. One can see that if $n$ is even, then $d = 2$ is a divisor of $2$, however, then $1^2$ and $2026^2$ both divide $2027$ with remainder $1$, which does not satisfy. Therefore, $n$ is odd. We see that $n = 1$ satisfies the given condition.\n\nNow consider $n > 1$ and set $n = p_1^{a_1} p_2^{a_2} \\dots p_k^{a_k}$ with $p_1, p_2, \\dots, p_k$ being distinct odd primes and $a_1, a_2, \\dots, a_k$ positive integers. Since $\\tau(n)^2 \\mid n$, $\\tau(n)$ is also odd.\n\nWe have $\\tau(n) = (a_1+1)(a_2+1)\\cdots(a_k+1)$ which is odd, so $a_1, a_2, \\dots, a_k$ are all even. Therefore, $n$ must be a perfect square. Note that if $k \\ge 3$ then $n \\ge (3 \\cdot 5 \\cdot 7)^2 > 2027$, is not satisfied. Hence $k \\in \\{1; 2\\}$. We consider the following cases.\n\n1. If $k = 1$ then $n = p^{2m}$ with $p$ being an odd prime and $m \\in \\mathbb{Z}^+$. Note that if $2m \\ge 6$ then $\\tau(n) \\ge 7$ so $p \\ge 7$ and $n \\ge 7^6 > 2027$, does not satisfy. Therefore, we reduce to considering if $2m = 2$ then $p = 3$ so $n = 9$; and if $2m = 4$ then $\\tau(n) = 5$ so $p = 5$, which implies $n = 5^4 = 625$. We can check that the numbers $n = 9, n = 625$ all satisfy condition (ii).\n\n2. If $k = 2$ then $n = p^{2a}q^{2b}$ with $p, q$ being distinct odd primes and $a, b \\in \\mathbb{Z}^+$. Note that if both $2a$ and $2b \\ge 4$ then $n \\ge (3 \\cdot 5)^4 > 2027$, is not satisfied. Therefore, we can assume $2b = 2$, which implies that $3|\\tau(n)$ so $p = 3$ or $q = 3$.\n\na. If $q = 3$ then $n = 9p^{2a}$, then if $2a = 2$ then $\\tau(n) = 9$ so $9^2|n$, is not satisfied; therefore $2a \\ge 4$ so $n \\ge 9 \\cdot 5^4 > 2027$, is also not satisfied.\n\nb. If $p = 3$ then $n = 3^{2a}q^2$; if $2a = 2$ then the same absurdity exists, so $2a > 2$. But if $2a \\ge 6$ then $n \\ge 3^6 \\cdot 5^2 > 2027$, is not satisfied. Therefore $2a = 4$ and $5|\\tau(n)$ so $q = 5$, implies $n = 2025$. We see that this number satisfies condition (ii).\n\nTherefore $n \\in \\{9; 625; 2025\\}$. We will prove that these numbers are good.\n\nConsider an arbitrary positive divisor $d$ of $n$ then $d$ is odd. Suppose there exist $u, v \\in \\{1; 2; \\dots; 2026\\}$ and $u \\neq v$ such that $u^d \\equiv v^d \\pmod{2027}$. Let $s$ be the inverse of $v$ modulo $2027$ then let $t = us$, we have $t^d \\equiv 1 \\pmod{2027}$. Next, let $h = \\text{ord}_{2027}(t)$ then $h \\nmid d$ and by Fermat's little theorem $h \\nmid 2026$ so $h \\nmid \\text{gcd}(d, 2026)$. Note that $2026 = 2 \\cdot 1013$ with $1013$ being a prime number, it is easy to see that $\\text{gcd}(d, 2026) = 1$ and $h = 1$. Then, we will have\n$$\nt \\equiv 1 \\pmod{2027} \\iff us \\equiv 1 \\pmod{2027}.\n$$\nMultiplying both sides of this congruence by $v$ gives $u \\equiv v \\pmod{2027}$, which is absurd. Thus, all positive divisors $d$ of these $n$ satisfy condition (i).\n\nSo all the good numbers are $1, 9, 625, 2025$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40797,
    "subject": "Mathematics (Multi-modal)",
    "question": "100 stones are placed next to each other in a row: black, white, black, white, ..., black, white. In each move, it is allowed to choose two black stones between which all stones are white and paint those white stones black, or choose two white stones with all stones between them black and paint those black stones white. Is it possible after several moves to get a situation where the first 50 stones in the row are black and the last 50 are white?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40798,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let a grid $2m \\times 2n$ is given. Laminates colors each cell of the grid either black or white such that the cells of each color form a polygon. Then Yamal covers the whole grid by $2mn$ dominoes of size $1 \\times 2$. The domino is called combined if it covers 1 black and 1 white cell. Find the maximal number of combined dominoes Lamin may guarantee regardless to the strategy of Yamal.",
    "options": [],
    "answer": "m + n - 1",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40799,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\nx^2 + a x + 1 = 0 \\text{ and } x^2 + b x + c = 0\n$$\nhave a common real root. In addition, the equations\n$$\nx^2 + x + a = 0 \\text{ and } x^2 + c x + b = 0\n$$\nhave a common real root. What values can the sum $a + b + c$ take?",
    "options": [],
    "answer": "(-∞, -3]",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40800,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define the function $S: \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ to be the sum of prime divisors (not necessarily distinct) of $n$ (if $v_p(n) = a$ then add $p \\cdot a$ to the sum). Prove that\n$$\n\\frac{\\tau(n!)}{\\tau((n-1)!)} \\le \\frac{4S(n)}{n}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40801,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. There are $2n$ knights sitting at a round table. They consist of $n$ pair of partners, each pair of which wishes to shake hands. A pair can shake hands only with next to each other. Every minute, one pair of adjacent knights swaps places. Find the minimum number of exchanges of adjacent knights such that, regardless of the initial arrangement, every knight can meet her partner and shake hands at some moment.",
    "options": [],
    "answer": "n(n−1)/2",
    "solution": "*Solution.* (Solution of Ahmad Alshehri, IMO 2025's team member)\nSAUDI ARABIAN IMO Booklet 2025\n---\n## Saudi Booklet 2025 — Page 46\n46\nSolution of IMO Team selection tests\nنلاحظ في المثال : ① ② ③ ④",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40802,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest real number $c > 0$ such that\n$$\n\\{x\\} + \\{y\\} < c + xy\n$$\nfor all real numbers $x, y > 0$ and $xy < 1$.",
    "options": [],
    "answer": "1",
    "solution": "First, choose $x = \\frac{n+1}{n+2}$, $y = \\frac{n}{n+1}$ with $n \\in \\mathbb{Z}^+$ then $xy = \\frac{n}{n+2} < 1$. Substituting into the given inequality, we have\n$$\n\\begin{cases}\n\\frac{n+1}{n+2} + \\frac{n}{n+1} < c + \\frac{n}{n+2} \\\\\n\\iff c > \\frac{n}{n+1} + \\frac{1}{n+2}.\n\\end{cases}\n$$\nHere, for $n \\to +\\infty$ we have $c \\ge 1$. Next, we will prove the inequality with $c = 1$: $\\{x\\} + \\{y\\} < 1 + xy$ for all $x, y > 0$ and $xy < 1$. Indeed, we have\n$$\n(1 - \\{x\\})(1 - \\{y\\}) > 0 \\implies \\{x\\} + \\{y\\} < 1 + \\{x\\}\\{y\\} \\le 1 + xy.\n$$\nThis inequality is true. Therefore, the minimum value we need to find is $c = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40803,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying the following conditions:\n(i) there is at most one number $a$ such that $f(a) = 0,$\n(ii) $f(f(x+y)f(x-y) + 25xy + y^2) = 24yf(x) + xf(x+y)$ for all real $x, y$.",
    "options": [],
    "answer": "f(x) = x for all real x; f(x) = -x for all real x",
    "solution": "First, setting $y = 0$ to give $f(f(x)^2) = x f(x)$.\n\nThen setting $x = 0$ to get $f(f(y)f(-y) + y^2) = 24y f(0)$. Now change $y$ by $-y$ then the LHS does not change, so we will get\n$$\n24y f(0) = -24y f(0), \\forall y \\implies f(0) = 0.\n$$\nFrom $f(f(y)f(-y) + y^2) = 0$, one can get $f(y)f(-y) = -y^2$.\n\nSetting $y = -x$ to get $f(-24x^2) = -24x f(x)$, change $x$ to $-x$ will implies that $f(x)$ is odd. Thus $f(y)^2 = y^2$. We have two solutions: $f(x) = x$ for all $x$ and $f(x) = -x$ for all $x$.\n\nNow we need to handle the point-wise trap. Let $A = \\{a \\in \\mathbb{R} \\setminus \\{0\\} \\mid f(a) = a\\}$ and $B = \\{a \\in \\mathbb{R} \\setminus \\{0\\} \\mid f(b) = -b\\}$, suppose that these sets are non-empty. Take $a \\in A$ and $b \\in B$.\n\nLet $x = a, y = b - a$ to give $f(x + y) = f(b) = -b$ and $f(x - y) = f(2a - b)$. Thus\n$$\n\\text{LHS} = f(-b f(2a - b) + 2a(b - a) + (b - a)^2) = \\pm(-b \\pm (2a - b) + b^2 - a^2).\n$$\n$$\n\\text{RHS} = (b - a)a + a(-b) = -a^2.\n$$\nThus $\\pm(\\pm(2a - b)(-b) + b^2 - a^2) = -a^2$. For each pair of $a, b$, there are 4 ways to choose the sign so for some fixed value $a$, there are at most 8 ways to choose $b$ which implies that $|B| \\le 8$. Similarly, $|A| \\le 8$, this is contradiction since $A \\cup B = \\mathbb{R} \\setminus \\{0\\}$.\n\nHence, $f(x) = x$ for all $x$ and $f(x) = -x$ for all $x$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40804,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find $n \\in \\mathbb{N}$ for which $5^n + n^4 + 20$ is a perfect square.",
    "options": [],
    "answer": "no such n",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40805,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a square board of $8 \\times 8$ and one colors exactly $7$ squares of the board. Prove that for every way of coloring, there always exists a sub-square size $2 \\times 2$ of the board on which exactly one square is colored.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We number rows numbered from $1$ to $8$ (from top to bottom), and columns numbered from $1$ to $8$ (from left to right). Suppose by contradiction that each $2 \\times 2$ square board does not contain exactly one colored square.\n\nObviously there must also be an empty column, let's say it is the $i$ column with $1 \\le i \\le 8$. Let's consider the left column (if any) of this $i$ column, that is, the $(i-1)$-th column. If on that $(i-1)$-th column, there is a colored square. Let's consider the corresponding $2 \\times 2$ square containing $2$ squares on $i$-th column (which are two uncolored squares) and also containing the $\\times$ cell. Thus, the remaining cell in $(i-1)$-th column must also be colored because otherwise the $2 \\times 2$ table contains exactly $1$ colored cell, which contradicts the assumption.\n\nContinuing like that, consider the squares containing $2$ cells in column $i$ and containing a colored cell in column $i-1$, then by similar reasoning, the remaining cell is also colored. This shows that the entire column $i-1$ must be colored, that is, there are $8$ colored cells, which is absurd. From there, we can deduce that column $i-1$ is an empty column. The same reasoning goes for column $i+1$. From there, we can continue to expand to columns $i-2, i-3, \\dots, 1$ and columns $i+2, i+3, \\dots, 8$, which are also empty columns. Therefore, the entire table is not colored, which is also absurd. This proves the given assertion. $\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40806,
    "subject": "Mathematics (Multi-modal)",
    "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": "Detailed solution",
    "solution": "We show the stronger result that the set of prime $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": 40807,
    "subject": "Mathematics (Multi-modal)",
    "question": "Bank is planning to produce and grant a special credit card to its loyal costumers. Each of these credit cards has a unique ID-number that has 16 digits and satisfies the following properties:\n(i) The first four digits of the ID-numbers is fixed and is equal to 2024.\n(ii) For each pair of cards their corresponding ID-numbers must differ in at least two digits.\nProve that the maximal possible number of cards that can be produced is $10^{11}$.",
    "options": [],
    "answer": "10^11",
    "solution": "Let the ID-number be $d_1d_2\\ldots d_{16}$, where each $d_i$ is a digit ($0$ to $9$). By condition (i), $d_1d_2d_3d_4 = 2024$ is fixed. So, the remaining $12$ digits $d_5, d_6, \\ldots, d_{16}$ can be chosen.\n\nLet $S$ be the set of all possible $12$-digit strings (from $0$ to $9$ in each position), i.e., $|S| = 10^{12}$.\n\nWe are to select a subset $A \\subset S$ such that for any two distinct elements $x, y \\in A$, $x$ and $y$ differ in at least $2$ positions (digits).\n\nWe claim that the maximal possible size of $A$ is $10^{11}$.\n\nProof:\n\nLet us construct such a set $A$ of size $10^{11}$.\n\nLet $A$ be the set of all $12$-digit numbers whose last digit is $0$. That is, $d_{16} = 0$. There are $10^{11}$ such numbers, since $d_5, d_6, \\ldots, d_{15}$ can be arbitrary ($11$ digits), and $d_{16} = 0$.\n\nNow, for any two distinct elements $x, y \\in A$, they must differ in at least one of the first $11$ digits (since the last digit is always $0$). But since they are distinct, there is at least one position among $d_5$ to $d_{15}$ where they differ. However, we need to ensure that they differ in at least $2$ digits.\n\nSo, instead, let us consider the set $A$ of all $12$-digit numbers whose last two digits are $00$. That is, $d_{15} = d_{16} = 0$. There are $10^{10}$ such numbers.\n\nBut the problem asks for $10^{11}$.\n\nLet us generalize. Consider the set $A$ of all $12$-digit numbers where $d_5$ is fixed (say $0$), and the remaining $11$ digits are arbitrary. Then $|A| = 10^{11}$.\n\nFor any two distinct elements $x, y \\in A$, they differ in at least one of $d_6$ to $d_{16}$. Since $d_5$ is fixed, the only way $x$ and $y$ can differ is in $d_6$ to $d_{16}$. But since they are distinct, there is at least one position where they differ. However, it is possible that they differ in only one digit, which does not satisfy the condition.\n\nTherefore, we need to find the largest possible subset of $S$ such that any two elements differ in at least $2$ digits.\n\nThis is a classical coding theory problem: the largest possible size of a code of length $12$ over an alphabet of size $10$ with minimal distance $2$.\n\nThe answer is given by the following construction:\n\nLet $A$ be the set of all $12$-digit numbers with no two the same, and such that for each $x \\in S$, at most one element of $A$ is at Hamming distance $0$ or $1$ from $x$.\n\nLet us count the number of $12$-digit numbers at Hamming distance $0$ or $1$ from a fixed $x$:\n- Hamming distance $0$: $x$ itself.\n- Hamming distance $1$: For each of the $12$ positions, there are $9$ choices (since each digit can be changed to $9$ other digits), so $12 \\times 9 = 108$.\n\nSo, for each $x$, there are $1 + 108 = 109$ numbers at Hamming distance $0$ or $1$ from $x$.\n\nIf $A$ is a code with minimal distance $2$, then the balls of radius $1$ around each codeword are disjoint and cover $|A| \\times 109$ numbers. Since the total number of $12$-digit numbers is $10^{12}$, we have:\n$$\n|A| \\leq \\frac{10^{12}}{109} < 10^{11}\n$$\n\nBut the problem claims that $10^{11}$ is possible.\n\nLet us construct $A$ as follows:\nLet $A$ be the set of all $12$-digit numbers where the sum of the digits is congruent to $0$ modulo $10$.\n\nFor any two distinct elements $x, y \\in A$, their digit sums are both $0$ modulo $10$. If $x$ and $y$ differ in only one digit, say at position $i$, then the sum of digits of $x$ and $y$ would differ by a nonzero amount, so their sums modulo $10$ would be different, which is a contradiction. Therefore, any two elements of $A$ differ in at least $2$ digits.\n\nHow many such numbers are there? For any choice of the first $11$ digits, the last digit is determined modulo $10$ to make the sum $0$ modulo $10$. So, $10^{11}$ such numbers.\n\nTherefore, the maximal possible number is $10^{11}$.\n\nThus, the answer is $10^{11}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40808,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two circles $(O), (O')$ with different radii and intersecting at $A, B$. Circle of center $A$, radius $AB$ intersects $(O), (O')$ again at $C, D$ respectively. Let $EF$ be the common tangent closer to $B$ of the two circles with $E \\in (O), F \\in (O')$. Rays $AE, AF$ intersect $BC, BD$ at $M, N$ respectively. Prove that the internal bisector of $\\angle CBD$ passes through the circumcenter of triangle $AMN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $AB = AC$, in circle $(O)$, we have $\\angle ABC = \\angle AEB$ so $ABM, AEB$ are two similar triangles. This implies that\n$$\n\\frac{AM}{AB} = \\frac{AB}{AE} \\implies AB^2 = AM \\cdot AE.\n$$\nSimilarly, one could get $AB^2 = AN \\cdot AF$ so $AM \\cdot AE = AN \\cdot AF$, thus $EMNF$ is cycle. Hence, $\\angle AMN = \\angle AFE$. On the other hand\n$$\n\\angle AFE = \\angle AFB + \\angle BFE = \\angle ABD + \\angle BAF = \\angle BNF\n$$\nso $\\angle AMN = \\angle BNF = \\angle AND$, this means that $BD$ in tangent to $(AMN)$. Similarly, $BC$ is also tangent to $(AMN)$, so the center of $(AMN)$ is equidistant from the two segments $BC$ and $BD$. In other words, the angle bisector of $\\angle CBD$ passes through the center of $(AMN)$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40809,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R}^+ \\to \\mathbb{R}$ such that for all $x, y > 0$ we have\n$$\nf(x^2 + x f(y) + y) = 2x + f(y).\n$$",
    "options": [],
    "answer": "All solutions are f(x) = sqrt(4x + C) for a constant C >= 0.",
    "solution": "Let $r > s$ be two positive real numbers. Then, there would be a positive real number $t$ such that $t^2 + f(s)t + s - r = 0$. Now, put $(x, y) = (t, s)$ it follows that\n$$\nf(r) = f(t^2 + f(s)t + s) = 2t + f(s) > f(s).\n$$\nSo, $f$ is strictly increasing and hence injective. Plugging $(x, y) = (\\frac{f(z)}{2}, y)$ it follows that\n$$\nf(z) + f(y) = f(f(z)^2/4 + f(z)f(y)/2 + y) = f(f(y)^2/4 + f(z)f(y)/2 + z)\n$$\nThat is, $f(z)^2/4 - z = C$. Hence, $f(z) = \\sqrt{4z + C}$, for some $C \\ge 0$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40810,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a non-isosceles triangle with centroid $G$ and inscribed in circle $(O)$. Let $M, N, P$ be the midpoints of $BC, CA, AB$ respectively. Let $D, E, F$ be the reflection points of the foot of the internal bisector of angles $A, B, C$ through $M, N, P$ respectively. Prove that the orthocenter of triangle $DEF$ lies on line $OG$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose $EF$ intersects $BC$ at $D'$. Let $(\\omega_A)$ be the circle with diameter $DD'$. By the angle bisector theorem, one can get\n$$\n\\frac{DB}{DC} = \\frac{AC}{AB}, \\quad \\frac{EC}{EA} = \\frac{AB}{BC}, \\quad \\frac{FA}{FB} = \\frac{BC}{CA}.\n$$\nBased on Ceva's theorem, $AD, BE, CF$ are concurrent and $(DD', BC) = -1$. If $AU, AV$ are the internal, external angle bisectors of triangle $ABC$, then $(UV, BC) = -1$. Through the reflection over the center $M$, then $(DV', CB) = -1$ with $V'$ being the symmetric point to $V$ through $M$. From there, $D' \\equiv V'$ so $(\\omega_A)$ is the reflection of the $A$-Apollonius circle through the perpendicular bisector of $BC$.\n\nSuppose that $A$-Apollonius circle intersects $(O)$ at $K$ then $\\frac{KB}{KC} = \\frac{AB}{AC}$ so $ABKC$ is harmonic, entailing $AK$ is the symmetric line of triangle $ABC$. Therefore, $(\\omega_A)$ intersects $(O)$ at $A', R$ will be the points symmetric to $A, K$ through the perpendicular bisector $BC$. Therefore, $KR \\parallel BC$ so $AK, AR$ are isogonal in angle $A$, which implies that $AR$ is the median of triangle $ABC$ or $R \\in AM$. Suppose $AR$ intersects $(\\omega_A)$ at $S$, then by the property of power of a point and Newton's formula, we have\n$$\n\\overline{MR} \\cdot \\overline{MS} = \\overline{MD} \\cdot \\overline{MD'} = MB^2 = -\\overline{MB} \\cdot \\overline{MC} = -\\overline{MR} \\cdot \\overline{MA}.\n$$\nFrom there, we deduce $\\overline{MS} = -\\overline{MA}$ or $M$ is the midpoint of $AS$. Here we have\n$$\n\\frac{\\mathcal{P}_{G/(O)}}{\\mathcal{P}_{G/(\\omega_A)}} = \\frac{\\overline{GR} \\cdot \\overline{GA}}{\\overline{GR} \\cdot \\overline{GS}} = \\frac{\\overline{GA}}{\\overline{GS}} = \\frac{2/3 \\cdot \\overline{MA}}{4/3 \\cdot \\overline{MS}} = -\\frac{1}{2}.\n$$\nTherefore $\\mathcal{P}_{G/(\\omega_A)} = -2\\mathcal{P}_{G/(O)}$ is equal between vertices $A, B, C$. Therefore we have $G$ has the same power to the three circles $(\\omega_A), (\\omega_B), (\\omega_C)$.\n\nOn the other hand, $OA$ is tangent to the Apollonius circle at vertex $A$ so $OA'$ is tangent to $(\\omega_A)$. From there we have\n$$\n\\mathcal{P}_{O/(\\omega_A)} = OA'^2 = R^2\n$$\nwith $R$ being the radius of $(O)$. Also because of this symmetry, $G$ has the same power to the three circles $(\\omega_A), (\\omega_B), (\\omega_C)$. Let $X$ be the projection of $D$ onto $EF$ then $X \\in DH$ and $X \\in (\\omega_A)$. Then, $\\mathcal{P}_{H/(\\omega_A)} = \\overline{HX} \\cdot \\overline{HD}$. Obviously, this quantity is also symmetric between the vertices $D, E, F$ in triangle $DEF$, so $H$ also has the same power to the three circles $(\\omega_A), (\\omega_B), (\\omega_C)$. From that we immediately have $H \\in OG$.\n\n$\\boxed{}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40811,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all triples $(a, b, c)$ of integers such that\n$$\na^3 + b^3 + c^3 = 25(abc + a^2b + b^2c + c^2a).\n$$",
    "options": [],
    "answer": "a = b = c = 0",
    "solution": "We will show that there is no solution other than the trivial $a = b = c = 0$. For any other triple, we can write $a = dx$, $b = dy$, $c = dz$, where $\\gcd(x, y, z) = 1$ and $d$ is a positive integer. After dividing out $d$, we obtain the same equation but for $x$, $y$, and $z$. We will demonstrate that each of $x$, $y$, and $z$ must be divisible by $3$, which leads to a contradiction.\n\nAssume first that none of $x$, $y$, or $z$ is divisible by $3$. Then $x^2 \\equiv y^2 \\equiv z^2 \\equiv 1 \\pmod{3}$, which implies that\n$$\nx + y + z \\equiv xyz + y + z + x \\pmod{3},\n$$\nmeaning $xyz \\equiv 3 \\pmod{3}$, a contradiction. Thus, we may assume that, for example, $z$ is divisible by $3$. Then $3$ also divides $x^3 + y^3 - 25x^2y$. Note that we always have $3 \\mid x^3 - x$, $3 \\mid y^3 - y$ thus we get $3 \\mid x + y(1 - x^2)$. Now, if $3$ does not divide $x$, then it would not divide $y(1 - x^2)$, which is a contradiction since $3 \\mid 1 - x^2$. Therefore, $3 \\mid x$, and consequently also $y$, as required. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40812,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a non-isosceles triangle with $M$ as the midpoint of $BC$. Take a point $D \\neq M$ on the segment $BC$ and the circumcircles of triangles $ABD$, $ACD$ intersect $AC$, $AB$ again at $E$, $F$ respectively. Line $BE$, $CF$ meet at $K$ and $AK$, $EF$ meet at $I$. Denote $O$ as the circumcenter of triangle $AEF$ and suppose that $(\\omega)$ is some circle passing through $B$, $C$ and external tangent to $(O)$. Prove that the common tangent of $(\\omega)$ and $(O)$ passes through the orthocenter of triangle $IMO$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40813,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence of positive integers $\\{a_n\\}_{n \\ge 1}$ is such that\n$$\na_n \\le 1000^{1000^{\\sqrt{\\log_2 n}}}, \\quad \\forall n = 1, 2, 3, \\dots\n$$\nIt is known that any number appears in the sequence at most $5000$ times. Prove that there are infinitely many primes, each of which divides at least one member of the sequence.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S$ be the set of all numbers that appear in the sequence $\\{a_n\\}_{n \\ge 1}$. Since each number appears at most $5000$ times, for $N$ terms of the sequence, the number of distinct elements in $S$ is at least $N/5000$.\n\nSuppose, for contradiction, that only finitely many primes divide the elements of $S$. Then all elements of $S$ are composed only of these finitely many primes, say $p_1, p_2, \\dots, p_k$.\n\nThus, every $a_n$ is of the form $p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_k^{\\alpha_k}$, where $\\alpha_i \\ge 0$ are integers.\n\nLet $M = 1000^{1000^{\\sqrt{\\log_2 N}}}$, so $a_n \\le M$ for $n \\le N$.\n\nThe number of possible $a_n$ is at most the number of $k$-tuples $(\\alpha_1, \\dots, \\alpha_k)$ such that $p_1^{\\alpha_1} \\cdots p_k^{\\alpha_k} \\le M$.\n\nFor each $i$, $\\alpha_i \\le \\log_{p_i} M$. So the total number of possible $a_n$ is at most\n$$\n\\prod_{i=1}^k (1 + \\log_{p_i} M) \\le (1 + \\log_2 M)^k\n$$\nsince $p_i \\ge 2$.\n\nBut $\\log_2 M = \\log_2 (1000^{1000^{\\sqrt{\\log_2 N}}}) = 1000^{\\sqrt{\\log_2 N}} \\cdot \\log_2 1000$.\n\nSo $1 + \\log_2 M \\le 2 \\cdot 1000^{\\sqrt{\\log_2 N}}$ for large $N$.\n\nTherefore, the number of possible $a_n$ is at most $\\left(2 \\cdot 1000^{\\sqrt{\\log_2 N}}\\right)^k$.\n\nBut the number of distinct $a_n$ among the first $N$ terms is at least $N/5000$.\n\nSo\n$$\n\\frac{N}{5000} \\le \\left(2 \\cdot 1000^{\\sqrt{\\log_2 N}}\\right)^k\n$$\nTake logarithms:\n$$\n\\log_2 N - \\log_2 5000 \\le k \\left(1 + \\sqrt{\\log_2 N} \\cdot \\log_2 1000\\right)\n$$\nFor large $N$, $\\log_2 N$ grows much faster than $\\sqrt{\\log_2 N}$, so for fixed $k$, this inequality fails for large $N$.\n\nTherefore, our assumption that only finitely many primes divide the sequence is false.\n\nThus, there are infinitely many primes, each of which divides at least one member of the sequence.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40814,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all non-negative real numbers $c \\ge 0$ such that there exists a function $f : (0, +\\infty) \\to (0, +\\infty)$ with the property\n$$\nf(y^2 f(x) + y + c) = x f(x + y^2)\n$$\nfor all $x, y > 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Solution.** (Solution of Yousif Bakheet, IMO 2025's team member)\n![](attached_image_1.png)\nSAUDI ARABIAN IMO Booklet 2025\n---\n## Saudi Booklet 2025 — Page 49\nSolution of IMO Team selection tests\n$$ \\therefore \\text{يوجد و يحقق التساوي } f(x+\\frac{1}{x}) \\ge f(x) \\text{ ($x \\ge 0$)} \\\\ \\text{لما } x > 0 \\text{ نريد } \\therefore \\text{ أن } f(x) \\ge 0 $$\n$$\n\\therefore \\text{如果 } \\left( \\frac{1-f(x)}{x} \\right)^{\\frac{1}{1-f(x)}} \\le \\frac{1}{4} \\text{ と } \\left( \\frac{1-f(x)}{x} \\right)^{\\frac{1}{1-f(x)}} > \\frac{1}{4} \\\\ \\text{那么 } f(x) \\ge \\frac{1+\\sqrt{1-4x(1-f(x))}}{\\left(\\frac{1-f(x)}{x}\\right)^{\\frac{1}{1-f(x)}}} \\text{ と } \\therefore \\text{ يحقق التساوي } f(x) \\ge 0\n$$\n$$\n\\text{لما } \\left( \\frac{1-f(x)}{x} \\right)^{\\frac{1}{1-f(x)}} \\le \\frac{1}{4} \\text{ と } \\left( \\frac{1-f(x)}{x} \\right)^{\\frac{1}{1-f(x)}} = \\frac{1}{4} \\text{ と } \\text{يحقق التساوي } f(x) = 0 \\text{ (لما نحقق قسمة f(x) في x)}\n$$\n$$\n\\text{لما } \\left( \\frac{1-f(x)}{x} \\right)^{\\frac{1}{1-f(x)}} > \\frac{1}{4} \\text{ と } \\left( \\frac{1-f(x)}{x} \\right)^{\\frac{1}{1-f(x)}} \\le \\frac{1}{4} \\text{ と } \\text{يحقق قسمة } f(x) \\text{ في x} \\text{، و نستنتج أن } C > 0 \\text{ يحقق}\n$$\n![](attached_image_2.png)\nالآن ليكن $C > 0$، لربما\n$$\nf(x) = f(x+\\frac{1}{x}) \\text{ يحقق قسمة f(x) في x} \\text{، و يحقق قسمة } f(x) = C \\text{ في x}\n$$\nنستعمل جدراً\nليكن\n$$\n\\text{إنما لو } f(x) = 1 \\text{ فن ن查閱 } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} \\le f(x) \\text{ يحقق قسمة } f(x) = 1 \\text{ بالعواملة}\n$$\nوما يrest في إيجاد العواملة و كن لك في باقي المات\n$$\n\\text{لما لو } f(x) = 1 \\text{ نريد } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} \\le \\frac{1}{4} \\text{ と } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} = \\frac{1}{4} \\text{ يحقق قسمة } f(x) = 0 \\text{ (لما نحقق قسمة } f(x) = 1 \\text{ في x)}\n$$\n$$\n\\text{لما مرتب على قسمات } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} \\le \\frac{1}{4} \\text{ と } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} = \\frac{1}{4} \\text{ يحقق قسمة } f(x) = 0 \\text{ (لما نحقق قسمة } f(x) = 1 \\text{ في x)}\n$$\n$$\n\\text{لما لو } f(x) = 1 \\text{ نريد } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} \\le \\frac{1}{4} \\text{ と } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} = \\frac{1}{4} \\text{ يحقق قسمة } f(x) = 0 \\text{ (لما نحقق قسمة } f(x) = 1 \\text{ في x)}\n$$\n$$\n\\text{و ما مرتب على قسمات } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} \\ge \\frac{1}{4} \\text{ と } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} = \\frac{1}{4} \\text{ يحقق قسمة } f(x) = 0 \\text{ (لما نحقق قسمة } f(x) = 1 \\text{ في x)}\n$$\n$$\n\\text{لما نستنتج أن } C > 0 \\text{ يحقق قسمة } f(x) = C \\text{ (لما نحقق قسمة } f(x) = 0 \\text{ في x)} \\\\ \\text{لما نستنتج أن } C < 0 \\text{ يحقق قسمة } f(x) = C \\text{ (لما نحقق قسمة } f(x) = 0 \\text{ في x)}\n$$\n![](attached_image_3.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40815,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $k \\ge 2$ such that for all $x, y, z \\ne 1$ satisfying $x + y + z = 3$ and $xyz = 1$ then the $k$-th root of numbers $|x|, |y|$ and $|z|$ are the side lengths of some triangle.",
    "options": [],
    "answer": "4",
    "solution": "Suppose that $x = \\max\\{x, y, z\\}$. We see that if $x, y, z > 0$ then based on the AM-GM inequality, we have $x + y + z \\ge 3\\sqrt[3]{xyz} = 3$ and the equality must occur, which means that $x = y = z = 1$, is not satisfied. \n\nNext, consider $y, z < 0$ and put $u = -y > 0$, $v = -z > 0$. One can rewrite the given conditions as $x = 3 + u + v$ and $xuv = 1$. Thus $x$ is always greater than the sum of the remaining two numbers, we need to find $k$ such that\n$$\n\\sqrt[k]{x} < \\sqrt[k]{u} + \\sqrt[k]{v}.\n$$\nChecking with $u = v$, one can solve the given condition to get $(x, u, v) = (4, \\frac{1}{2}, \\frac{1}{2})$. Then, by substituting this tuple into the above inequality, one can get\n$$\n\\sqrt[6]{4} < \\frac{2}{\\sqrt[6]{2}} \\implies \\sqrt[6]{8} < 2.\n$$\nFrom this we have $k > 3$ or $k \\ge 4$. Now we will prove that $k = 4$ satisfies the given condition, that is,\n$$\n\\sqrt[4]{x} < \\sqrt[4]{u} + \\sqrt[4]{v} \\iff \\sqrt{x} < \\sqrt{u} + \\sqrt{v} + 2\\sqrt[4]{uv} = \\sqrt{u} + \\sqrt{v} + \\frac{2}{\\sqrt[4]{x}}\n$$\nWe will prove the stronger inequality\n$$\n\\sqrt{x} < \\sqrt{u} + \\sqrt{v} + \\frac{2}{\\sqrt{x}} \\iff \\sqrt{x} - \\frac{2}{\\sqrt{x}} < \\sqrt{u} + \\sqrt{v}.\n$$\nSince $x \\ge 4$ then $\\sqrt{x} - \\frac{2}{\\sqrt{x}} > 0$ and squaring both sides gives\n$$\nx + \\frac{4}{x} - 4 < u + v + 2\\sqrt{uv} \\iff \\frac{4}{x} < 1 + \\frac{2}{\\sqrt{x}}.\n$$\nThe final inequality is true because $x > 3$. Therefore $k_{\\min} = 4$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40816,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each student in the class wrote down in his notebook a geometric progression consisting of different positive numbers. Let $A$ be the sum of the first terms of all these progressions, $B$ the sum of their second terms, and $C$ the sum of their third terms. Prove that the sequence $A$, $B$, $C$ is not an arithmetic progression.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40817,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $G$ is a simple planar graph with minimum degree at least $5$. Prove that $G$ contains an edge $\\{x, y\\}$ such that $\\deg(x) + \\deg(y) \\le 11$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By adding edges we can assume that $G$ is a triangulation (note that if we add edges the minimum degree only increases and if we find an edge $\\{x, y\\}$ such that $\\deg(x) + \\deg(y) \\le 11$ in the new graph, this edge will also be present in the original graph, since otherwise one of its endpoints would have degree smaller than $5$ in the original graph).\n\nLet $V$ be the set of vertices, $E$ the set of edges, and $F$ the set of faces of our triangulation. Let $d(v)$ be the degree of $v \\in V$ and let $d(f)$ be the number of edges on the boundary of $f$. Here, if $e \\in E$ is on the boundary of only one face, then we count it two times. Hence,\n$$\n\\sum_{f} d(f) = 2|E| = \\sum_{v \\in V} d(v)\n$$\nThus, by Euler's formula, we have\n$$\n\\sum_{v \\in V} (6 - d(v)) = \\sum_{f} (6 - 2d(f)) + \\sum_{v \\in V} (6 - d(v)) = 6|F| - 4|E| + 6|V| - 2|E| = 12\n$$\nsince for every face $d(f) = 3$. Let us give a charge $6 - d(v)$ to every vertex $v \\in V$. The total charge is $12$. The only vertices with positive initial charge are vertices satisfying $d(v) = 5$. Now we discharge the system using a single rule: every vertex of degree $5$ gives charge $\\frac{1}{5}$ to each of its neighbors. Clearly the total final charge is still $12$. Thus there are vertices with positive final charge. Suppose $v$ is such a vertex. Then its final charge $c(v)$ satisfies\n$$\n0 < c(v) \\le 6 - d(v) + \\frac{1}{5}d(v) = 6 - \\frac{4}{5}d(v).\n$$\nThus $d(v) \\le 7$. Now we consider three cases.\n\ni. Suppose $v$ with positive final charge satisfies $d(v) = 6$. Then its initial charge was $0$ and thus this vertex must have gained some charge. So, one of its neighbors $u$ has degree $5$ and $\\{u, v\\}$ is the desired edge.\n\nii. Suppose $v$ with positive final charge satisfies $d(v) = 5$. Then in the discharging process this vertex gave all its charge to its five neighbors. But since the final charge is positive, it must have gained some charge from a vertex of degree $5$. Thus, one of its neighbors $u$ has degree $5$ and so $\\{v, u\\}$ is the desired edge.\n\niii. Suppose $v$ with positive final charge satisfies $d(v) = 7$. Since the initial charge of $v$ was $-1$, $v$ gained charge from at least $6$ of its neighbors, so it has at least $6$ neighbors with degree $5$. There is one more neighbor of $v$ whose degree we do not control. Since $G$ is a triangulation, the neighbors of $v$ form a cycle and clearly on this cycle there are two adjacent vertices $u_1$ and $u_2$ of degree $5$. Thus $\\{u_1, u_2\\}$ is the desired edge.\n\n$\\boxed{}$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 40818,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. Point $D$ lies on side $BC$, such that the incircles of triangles $ABD$ and $ACD$ are congruent. Let $\\Omega_B$ be the circle with diameter $AB$, and let $\\Omega_C$ be the circle with diameter $AC$. Prove that line $AD$ is perpendicular to one of the common tangents to the circles $\\Omega_B$ and $\\Omega_C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $AH$ as the altitude of triangle $ABC$ then clearly $H \\in \\Omega_B, \\Omega_C$ so $AH$ is the common chord of the two circles. Let $EF$ be the common tangent near $A$ of the two circles with $E \\in \\Omega_B, F \\in \\Omega_C$. According to the familiar property, $AH$ passes through the midpoint $K$ of $EF$. Draw $Ax \\parallel EF$ then $A(Kx, EF) = -1$.\n\nWe redefine $D$ as a point on $BC$ such that $AD \\perp EF$. Let $I, J$ be the two incenters of $ABD, ACD$ respectively and let $AI, AJ$ intersect $BC$ at $M, N$ respectively. Let $R, S$ be the centers of the circles $\\Omega_B, \\Omega_C$ (which are also the midpoints of $AB, AC$) then $RE \\parallel AM$ (because they are both perpendicular to $EF$) so\n$$\n\\angle ABE = \\frac{1}{2}\\angle ARE = \\frac{1}{2}\\angle DAB = \\angle MAB,\n$$\nwhich implies $BE \\parallel AM$ or $AE \\perp AM$. Similarly, $AF \\perp AN$.\n\n![](attached_image_1.png)\n\nNow draw $Ay \\parallel MN$ then $AK \\perp Ay$. From there, according to the orthogonality property of two harmonic bundles, we immediately have $A(Dy, MN) = -1$. Thus, $DM = DN$. Then, according to the property of the angle bisector,\n$$\n\\frac{IM}{IA} = \\frac{DM}{DA} = \\frac{DN}{DA} = \\frac{JN}{JA} \\implies IJ \\parallel MN.\n$$\nTherefore, the distances from $I, J$ to the line $BC$ are equal implies that the radius of the circle inscribed in triangles $ABD, ACD$ are equal. So in other words, $D$ is the given point in the problem. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40819,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and $A = \\{n, n+1, \\dots, 2n-1\\}$. Prove that there is a number $a \\in A$ such that the sum of elements of the set $A \\setminus \\{a\\}$ is not divisible by any element of $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40820,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N$ be a positive integer. Ali and Hadi play a game in which they start by writing the numbers $1, 2, \\dots, N$ on a board. They then take turns to make a move, starting with Ali. Each move consists of choosing a pair of integers $(k, n)$, where $k \\ge 0$ and $n$ is one of the integers on the board, and then erasing every integer $s$ on the board such that $2^k|n - s$. The game continues until the board is empty. The player who erases the last integer on the board loses.\nDetermine all values of $N$ for which Ali can ensure that he wins, no matter how Hadi plays.",
    "options": [],
    "answer": "Ali can force a win exactly when N = 2^n with n odd, or when N = t·2^n with n even and odd t > 1.",
    "solution": "The answer is that Geoff wins when $N$ is of the form $2^n$ for $n$ odd or of the form $t2^n$ for $n$ even and $t > 1$ odd.\n\nCommon remarks. We will say that a set $S$ wins if the current player wins given $S$ as the current set of integers on the board. Otherwise, we will say that $S$ loses. We will let $J(S, \\mathcal{T}) = (2S - 1) \\cup (2\\mathcal{T})$. Note that every subset of $\\mathbb{Z}$ can be written as $J(S, \\mathcal{T})$ for some unique pair $(S, \\mathcal{T})$ of subsets of $\\mathbb{Z}$. We will let $[n]$ denote the set $\\{1, 2, \\dots, n\\}$.\n\n**Lemma 1.** For any set $S, \\mathcal{T}$ wins if and only if $J(S, \\emptyset)$ wins. Similarly, $S$ wins if and only if $J(\\emptyset, S)$ wins.\n*Proof.* Let $(k, m)$ be a move on $S$, and let $\\mathcal{T}$ be the result of applying the move. Then we can reduce $J(S, \\emptyset)$ to $J(\\mathcal{T}, \\emptyset)$ by applying the move $(k + 1, 2m - 1)$.\nConversely, let $(k, m)$ be a move on $J(S, \\emptyset)$. We can express the result of this move as $J(\\mathcal{T}, \\emptyset)$ for some $\\mathcal{T}$. Then we can reduce $S$ to $\\mathcal{T}$ by applying the move $(\\max(k - 1, 0), (k + 1)/2)$.\nThis gives us a natural bijection between games starting with $S$ and games starting with $J(S, \\emptyset)$ and thus proves the first part of the lemma. The second part follows by a similar argument.\n\n**Lemma 2.** If $S, \\mathcal{T}$ are nonempty and at least one of them loses, then $J(S, \\mathcal{T})$ wins.\n*Proof.* If $S$ is losing, then we can delete $J(\\emptyset, \\mathcal{T})$ using the move $(1, t)$ for some $t \\in J(\\emptyset, \\mathcal{T})$, which leaves the losing set $J(S, \\emptyset)$. Similarly, if $\\mathcal{T}$ is losing, then we can delete $J(S, \\emptyset)$ using the move $(1, s)$ for some $s \\in J(S, \\emptyset)$, leaving the losing set $J(\\emptyset, \\mathcal{T})$.\n\n**Lemma 3.** If $S$ is nonempty and wins, then $J(S, S)$ loses.\n*Proof.* From this position, we can convert any sequence of moves into another valid sequence of moves by replacing $(k, 2n - 1)$ with $(k, 2n)$, and vice versa. Thus we may assume that the initial move $(k, m)$ has $m$ odd. We want to show that any such move results in a winning position for the other player.\nThe move $(0, m)$ loses immediately. Otherwise, the move results in the set $J(\\mathcal{T}, S)$ for some set $\\mathcal{T}$. There are three cases. If $\\mathcal{T}$ is empty then the other player gets the winning set $J(\\emptyset, S)$. If $\\mathcal{T}$ is losing then the other player can choose the move $(1, s)$ for some $s \\in J(\\emptyset, S)$, which leaves the losing set $J(\\mathcal{T}, \\emptyset)$. If $\\mathcal{T}$ is nonempty winning then the other player can choose the move $(k, m + 1)$, which results in the position $J(\\mathcal{T}, \\mathcal{T})$. We can do induction on $|S|$ to show that this is a losing set.\n\n**Lemma 4.** $[2n]$ wins if and only if $[n]$ loses.\n*Proof.* Note that $[2n] = J([n], [n])$, the result follows directly from Lemma 2, 3.\n\n**Lemma 5.** For any integer $n \\ge 1$, $[2n + 1]$ wins.\n*Proof.* By Lemma 4, either $[n]$ or $[2n]$ loses. If $[n]$ loses, then by Lemma 2 we have that $[2n+1] = J([n+1], [n])$ wins. Otherwise, $[2n]$ loses, and therefore $[2n+1]$ wins by choosing the move $(k, 2n+1)$ for $k$ big enough so only $2n+1$ is eliminated.\n\nIt remains to verify the original answer. We have two cases to consider:\n* Suppose $N = 2^n$ for some $n$. For $N = 1$, every move is an instant loss for Geoff. Then by Lemma 4, Geoff wins for $N = 2^n$ if and only if Geoff loses for $N = 2^{n-1}$, and thus by induction we have that Geoff wins for $N = 2^n$ if and only if $n$ is odd.\n* Otherwise, $N = t2^n$, for some $n$ and some $t > 1$ with $t$ odd. By Lemma 5, Geoff wins when $n = 0$. Then by Lemma 4, Geoff wins for $N = t2^n$ if and only if Geoff loses for $N = t2^{n-1}$, and thus by induction on $n$ we have that Geoff wins for $N = t2^n$ if and only if $n$ is even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40821,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $D$ тачка на страници $A C$ троугла $A B C$ у коме је $A B < B C$ таква да је $A B = B D$. Круг уписан у $\\triangle A B C$ додирује $A B$ у $K$ и $A C$ у $L$, а $J$ је центар уписаног круга троугла $B C D$. Доказати да $K L$ полови дуж $A J$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nНека је $M$ тачка на $A C$ таква да је $J M \\parallel K L$. Довољно је доказати да је $A M = 2 A L$.\n\nИз $\\angle B D A = \\alpha$ добијамо $\\angle J D M = 90^\\circ - \\frac{\\alpha}{2} = \\angle K L A = \\angle J M D$, па је $J M = J D$, а додирна тачка уписаног круга $\\triangle B C D$ са $C D$ је средиште $T$ дужи $M D$. Према томе, $D M = 2 D T = B D + C D - B C = A B - B C + C D$, одакле је\n$$\nA M = A D + D M = A C + A B - B C = 2 A L\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40822,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nОдредити све парове природних бројева $(x, n)$ који су решења једначине\n\n$$\nx^{3}+2x+1=2^{n}\n$$",
    "options": [],
    "answer": "(1, 2)",
    "solution": "Solution:\n\nПровером се добија да је за $n \\leqslant 2$ једино решење пар $(1,2)$. Докажимо да за $n \\geqslant 3$ нема решења.\n\nБрој $x$ мора бити непаран, па је $x^{2}+2 \\equiv 3 \\pmod{8}$. Сада из $x\\left(x^{2}+2\\right) \\equiv -1 \\pmod{8}$ следи да је $x \\equiv 5 \\pmod{8}$. Шта више, како $3 \\mid x\\left(x^{2}+2\\right)$ (уколико $3 \\nmid x$ тада $3 \\mid x^{2}+2$), мора бити $2^{n} \\equiv 1 \\pmod{3}$, па је $n$ паран број.\n\nДодавањем обема странама једнакости броја 2 добија се\n$$\n(x+1)\\left(x^{2}-x+3\\right)=2^{n}+2\n$$\nКако је $n$ паран, $2^{n}$ је потпун квадрат, па је број $-2$ квадратни остатак по сваком непарном простом делиоцу $p$ броја $(x+1)\\left(x^{2}-x+3\\right)$. Зато је\n$$\n1=\\left(\\frac{-2}{p}\\right)=\\left(\\frac{-1}{p}\\right)\\left(\\frac{2}{p}\\right)=(-1)^{\\frac{p-1}{2}}(-1)^{\\frac{p^{2}-1}{8}}=(-1)^{\\frac{(p-1)(p+5)}{8}}\n$$\nодакле следи да је $p$ облика $8k+1$ или $8k+3$. Будући производ таквих простих бројева, и сам број $x^{2}-x+3$ мора бити тог облика. Међутим, како је $x \\equiv 5 \\pmod{8}$, важи $x^{2}-x+3 \\equiv 7 \\pmod{8}$ што је контрадикција.\n\nДакле, једино решење дате једначине је $(x, n)=(1,2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40823,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $k$ природан број. За сваку функцију $f: \\mathbb{N} \\rightarrow \\mathbb{N}$, нека је низ функција $\\left(f_{m}\\right)_{m \\geqslant 1}$ дефинисан са $f_{1}=f$ и $f_{m+1}=f \\circ f_{m}$ за $m \\geqslant 1$. Функција $f$ је $k$-фина уколико за све $n \\in \\mathbb{N}$ важи\n$$\nf_{k}(n)=f(n)^{k}\n$$\n\na) За које $k$ постоји $1$-$1$ $k$-фина функција $f$?\n\nб) За које $k$ постоји на $k$-фина функција $f$?",
    "options": [],
    "answer": "a) For all k. b) Only for k equal to 1.",
    "solution": "Solution:\n\nСвака функција је $1$-фина, па је одговор на оба дела задатка потврдан. Нека је надаље $k \\geqslant 2$. Свака $k$-фина функција је $1$-$1$ јер из $f(m)=f(n)$ следи $m^{k}=f_{k}(m)=f_{k}(n)=n^{k}$, тј. $m=n$.\n\na) Одговор: ДА. Конструишимо функцију $f$ индуктивно на следећи начин. Нека је $n$ најмањи природан број чија слика није одређена.\n\n(1) ако је $n=1$, онда је $f(n)=1$;\n\n(2) ако је $n=a^{k}$ за неко цело $a>1$, дефинишемо $f(n)=f(a)^{k}$;\n\n(3) ако $n$ није потпун $k$-ти степен изаберемо најмањих $k-1$ природних бројева $n_{1}, n_{2}, \\ldots, n_{k-1}$ који нису потпуни $k$-ти степени и за које до сада нису одређене слике, и дефинишемо $f\\left(n_{1}\\right)=n_{2}, f\\left(n_{2}\\right)=n_{3}, \\ldots, f\\left(n_{k-1}\\right)=n_{1}^{k}$.\n\nНа овај начин функција $f$ је добро дефинисана. Покажимо да је она $k$-фина. За свако $n \\in \\mathbb{N}$ које није $k$-ти степен постоје бројеви $n_{1}, \\ldots, n_{k-1}$ из услова (3) такви да је $n_{i}=n$ за неко $1 \\leqslant i \\leqslant k-1$. Тада важи $f_{k}\\left(n_{i}\\right)=f_{i}\\left(n_{1}^{k}\\right)=f_{i}\\left(n_{1}\\right)^{k}=f\\left(n_{i}\\right)^{k}$. Такође, ако је $n$ потпун $k$-ти степен, тада је $n=n_{i}^{k^{s}}$ за неко $i$ и $s$, па према (2) важи $f_{k}(n)=f_{k}\\left(n_{i}\\right)^{k^{s}}=n_{i}^{k^{s+1}}=n^{k}$, што доказује наше тврђење.\n\nб) Одговор: НЕ. Заиста, ако је $f$ на и $k$-фина, за свако $a_{0}$ постоји низ природних бројева $a_{1}, a_{2}, \\ldots$ таквих да је $f\\left(a_{k+1}\\right)=a_{k}$ за све $k$, одакле је $a_{k}^{k}=f_{k}\\left(a_{k}\\right)=a_{0}$, што је немогуће ако $a_{0}$ није $k$-ти степен.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40824,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nДат је неједнакокраки троугао $ABC$. Нека су $AD$, $BE$, $CF$ симетрале углова овог троугла ($D \\in BC$, $E \\in AC$, $F \\in AB$). Нека су $K_{a}$, $K_{b}$, $K_{c}$ тачке на уписаном кругу троугла $ABC$ такве да су $DK_{a}$, $EK_{b}$, $FK_{c}$ тангенте на уписани круг и да $K_{a} \\notin BC$, $K_{b} \\notin AC$, $K_{c} \\notin AB$. Нека су $A_{1}$, $B_{1}$, $C_{1}$ средишта страница $BC$, $CA$, $AB$. Доказати да се праве $A_{1}K_{a}$, $B_{1}K_{b}$, $C_{1}K_{c}$ секу на уписаном кругу троугла $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nДокажимо да су троуглови $K_{a}K_{b}K_{c}$ и $A_{1}B_{1}C_{1}$ хомотетични. Да бисмо то доказали, довољно је да докажемо да је $K_{a}K_{b} \\parallel A_{1}B_{1}$, односно $K_{a}K_{b} \\parallel AB$ (аналогно ће следити и за друге парове страница).\n\nОзначимо $M = K_{a}K_{b} \\cap BC$, са $S$ означимо центар уписаног круга, и са $T$ означимо произвољну тачку на уписаном кругу. Означимо $\\alpha = \\angle BAS$, $\\beta = \\angle CBS$, $\\gamma = \\angle ACS$. Користећи оријентисане углове (по модулу $180^\\circ$), добијамо $\\angle B'EB = \\beta + 2\\gamma$, и аналогно $\\angle A'DA = \\alpha + 2\\beta$, а одатле и $\\angle A'DK_{a} = 2\\alpha + 4\\beta$. Затим, $\\angle B'TK_{b} = \\angle B'SE = 90^\\circ + \\angle B'ES = \\gamma - \\alpha$ и аналогно $\\angle A'TK_{a} = \\beta - \\gamma$. Затим, $\\angle A'TB' = \\angle A'SC = 90^\\circ + \\angle A'CS = \\alpha + \\beta$. И на крају добијамо $\\angle K_{a}TK_{b} = \\angle K_{a}TA' + \\angle A'TB' + \\angle B'TK_{b} = 2\\gamma$.\n\nТакође, из троугла $DK_{a}M$ добијамо $\\angle CMK_{a} = \\angle CDK_{a} + \\angle DK_{a}M = \\angle A'DK_{a} + \\angle DK_{a}K_{b} = (2\\alpha + 4\\beta) + \\angle K_{a}TK_{b} = (2\\alpha + 4\\beta) + 2\\gamma = 2\\beta$. Дакле, $\\angle CMK_{a} = \\angle CBA$, одакле следи да је $K_{a}K_{b} \\parallel AB$, што је требало доказати. Дакле, троуглови $K_{a}K_{b}K_{c}$ и $A_{1}B_{1}C_{1}$ су хомотетични.\n\nПриметимо такође да је коефицијент хомотетије позитиван: ако би био негативан, дужи $K_{a}A_{1}$, $K_{b}B_{1}$, $K_{c}C_{1}$ би се секле у једној тачки. Ако је $\\alpha > \\beta$, онда се тачке $C_{1}$ и $K_{c}$, па и цела дуж $K_{c}C_{1}$, налазе унутар четвороугла $SFB D$. Зато, ако без умањења општости претпоставимо $\\alpha > \\beta > \\gamma$, онда $K_{c}C_{1} \\subset SFB D$, али $K_{a}A_{1} \\subset SDCE$, па су ове две дужи дисјунктне.\n\nКако су троуглови $K_{a}K_{b}K_{c}$ и $A_{1}B_{1}C_{1}$ хомотетични, њихови описани кругови су такође хомотетични. Али то су Ојлеров и уписани круг троугла $ABC$, респективно, и познато је да се та два круга додирују изнутра у Фојербаховој тачки троугла $ABC$. Стога (уз чињеницу да је коефицијент хомотетије позитиван), центар хомотетије је управо Фојербахова тачка. Одавде следи да се $A_{1}K_{a}$, $B_{1}K_{b}$, $C_{1}K_{c}$ секу у Фојербаховој тачки троугла $ABC$, која припада уписаном кругу троугла $ABC$, чиме је тврдње доказано.\n\n![](attached_image_1.png)\nSolution:\n\nАлтернативно решење. Нека је уписани круг троугла $ABC$ јединична кружница у комплексној равни. Тада је $a = \\frac{2b'c'}{b'+c'}$, $b = \\frac{2a'c'}{a'+c'}$, $c = \\frac{2a'b'}{a'+b'}$. Затим је\n$$\na_{1} = \\frac{b+c}{2} = \\frac{a'^{2}b' + a'^{2}c' + 2a'b'c'}{(a'+b')(a'+c')}\n$$\nВредност $k_{a}$ налазимо из услова $\\frac{k_{a}}{a} = \\overline{\\left(\\frac{a'}{a}\\right)}$, одакле је $k_{a} = \\frac{1}{a'}\\overline{\\bar{a}} = \\frac{b'c'}{a'}$. Сада налазимо пресечну тачку $z$ уписаног круга (одакле $|z| = 1$) и праве $K_{a}A_{1}$ (одакле $\\frac{z-k_{a}}{a_{1}-k_{a}} = \\overline{\\left(\\frac{z-k_{a}}{a_{1}-k_{a}}\\right)}$). Други услов можемо трансформисати у облик\n$$\n\\overline{\\left(a_{1}-k_{a}\\right)}(z-k_{a}) = \\left(\\frac{1}{z} - \\frac{1}{k_{a}}\\right)(a_{1}-k_{a})\n$$\nодакле (због $z \\neq k_{a}$) следи $\\overline{\\left(a_{1}-k_{a}\\right)} = -\\frac{1}{zk_{a}}(a_{1}-k_{a})$, па је\n$$\nz = -\\frac{1}{k_{a}} \\frac{a_{1}-k_{a}}{\\overline{\\left(a_{1}-k_{a}\\right)}} = -\\frac{(a'^{2}-b'c')(a'b'+a'c'+b'c')}{(b'c'-a'^{2})(a'+b'+c')} = \\frac{a'b'+a'c'+b'c'}{a'+b'+c'}\n$$\nКако је наведени израз симетричан по $a'$, $b'$, $c'$, аналогно се доказује да ће праве $K_{b}B_{1}$ и $K_{c}C_{1}$ сећи уписани круг у истој тачки, чиме је тврђење доказано.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40825,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $k$ природан број. Доказати да за позитивне реалне бројеве $x, y, z$ чији је збир једнак $1$, важи неједнакост\n\n$$\n\\frac{x^{k+2}}{x^{k+1}+y^{k}+z^{k}}+\\frac{y^{k+2}}{y^{k+1}+z^{k}+x^{k}}+\\frac{z^{k+2}}{z^{k+1}+x^{k}+y^{k}} \\geqslant \\frac{1}{7}\n$$\n\nКада важи једнакост?",
    "options": [],
    "answer": "Equality holds if and only if x = y = z = 1/3.",
    "solution": "Solution:\n\nДати израз је симетричан, па се без губљења општости може претпоставити да је $x \\geqslant y \\geqslant z$. Тада је\n$$\nx^{k+1}+y^{k}+z^{k} \\leqslant y^{k+1}+z^{k}+x^{k} \\leqslant z^{k+1}+x^{k}+y^{k}\n$$\nЗаиста, довољно је доказати прву неједнакост, тј. да је $x^{k+1}+y^{k} \\leqslant y^{k+1}+x^{k}$. Ова неједнакост је еквивалентна са $\\left(\\frac{y}{x}\\right)^{k} \\leqslant \\frac{1-x}{1-y}$. Како је $y \\leqslant x$, довољно је доказати да је $\\frac{y}{x} \\leqslant \\frac{1-x}{1-y}$, што је еквивалентно тачној неједнакости $0 \\leqslant x-x^{2}-y+y^{2}=(x-y)(1-x-y)=(x-y) z$.\n\nПрименом неједнакости Чебишева на тројке $\\left(x^{k+2}, y^{k+2}, z^{k+2}\\right)$ и $\\left(\\frac{1}{x^{k+1}+y^{k}+z^{k}}, \\frac{1}{y^{k+1}+z^{k}+x^{k}}, \\frac{1}{z^{k+1}+x^{k}+y^{k}}\\right)$ добија се\n$$\n\\sum_{\\text{cyc}} \\frac{x^{k+2}}{x^{k+1}+y^{k}+z^{k}} \\geqslant \\frac{1}{3} \\sum_{\\text{cyc}} x^{k+2} \\sum_{\\text{cyc}} \\frac{1}{x^{k+1}+y^{k}+z^{k}} = L\n$$\nУколико се у $L$ поново примени неједнакост Чебишева на тројке $(x, y, z)$ и $\\left(x^{k+1}, y^{k+1}, z^{k+1}\\right)$ добија се\n$$\nL \\geqslant \\frac{1}{3} \\cdot \\frac{1}{3} \\sum_{\\text{cyc}} x \\sum_{\\text{cyc}} x^{k+1} \\sum_{\\text{cyc}} \\frac{1}{x^{k+1}+y^{k}+z^{k}} = L'\n$$\nИз неједнакости Коши-Шварц-Буњаковског следи\n$$\n\\sum_{\\text{cyc}} \\frac{1}{x^{k+1}+y^{k}+z^{k}} \\sum_{\\text{cyc}}\\left(x^{k+1}+y^{k}+z^{k}\\right) \\geqslant 9\n$$\nпа је\n$$\nL' \\geqslant \\frac{x^{k+1}+y^{k+1}+z^{k+1}}{x^{k+1}+y^{k+1}+z^{k+1}+2\\left(x^{k}+y^{k}+z^{k}\\right)}\n$$\nи самим тим довољно је доказати да је\n$$\n3\\left(x^{k+1}+y^{k+1}+z^{k+1}\\right) \\geqslant x^{k}+y^{k}+z^{k}\n$$\nПоследња неједнакост се добија поновном применом неједнакости Чебишева на тројке $(x, y, z)$ и $\\left(x^{k}, y^{k}, z^{k}\\right)$.\n\nЈеднакост у свим примењеним неједнакостима важи ако и само ако је $x=y=z$, тј. ако и само ако је $x=y=z=\\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40826,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nУ скупу целих бројева решити једначину\n$$\n12^{x} + y^{4} = 2008^{z}.\n$$",
    "options": [],
    "answer": "(0, 0, 0)",
    "solution": "Solution:\n\nЗа $x < 0$ или $z \\leq 0$ једино решење је тривијално $(0, 0, 0)$. Надаље је $z > 0$. Како је $2008 = 2^{3} \\cdot 251$, обе стране једначине су дељиве са $251$. Претпоставимо да је $x = 2x_{1}$ парно. Тада $\\left(2^{x_{1}}\\right)^{2} \\equiv -\\left(y^{2}\\right)^{2} \\pmod{251}$, што дизанјем на $125$-ти степен даје $1 \\equiv \\left(2^{x_{1}}\\right)^{250} \\equiv -\\left(y^{2}\\right)^{250} \\equiv -1$ по малој Фермаовој теореми, а то је немогуће. Према томе, $x$ мора бити непарно.\n\nОчигледно је $y$ парно. Напишимо $y = 2^{u} y_{1}$ за непарно $y_{1}$. Имамо\n$$\n2^{2x} 3^{x} + 2^{4u} y_{1}^{4} = 2^{3z} 251^{z}\n$$\nКако је $2x \\neq 4u$ јер је $x$ непарно, највећи степен двојке који дели леву страну је $2^{2x}$ или $2^{4u}$, док је највећи степен двојке који дели десну страну једнак $2^{3z}$, одакле је $3z = 2x$ или $3z = 4u$. Показаћемо да ни у једном од ова два случаја дата једначина нема решења.\n\ni. $3z = 2x < 4u$; дакле, $2 \\mid z$. Скраћивање са $2^{2x}$ даје $3^{x} + 2^{4u-2x} y_{1}^{4} = 251^{z}$ што је немогуће јер је лева страна облика $4k+3$ (јер $2 \\nmid x$), а десна облика $4k+1$ ($k \\in \\mathbb{N}$).\n\nii. $3z = 4u < 2x$; опет $2 \\mid z$. Скраћивање са $2^{4u}$ даје $2^{2x-4u} 3^{x} + y_{1}^{4} = 251^{z}$. Десна страна је облика $5k+1$, па за $5 \\nmid y_{1}$ имамо $y_{1}^{4} \\equiv 1$ и $2^{2x-4u} 3^{x} \\equiv 0 \\pmod{5}$ што је немогуће, док за $5 \\mid y_{1}$ имамо $1 \\equiv 2^{2x-4u} 3^{x} \\equiv \\pm 3^{x} \\equiv \\pm 3 \\pmod{5}$ јер $2 \\nmid x$, опет немогуће.\nSolution:\n\nЗа парно $x$ лева страна једначине је облика $a^{2} + b^{2}$, а за непарно $x$ лева страна је облика $a^{2} + 3b^{2}$. Међутим, како су $-1$ и $-3$ квадратни неостаци по модулу $251$, ни $a^{2} + b^{2}$ ни $a^{2} + 3b^{2}$ не могу да буду дељиви са $251$ ако $251 \\nmid a, b$. Према томе, дата једначина нема целобројних решења за $z \\geq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40827,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nДат је троугао $A B C$. Нека су тачке $D$ и $E$ на правој $A B$ такве да је $D-A-B-E$, $A D = A C$ и $B E = B C$. Симетрале унутрашњих углова код темена $A$ и $B$ секу наспрамне странице у тачкама $P$ и $Q$, редом, а описану кружницу око троугла $A B C$ у тачкама $M$ и $N$, редом. Права која спаја тачку $A$ са центром кружнице описане око троугла $B M E$ и права која спаја тачку $B$ са центром кружнице описане око троугла $A N D$ секу се у тачки $X$, $X \\neq C$. Доказати да је $C X \\perp P Q$. (Дуиан Ђукић)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nОзначимо са $U$ центар описаног круга $\\triangle B M E$. Применимо инверзију са центром $A$ и квадратом полупречника $A B \\cdot A C$. Тачке $B$ и $C$ се сликају у тачке $B'$ и $C'$ симетричне тачкама $C$ и $B$ у односу на $A P$, тачке $P$ и $M$ се сликају једна у другу, а $E$ се слика у тачку $E'$ симетричну $Q$ у односу на $A P$. Према томе, права $A U$ се поклапа са правом која спаја $A$ са центром круга $B' P E'$ (наравно, центри се не сликају један у други!). Видимо да је та права симетрична\n\n![](attached_image_1.png)\n\nправој $A Z$ у односу на симетралу угла $A$, где је $Z$ центар круга описаног око $\\triangle C P Q$.\n\nАналогно се добија да је права $B Z$ симетрична правој која спаја $B$ са центром $V$ круга $A N D$ у односу на симетралу угла $B$. По Чевиној теореми у тригонометријском облику (или по тврђењу о изогонално спрегнутим тачкама), праве симетричне правим $A U, B V, C X$ у односу на симетрале углова $A, B, C$ редом се такође секу у једној тачки, што значи да је права $C Z$ симетрична $C X$ у односу на симетралу угла $C$. Али $Z$ је центар круга $C P Q$, одакле следи да права $C X$ садржи висину троугла $C P Q$, а то смо и желели да докажемо.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40828,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nДоказати да за позитивне реалне бројеве $a, b$ и $c$, такве да је $a+b+c=1$, важи неједнакост\n$$\n\\frac{1}{b c+a+\\frac{1}{a}}+\\frac{1}{c a+b+\\frac{1}{b}}+\\frac{1}{a b+c+\\frac{1}{c}} \\leqslant \\frac{27}{31}\n$$\n(Марко Радовановић са сарадницима)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nТражена неједнакост је очигледно еквивалентна неједнакости\n$$\n\\frac{a}{p+a^{2}}+\\frac{b}{p+b^{2}}+\\frac{c}{p+c^{2}} \\leq \\frac{27}{31}\n$$\nгде је $a+b+c=1$ и $p=a b c+1$. Посматраћемо функцију\n$$\nf(x)=\\frac{3(a+b+c)}{3 x+a^{2}+b^{2}+c^{2}}-\\frac{a}{x+a^{2}}-\\frac{b}{x+b^{2}}-\\frac{c}{x+c^{2}}\n$$\nДоказаћемо да важи $f(x) \\geq 0$ за све $x \\geq a b+b c+c a$. Свођење израза за $f(x)$ на заједнички именилац даје\n$$\nf(x)=\\frac{A x^{2}+B x+C}{\\left(x+a^{2}\\right)\\left(x+b^{2}\\right)\\left(x+c^{2}\\right)\\left(3 x+a^{2}+b^{2}+c^{2}\\right)}\n$$\nпри чему је $A \\geq 0 \\geq C$. Заправо, лако се добија\n$$\n\\begin{gathered}\nA=2 a^{3}+2 b^{3}+2 c^{3}-a b(a+b)-a c(a+c)-b c(b+c) \\geq 0 \\\\\nC=-a b c\\left[a\\left(b^{3}+c^{3}\\right)+b\\left(c^{3}+a^{3}\\right)+c\\left(a^{3}+b^{3}\\right)-2 a b c(a+b+c)\\right] \\leq 0\n\\end{gathered}\n$$\nПриметимо да није важно колико је $B$. Према томе, полином $P(x)=A x^{2}+B x+C$ (ако није константно 0 ) има две реалне нуле, једну позитивну (рецимо $x=x_{0}$ ) и једну негативну, и важи $P(x) \\leq 0$ за $0 \\leq x \\leq x_{0}$ и $P(x) \\geq 0$ за $x \\geq x_{0}$. Тврдимо да је $f(a b+b c+c a) \\geq 0$. Заиста,\n$$\n\\begin{aligned}\n& f(a b+b c+c a) \\\\\n& =\\frac{3(a+b+c)}{a^{2}+b^{2}+c^{2}+3(a b+b c+c a)}-\\frac{a}{(a+b)(a+c)}-\\frac{b}{(b+c)(b+a)}-\\frac{c}{(c+a)(c+b)} \\\\\n& =\\frac{3(a+b+c)}{a^{2}+b^{2}+c^{2}+3(a b+b c+c a)}-\\frac{2(a b+b c+c a)}{(a+b)(b+c)(c+a)} \\geq 0 \\quad \\text{jep je} \\\\\n& \\quad \\frac{3(a+b+c)}{a^{2}+b^{2}+c^{2}+3(a b+b c+c a)} \\geq \\frac{9}{4(a+b+c)} \\geq \\frac{2(a b+b c+c a)}{(a+b)(b+c)(c+a)},\n\\end{aligned}\n$$\nшто смо и желели. Према томе, $P(a b+b c+c a) \\geq 0$, тј. $x_{0} \\leq a b+b c+c a$, одакле следи да је и $P(x) \\geq 0$ и $f(x) \\geq 0$ за све $x \\geq a b+b c+c a$. Између осталог, $f(1+a b c) \\geq 0$ јер је $1+a b c>1>a b+b c+c a$. Тако смо доказали\n$$\n\\frac{a}{1+a b c+a^{2}}+\\frac{b}{1+a b c+b^{2}}+\\frac{c}{1+a b c+c^{2}} \\leq \\frac{3}{3+a^{2}+b^{2}+c^{2}+3 a b c}\n$$\nОстаје још само да докажемо да је $a^{2}+b^{2}+c^{2}+3 a b c \\geq \\frac{4}{9}$, што ће заједно са (1) дати тражену неједнакост. Хомогенизација даје $9(a+b+c)\\left(a^{2}+b^{2}+c^{2}\\right)+27 a b c \\geq 4(a+b+c)^{3}$, што је еквивалентно са\n$$\n5\\left(a^{3}+b^{3}+c^{3}\\right)+3 a b c \\geq 3(a b(a+b)+a c(a+c)+b c(b+c))\n$$\nПоследња неједнакост одмах следи из Шурове неједнакости. Овим је доказ тврђења задатка коначно завршен.\n\n\nДруго решење. Након хомогенизације, свођења на заједнички именилац и скраћивања неједнакост се своди на симетричну неједнакост која се директно доказује Мјурхедовом неједнакошћу:\n$$\n\\begin{aligned}\n& \\frac{23}{2} T_{900}+122 T_{810}+260 T_{720}+282 T_{630}+193 T_{540}+\\frac{547}{2} T_{711}+807 T_{620}+284 T_{531} \\\\\n&+91 T_{522}-98 T_{441}-1669 T_{432}-557 T_{333} \\geq 0\n\\end{aligned}\n$$\nгде је $T_{ijk}$ симетрична сума $x^{i} y^{j} z^{k}+\\cdots$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40829,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је низ $\\left(a_{n}\\right)_{n \\geqslant 1}$ дефинисан са $a_{1}=3, a_{2}=11$ и $a_{n}=4 a_{n-1}-a_{n-2}$, за $n \\geqslant 3$. Доказати да је сваки члан овог низа облика $a^{2}+2 b^{2}$ за неке природне $a$ и $b$.\n\n(Ђорђе Баралић)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nИмамо $a_{1}=1+2 \\cdot 1^{2}, a_{2}=3^{2}+2 \\cdot 1^{2}, a_{3}=3^{2}+2 \\cdot 4^{2}, a_{4}=11^{2}+2 \\cdot 4^{2}$, итд. Доказаћемо индукцијом по $n$ да важи\n$$\na_{2 n-1}=a_{n-1}^{2}+2\\left(\\frac{a_{n}-a_{n-1}}{2}\\right)^{2} \\quad \\text { и } \\quad a_{2 n}=a_{n}^{2}+2\\left(\\frac{a_{n}-a_{n-1}}{2}\\right)^{2},\n$$\nпри чему је $a_{0}=1$. Претпоставимо да тврђење важи за $n$. Тада је\n$$\n\\begin{aligned}\na_{2 n+1} & =4 a_{2 n}-a_{2 n-1}=4 a_{n}^{2}+8\\left(\\frac{a_{n}-a_{n-1}}{2}\\right)^{2}-a_{n-1}^{2}-2\\left(\\frac{a_{n}-a_{n-1}}{2}\\right)^{2} \\\\\n& =\\frac{11}{2} a_{n}^{2}-3 a_{n} a_{n-1}+\\frac{1}{2} a_{n-1}^{2}=\\frac{11}{2} a_{n}^{2}-3 a_{n}\\left(4 a_{n}-a_{n+1}\\right)+\\frac{1}{2}\\left(4 a_{n}-a_{n+1}\\right)^{2} \\\\\n& =\\frac{3}{2} a_{n}^{2}-a_{n} a_{n+1}+\\frac{1}{2} a_{n+1}^{2}=a_{n}^{2}+2\\left(\\frac{a_{n+1}-a_{n}}{2}\\right)^{2} \\\\\na_{2 n+2} & =4 a_{2 n+1}-a_{2 n}=4 a_{n}^{2}+8\\left(\\frac{a_{n+1}-a_{n}}{2}\\right)^{2}-a_{n}^{2}-2\\left(\\frac{a_{n}-a_{n-1}}{2}\\right)^{2} \\\\\n& =3 a_{n}^{2}+8\\left(\\frac{a_{n+1}-a_{n}}{2}\\right)^{2}-2\\left(\\frac{a_{n+1}-3 a_{n}}{2}\\right)^{2}=\\frac{3}{2} a_{n+1}^{2}-a_{n} a_{n+1}+\\frac{1}{2} a_{n}^{2} \\\\\n& =a_{n+1}^{2}+2\\left(\\frac{a_{n+1}-a_{n}}{2}\\right)^{2},\n\\end{aligned}\n$$\nчиме је доказ завршен.\n\n\nДруго решење. Познато је да се непаран природан број $m>1$ може представити у облику $a^{2}+2 b^{2}$ за неке узајамно просте $a, b \\in \\mathbb{N}$ ако и само ако су сви прости делиоци $m$ облика $8 k+1$ или $8 k+3, k \\in \\mathbb{N}_{0}$. Лако се види да су сви чланови низа ( $a_{n}$ ) непарни; остаје да покажемо да ако прост број $p$ дели $a_{n}$, онда је $p=8 k+1$ или $8 k+3$ за неко $k \\in \\mathbb{N}_{0}$.\nПоказује се индукцијом по $n$ да је $a_{n} a_{n+2}=a_{n+1}^{2}+2$. Заиста, ово важи за $n \\leq 2$, а за $n>2$, уз претпоставку да важи за $n-2$, имамо\n$$\n\\begin{aligned}\n\\frac{a_{n+1}^{2}+2}{a_{n}} & =\\frac{\\left(4 a_{n}-a_{n-1}\\right)^{2}+2}{a_{n}}=16 a_{n}-8 a_{n-1}+\\frac{a_{n-1}^{2}+2}{a_{n}} \\\\\n& =16 a_{n}-8 a_{n-1}+a_{n-2}=4 a_{n+1}-a_{n}=a_{n+2}\n\\end{aligned}\n$$\nОдавде следи да је $-2$ квадратни остатак по сваком простом делиоцу $p$ броја $a_{n}$, па је $p \\equiv 1$ или $p \\equiv 3(\\bmod 8)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40830,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $ABCDE$ конвескан петоугао у коме је $AB=1$, $\\angle BAE=\\angle ABC=120^\\circ$, $\\angle CDE=60^\\circ$ и $\\angle ADB=30^\\circ$. Доказати да је површина петоугла $ABCDE$ мања од $\\sqrt{3}$.\n\n(Милош Милосављевић)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nНека је $k$ круг описан око троугла $ABD$, и $l$ права кроз $D$ паралелна са $AB$. Полупречник круга $k$ је $1$. Полуправе $BC$ и $AE$ секу $k$ у тачкама $H$ и $I$, а праву $l$ у $F$ и $G$, редом. Троуглови $FCD$ и $GDE$ су слични јер је $\\angle CFD=\\angle DGE=60^\\circ$ и $\\angle FCD=120^\\circ-\\angle CDF=\\angle GDE$. Означимо са $k=\\frac{FC}{GD}=\\frac{FD}{GE}$ коефицијент сличности, са $h$ растојање тачке $D$ од $HI$, и $x=FD$, $y=GD$. Лако се налази да је $x+y=2+\\frac{2}{\\sqrt{3}} h$ и $xy=\\frac{4}{3} h^2+\\frac{2}{\\sqrt{3}} h$ (производ $xy$ је потенција тачке $F$ у односу на $k$ и једнак је $OF^2-1$, где је $O$ центар круга $k$). Тако добијамо\n$$\n\\begin{aligned}\n& P_{ABFG}=\\frac{1}{2}(1+x+y)\\left(\\frac{\\sqrt{3}}{2}+h\\right)=\\frac{1}{\\sqrt{3}} h^2+2h+\\frac{3\\sqrt{3}}{4} \\\\\n& P_{FCD}+P_{GDE}=\\frac{1}{2}(x \\cdot FC+y \\cdot GE) \\sin 60^\\circ=\\frac{\\sqrt{3}}{4} xy\\left(k+\\frac{1}{k}\\right) \\geq \\frac{\\sqrt{3}}{2} xy=\\frac{2}{\\sqrt{3}} h^2+h\n\\end{aligned}\n$$\nпа је\n$$\nP_{ABCDE}=P_{ABFG}-(P_{FCD}+P_{GDE}) \\leq -\\frac{1}{\\sqrt{3}} h^2+h+\\frac{3\\sqrt{3}}{4}=f(h)\n$$\nКвадратна функција $f(h)$ достиже максимум за $h=\\frac{\\sqrt{3}}{2}$, чиме је доказано да је $P_{ABCDE} \\leq \\sqrt{3}$. Једнакост би важила само ако је $h=\\frac{\\sqrt{3}}{2}$ и $k=1$; тада би (ако без смањења општости претпоставимо $DA \\geq DB$) тачка $D$ била симетрична тачки $B$ у односу на $HI$, па би троугао $ADG$ био једнакостраничан и $FC=GD=GA=FB$ што је немогуће јер би се $B$ и $C$ поклапале. Зато је горња неједнакост строга.\n\n![](attached_image_1.png)\n\nДруго решење.\n\nПретпоставимо да троугао $ABD$ није тупоугли. Тада се тачке $C'$ и $E'$ симетричне тачкама $C$ и $E$ у односу на праве $BD$ и $AD$ редом налазе унутар троугла $ABD$, на истој правој кроз $D$, и важи\n$S_{ABCDE}=S_{ABD}+S_{ADE}+S_{BDC}=S_{ABD}+S_{ADE'}+S_{BDC'} \\leq 2S_{ABD}-S_{ABF}$, при чему $F$ тачка пресека правих $AE'$ и $BC'$. Једнакост важи ако и само ако је $F \\equiv C' \\equiv E'$. Означимо $\\angle BAD=\\alpha$. Тада је $\\angle ABD=150^\\circ-\\alpha$, $\\angle BAF=2\\alpha-120^\\circ$, $\\angle ABF=180^\\circ-2\\alpha$, па важи\n$$\n\\begin{aligned}\n& S_{ABD}=\\sin \\alpha \\sin (150^\\circ-\\alpha)=\\frac{\\sqrt{3}}{4}+\\frac{1}{2} \\cos (150^\\circ-2\\alpha)=\\frac{\\sqrt{3}}{4}+\\frac{1}{2}u \\\\\n& S_{ABF}=\\frac{1}{\\sqrt{3}} \\sin (2\\alpha-120^\\circ) \\sin 2\\alpha=-\\frac{\\sqrt{3}}{12}+\\frac{\\sqrt{3}}{6} \\cos (300^\\circ-4\\alpha)=-\\frac{\\sqrt{3}}{4}+\\frac{1}{\\sqrt{3}}u^2\n\\end{aligned}\n$$\nгде је $\\cos (150^\\circ-2\\alpha)=u$ и одатле $\\cos (300^\\circ-4\\alpha)=2u^2-1$. Сада имамо\n$$\nS_{ABCDE} \\leq 2S_{ABD}-S_{ABF}=\\frac{3\\sqrt{3}}{4}+u-\\frac{u^2}{\\sqrt{3}} \\leq \\sqrt{3}\n$$\nуз једнакост која би важила за $u=\\frac{\\sqrt{3}}{2}$, тј. $\\alpha \\in \\{60^\\circ, 90^\\circ\\}$, и $F \\equiv C' \\equiv E'$, али се никада не достиже јер се за ове вредности $\\alpha$ тачка $F$ налази у темену правог угла, па је петоугао дегенерисан.\n\nУ случају тупоуглог троугла $ABD$, уз исте ознаке, тачка $F$ се налази изван троугла $ABD$, али горњи израз за површину $\\triangle ABF$ узима негативне вредности, па се опет добија $S_{ABCDE} \\leq \\frac{3\\sqrt{3}}{4}+u-\\frac{u^2}{\\sqrt{3}}<\\sqrt{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40831,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека су $\\alpha$ и $\\beta$ углови неједнакокраког троугла $A B C$ код темена $A$ и $B$, редом. Нека симетрале ових углова секу наспрамне странице троугла у $D$ и $E$, редом. Доказати да оштар угао између правих $D E$ и $A B$ није већи од $\\frac{|\\alpha-\\beta|}{3}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nКао и обично, означимо $\\varangle A C B=\\gamma$ и $B C=a, C A=b, A B=c$, при чему је без смањења општости $a>b$ и $\\alpha>\\beta$. Нека је $F$ тачка пресека правих $D E$ и $A B$, а $\\varphi$ угао између ових правих. Из односа $\\frac{B D}{D C}=\\frac{c}{b}$ и $\\frac{C E}{E A}=\\frac{a}{c}$ лако налазимо $B D=\\frac{a c}{b+c}, D C=\\frac{a b}{b+c}$, $C E=\\frac{a b}{a+c}$ и $E A=\\frac{b c}{a+c}$. Менелајева теорема за праву $D E$ и троугао $A B C$ даје $A F=\\frac{b c}{a-b}$ и $F B=\\frac{a c}{a-b}$.\n\n![](attached_image_1.png)\n\nСада на основу синусне теореме у троугловима $F E A$ и $F D B$ имамо\n$$\n\\begin{aligned}\n& \\frac{\\sin (\\alpha-\\varphi)}{\\sin \\varphi}=\\frac{\\sin \\varangle F E A}{\\sin \\varangle E F A}=\\frac{F A}{E A}=\\frac{\\frac{b c}{a-b}}{\\frac{b c}{a+c}}=\\frac{a+c}{a-b} \\\\\n& \\frac{\\sin (\\beta+\\varphi)}{\\sin \\varphi}=\\frac{\\sin \\varangle F D B}{\\sin \\varangle D F B}=\\frac{F B}{D B}=\\frac{\\frac{a c}{a-b}}{\\frac{a c}{b+c}}=\\frac{b+c}{a-b}\n\\end{aligned}\n$$\nиз чега добијамо $\\sin \\varphi=\\sin (\\alpha-\\varphi)-\\sin (\\beta+\\varphi)=2 \\sin \\frac{\\alpha-\\beta-2 \\varphi}{2} \\cos \\frac{\\alpha+\\beta}{2}<$ $\\sin (\\alpha-\\beta-2 \\varphi)$. Одавде је $\\varphi<\\alpha-\\beta-2 \\varphi$, тј. $3 \\varphi<\\alpha-\\beta$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40832,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nОдредити најмањи природан број који је дељив са $2009$ и коме је збир цифара једнак $2009$.",
    "options": [],
    "answer": "The number with decimal form 4998 followed by 170 nines, then 8, then 49 nines (equivalently 5*10^223 − 10^220 − 10^49 − 1).",
    "solution": "Solution:\n\nПошто је $2009 = 223 \\cdot 9 + 2$, тражени број има бар $224$ цифре. Посматраћемо $224$-цифрене бројеве $x = \\overline{c_{223} c_{222} \\ldots c_{1} c_{0}}$. Јасно је да је $c_{223} \\geq 2$. Притом, ако је $c_{223} = 2$ онда је $c_{222} = \\cdots = c_{0} = 9$ и $x = 3 \\cdot 10^{223} - 1 \\equiv 3 \\cdot 10 - 1$, а то није дељиво са $2009 = 7^{2} \\cdot 41$ јер је $x \\equiv 1 \\pmod{7}$.\n\nНека је сада $c_{223} = 3$. Тада број $x$ има облик $399\\ldots 9899\\ldots 9 = 4 \\cdot 10^{223} - 10^{i} - 1$ за неко $i$. Како је $10^{5} \\equiv 1 \\pmod{41}$, имамо $10^{i} \\equiv 1, 10, 18, 16$ или $37 \\pmod{41}$ за $i = 0, 1, 2, 3, 4 \\pmod{5}$ редом, и одатле $x \\equiv 4 \\cdot 10^{3} - 10^{i} - 1 \\equiv 22 - 10^{i} \\pmod{41}$ никад није дељиво са $41$.\n\nНека је $c_{223} = 4$. Међу цифрама $c_{222}, \\ldots, c_{0}$ налазе се две осмице или једна седмица, док су све остале деветке; у сваком случају, $x = 5 \\cdot 10^{223} - 10^{i} 10^{j} - 1 \\equiv 38 - (10^{i} + 10^{j}) \\pmod{41}$, где $i$ и $j$ нису обавезно различити. По претходном је $10^{i} + 10^{j} \\equiv 38 \\pmod{41}$ ако и само ако је $(i, j) \\equiv (0, 4)$ или $(4, 0) \\pmod{5}$. Између осталог, $i \\neq j$ и $i, j \\leq 220$.\n\nПробајмо да ставимо $j = 220$ и $i \\equiv 4 \\pmod{5}$. Треба одабрати $i$, ако постоји, тако да $7^{2} \\mid x = 5 \\cdot 10^{223} - 10^{220} - 10^{i} - 1 \\equiv 5 \\cdot 10^{13} - 10^{10} - 10^{i} - 1 \\equiv 31 - 10^{i} \\pmod{49}$. Лако налазимо да је $10^{i} \\equiv 31 \\pmod{49}$ ако и само ако је $i \\equiv 7 \\pmod{42}$, што заједно са $i \\equiv 4 \\pmod{5}$ даје као једину могућност $i = 49$. Према томе, тражени број је\n\n$$\n4998\\underbrace{9 \\ldots 99}_{170}8\\underbrace{99 \\ldots 9}_{49}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40833,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nОдредити највећи природан број $n$ за који постоје различити скупови $S_{1}, S_{2}, \\ldots, S_{n}$ такви да је:\n\n$1^{\\circ}\\left|S_{i} \\cup S_{j}\\right| \\leqslant 2004$ за свака два цела броја $1 \\leqslant i, j \\leqslant n$, и\n\n$2^{\\circ} S_{i} \\cup S_{j} \\cup S_{k}=\\{1,2, \\ldots, 2008\\}$ за свака три цела броја $1 \\leqslant i<j<k \\leqslant n$.",
    "options": [],
    "answer": "32",
    "solution": "Solution:\n\nСваки скуп $S_{i}$ има највише 2003 елемената. Заиста, ако је $\\left|S_{i}\\right|=2004$, из услова $1^{\\circ}$ следи да је $S_{j} \\subset S_{i}$ за све $j$, противно услову $2^{\\circ}$. Посматрајмо скупове\n$$\nG_{\\{i, j\\}}=\\{1,2, \\ldots, 2008\\} \\backslash\\left(S_{i} \\cup S_{j}\\right) \\quad \\text{ за } 1 \\leq i, j \\leq n\n$$\nТада је $\\left|G_{\\{i, j\\}}\\right| \\geqslant 4$ и свих $\\binom{n}{2}$ скупова $G_{\\{i, j\\}}$ су међусобно дисјунктни (у супротном, ако $x \\in G_{\\{i, j\\}} \\cap G_{\\{k, l\\}}$, онда $x \\notin S_{i} \\cup S_{j} \\cup S_{k} \\cup S_{l}$, што је немогуће ако су бар три међу $i, j, k, l$ различита). Следи да је $4\\binom{n}{2} \\leqslant 2008$, одакле je $n \\leqslant 32$.\n\nКонструисаћемо 32 скупа који задовољавају $1^{\\circ}$ и $2^{\\circ}$. Разложимо скуп $\\{1,2, \\ldots, 2008\\}$ произвољно на $\\binom{32}{2}=496$ (дисјунктних) скупова $G_{\\{i, j\\}}$, при чему је $\\left|G_{\\{i, j\\}}\\right| \\geqslant 4$ за $1 \\leqslant i, j \\leqslant 32$, и дефинишимо\n$$\nS_{i}=\\{1,2, \\ldots, 2008\\} \\backslash \\bigcup_{j \\neq i} G_{\\{i, j\\}} \\quad \\text{ за } i=1, \\ldots, 32\n$$\nУслов $1^{\\circ}$ је аутоматски задовољен. Осим тога, свако $s \\in\\{1,2, \\ldots, 2008\\}$ припада највише једном од скупова $G_{\\{p, q\\}}$, што значи да постоје највише два скупа $S_{i}$ који га не садрже (то су $S_{p}$ и $S_{q}$ ), па је и услов $2^{\\circ}$ задовољен. Према томе, одговор је $n=32$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40834,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $n \\in \\mathbb{N}$ и $A_{n}$ скуп свих пермутација $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ скупа $\\{1,2, \\ldots, n\\}$ таквих да важи\n$$\nk \\mid 2\\left(a_{1}+a_{2}+\\cdots+a_{k}\\right) \\quad \\text{ за свако } 1 \\leqslant k \\leqslant n\n$$\nОдредити број елемената скупа $A_{n}$.",
    "options": [],
    "answer": "F1 = 1, F2 = 2, and for n ≥ 3: Fn = 3 · 2^(n−2)",
    "solution": "Solution:\n\nОзначимо са $F_{n}$ број елемената скупа $A_{n}$. Имамо $F_{1}=1, F_{2}=2$ и $F_{3}=6$. За $n>3$, посматрајмо било коју пермутацију $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ у $A_{n}$. Како $n-1$ дели $2\\left(a_{1}+\\cdots+a_{n-1}\\right)=n(n+1)-2 a_{n} \\equiv 2-2 a_{n}(\\bmod n-1)$, следи да је $a_{n}$ једнако $1, \\frac{n+1}{2}$ или $n$.\n\nПретпоставимо да је $a_{n}=\\frac{n+1}{2}$. Тада $n-2$ дели $2\\left(a_{1}+\\cdots+a_{n-2}\\right)=n^{2}-1-2 a_{n-1} \\equiv 3-2 a_{n-1}(\\bmod n-2)$. Зато мора бити $2 a_{n-1}-3=n-2$, али тада је $a_{n-1}=\\frac{n+1}{2}=a_{n}$, контрадикција.\n\nАко је $a_{n}=n$, онда је $\\left(a_{1}, \\ldots, a_{n}\\right) \\rightarrow\\left(a_{1}, \\ldots, a_{n-1}\\right)$ бијективно пресликавање у скуп $A_{n-1}$, па оваквих пермутација има $F_{n-1}$.\n\nАко је $a_{n}=1$, онда је $\\left(a_{1}-1, \\ldots, a_{n-1}-1\\right)$ пермутација на $\\{1, \\ldots, n-1\\}$ која припада скупу $A_{n-1}$ јер је $2\\left(\\left(a_{1}-1\\right)+\\cdots+\\left(a_{k}-1\\right)\\right)=2\\left(a_{1}+\\cdots+a_{k}\\right)-2 k$ дељиво са $k$ за $1 \\leqslant k \\leqslant n-1$. Као и у претходном случају, оваквих пермутација има $F_{n-1}$.\n\nЗакључујемо да је $F_{n}=2 F_{n-1}$ за $n>3$, што заједно са $F_{3}=6$ даје $F_{n}=3 \\cdot 2^{n-2}$ за $n \\geqslant 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40835,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека су $x, y, z$ позитивни реални бројеви такви да је $x y + y z + z x = x + y + z$. Доказати неједнакост\n$$\n\\frac{1}{x^{2} + y + 1} + \\frac{1}{y^{2} + z + 1} + \\frac{1}{z^{2} + x + 1} \\leqslant 1\n$$\nКада се у претходној неједнакости достиже знак једнакости?",
    "options": [],
    "answer": "Equality holds only when x = y = z = 1",
    "solution": "Solution:\n\nКоши-Шварцова неједнакост за тројке $(x, \\sqrt{y}, 1)$ и $(1, \\sqrt{y}, z)$ даје $\\frac{1}{x^{2} + y + 1} \\leqslant \\frac{1 + y + z^{2}}{(x + y + z)^{2}}$. Аналогно важи $\\frac{1}{y^{2} + z + 1} \\leqslant \\frac{1 + z + x^{2}}{(x + y + z)^{2}}$ и $\\frac{1}{z^{2} + x + 1} \\leqslant \\frac{1 + x + y^{2}}{(x + y + z)^{2}}$. Сабиранјем ових неједнакости добијамо\n$$\n\\frac{1}{x^{2} + y + 1} + \\frac{1}{y^{2} + z + 1} + \\frac{1}{z^{2} + x + 1} \\leqslant \\frac{3 + x + y + z + x^{2} + y^{2} + z^{2}}{(x + y + z)^{2}} = S\n$$\nОстаје да се докаже да је $S \\leqslant 1$, а то је еквивалентно са $3 + x + y + z \\leqslant 2(x y + y z + z x) = 2(x + y + z)$ по услову задатка, тј. $x + y + z \\geqslant 3$. Ово, међутим, следи из $x + y + z = x y + y z + z x \\leqslant \\frac{(x + y + z)^{2}}{3}$.\nЈеднакост важи само за $x = y = z = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40836,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $k$ уписана кружница неједнакокраког $\\triangle A B C$, чији је центар $S$. Кружница $k$ додирује странице $B C, C A, A B$ у тачкама $P, Q, R$, редом. Права $Q R$ сече праву $B C$ у тачки $M$. Нека кружница која садржи тачке $B$ и $C$ додирује $k$ у тачки $N$. Описана кружница $\\triangle M N P$ сече праву $A P$ у тачки $L$, различитој од $P$. Доказати да су тачке $S, L$ и $M$ колинеарне.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nПосматрајмо хомотетију са центром $N$ која слика круг $k$ у круг $B C N$; нека она слика тачку $P$ у $P_{1}$. Тангента на круг $B C N$ у $P_{1}$ је паралелна тангенти на $k$ у $P$, тј. правој $B C$, што значи да је $P_{1}$ средиште лука $B C$ круга $B C N$. Дакле, $N P$ је симетрала угла $C N B$, па је $\\frac{B N}{C N}=\\frac{B P}{C P}$. Шта више, по Менелајевој теореми је $\\frac{B M}{M C}=\\frac{B R}{R A} \\cdot \\frac{A Q}{Q C}=\\frac{B P}{P C}=\\frac{B N}{N C}$, па је $N M$ спољна симетрала угла $C N B$.\n\n![](attached_image_1.png)\n\nПрема томе, $N$ лежи на кругу над пречником $M P$, а $L$ је подножје нормале из $M$ на $A P$. Остаје да се докаже да је $M S \\perp A P$.\n\nНека је $L^{\\prime}$ подножје нормале из $S$ на $A P$. Тачке $A, L^{\\prime}, Q, R, S$ леже на кругу $\\omega$ над пречником $A S$. Инверзија у односу на $k$ слика кругове $\\omega$ и $S P L^{\\prime}$ редом у праве $Q R$ и $B C$, па слика тачку $L^{\\prime}$ у $M$. Дакле, $M$ лежи на правој $S L^{\\prime}$, одакле следи тврђење (и $L^{\\prime} \\equiv L$ ).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40837,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНеки од $n$ градова су повезани авионским линијама (све линије су двосмерне). Постоји тачно $m$ линија. Нека је $d_{i}$ број линија које полазе из града $i$, за $i=1,2, \\ldots, n$. Ако је $1 \\leqslant d_{i} \\leqslant 2010$, за свако $i=1,2, \\ldots, n$, доказати да важи\n$$\n\\sum_{i=1}^{n} d_{i}^{2} \\leqslant 4022 m-2010 n\n$$\nОдредити све $n$ за које може да се достигне једнакост.",
    "options": [],
    "answer": "Equality holds if and only if n is even, or n is odd and at least 2011.",
    "solution": "Solution:\n\nУслов задатка нам даје $0 \\leqslant (d_{i}-1)(2010-d_{i})$ за све $i$, тј. $d_{i}^{2} \\leqslant 2011 d_{i}-2010$. Користећи услов $\\sum_{i=1}^{n} d_{i}=2 m$, сабирањем ових неједнакости добијамо\n$$\n\\sum_{i=1}^{n} d_{i}^{2} \\leqslant 2011 \\cdot \\sum_{i=1}^{n} d_{i}-2010 n=4022 m-2010 n\n$$\nа једнакост важи ако и само ако је $d_{i} \\in\\{1,2010\\}$ за свако $i \\in\\{1,2, \\ldots, n\\}$.\n\n$1^{\\circ}$ Нека је $n=2 k, k \\in \\mathbb{N}$. Ако успоставимо авиолинију између градова $i$ и $j$ ако и само ако је $|j-i|=k$, имамо $d_{i}=1$ за све $i$.\n\n$2^{\\circ}$ Нека је $n=2 k-1, k \\in \\mathbb{N}$. Не може да важи $d_{i}=1$ за све $i$ јер би иначе било $2 m=n=2 k-1$. Зато мора да буде $d_{j}=2010$ за неко $j$; отуда је $n \\geqslant 2011$. С друге стране, успостављањем авиолиније између градова 1 и $i$ $(1 \\leq i \\leq 2010)$ и између градова $2 i$ и $2 i+1$ $(i=1006, \\ldots, k)$ даје мрежу у којој је $d_{1}=2010$ и $d_{i}=1$ за $2 \\leqslant i \\leqslant n$.\n\nПрема томе, једнакост се може достићи ако $2 \\mid n$, или $2 \\nmid n$ и $n \\geq 2011$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40838,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nУ оштроуглом $\\triangle ABC$ тачка $M$ је средиште странице $BC$, а тачке $D, E$ и $F$ су подножја висина из темена $A, B$ и $C$, редом. Нека је $H$ ортоцентар $\\triangle ABC$, $S$ средиште дужи $AH$, а $G$ пресек дужи $FE$ и $AH$. Ако је $N$ тачка пресека тежишне дужи $AM$ и описане кружнице $\\triangle BCH$, доказати да је $\\varangle HMA = \\varangle GNS$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nНека је $A'$ тачка таква да је $ABA'C$ паралелограм. Тада важи $\\varangle BA'C = \\varangle BAC = 180^{\\circ} - \\varangle BHC = 180^{\\circ} - \\varangle BNC$, па су тачке $A', B, C, H, N$ на истом кругу, тј. кругу над пречником $HA'$. Одавде је $\\varangle ANH = 90^{\\circ}$, дакле $N$ је на описаном кругу троугла $AEF$ чији је центар у $S$.\n\nСада имамо $\\varangle SFG = 90^{\\circ} - \\varangle EAF = \\varangle ACF = \\varangle ADF$. Следи да су троуглови $SFG$ и $SDF$ слични, и одатле $SG \\cdot SD = SF^{2} = SN^{2}$. Ово повлачи да је и $\\triangle SNG \\sim \\triangle SDN$, и најзад $\\varangle GNS = \\varangle SDN = \\varangle HMN$ јер је четвороугао $HDMN$ тетиван.\n\n![](attached_image_1.png)\n\nДруго решење. Четвороуглови $BDHF$ и $DCEH$ су тетивни и $AF \\cdot AB = \\overline{AH \\cdot AD = AE} \\cdot AC$. Применимо инверзију $\\mathcal{I}$ са центром $A$ и потенцијом $AF \\cdot AB$. Очигледно је $\\mathcal{I}(F) = B$, $\\mathcal{I}(H) = D$, $\\mathcal{I}(E) = C$, па $\\mathcal{I}$ слика праву $BC$ у описани круг $\\triangle FHE$, тј. у круг над пречником $AH$; даље, $\\mathcal{I}$ слика круг $BCH$ у описани круг $\\omega$ троугла $FDE$, а то је Ојлеров круг у $\\triangle ABC$. Пошто је $M \\in \\omega \\cap AM$ и $\\mathcal{I}$ чува праву $AM$, имамо $\\mathcal{I}(M) = N$.\n\nНека је $\\mathcal{I}(G) = G^{*}$ и $\\mathcal{I}(S) = S^{*}$. Како је $\\mathcal{I}(EF)$ описани круг $\\triangle ABC$, $S \\in \\omega$ и $\\mathcal{I}(AH) = AH$, тачке $G^{*}$ и $S^{*}$ су други пресеци праве $AH$ са круговима $ABC$ и $HBC$, редом. Према томе, $\\varangle GNS = \\varangle G^{*}MS^{*} = \\varangle HMA$ јер су $G^{*}$ и $S^{*}$ симетричне тачкама $H$ и $A$ у односу на $BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40839,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $A$ бесконачан подскуп скупа природних бројева. Одредити све природне бројеве $n$ такве да за свако $a \\in A$ важи\n$$\na^{n}+a^{n-1}+\\ldots+a^{1}+1 \\mid a^{n!}+a^{(n-1)!}+\\ldots+a^{1!}+1\n$$",
    "options": [],
    "answer": "[1, 2]",
    "solution": "Solution:\n\nОзначимо $P(x)=x^{n}+x^{n-1}+\\cdots+1$ и $Q(x)=x^{n!}+\\cdots+x^{1!}+1$; нека је $Q(x)=C(x) P(x)+R(x)$, где су $C$ и $R$ полиноми са целим коефицијентима и $\\operatorname{deg} R<\\operatorname{deg} P$. По услову задатка $P(a) \\mid Q(a)$, и самим тим $P(a) \\mid R(a)$, за бесконачно много целих бројева $a$. Како за довољно велико $a$ важи $|R(a)|<|P(a)|$, мора бити $R(a)=0$; дакле, $R(x)$ има бесконачно много нула, па је $R(x) \\equiv 0$ и $P(x) \\mid Q(x)$.\n\nЛема. Нека су $k_{0}, k_{1}, \\ldots, k_{n} \\in \\mathbb{N}_{0}$. Полином $P(x)=x^{n}+x^{n-1}+\\cdots+1$ дели $Q(x)=x^{k_{n}}+\\cdots+x^{k_{1}}+x^{k_{0}}$ ако и само ако је $\\left\\{k_{0}, k_{1}, \\ldots, k_{n}\\right\\}$ потпун систем остатака по модулу $n+1$.\n\nДоказ. Нека је $r_{i}$ остатак при дељењу $k_{i}$ са $n+1$. Пошто $x^{n+1}-1$ дели $x^{k_{i}}-x^{r_{i}}$ за све $i$, следи да $P(x)=\\frac{x^{n+1}-1}{x-1}$ дели $Q(x)-Q_{1}(x)$, где је $Q_{1}(x)=x^{r_{0}}+x^{r_{1}}+\\cdots+x^{r_{n}}$ и притом $\\operatorname{deg} Q_{1} \\leqslant n$. Ако $P(x) \\mid Q(x)$, онда $P(x) \\mid Q_{1}(x)$, тј. $Q_{1}(x)=c P(x)$ за неку константу $c$, а то важи ако и само ако је $c=1$ и $\\left\\{r_{0}, r_{1}, \\ldots, r_{n}\\right\\}=\\{0,1, \\ldots, n\\}$.\n\nИз леме следи да су тражени бројеви $n$ они за које је $\\{0,1!, \\ldots, n!\\}$ потпун систем остатака по модулу $n+1$.\n\nАко је $n>3$ и $n+1$ је сложен број, онда је $n!\\equiv 0(\\bmod n+1)$, па услов није задовољен. Ако је $n+1=p>3$ прост, по Вилсоновој теореми је $(p-1)!\\equiv-1(\\bmod p)$, одакле је $(p-2)!\\equiv 1=1!(\\bmod p)$, и опет услов није задовољен. Остају случајеви $n \\leqslant 3$; директном провером се добија да $n=1$ и $n=2$ задовољавају услове.\nSolution:\n\nДруго решење. Доказаћемо јаче тврђење: ако $A=a^{n}+\\cdots+a+1$ дели $a^{k_{n}}+\\cdots+a^{k_{1}}+a^{k_{0}}$ за неко $a \\in \\mathbb{N} \\backslash\\{1\\}$, онда је $\\left\\{k_{0}, k_{1}, \\ldots, k_{n}\\right\\}$ потпун систем остатака по модулу $n+1$.\n\nНека $A \\mid B=a^{k_{n}}+\\cdots+a^{k_{1}}+a^{k_{0}}$. Означимо са $r_{i, j}$ остатак при дељењу $k_{i}+j$ са $n+1$ и посматрајмо бројеве $B_{j}=a^{r_{n, j}}+\\cdots+a^{r_{1, j}}+a^{r_{0, j}}$. Тада је $B_{j} \\equiv a^{j} B(\\bmod A)$, одакле следи да $A \\mid B_{j}$ за $j=0,1, \\ldots, n$. С друге стране, $B_{0}+B_{1}+\\cdots+B_{n}=\\sum_{j} a^{r_{n, j}}+\\cdots+\\sum_{j} a^{r_{1, j}}+\\sum_{j} a^{r_{0, j}}=(n+1) A$, па како је $B_{j}>0$, мора бити $B_{j}=A$ за све $j$. Из неједнакости $A<2 a^{n}$ закључујемо да је, за свако $j$, највише један од остатака $r_{i, j}$ једнак $n$. Али ако је $k_{i} \\equiv k_{i^{\\prime}} \\equiv n-j(\\bmod n+1)$, онда је $r_{i, j}=r_{i^{\\prime}, j}=n$, што је немогуће. Према томе, $k_{0}, k_{1}, \\ldots, k_{n}$ су међусобно различити по модулу $n+1$, што је и требало доказати.\n\nИз леме следи да су тражени бројеви $n$ они за које је $\\{0,1!, \\ldots, n!\\}$ потпун систем остатака по модулу $n+1$.\nАко је $n>3$ и $n+1$ је сложен број, онда је $n!\\equiv 0(\\bmod n+1)$, па услов није задовољен. Ако је $n+1=p>3$ прост, по Вилсоновој теореми је $(p-1)!\\equiv-1(\\bmod p)$, одакле је $(p-2)!\\equiv 1=1!(\\bmod p)$, и опет услов није задовољен. Остају случајеви $n \\leqslant 3$; директном провером се добија да $n=1$ и $n=2$ задовољавају услове.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40840,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $O$ центар описане кружнице $\\triangle ABC$. Права кроз $O$ сече странице $CA$ и $CB$ у тачкама $D$ и $E$, редом, и описану кружницу $\\triangle ABO$ у тачки $P$ унутар троугла (различитој од $O$). Тачка $Q$ на страници $AB$ је таква да је $\\frac{AQ}{QB} = \\frac{DP}{PE}$. Доказати да је $\\angle APQ = 2 \\cdot \\angle CAP$. (Душан Ђукић)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nНека је $X$ тачка на полуправој $AP$ таква да је $EX \\parallel AC$. По Талесовој теореми је $AP : PX = DP : PE = AQ : QB$, одакле следи $BX \\parallel QP$.\n\nПрава $PE$ је спољашња симетрала угла $APB$ и полови угао $BPX$. Такође, пошто је $\\angle BEX = 180^{\\circ} - \\angle ACB$ и $\\angle BPX = 180^{\\circ} - \\angle APB = 180^{\\circ} - 2\\angle ACB$, добијамо\n\n![](attached_image_1.png)\n\n$90^{\\circ} + \\frac{1}{2} \\angle BPX$. Следи да је $E$ центар уписаног круга троугла $BPX$ и одатле $\\angle APQ = \\angle PXB = 2 \\angle PXE = 2 \\angle CAP$.\n\n\nДруго решење. Конструкција из задатка је могућа само ако је $\\triangle ABC$ оштроугли. Означимо $\\angle PAD = \\varphi$, $\\angle QPA = \\psi$ и $\\angle BCA = \\gamma$. Из $\\angle APB = 2\\gamma$ и $\\angle DAP + \\angle EBP = \\angle APB - \\angle ACB = \\gamma$ следи $\\angle PBE = \\gamma - \\varphi$ и $\\angle BPQ = 2\\gamma - \\psi$. Како је $\\angle APD = \\angle BPE = 90^{\\circ} - \\gamma$, такође имамо $\\angle ADP = 90^{\\circ} + \\gamma - \\varphi$ и $\\angle BEP = 90^{\\circ} + \\varphi$.\n\nСинусне теореме у троугловима $APD$ и $PBE$ дају\n$$\n\\frac{DP}{PE} = \\frac{DP}{PA} \\cdot \\frac{PA}{PB} \\cdot \\frac{PB}{PE} = \\frac{\\sin \\varphi \\cos \\varphi}{\\sin(\\gamma - \\varphi) \\cos(\\gamma - \\varphi)} \\cdot \\frac{PA}{PB} = \\frac{\\sin 2\\varphi}{\\sin(2\\gamma - 2\\varphi)} \\cdot \\frac{PA}{PB}.\n$$\nС друге стране,\n$$\n\\frac{AQ}{QB} = \\frac{AQ}{AP} \\cdot \\frac{AP}{BP} \\cdot \\frac{BP}{QB} = \\frac{\\sin \\psi}{\\sin(2\\gamma - \\psi)} \\cdot \\frac{AP}{PB},\n$$\nпа се услов $\\frac{AQ}{QB} = \\frac{DP}{PE}$ своди на $f(2\\varphi) = f(\\psi)$, где је\n$$\nf(x) = \\frac{\\sin(2\\gamma - x)}{\\sin x} = \\sin 2\\gamma \\cot x - \\cos 2\\gamma.\n$$\nЈасно је да је $f$ строго опадајућа функција на $(0, \\pi)$, па мора бити $\\psi = 2\\varphi$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40841,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nТаблица димензија $n \\times n$, на чијим пољима су бројеви $1,2, \\ldots, n^{2}$ (на сваком пољу тачно један број и сваки број на тачно једном пољу) назива се нишка ако сви производи од по $n$ бројева који се налазе на $n$ „разбацаних\" поља дају исти остатак при дељењу са $n^{2}+1$. Да ли постоји нишка таблица за:\n\na. $n=8$;\n\nб. $n=10$ ?\n\n($n$ поља су „разбацана\" ако никоја два нису у истој врсти или у истој колони.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nПретпоставимо да постоји нишка таблица $8 \\times 8$ и да производ ма којих 8 разбацаних бројева даје остатак $r$ по модулу $8^{2}+1=65=5 \\cdot 13$. Сви бројеви у таблици се могу поделити на 8 дисјунктних осморки разбацаних бројева. Међу овим осморкама постоји једна која садржи умножак броја 13 и једна која не садржи такав умножак. Производ бројева у првој осморци је дељив са 13, а у другој није, контрадикција. Закључујемо да нишка таблица $8 \\times 8$ не постоји.\n\nб.\nБрој $n^{2}+1=101$ је прост. Попунимо таблицу као на слици, где је $g$ примитиван корен по модулу 101. Лако се види да је производ бројева у ма којих 10 разбацаних поља конгруентан са $g^{495}(\\bmod 101)$, па је ово пример нишке таблице.\n\n| $g^{0}$ | $g^{1}$ | $g^{2}$ | $\\cdots$ | $g^{9}$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $g^{10}$ | $g^{11}$ | $g^{12}$ | $\\cdots$ | $g^{19}$ |\n| $g^{20}$ | $g^{21}$ | $g^{22}$ | $\\cdots$ | $g^{29}$ |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |  | $\\vdots$ |\n| $g^{90}$ | $g^{91}$ | $g^{92}$ | $\\cdots$ | $g^{99}$ |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40842,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека су $a_{0}$ и $a_{n}$ различити делиоци природног броја $m>1$, а низ природних бројева $a_{0}, a_{1}, a_{2}, \\ldots, a_{n}$ такав да задовољава\n$$\na_{i+1}=\\left|a_{i} \\pm a_{i-1}\\right| \\quad \\text{ за } 0<i<n\n$$\nАко је НЗД $\\left(a_{0}, \\ldots, a_{n}\\right)=1$, доказати да у низу постоји члан који је мањи од $\\sqrt{m}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nПосматрајмо два најмања (различита) члана низа, $p$ и $q$. Ако је $\\min \\{p, q\\}= 1$, тврђење тривијално важи; зато надаље претпостављамо да је $p, q>1$.\n\nЛема 1. Постоје индекси $k$ и $l$ за које је $a_{k}=p, a_{l}=q$ и $|k-l| \\leq 2$.\n\nДоказ. Нека је $a_{k}=p$ и $a_{l}=q$ ($k<l$). Претпоставимо да је $r=l-k>2$. Доказаћемо индукцијом по $r$ да за неко $i$, $k<i<l$, важи $a_{i} \\in\\{p, q\\}$. Како је $a_{k+3} \\neq\\left|a_{k+2}-a_{k+1}\\right|=p$, имамо $a_{k+3}=a_{k+1}+a_{k+2}$; слично је и $a_{l-3}=a_{l-2}+a_{l-1}$. Нека је $a_{m}=\\max _{p<i<q} a_{i}$. Јасно је да је $k+2<m<l-2$ (дакле, $l-k \\geqslant 6$ ), а такође и $a_{m+2}=a_{m}-a_{m+1}=a_{m-1}$ и $a_{m+1}=a_{m}-a_{m-1}=a_{m-2}$. То значи да низ $\\left(a_{i}^{\\prime}\\right)$, дефинисан са $a_{i}^{\\prime}=a_{i}$ за $i<m$ и $a_{i}^{\\prime}=a_{i+3}$ за $i \\geqslant m$, задовољава услове задатка; притом је $a_{k}^{\\prime}=p$ и $a_{l-3}^{\\prime}=q$, па по индуктивној претпоставци (јер је $(l-3)-k \\geqslant 3$ ) важи $a_{i}^{\\prime} \\in\\{p, q\\}$ за неко $i$ ($k<i<l-3$). Тада је и $a_{i} \\in\\{p, q\\}$ или $a_{i+3} \\in\\{p, q\\}$, чиме је индукција готова.\n\nЛема 2. За свако $i$ ($0 \\leqslant i \\leqslant n$) постоје $x_{i}, y_{i} \\in \\mathbb{N}_{0}$ такви да је $a_{i}=x_{i} p+y_{i} q$ и $\\left(x_{i}, y_{i}\\right)=1$.\n\nДоказ. Посматрајмо векторе $v_{k}=(1,0)$ и $v_{l}=(0,1)$ и, за свако $i$, дефинишимо $v_{i+2}=\\varepsilon_{i} v_{i}+\\varepsilon_{i}^{\\prime} v_{i+1}$ ако је $a_{i+2}=\\varepsilon_{i} a_{i}+\\varepsilon_{i}^{\\prime} a_{i+1}$ ($\\varepsilon_{i}, \\varepsilon_{i}^{\\prime} \\in\\{-1,1\\}$). Једноставном индукцијом добијамо да за $v_{i}=(x_{i}, y_{i})$ важи\n$$\na_{i}=x_{i} p+y_{i} q\n$$\nКако је $x_{i+1} y_{i+2}-x_{i+2} y_{i+1}=x_{i+1}\\left(\\varepsilon_{i} y_{i}+\\varepsilon_{i}^{\\prime} y_{i+1}\\right)-\\left(\\varepsilon_{i} x_{i}+\\varepsilon_{i}^{\\prime} x_{i+1}\\right) y_{i+1}=-\\varepsilon_{i}\\left(x_{i} y_{i+1}-x_{i+1} y_{i}\\right)$ и слично $x_{i} y_{i+2}-x_{i+2} y_{i}=\\varepsilon_{i}^{\\prime}\\left(x_{i} y_{i+1}-x_{i+1} y_{i}\\right)$, имамо\n$$\nx_{i} y_{i+1}-x_{i+1} y_{i}, x_{i} y_{i+2}-x_{i+2} y_{i} \\in\\{-1,1\\} \\text{ за } 0 \\leqslant i<n\n$$\nи одатле $\\left(x_{i}, y_{i}\\right)=1$. Остаје још да покажемо да су $x_{i}, y_{i} \\geqslant 0$ за све $i$.\n\nПретпоставимо да је $x_{i}<0$ за неко $i<k$ (случај $y_{i}<0$ и/или $i>l$ је аналоган) и посматрајмо највеће такво $i$. Због $a_{i}>0$ и (1) имамо $y_{i}>0$. Из (2) и $x_{i} y_{i+1}, x_{i} y_{i+2} \\leqslant 0 \\leqslant x_{i+1} y_{i}, x_{i+2} y_{i}$ следи да су $v_{i+1}$, $v_{i+2} \\in\\{(0,1),(1,0)\\}$, а тада мора бити $v_{i}= \\pm(1,-1)$, тј. $a_{i}=|p-q|$. Међутим, како су $p$ и $q$ узајамно прости и већи од 1, важи $\\max \\{p, q\\}>|p-q| \\notin\\{p, q\\}$, противно избору $p$ и $q$.\n\nНека је $m=d a_{0}=e a_{n}$. По леми 2 је $m=d x_{0} p+d y_{0} q=e x_{n} p+e y_{n} q$, при чему је $\\left(d x_{0}, d y_{0}\\right) \\neq\\left(e x_{n}, e y_{n}\\right)$ јер због $a_{0} \\neq a_{n}$ важи $\\frac{x_{0}}{y_{0}} \\neq \\frac{x_{n}}{y_{n}}$. Одатле следи да $p \\mid d y_{0}-e y_{n}$, па је $d y_{0}>p$ или $e y_{n}>p$; коначно, $m>p q$ и према томе $\\min (p, q)<\\sqrt{m}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40843,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $n \\geqslant 2$ природан број и нека позитивни реални бројеви $a_{0}, a_{1}, \\ldots, a_{n}$ задовољавају једнакост\n$$\n\\left(a_{k-1}+a_{k}\\right)\\left(a_{k}+a_{k+1}\\right)=a_{k-1}-a_{k+1} \\quad \\text{ за свако } k=1,2, \\ldots, n-1 \\text{. }\n$$\nДоказати да је $a_{n}<\\frac{1}{n-1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nДата једнакост је еквивалентна са\n$$\n\\frac{1}{a_{k}+a_{k+1}}=1+\\frac{1}{a_{k-1}+a_{k}}\n$$\nза свако $k>0$. Индукцијом следи $\\frac{1}{a_{k}+a_{k+1}}=k+\\frac{1}{a_{0}+a_{1}}$ за $k>0$, одакле добијамо да је $\\frac{1}{a_{n-1}+a_{n}} > n-1$ и према томе $a_{n}<\\frac{1}{n-1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40844,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $n$ непаран природан број такав да су бројеви $\\varphi(n)$ и $\\varphi(n+1)$ степени броја два ($\\varphi(n)$ је број природних бројева не већих од $n$ и узајамно простих са $n$). Доказати да је $n+1$ степен броја два или је $n=5$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nАко је $n=\\prod_{i=1}^{k} p_{i}^{r_{k}}$ канонска факторизација $n$, важи $\\varphi(n)=\\prod_{i=1}^{k} p_{i}^{r_{k}-1}\\left(p_{i}-1\\right)$, па пошто $n$ нема других простих чинилаца осим двојке, мора бити $a_{i}=1$ и $p_{i}-1=2^{b_{i}}$ за свако $i$ и неке $b_{i}$. Како $2^{b_{i}}+1$ може бити прост само ако је $b_{i}$ степен двојке, имамо $p_{i}=2^{2^{c_{i}}}+1$ за неке различите $c_{i}$.\n\nПретпоставимо да $n+1$ није степен двојке. Из чињенице да је $\\varphi(n+1)$ степен двојке добијамо да су сви непарни прости делиоци броја $n+1$ облика $2^{2^{d_{i}}}+1$. Према томе,\n$$\nn=\\prod_{i=1}^{k}\\left(2^{2^{c_{i}}}+1\\right), \\quad n+1=2^{t} \\prod_{j=1}^{l}\\left(2^{2^{d_{j}}}+1\\right)\n$$\nпри чему су сви $c_{i}$ и $d_{j}$ међусобно различити. Можемо узети без смањења општости да је $c_{1}<\\cdots<c_{k}$ и $d_{1}<\\cdots<d_{l}$.\n\nЗа свако $m, M \\in \\mathbb{N}, m \\leq M$, једноставном индукцијом се показује да важи\n$$\n\\frac{2^{2^{m}}+1}{2^{2^{m}}}<\\prod_{i=m}^{M} \\frac{2^{2^{i}}+1}{2^{2^{i}}}=\\frac{2^{2^{m}}}{2^{2^{m}}-1} \\cdot \\frac{2^{2^{M+1}}-1}{2^{2^{M+1}}}<\\frac{2^{2^{m}}}{2^{2^{m}}-1}\n$$\nОдавде добијамо\n$$\n\\frac{2^{2^{c_{1}}}+1}{2^{2^{c_{1}}}} 2^{c} \\leqslant n<\\frac{2^{2^{c_{1}}}}{2^{2^{c_{1}}}-1} 2^{c} \\quad \\text{и} \\quad \\frac{2^{2^{d_{1}}}+1}{2^{2^{d_{1}}}} 2^{d} \\leqslant n+1<\\frac{2^{2^{d_{1}}}}{2^{2^{d_{1}}}-1} 2^{d}\n$$\nгде је $c=\\sum_{i} 2^{c_{i}}$ и $d=t+\\sum_{j} 2^{d_{i}}$. Следи да је $c=d$. Ако је $d_{1}>c_{1}$, важи $\\frac{2^{2^{d_{1}}}}{2^{2^{d_{1}}}-1}<\\frac{2^{2^{c_{1}}}+1}{2^{2^{c_{1}}}}$, па је $n+1<n$, контрадикција. Према томе, $d_{1}<c_{1}$, а тада је $n+1 \\geqslant \\frac{2^{2^{d_{1}}}+1}{2^{2^{d_{1}}}} 2^{c}>\\frac{2^{2^{c_{1}}}}{2^{2^{c_{1}}}-1} 2^{c}>n$, па је $\\frac{n+1}{n}>\\frac{2^{2^{d_{1}}}+1}{2^{2^{d_{1}}}} \\cdot \\frac{2^{2^{c_{1}}}-1}{2^{2^{c_{1}}}}$ и, због $n \\geqslant 2^{2^{c_{1}}}+1 \\geqslant a^{2}+1$ за $2^{2^{d_{1}}}=a, \\frac{n+1}{n}>\\frac{(a+1)\\left(a^{2}-1\\right)}{a^{3}}=1+\\frac{a^{2}-a-1}{a^{3}}$, одакле закључујемо $a^{2}+1 \\leqslant n<\\frac{a^{3}}{a^{2}-a-1}$. Једина могућност је $a=2$ и $n=5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40845,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $H$ ортоцентар, а $O$ центар описане кружнице оштроуглог троугла $ABC$. Тачке $D$ и $E$ су подножја висина из $A$ и $B$, редом. Обележимо са $K$ пресечну тачку правих $OD$ и $BE$, а са $L$ пресечну тачку правих $OE$ и $AD$. Нека је $X$ друга пресечна тачка кружница описаних око троуглова $HKD$ и $HLE$, а $M$ средиште странице $AB$. Доказати да су тачке $K, L$ и $M$ колинеарне ако и само ако је $X$ центар описане кружнице троугла $EOD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nАко је $X$ центар описаног круга $\\triangle ODE$, онда је $90^\\circ - \\angle KDE = 90^\\circ - \\angle ODE = \\angle XEO = \\angle XEL = \\angle XHD = \\angle XKD$ (сви углови су оријентисани), одакле следи да је $XK \\perp DE$; аналогно $XL \\perp DE$, тј. $K$ и $L$ леже на симетрали дужи $DE$, па је $DEHO$ једнакокраки трапез, дакле $D, E, O, H$ леже на кругу.\n\nС друге стране, ако $O$ лежи на кругу $HDE$, дакле на кругу над пречником $CH$, периферијски углови над $EH$ и $OD$ су једнаки ($\\angle ECH = \\angle OCD$), па је $DEHO$ једнакокраки трапез и одатле $DL = EL$. Сада имамо $\\angle EXH = \\angle ELH = 2 \\angle EDH$\n\n![](attached_image_1.png)\n\nи аналогно $\\angle DXH = 2 \\angle DEH$, па следи да је $X$ центар круга $DEOH$. Према томе, $X$ је центар круга $ODE$ ако и само ако $D, E, O$ и $H$ леже на кругу.\n\nАко су тачке $D, E, O, H$ на кругу, тада $K, L$ и $M$ леже на симетрали дужи $DE$, чиме је један смер задатка доказан. Претпоставимо сада да $O$ лежи ван круга $CDHE$ (случај када је $O$ унутар круга се разматра на исти начин). Како је $CO \\perp DE$, важи $DL > LE$ и $EK > KD$, тј. $K$ и $L$ леже на разним странама симетрале дужи $DE$, а $M$ припада овој симетрали. Према томе, ако су $K, L$ и $M$ колинеарне, $M$ се мора налазити између $K$ и $L$. Следи да је једна од тачака $K$ и $L$ ван троугла $ABC$, а друга унутар троугла. Међутим, када је $O$ изван четвороугла $ABDE$, обе тачке $K$ и $L$ су ван троугла, а у супротном су обе унутар троугла. То је контрадикција са претпоставком да $M$ лежи на правој $KL$, што доказује други смер.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40846,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНа страницама $AB$, $AC$ и $BC$ троугла $ABC$ дате су, редом, тачке $M$, $X$ и $Y$ тако да је $AX = MX$ и $BY = MY$. Нека су $K$ и $L$, редом, средишта дужи $AY$ и $BX$, а $O$ центар описане кружнице троугла $ABC$. Ако су $O_{1}$ и $O_{2}$ тачке симетричне тачки $O$ у односу на $K$ и $L$, редом, доказати да тачке $X$, $Y$, $O_{1}$ и $O_{2}$ леже на истој кружници.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nПоставимо координатни систем са почетком у тачки $M$ и $x$-осом дуж праве $AB$. Нека тачке $X$ и $Y$ имају координате $(a, b)$ и $(c, d)$ редом. Због $AX = XM$ и $BY = YM$, координате тачака $A$ и $B$ су $(2a, 0)$ и $(2c, 0)$, а тачака $K$ и $L$ су $\\left(a + \\frac{c}{2}, \\frac{d}{2}\\right)$ и $\\left(c + \\frac{a}{2}, \\frac{b}{2}\\right)$, редом. Тачка $O$ има координате $(a + c, e)$ за неко $e$, одакле добијамо $O_{1}(a, d - e)$ и $O_{2}(c, b - e)$. Према томе, тачке $O_{1}$ и $O_{2}$ су симетричне тачкама $X$ и $Y$ у односу на праву $y = \\frac{b + d - e}{2}$, па су $X$, $Y$ и $O_{1}$, $O_{2}$ темена (могуће дегенерисаног) једнакокраког трапеза, и зато леже на кругу.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40847,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nДа ли постоје природни бројеви $a, b$ и $c$, већи од 2011, такви да у децималном запису важи једнакост\n$$\n(a+\\sqrt{b})^{c}=\\ldots 2010,2011 \\ldots ?\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nПоказаћемо да такви бројеви $a, b$ и $c$ постоје. Број $x=(a+\\sqrt{b})^{c}+(a-\\sqrt{b})^{c}$ је цео. Довољно је одабрати $a, b, c$ тако да $x$ буде дељиво са $10^{4}$ и $7989,7989> (a-\\sqrt{b})^{c}>7989,7988$.\n\nЗа непарно $c$, број $x=2 a^{c}+2\\binom{c}{2} a^{c-2}+\\cdots+2\\binom{c}{c-1} a$ је дељив са $a$, па је довољно узети $a$ које је дељиво са $10^{4}$. Други услов постижемо избором $a$ и $b$ тако да је $1<a-\\sqrt{b}<\\sqrt{\\frac{7989,7989}{7989,7988}}$ - на пример, $a=10^{8}$ и $b=(a-1)^{2}-1$. Заиста, нека је $c$ најмањи непаран природан број за који је $(a-\\sqrt{b})^{c}>7989,7988$. Такво $c$ је очигледно веће од 2011 (у овом случају $c=1797184159)$ и $(a-\\sqrt{b})^{c}<7989,7989$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40848,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $P$ тачка на дијагонали $BD$ паралелограма $ABCD$ таква да је $\\varangle PCB = \\varangle ACD$. Кружница описана око троугла $ABD$ сече праву $AC$ у тачкама $E$ и $A$. Доказати да је\n$$\n\\varangle AED = \\varangle PEB. \\quad \\text{(Марко Ђикић)}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nДоказ изводимо у случају када је $\\angle BAC \\leq 90^\\circ$. Други случај је аналоган.\nНека се праве $DE$ и $BC$ секу у $L$. Четвороугао $CDPL$ је тетиван јер је $\\angle PDL = \\angle PCL$, одакле имамо $\\angle PLE = \\angle PCD = \\angle BCA = \\angle DAC = \\angle DBE = \\angle PBE$, па је и четвороугао $BPEL$ тетиван. Из ове две тетивности коначно добијамо $\\angle PEB = \\angle PLB = \\angle PDC = \\angle DBA = \\angle DEA$.\n\n![](attached_image_1.png)\nSolution:\n\nДруго решење. Нека је $P'$ тачка на $BD$ таква да је $\\angle DEA = \\angle PEB$.\nПо синусној теореми је $\\frac{BP}{DP} = \\frac{BP}{CP} \\cdot \\frac{CP}{DP} = \\frac{\\sin \\angle BCP}{\\sin \\angle CBD} \\cdot \\frac{\\sin \\angle CDB}{\\sin \\angle PCD}$. Аналогно је $\\frac{BP'}{DP'} = \\frac{\\sin \\angle BEP'}{\\sin \\angle EBD} \\cdot \\frac{\\sin \\angle EDB}{\\sin \\angle P'ED}$. Како је $\\angle BCP = \\angle EDB$, $\\angle CBD = \\angle P'ED$, $\\angle CDB = \\angle BEP'$, и $\\angle PCD = \\angle EBD$, следи $\\frac{BP}{DP} = \\frac{BP'}{DP'}$, дакле $P \\equiv P'$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40849,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНаћи све природне бројеве $a$ и $b$ такве да\n$$\na \\mid b^{2}, \\quad b \\mid a^{2} \\quad \\text{и} \\quad a+1 \\mid b^{2}+1.\n$$",
    "options": [],
    "answer": "All pairs (a, b) of the forms (t^2, t), (t^2, t^3), and (t^2(t^2 − 1), t(t^2 − 1)^2), where t is a natural number (with t ≥ 2 in the third family).",
    "solution": "Solution:\n\nНека је $b^{2}=c a$. Услови задатка дају $b^{2}=c a \\mid a^{4}$ и $a+1 \\mid c a+1$, а то је еквивалентно са\n$$\nc \\mid a^{3} \\quad \\text{и} \\quad a+1 \\mid c-1.\n$$\nНека је $c=d(a+1)+1,\\ d \\in \\mathbb{N}_{0}$. Како је $a^{3} \\equiv -1 \\pmod{a+1}$, имамо $\\frac{a^{3}}{c} \\equiv -1 \\pmod{a+1}$, тј. $\\frac{a^{3}}{c}=e(a+1)-1$ за неко $e \\in \\mathbb{N}$. Следи да је $a^{3}=(d(a+1)+1)(e(a+1)-1)$, што након множења и скраћивања са $a+1$ постаје $a^{2}-a+1=d e(a+1)+(e-d)$. Одавде имамо $e-d \\equiv a^{2}-a+1 \\equiv 3 \\pmod{a+1}$, дакле\n$$\ne-d=k(a+1)+3 \\quad \\text{и} \\quad d e=a-2-k \\quad (k \\in \\mathbb{Z}).\n$$\nРазликујемо следеће случајеве:\n\n(1) $k \\notin\\{-1,0\\}$. У овом случају (*) повлачи $d e<|e-d|-1$, што је могуће само за $d=0$. Сада је $c=1$ и $b^{2}=a$, дакле $(a, b)=\\left(t^{2}, t\\right)$.\n\n(2) $k=-1$. Из (*) добијамо $a=d+1$. Сада је $c=a^{2}$ и $b^{2}=a^{3}$, дакле $(a, b)=\\left(t^{2}, t^{3}\\right)$.\n\n(3) $k=0$. Из (*) добијамо $a=d^{2}+3 d+2$. Сада је $c=d(a+1)+1=(d+1)^{3}$ и $b^{2}=c a=(d+1)^{4}(d+2)$. Следи да је $d+2=t^{2}$ за неко $t \\in \\mathbb{N}$, што даје $(a, b)=\\left(t^{2}(t^{2}-1), t(t^{2}-1)^{2}\\right),\\ t \\geq 2$.\n\nОдговор: То су парови $(a, b)$ облика $\\left(t^{2}, t\\right),\\ \\left(t^{2}, t^{3}\\right)$ и $\\left(t^{2}(t^{2}-1), t(t^{2}-1)^{2}\\right)$, где је $t \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40850,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nУ неким чворовима квадратне решетке $2012 \\times 2012$ налази се мува и $k$ паукова. Један потез састоји се у следећем: мува се помера на суседан чвор или остаје на истом месту, а након тога се сваки од $k$ паукова помера на неки суседан чвор или остаје на истом месту (у једном чвору може бити више паукова). У сваком тренутку сваки паук и мува имају увид у позиције осталих.\na) Наћи најмање $k$ тако да пауци могу ухватити муву у коначном броју потеза, без обзира на почетну позицију муве и паукова.\nб) Одговорити на исто питање за кубну решетку $2012 \\times 2012 \\times 2012$.\n(Чворови су суседни ако се разликују на тачно једној координати, и то за 1. Паук хвата муву уколико се налазе у истом чвору.)\n(Никола Милосављевић)",
    "options": [],
    "answer": "a) 2; b) 2",
    "solution": "Solution:\n\nЈедан паук не може да ухвати муву. Довољно је да мува не мрда ако паук није на суседном пољу, односно да се помери на поље дијагонално супротно пауковом ако јесте.\nДоказаћемо да су два паука довољна у оба дела задатка - означимо их са $P$ и $Q$, муву са $M$, а $x$- и $y$-координату тачке $A$ са $A_{x}$ и $A_{y}$.\n\na) Поставимо координатни почетак у доњи леви угао мреже. На почетку, крећући се дуж $x$-осе, $P$ у коначном броју потеза постиже да буде $P_{x}=M_{x}$. Аналогно, $Q$ постиже да буде $Q_{y}=M_{y}$. Потом се пауци крећу на следећи начин: кад год мува промени $x$-координату, $P$ учини исто тако да остане $P_{x}=M_{x}$, у супротном приђе муви један корак дуж $y$-осе; кретање $Q$ је аналогно. На овај начин величина $|P_{y}-M_{y}|+|Q_{x}-M_{x}|$ се или смањује или остаје иста, при чему може остати иста кроз највише $2 \\cdot 2010$ потеза (када се мува повлачи). Према томе, после коначног броја потеза бар један сабирак ће постати нула, тј. мува ће бити ухваћена.\n\nб) Занемарујући $z$-осу, на основу дела под (a), један од паукова, рецимо $P$, може постићи да буде $P_{x}=M_{x}$ и $P_{y}=M_{y}$. Надаље, кад год мува направи корак дуж $z$-осе, паук $P$ иде ка њој, а супротном се креће тако да увек буде тачно испод муве. Јасно је да мува може само коначно много пута да направи корак дуж $z$-осе или остане у месту, а да је паук $P$ не ухвати. Према томе, почев од неког момента, мува мора да се креће искључиво по својој $xy$-равни, без стајања у месту.\nСада паук $Q$ у коначно много потеза долази до $xy$-равни по којој се креће мува. Такође, стајањем у месту по потреби, $Q$ постиже да $f=|Q_{x}-M_{x}|+|Q_{y}-M_{y}|$ буде парно. У сваком следећем потезу, паук $Q$ се приближава муви дуж $x$-осе ако је $|Q_{x}-M_{x}|>|Q_{y}-M_{y}|$, а дуж $y$-осе у супротном. После сваког потеза, величина $f$ се не повећава и не мења парност, при чему само коначно много пута може да остане иста (када мува бежи од паука према крају решетке). Према томе, у неком тренутку ће бити $f=0$, и мува ће бити ухваћена.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40851,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНаћи све природне бројеве $n$ за које постоји пермутација $(p_{1}, p_{2}, \\ldots, p_{n})$ бројева $(1,2, \\ldots, n)$ таква да скупови $\\{p_{i}+i \\mid 1 \\leqslant i \\leqslant n\\}$ и $\\{p_{i}-i \\mid 1 \\leqslant i \\leqslant n\\}$ чине потпуне системе остатака по модулу $n$. (Марко Ђикић)",
    "options": [],
    "answer": "All natural numbers n with gcd(n, 6) = 1",
    "solution": "Solution:\n\nПретпоставимо да таква пермутација постоји. Како је $\\{p_{i}+i \\mid 1 \\leq i \\leq n\\}$ потпун систем остатака по модулу $n$, важи $\\sum_{k=1}^{n} k \\equiv \\sum_{i=1}^{n}(p_{i}+i) \\equiv \\sum_{i=1}^{n} i+\\sum_{i=1}^{n} p_{i} \\equiv 2 \\sum_{k=1}^{n} k\\ (\\bmod\\ n)$, дакле $\\sum_{k=1}^{n} k=\\frac{n(n+1)}{2} \\equiv 0\\ (\\bmod\\ n)$, одакле следи $2 \\nmid n$.\n\nШта више, важи $2 \\sum_{k=1}^{n} k^{2} \\equiv \\sum_{k=1}^{n}\\left((p_{i}+i)^{2}+(p_{i}-i)^{2}\\right) \\equiv \\sum_{k=1}^{n}\\left(2 p_{i}^{2}+2 i^{2}\\right) \\equiv 4 \\sum_{k=1}^{n} k^{2}$, одакле је $2 \\sum_{k=1}^{n} k^{2}=\\frac{n(n+1)(2 n+1)}{3} \\equiv 0\\ (\\bmod\\ n)$, дакле $3 \\nmid n$.\n\nПрема томе, мора бити $(n, 6)=1$.\n\nС друге стране, ако је $(n, 6)=1$ и $p_{i} \\equiv 2 i\\ (\\bmod\\ n),\\ p_{i} \\in\\{1, \\ldots, n\\}$, тада је $(p_{1}, p_{2}, \\ldots, p_{n})$ пермутација скупа $\\{1, \\ldots, n\\}$ и задовољава услове, јер су $\\{p_{i}+i \\mid 1 \\leq i \\leq n\\} \\equiv\\{3 i \\mid 1 \\leq i \\leq n\\}$ и $\\{p_{i}-i \\mid 1 \\leq i \\leq n\\} \\equiv\\{i \\mid 1 \\leq i \\leq n\\}$ $(\\bmod\\ n)$ потпуни системи остатака по модулу $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40852,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $\\mathcal{K}$ целобројна решетка. Да ли постоји бијекција $f: \\mathbb{N} \\rightarrow \\mathcal{K}$ таква да за све међусобно различите $a, b, c \\in \\mathbb{N}$ важи\n$$\n\\text{НЗД}(a, b, c)>1 \\quad \\Longrightarrow \\quad f(a), f(b), f(c) \\text{ нису колинеарне? }\n$$\n(Целобројна решетка је скуп тачака у равни са целобројним координатама у Декартовом координатном систему.)\n(Стеван Гајовић)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nПоређајмо све тачке решетке у низ $A_{1}, A_{2}, \\ldots$. Ово се може урадити нпр. спирално: $(0,0), (0,1), (1,1), (1,0), (1,-1), (0,-1), (-1,-1), \\ldots$ Конструисаћемо индуктивно пример бијекције са траженим својством.\n\nСтавимо $f(1)=A_{1}$. Претпоставимо да су $f(1), \\ldots, f(n-1)$ одређене и узмимо за $f(n)$ тачку $A_{m}$ са најмањим $m$ такву да, ни за које природне $i, j \\leq n$, $(i, j, n)>1$, $A_{m}$ не лежи на правој $f(i) f(j)$. Правих $f(i) f(j)$ има коначно много, па заиста постоји тачка решетке која не лежи ни на једној од њих. Приметимо да за просто $p$ нема ограничења при избору $f(p)$, што гарантује да ће свака тачка бити слика неког природног броја.\n\nОвако дефинисано пресликавање $f$ задовољава све услове задатка.\n\n\nДруго решење. Ако је $n$ сложен број, узмимо за $f(n)$ тачку $(n, n^{2})$. За прост број $p$, нека је $f(p)$ било која од дозвољених тачака на минималном растојању од тачке $(0,0)$. Покажимо да је ово пресликавање добро дефинисана бијекција са траженим својством.\n\nЗа свако $p$ постоји дозвољена тачка - нпр. нека од неупотребљених тачака на параболи $y=x^{2}$. Заиста, произволна права кроз ту тачку сече ову параболу у још највише једној тачки, тј. садржи слику највише једног сложеног броја.\n\nС друге стране, за произвољну тачку $A \\in \\mathcal{K} \\setminus \\{(n, n^{2}) \\mid n$ је сложен$\\}$ постоји прост број $p$ за који је $A$ дозвољена тачка. Наиме, постоји само коначно много правих кроз $A$ које садрже две целобројне тачке на параболи $y=x^{2}$ (различите од $A$); означимо те праве са $p_{1}, \\ldots, p_{k}$ и целобројне тачке у пресеку $p_{i}$ са параболом $y=x^{2}$ са $A_{i}, B_{i}$. Довољно је узети $p$ које не дели $(f^{-1}(A_{i}), f^{-1}(B_{i}))$ ни за једно $i$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40853,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека композиција садржи $n>1$ вагона са златницима. Постоје две врсте наизглед истих златника: прави и лажни. У сваком вагону се налазе златници само једне врсте. Златници исте врсте су исте масе, док златници различитих врста немају исту масу. Маса правог златника је позната.\nОдредити минималан број мерења на дигиталној ваги којима је могуће утврдити који све вагони садрже лажне златнике, као и која је маса лажног златника.\n(Претпоставља се да се из сваког вагона може узети било који број златника.)\n(Милош Милосављевић)",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nДоказаћемо да је минималан број мерења једнак 2. Означимо тежине правог и лажног златника са $x$ и $y$ редом, и нека је $a_{i}=1$ ако су златници у $i$-том вагону лажни, а $a_{i}=0$ у супротном.\n\nУзмимо у првом мерењу по један златник из сваког вагона. Тада је $a_{1}+a_{2}+\\cdots+a_{n} = \\frac{n x - m_{1}}{x-y}$, где је $m_{1}$ добијена маса. Претпостављамо да је $m_{1} \\neq n x$, јер у супротном нема лажних златника.\n\nУ другом мерењу, за неко $q \\in \\mathbb{N}$, узимамо из $i$-тог вагона $q^{i-1}$ златника. Ако је добијена маса $m_{2}$, имамо $a_{1} + q a_{2} + \\cdots + q^{n-1} a_{n} = \\frac{(1+q+\\cdots+q^{n-1}) x - m_{2}}{x-y}$. Одавде добијамо\n$$\nf\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right) = \\frac{a_{1} + q a_{2} + \\cdots + q^{n-1} a_{n}}{a_{1} + a_{2} + \\cdots + a_{n}} = \\frac{(1+q+\\cdots+q^{n-1}) x - m_{2}}{n x - m_{1}}\n$$\nЖелимо да нам вредност $f$ једнозначно одреди $a_{1}, \\ldots, a_{n}$. Дакле, довољно је показати да постоји природан број $q$ такав да је функција $f:\\{0,1\\}^{n} \\setminus \\{(0,0, \\ldots, 0)\\} \\rightarrow \\mathbb{R}$ инјективна.\n\nЗа фиксиране $\\mathfrak{a} = (a_{1}, a_{2}, \\ldots, a_{n})$ и $\\mathfrak{b} = (b_{1}, b_{2}, \\ldots, b_{n})$, једнакост $f(\\mathfrak{a}) = f(\\mathfrak{b})$ је еквивалентна са $P_{\\mathfrak{a}, \\mathfrak{b}}(q) = (a b_{n} - b a_{n}) q^{n-1} + \\cdots + (a b_{2} - b a_{2}) q + (a b_{1} - b a_{1}) = 0$, где је $a = a_{1} + a_{2} + \\ldots + a_{n} \\neq 0 \\neq b = b_{1} + b_{2} + \\ldots + b_{n}$. Према томе, ако функција $f$ није инјективна, онда је $q$ нула полинома\n$$\nP(q) = \\prod_{\\mathfrak{a}, \\mathfrak{b}} P_{\\mathfrak{a}, \\mathfrak{b}}(q)\n$$\nКако ниједан од полинома $P_{\\mathfrak{a}, \\mathfrak{b}}$ није идентички једнак 0, постоји само коначно много бројева $q$ за које је $P(q) = 0$, па је могуће одабрати $q$ за које је $f$ инјективна функција.\n\nОвако у два мерења можемо да одредимо $a_{1}, \\ldots, a_{n}$, тј. вагоне са лажним златницима. Најзад, $y$ одређујемо из једнакости $y = x - \\frac{n x - m_{1}}{a_{1} + a_{2} + \\cdots + a_{n}}$.\n\nС друге стране, једно мерење није довољно, јер узимањем $k_{i}$ златника из $i$-тог вагона добијамо једначину $k_{1} a_{1} + \\cdots + k_{n} a_{n} = \\frac{k x - m}{x-y}$ (где је $k = k_{1} + \\cdots + k_{n}$) која у општем случају има више решења. На пример, два могућа решења за $(a_{1}, a_{2}, \\ldots, a_{n}, y)$ су $(1, 0, \\ldots, 0, x - \\frac{k x - m}{k_{1}})$ и $(0, 0, \\ldots, 1, x - \\frac{k x - m}{k_{n}})$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40854,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nДат је природан број $k$. Нека је $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ бијекција таква да за свака два цела броја $i$ и $j$ за које је $|i-j| \\leqslant k$ важи $|f(i)-f(j)| \\leqslant k$. Доказати да за све $i, j \\in \\mathbb{Z}$ важи\n$$\n|f(i)-f(j)|=|i-j|\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nЗа $k=1$ тврђење је тривијално. Нека је зато $k>2$. Интервалом дужине $k$ зовемо скуп облика $\\{x, x+1, \\ldots, x+k\\}, x \\in \\mathbb{Z}$. Два цела броја $x$ и $y$ ће бити узастопна ако и само ако постоје интервали $I_{1}$ и $I_{2}$ дужине $k$ за које је $I_{1} \\cap I_{2}=\\{x, y\\}$. Међутим, по услову задатка су $f\\left(I_{1}\\right)$ и $f\\left(I_{2}\\right)$ такође интервали дужине $k$, па како је $\\{f(x), f(y)\\}=f\\left(I_{1}\\right) \\cap f\\left(I_{2}\\right)$, следи да су и $f(x)$ и $f(y)$ узастопни бројеви. Одавде је $|f(x+1)-f(x)|=1$ за $x \\in \\mathbb{Z}$. Коначно, користећи инјективност пресликавања, једноставном индукцијом по $n$ добијамо да је $|f(x+n)-f(x)|=n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40855,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је\n$$\nS_{n} = \\left\\{ \\binom{n}{n}, \\binom{2n}{n}, \\binom{3n}{n}, \\ldots, \\binom{n^{2}}{n} \\right\\}, \\quad \\text{за } n \\in \\mathbb{N}\n$$\n\na) Доказати да постоји бесконачно много сложених природних бројева $n$ таквих да $S_{n}$ није потпун систем остатака по модулу $n$.\n\nб) Доказати да постоји бесконачно много сложених природних бројева $n$ таквих да $S_{n}$ јесте потпун систем остатака по модулу $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nа. Доказаћемо да $n = 2p$ задовољава услове, где је $p > 2$ прост број. Имамо\n$$\n\\binom{2kp}{2p} = k \\prod_{i=1}^{p-1} \\frac{2kp - i}{2p - i} \\cdot (2k-1) \\prod_{i=1}^{p-1} \\frac{2kp - p - i}{p - i} \\equiv k(2k-1) \\pmod{p}\n$$\nКонкретно, одавде је $\\binom{2kp}{2p}$ дељиво са $p$ за $k \\in \\left\\{ \\frac{p+1}{2}, p, 2p \\right\\}$, тј. $S_{2p}$ има три елемента дељива са $p$, па није потпун систем остатака.\n\nб. Доказаћемо да $n = p^{2}$ задовољава услове, где је $p > 2$ прост број. Имамо\n$$\n\\binom{k p^{2}}{p^{2}} = \\prod_{i=0}^{p^{2}-1} \\frac{k p^{2} - i}{p^{2} - i} = k \\prod_{j=1}^{p-1} \\frac{k p^{2} - j p}{j p} \\cdot \\prod_{p \\nmid j} \\frac{k p^{2} - i}{p^{2} - i},\n$$\nпа је по модулу $p^{2}$\n$$\n\\binom{k n}{n} \\equiv k \\prod_{j=1}^{p-1} \\frac{k p - j}{j} = k \\prod_{j=1}^{p-1} \\left(1 - \\frac{k p}{j}\\right) \\equiv k - k^{2} p \\sum_{j=1}^{p-1} \\frac{1}{j}\n$$\nКако је $\\sum_{j=1}^{p-1} \\frac{1}{j} = \\sum_{j=1}^{\\frac{p-1}{2}} \\left( \\frac{1}{j} + \\frac{1}{p-j} \\right ) = \\sum_{j=1}^{\\frac{p-1}{2}} \\frac{p}{j(p-j)} \\equiv 0 \\pmod{p}$, коначно следи да је $\\binom{k p^{2}}{p^{2}} \\equiv k \\pmod{p^{2}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40856,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека су $M$, $N$ и $P$ средишта страница $BC$, $AC$ и $AB$, редом, а $O$ центар описане кружнице оштроуглог троугла $ABC$. Кружнице описане око троуглова $BOC$ и $MNP$ секу се у различитим тачкама $X$ и $Y$ унутар троугла $ABC$. Доказати да је\n$$\n\\varangle BAX = \\varangle CAY\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nОбележимо са $k_{1}$ и $k_{2}$ редом кругове $MNP$ и $BOC$. Круг $k_{1}$ је Ојлеров круг у $\\triangle ABC$ и пролази кроз подножја висина $D$, $E$ из $B$, $C$ и средиште $O_{1}$ дужи $AH$, где је $H$ ортоцентар $\\triangle ABC$.\n\nПокажимо да друга пресечна тачка $Z$ праве $AY$ и круга $k_{1}$ лежи на Ојлеровом кругу $k_{3}$ троугла $ADE$. Сматраћемо да\n\n![](attached_image_1.png)\n\nје $Z$ између $A$ и $Y$; доказ у другом случају је аналоган. Нека су $D_{1}$ и $E_{1}$ редом средишта дужи $AD$ и $AE$. Како је $AY \\cdot AZ = AD \\cdot AN = AD_{1} \\cdot AC$, тачке $Y$, $Z$, $C$, $D_{1}$ су концикличне, па је $\\varangle AZD_{1} = \\varangle ACY$.\n\nАналогно је $\\varangle AZE_{1} = \\varangle ABY$, па је $\\varangle D_{1}ZE_{1} = \\varangle AZD_{1} + \\varangle AZE_{1} = \\varangle ACY + \\varangle ABY = \\varangle BYC - \\varangle BAC = \\varangle BAC$. Одавде следи да је $Z$ на $k_{3}$.\n\nПошто је $O_{1}$ центар описаног круга $\\triangle ADE$, трансформација сличности која слика $\\triangle ABC$ у $\\triangle ADE$ такође слика $k_{1}$ у $k_{2}$ и $k_{2}$ у $k_{3}$, па је слика тачке $X \\in k_{1} \\cap k_{2}$ тачка $Z \\in k_{2} \\cap k_{3}$. Према томе, $\\varangle BAX = \\varangle DAZ = \\varangle CAY$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40857,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nОдредити све $n \\in \\mathbb{N}$ за које је могуће поделити скуп $\\{1,2, \\ldots, 3 n\\}$ на $n$ дисјунктних трочланих подскупова облика $\\{a, b, c\\}$ у којима су $b-a$ и $c-b$ различити бројеви из скупа $\\{n-1, n, n+1\\}$.",
    "options": [],
    "answer": "All even n",
    "solution": "Solution:\n\nТражена партиција скупа $\\{1,2, \\ldots, 3 n\\}$ одговара партицији темена правилног $3 n$-угла $P_{1} P_{2} \\ldots P_{3 n}$ на тројке $\\left\\{A_{i}, B_{i}, C_{i}\\right\\}$ такве да су углови сваког од троуглова $A_{i} B_{i} C_{i}$ једнаки $\\frac{n-1}{3 n} \\pi, \\frac{n}{3 n} \\pi$ и $\\frac{n+1}{3 n} \\pi$. Погодним обележавањем темена $3 n$-угла можемо постићи да темена $A_{1}, B_{1}, C_{1}$ буду управо $P_{n}, P_{2 n-1}, P_{3 n}$. Другим речима, не смањујемо општост ако претпоставимо да се међу тројкама $\\{a, b, c\\}$ на које је скуп $\\{1,2, \\ldots, 3 n\\}$ подељен налази и тројка $\\{n, 2 n-1,3 n\\}$.\n\nЈедна од преосталих $n-1$ тројки мора да садржи два броја из интервала $[2 n, 3 n-1]$, а то могу да буду једино $2 n$ и $3 n-1$. Једина тројка која садржи ове бројеве и не садржи $n$ је $\\{n-1,2 n, 3 n-1\\}$.\n\nСве остале тројке садрже тачно по један број из сваког од интервала $[1, n-2],[n+1,2 n-2]$ и $[2 n+1,3 n-2]$. Пресликавањем $(a, b, c) \\rightarrow(a, b-2, c-4)$ за $a<b<c$ добијамо одговарајуће растављање скупа $\\{1,2, \\ldots, 3(n-2)\\}$ на тројке. Како за $n=1$ овакво растављање није могуће, једноставном индукцијом показујемо да оно није могуће ни за једно непарно $n$.\n\nС друге стране, за парно $n=2 m$ тројке ( $2 i-1,2 i+n, 2 i+2 n-1$ ) и $(2 i, 2 i+n-1,2 i+2 n)$ за $i=1, \\ldots, m$ задовољавају услове.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40858,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека су $A' $ и $B'$ подножја висина из темена $A$ и $B$, редом, оштроуглог троугла $ABC$ ($AC \\neq BC$). Кружница $k$ садржи тачке $A'$ и $B'$ и додирује страницу $AB$ у тачки $D$. Ако троуглови $ADA'$ и $BDB'$ имају једнаке површине, доказати да је\n$$\n\\varangle A' D B' = \\varangle ACB.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nНека је без смањења општости $BC > AC$. Тада се праве $A'B'$ и $AB$ секу у тачки $P$, при чему је $A$ између $P$ и $B$. Из једнакости површина $ADA'$ и $BDB'$ следи да је $\\frac{AD}{DB} = \\frac{PB'}{PA'}$. С друге стране, важи и $PD^2 = PA' \\cdot PB' = PA \\cdot PB$, одакле је $\\frac{PD}{PB} = \\frac{PA}{PD} = \\frac{AD}{DB} = \\frac{PB'}{PA'}$. Из последњих једнакости\n\n![](attached_image_1.png)\n\nследи да је $B'D \\parallel BC$ и $A'D \\parallel AC$, и зато је $\\varangle A' D B' = \\varangle ACB$.\n\n\nДруго решење. Како је $\\varangle CB' D = \\alpha + \\varangle A D B' = \\alpha + \\varangle B' A' D = \\varangle C A' D = x$, синусна теорема у $\\triangle A' B' D$ и $\\triangle A B' D$ нам даје $BD = \\frac{BA' \\sin x}{\\sin (\\beta + x)}$ и $AD = \\frac{AB' \\sin x}{\\sin (\\alpha + x)}$, па је $\\frac{AD}{BD} = \\frac{AB' \\sin (\\beta + x)}{BA' \\sin (\\alpha + x)} = \\frac{\\sin (\\beta + x) \\cos \\alpha}{\\sin (\\alpha + x) \\cos \\beta}$. С друге стране, из услова $[ADA'] = [BDB']$ добијамо $\\frac{AD}{DB} = \\frac{d(B', AB)}{d(A', AB)} = \\frac{\\sin \\alpha \\cos \\alpha}{\\sin \\beta \\cos \\beta}$, тј. $\\frac{\\sin (\\beta + x)}{\\sin (\\alpha + x)} = \\frac{\\sin \\alpha}{\\sin \\beta}$. Одавде лако добијамо $\\varangle AB' D = x = \\gamma$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40859,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНаћи највећу константу $K \\in \\mathbb{R}$ са следећим својством: ако су $a_{1}, a_{2}, a_{3}, a_{4}>0$ такви да за све $i, j, k \\in \\mathbb{N}, 1 \\leqslant i<j<k \\leqslant 4$, важи $a_{i}^{2}+a_{j}^{2}+a_{k}^{2} \\geqslant 2\\left(a_{i} a_{j}+a_{j} a_{k}+a_{k} a_{i}\\right)$, онда је\n$$\na_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2} \\geqslant K\\left(a_{1} a_{2}+a_{1} a_{3}+a_{1} a_{4}+a_{2} a_{3}+a_{2} a_{4}+a_{3} a_{4}\\right)\n$$",
    "options": [],
    "answer": "11/7",
    "solution": "Solution:\n\nНека је $\\max \\left\\{a_{1}, a_{2}\\right\\} \\leqslant a_{3} \\leqslant a_{4}$. Означимо $a_{2}=\\beta^{2}$ и $a_{3}=\\gamma^{2}, \\beta, \\gamma \\geqslant 0$. Из услова задатка следи $a_{1} \\leqslant(\\gamma-\\beta)^{2}$ и $a_{4} \\geqslant(\\gamma+\\beta)^{2}$.\nПретпоставимо да су обе ове неједнакости заправо једнакости. Тада имамо $a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}=3\\left(\\beta^{4}+4 \\beta^{2} \\gamma^{2}+\\gamma^{4}\\right)$ и $a_{1} a_{2}+a_{1} a_{3}+a_{1} a_{4}+a_{2} a_{3}+a_{2} a_{4}+a_{3} a_{4}=3\\left(\\beta^{4}+\\beta^{2} \\gamma^{2}+\\gamma^{4}\\right)$. При том је $\\gamma \\leqslant 2 \\beta$, па је\n$$\n\\frac{\\beta^{4}+4 \\beta^{2} \\gamma^{2}+\\gamma^{4}}{\\beta^{4}+\\beta^{2} \\gamma^{2}+\\gamma^{4}}=1+\\frac{3 \\beta^{2} \\gamma^{2}}{\\beta^{4}+\\beta^{2} \\gamma^{2}+\\gamma^{4}}=1+\\frac{3}{1+\\frac{\\beta^{2}}{\\gamma^{2}}+\\frac{\\gamma^{2}}{\\beta^{2}}} \\geqslant \\frac{11}{7}\n$$\nуз једнакост за $\\gamma=2 \\beta$. Према томе, у овом случају је $K \\geqslant \\frac{11}{7}$, а једнакост се достиже за $a_{1}: a_{2}: a_{3}: a_{4}=1: 1: 4: 9$.\n\nПокажимо још да можемо да узмемо $a_{1}=(\\gamma-\\beta)^{2}$ и $a_{4}=(\\gamma+\\beta)^{2}$. Посматрајмо израз\n$$\nF=a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}-\\frac{11}{7}\\left(a_{1} a_{2}+a_{1} a_{3}+a_{1} a_{4}+a_{2} a_{3}+a_{2} a_{4}+a_{3} a_{4}\\right)\n$$\nЗа фиксирано $a_{2}, a_{3}, a_{4}, F$ је опадајућа функција по $a_{1}$ за $a_{1}<\\frac{11}{14}\\left(a_{2}+a_{3}+a_{4}\\right)$, при чему је $\\frac{11}{14}\\left(a_{2}+a_{3}+a_{4}\\right) \\geqslant \\frac{11}{14}\\left(\\beta^{2}+\\gamma^{2}+(\\beta+\\gamma)^{2}\\right) \\geqslant(\\gamma-\\beta)^{2} \\geqslant a_{1}$, па се $F$ не повећава ако $a_{1}$ заменимо са $(\\gamma-\\beta)^{2}$. Сада можемо да ставимо без смањења општости да је $a_{1} \\leqslant a_{2}$, тј. $\\beta \\leqslant \\gamma \\leqslant 2 \\beta$. Слично као горе, за фиксирано $a_{1}, a_{2}, a_{3}, F$ је растућа функција по $a_{4}$ за $a_{4}>\\frac{11}{14}\\left(a_{1}+a_{2}+a_{3}\\right)$, и при том је $\\frac{11}{14}\\left(a_{1}+a_{2}+a_{3}\\right) \\leqslant \\frac{11}{14}\\left(\\beta^{2}+\\gamma^{2}+(\\gamma-\\beta)^{2}\\right) \\leqslant(\\gamma+\\beta)^{2} \\leqslant a_{4}$, дакле $F$ се не повећава ако $a_{4}$ заменимо са $(\\gamma+\\beta)^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40860,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nОдредити све функције $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ такве да за све $x, y \\in \\mathbb{R}$ важи\n$$\nf(x f(y)-y f(x))=f(x y)-x y .\n$$",
    "options": [],
    "answer": "f(x) = x and f(x) = |x|",
    "solution": "Solution:\n\nЗамена $y=0$ даје $f(x f(0))=f(0)$. Ако је $f(0) \\neq 0$, израз $x f(0)$ узима све реалне вредности, па је $f$ константна функција, а она не задовољава услове. Према томе, $f(0)=0$.\n\nСтављањем $y=x$ добијамо $f(0)=f\\left(x^{2}\\right)-x^{2}$, тј. $f\\left(x^{2}\\right)=x^{2}$. Дакле, $f(x)=x$ за све $x \\geqslant 0$.\n\nУбацимо сада произвољне $x, y<0$. Како је тада $f(x y)=x y$, имамо $f(x f(y)-y f(x))=0$, што може да важи само за $x f(y)-y f(x) \\leqslant 0$. Аналогно је и $y f(x)-x f(y) \\leqslant 0$, па је $y f(x)=x f(y)$, тј. $f(x) / x=f(y) / y$. Следи да је $f(x)=c x$ за све $x<0$, где је $c$ нека константа.\n\nСада за $x<0<y$ добијамо $f((1-c) x y)=f(x y)-x y=(c-1) x y$, тј. $f(z)=-z$ за $z=(1-c) x y$. Ако је $c=1$, онда је $f(x) \\equiv x$, што је очигледно решење. С друге стране, за $c \\neq 1$ је $z \\neq 0$ и отуда $f(z) \\in\\{c z, z\\}$, па тада мора бити $c=-1$, што даје функцију $f(x)=|x|$ за све $x$. И ова функција је решење: све случајеве осим $x>0>y$ смо већ проверили, а за $x>0>y$ је $-2 x y=f(-2 x y)=f(x f(y)-y f(x))=f(x y)-x y=-2 x y$.\n\nПрема томе, решења су функције $f(x)=x$ и $f(x)=|x|$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40861,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНа страницама $BC$ и $AC$ троугла $ABC$ дате су тачке $D$ и $E$, редом. Нека је $F$ ($F \\neq C$) тачка пресека кружнице описане око троугла $CED$ и праве која садржи тачку $C$ и паралелна је са правом $AB$. Нека је $G$ тачка пресека праве $FD$ и странице $AB$, а $H$ тачка на правој $AB$ таква да је $\\varangle HDA = \\varangle GEB$ и $H-A-B$. Ако је $DG = EH$, доказати да тачка пресека дужи $AD$ и $BE$ припада симетрали угла $ACB$.\n(Милош Милосављевић)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nКако је $\\varangle AGD = 180^\\circ - \\varangle CFD = \\varangle CED = 180^\\circ - \\varangle AED$, тачке $A$, $E$, $D$, $G$ су концикличне. Одавде је $\\varangle DAG = \\varangle DEG$.\n\nМеђутим, тада је $\\varangle DHB = \\varangle DAG - \\varangle HDA = \\varangle DEG - \\varangle BEG = \\varangle DEB$, па су и тачке $H$, $E$, $D$, $B$ концикличне. Следи да је $BHE = 180^\\circ - \\varangle BDE = \\varangle CDE = \\varangle CFE$, што\n\n![](attached_image_1.png)\n\nзначи да су тачке $F$, $E$ и $H$ колинеарне.\n\nСада је $EH = EF \\cdot \\frac{AE}{EC}$ и $DG = DF \\cdot \\frac{DB}{CD}$, па услов $EH = DG$ постаје $\\frac{AE}{EC} \\cdot \\frac{CD}{DB} = \\frac{DF}{EF}$. С друге стране, из $\\varangle DEF = \\varangle DCF = \\varangle ABC$ и $\\varangle DFE = \\varangle ACB$ следи да је $\\triangle DEF \\sim \\triangle ABC$, па је $\\frac{DF}{EF} = \\frac{AC}{CB} = \\frac{AM}{MB}$, где је $M$ пресечна тачка симетрале угла $ACB$ и странице $AB$.\n\nПрема томе, $\\frac{AE}{EC} \\cdot \\frac{CD}{DB} \\cdot \\frac{BM}{MA} = 1$, па се по Чевиној теореми праве $AD$ и $BE$ секу на $CM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40862,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nДва играча играју следећу игру. Играчи наизменично записују по један природан број већи од један, при чему није дозвољено записати линеарну комбинацију претходно записаних бројева са ненегативним целим коефицијентима. Игру губи играч који не може да запише нови број. Да ли неко од играча има победничку стратегију и, ако има, који? (Александар Илић / \"Квант”)",
    "options": [],
    "answer": "First player",
    "solution": "Solution:\n\nЗа $a, b \\in \\mathbb{N}$ дефинишимо $\\mathcal{L}\\{a, b\\} = \\left\\{ a x + b y \\mid x, y \\in \\mathbb{N}_0 \\right\\}$. Прво ћемо доказати једно помоћно тврђење.\n\nЛема. Нека су $a > 1$ и $b > 1$ узајамно прости природни бројеви.\n\n(a) $N = a b - a - b$ је највећи природан број ван скупа $\\mathcal{L}\\{a, b\\}$.\n\n(б) За свако $z \\in \\mathbb{Z}$, $z \\in \\mathcal{L}\\{a, b\\}$ ако и само ако $N - z \\notin \\mathcal{L}\\{a, b\\}$.\n\nДоказ. (a) Ако је $N = (b-1)a - b = a x + b y$ за неке $x \\in \\{0, \\ldots, b-1\\}$ и $y \\in \\mathbb{Z}$, онда је $x \\equiv b-1 \\pmod{b}$, па је $x \\geqslant b-1$ и одатле $y < 0$; дакле, $N \\notin \\mathcal{L}\\{a, b\\}$.\n\n(б) Јасно је да из $z \\in \\mathcal{L}\\{a, b\\}$ следи $N-z \\notin \\mathcal{L}\\{a, b\\}$ (у супротном би важило $N = z + (N-z) \\in \\mathcal{L}\\{a, b\\}$). Посматрајмо сада неко $z \\in \\mathbb{Z} \\setminus \\mathcal{L}\\{a, b\\}$. Ако је $x \\in \\{0, \\ldots, b-1\\}$ такво да је $a x \\equiv z \\pmod{b}$, онда је $z < a x$, тј. $a x - b y = z$ за неко $y \\in \\mathbb{N}$. Тада је $N - x = (b-1-x)a + b(y-1) \\in \\mathcal{L}\\{a, b\\}$.\n\nНека играч А започне игру уписујући прост број $a \\geqslant 5$, након чега играч Б уписује природан број $b$ такав да $a \\nmid b$. Само коначно много природних бројева не припада скупу $\\mathcal{L}\\{a, b\\}$, па је игра коначна. Дакле, један од играча има победничку стратегију.\n\nРазмотримо игру у којој је други потез играча А број $N$. Ако је то губитнички потез, сада Б има победнички одговор у виду неког дозвољеног броја $c$. Међутим, онда играч А може да у другом потезу, уместо броја $N$, запише број $c$ и надаље следи добитничку стратегију другог играча. Заиста, по леми, број $N-c$ није дозвољен, па након потеза $c$ ни број $N = (N-c) + c$ није дозвољен, те Б остаје управо у позицији која би за А била губитничка.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40863,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nЗа природан број $n$ кажемо да је шашав ако и само ако постоје природни бројеви $a>1$ и $b>1$ такви да је $n=a^{b}+b$. Да ли постоји 2014 узастопних природних бројева међу којима је тачно 2012 шашавих бројева?\n\n(Милош Милосављевић)",
    "options": [],
    "answer": "Yes",
    "solution": "Solution:\n\nПрво ћемо дати пример 2012 узастопних шашавих бројева. Довољно је узети бројеве $N+2, N+3, \\ldots, N+2013$, где је $N=2^{2013!}$.\nЗа природан број $n$, означимо са $f(n)$ број шашавих бројева међу $n, n+1, \\ldots, n+2013$. Како је $f(1)<2012$ (бројеви $1,2,3,4,5$ нису шашави), $f(N) \\geqslant 2012$ и $|f(n+1)-f(n)| \\leqslant 1$ за свако $n$, постоји $n$ такво да је $f(n)=2012$.\n\n\nДруго решење. Означимо $N=\\frac{2014!}{2011}$. Бројеви $2^{N}+i$ за $2 \\leqslant i \\leqslant 2010$ и $2012 \\leqslant i \\leqslant 2014$ су шашави; докажимо да $2^{N}+1$ и $2^{N}+2011$ то нису.\nПретпоставимо да је $2^{N}+2011=a^{b}+b$ (слично се испитује и број $2^{N}+1$). Из $2^{b} \\leqslant a^{b}<2^{N}$ следи $b<N$. Такође је $b>2011$, јер би у супротном било $2011-b=a^{b}-2^{N}=\\left(a-2^{\\frac{N}{b}}\\right)\\left(\\cdots+2^{\\frac{(b-1) N}{b}}\\right)>2^{\\frac{N}{b}}>2011$. Даље, ако $2 \\mid a$, имамо $2^{b} \\mid 2^{N}-a^{b}=b-2011<2^{b}$, што је немогуће. Најзад, за $2 \\nmid a$ и $2 \\mid b$ имамо $b-2011=\\left(2^{\\frac{N}{2}}-a^{\\frac{b}{2}}\\right)\\left(2^{\\frac{N}{2}}+a^{\\frac{b}{2}}\\right)>2^{\\frac{N}{2}}$, па је $2^{N}+2011=a^{b}+b>2^{2^{N / 2}}$, опет немогуће јер је $2^{\\frac{N}{2}}>N$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40864,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nПравилан $n$-тоугао подељен је на троуглове помоћу $n-3$ дијагонале од којих никоје две немају заједничких унутрашњих тачака. Колико највише међу овим троугловима може бити међусобно неподударних?\n(Душан Ђукић)",
    "options": [],
    "answer": "For n > 3, the maximum is floor((3n − 7)/4); for n = 3, it is 1.",
    "solution": "Solution:\n\nОдговор је $\\left[\\frac{3 n-7}{4}\\right]$ за $n>3$, односно 1 за $n=3$.\n\nТроуглове са по две, једном или ниједном страницом која је уједно страница $n$-тоугла ( $n>3$ ) зовемо редом ушиима, танким и дебелим троугловима. Нека у подели има $a$ дебелих троуглова, $b$ танких и $c$ ушију. Број страница $n$-тоугла које они заузимају је $b+2c=n$. С друге стране, укупан број троуглова је $a+b+c=n-2$. Из ове две релације добијамо $c=a+2$.\n\nКако троуглова који нису дебели има највише $\\left[\\frac{n-1}{2}\\right]$ различитих, укупан број $N$ неподударних троуглова у подели није већи од $\\frac{n-1}{2}+a$. С друге стране, у таквој подели има $a+2$ ушију, а све уши су подударне, па је $N \\leqslant n-2-(a+1)=n-a-3$. Сабирањем добијамо $2N \\leqslant \\left(\\frac{n-1}{2}+a\\right)+(n-a-3)=\\frac{3n-7}{2}$, тј. $N \\leqslant \\left[\\frac{3n-7}{4}\\right]$.\n\nНајзад, повлачењем дијагонала $A_{0}A_{2i}$ и $A_{2i-2}A_{2i}\\left(1 \\leqslant i \\leqslant \\left[\\frac{n}{4}\\right]\\right)$ и $A_{0}A_{j}$ $\\left(2\\left[\\frac{n}{4}\\right]<j \\leqslant n-2\\right)$ добијамо пример са тачно $\\left[\\frac{n-1}{2}\\right]+\\left[\\frac{n}{4}\\right]-1=\\left[\\frac{3n-7}{4}\\right]$ неподударних троуглова.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40865,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nСиметрале унутрашњих углова код темена $A$ и $B$ троугла $ABC$ секу наспрамне странице у тачкама $D$ и $E$, редом. Ромб је уписан у четвороугао $ABDE$ тако да се на свакој страници четвороугла налази тачно једно теме ромба. Ако је $\\angle BAC=\\alpha$ и $\\angle ABC=\\beta$, доказати да је барем један угао ромба не већи од $\\max\\{\\alpha, \\beta\\}$.\n\n(Душан Ђукић)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nНека је $KLMN$ ромб са $K \\in AB$, $L \\in BD$ и $N \\in EA$. Посматрајмо трапез $PQRS$ са $P, Q \\in AB$, $PQ \\parallel RS$ и $\\angle PQR=\\angle QPS=\\angle KNM=\\varphi$ такав да су $K, L, M, N$ на $PQ, QR, RS, SP$ редом. Претпоставимо да је $\\varphi>\\alpha, \\beta$. Тада $R$ и $S$ леже ван $\\triangle ABC$.\n\nКако је $\\angle SNM=180^\\circ-\\varphi-\\angle NMS=\\angle RML$ и $\\angle LRM=\\angle MSN$, троуглови $LRM$ и $MSN$ су подударни, па је $LR=MS$. Слично, ако је $Q'$ тачка на $AB$ таква да је $\\angle LQ'B=\\varphi$, троуглови $LRM$ и $KQ'L$ су подударни, па је $MR=LQ'=LQ$. Следи да је $RQ=RL+LQ=SM+MR=RS$.\n\n![](attached_image_1.png)\n\nСада је $d(R, AB)=RQ \\sin \\varphi>RS \\sin \\alpha>d(R, AC)$ и аналогно $d(S, AB)>d(S, BC)$, што значи да тачке $R$ и $S$ леже изнад правих $AD$ и $BE$ редом (тј. у полуравнима у којима је $C$), одакле следи да обе тачке леже изнад праве $DE$. Ово је немогуће јер се дужи $RS$ и $DE$ секу у $M$.\n\n\nДруго решење. Растојање од тачке $X$ до праве $p$ означавамо $d(X, p)$.\n\nЛема. За произвољну тачку $M$ на дужи $DE$ важи $d(M, AB)=d(M, AC)+d(M, BC)$.\n\nДоказ. Ако је $\\frac{DM}{DE}=k$, важи $d(M, AB)=k d(E, AB)+(1-k) d(D, AB)=k d(E, BC)+(1-k) d(D, AC)=d(M, BC)+d(M, AC)$.\n\nКористимо исте ознаке као у првом решењу. Нека је $a$ страница ромба и $O$ његов центар. Имамо\n\n$$\n\\begin{aligned}\nd(M, AC)+d(M, BC) & =a(\\sin \\angle MNC+\\sin \\angle MLC), \\\\\n\\text{и}\\quad d(M, AB) & =2 d(O, AB)=d(L, AB)+d(N, AB) \\\\\n& =a(\\sin \\angle NKA+\\sin \\angle LKB) .\n\\end{aligned}\n$$\n\nМеђутим, ако је $\\varphi>\\alpha, \\beta$, онда је $\\angle NKA=\\angle MNC+\\varphi-\\alpha>\\angle MNC$ и аналогно $\\angle LKB>\\angle MLC$, па из горњих једначина следи $d(M, AB)>d(M, AC)+d(M, BC)$, контрадикција.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40866,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nДат је тетивни четвороугао $A B C D$. Тачке $M, N, P$ и $Q$ су средишта страница $D A, A B, B C$ и $C D$, редом, а тачка $E$ је пресек дијагонала $A C$ и $B D$. Кружнице описане око $\\triangle E M N$ и $\\triangle E P Q$ секу се у тачки $F \\neq E$. Доказати да важи $E F \\perp A C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nТроуглови $E A B$ и $E D C$ су слични, па су то и троуглови $E B N$ и $E C Q$. Зато у оријентисаним угловима важи $\\varangle M F E=\\varangle M N E=\\varangle B E N=\\varangle Q E C=\\varangle E Q M$. Аналогно важи $\\varangle Q F E=\\varangle E M Q$, одакле следи да је $F$ ортоцентар троугла $E M Q$. Дакле, $E F \\perp Q M \\| A C$.\n\n\nДруго решење. Посматрајмо транслацију $\\mathcal{T}$ за вектор $\\frac{1}{2} \\overrightarrow{A C}$. Важи $\\mathcal{T}(M)=$\n![](attached_image_1.png)\n$Q$ и $\\mathcal{T}(N)=P$; означимо $\\mathcal{T}(E)=E^{\\prime}$. Из сличности троуглова $A E D$ и $B E C$ следи $\\triangle A E M \\sim \\triangle B E P$, одакле је $\\varangle Q E^{\\prime} E=\\varangle E M Q=\\varangle M E A=\\varangle B E P=\\varangle Q P E$, па тачка $E^{\\prime}$ лежи на кругу $P E Q$. Према томе, транслација $\\mathcal{T}$ слика круг $M E N$ у круг $P E Q$, па је права која спаја њихове центре паралелна са $A C$, одакле следи тврђење.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40867,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $k$ природан број. За $n \\in \\mathbb{N}$ означимо са $f_{k}(n)$ најмањи природан број већи од $k n$ такав да је $n f_{k}(n)$ потпун квадрат природног броја. Ако је испуњено $f_{k}(m)=f_{k}(n)$, доказати да важи $m=n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nПретпоставимо да је $f_{k}(m)=f_{k}(n)=q$. Напишимо број $q$ у облику $q=a u^{2}$, где су $a, u \\in \\mathbb{N}$ и $a$ није дељиво ниједним потпуним квадратом већим од 1. Како је $m q=a m u^{2}$ потпун квадрат, то је и $a m$, па следи да је $m=a v^{2}$ за неко $v \\in \\mathbb{N}$. Слично је $n=a w^{2}$ за неко $w \\in \\mathbb{N}$.\nКако је $f_{k}\\left(a v^{2}\\right)=a u^{2}$, $u$ је најмањи природан број већи од $v \\sqrt{k}$. Аналогно, $u$ је најмањи природан број већи од $w \\sqrt{k}$, па мора да важи $|v \\sqrt{k}-w \\sqrt{k}|<1$. Међутим, одавде следи $|v-w|<\\frac{1}{\\sqrt{k}}<1$, па мора бити $v=w$, тј. $m=n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40868,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nСтражар предлаже затвореницима следећу игру. Сви ће бити изведени у двориште, где ће свакоме од њих бити стављен на главу шешир у једној од 5 могућих боја. Стражар ће их потом поређати у врсту тако да сваки затвореник види све шешире осим сопственог и питати првог затвореника у врсти да ли зна боју свог шешира. Затвореник гласно одговара “да\" или \"не\". Ако одговори \"не\", биће одмах закључан у самицу. Ако одговори “да\", стражар ће га питати које је боје његов шешир, на шта затвореник треба да одговори на такав начин да остали затвореници не чују одговор. Уколико је одговор погрешан, тај затвореник биће одмах закључан у самицу пред свима, а ако је одговор тачан, тај затвореник биће одмах ослобођен пред свима. Стражар потом прилази следећем затворенику у реду и понавља исти поступак, и тако све до последњег затвореника. Затвореници имају могућност да осмисле стратегију пре почетка игре, али кад игра почне, никаква комуникација међу затвореницима није дозвољена. Ако у затвору има 2015 затвореника, који је максималан број затвореника који ће загарантовано бити ослобођени уколико затвореници примењују оптималну стратегију?",
    "options": [],
    "answer": "2013",
    "solution": "Solution:\n\nДоделимо бојама вредности $0,1,2,3,4$ и означимо са $B$ боју шешира другог затвореника, а са $S$ збир боја шешира од трећег до последњег затвореника по модулу 5. Описаћемо стратегију прва два затвореника након које ће сви остали затвореници знати $S$, те ће, знајући боје осталих шешира, моћи да одреде своју. Овако ће за бар 2013 затвореника бити обезбеђена слобода. Први затвореник одговара \"не\" ако је $S \\in\\{0, B, B+1\\}$ или $(B, S)=(4,1)$. На то други одговара \"не\" ако је $S=0$, а у супротном каже \"да\" и стражару даје одговор $S$. На овај начин, ако чују два одговора \"не\", остали знају да је $S=0$. Ако други каже \"да\" и буде ослобођен, остали знају да је $S=B$, док у супротном знају да је $S=B+1$ ако $B \\neq 4$, односно $S=1$ ако $B=4$.\n\nУ осталим случајевима први затвореник каже “да\" (небитно је шта ће бити с њим). На то други каже \"да\" ако $S \\in\\{2,4\\}$ и стражару даје одговор $S-2$. Ако је $B=0$, онда $S \\notin\\{0,1\\}$, па остали знају да је $S=2$ ако други буде ослобођен, $S=3$ ако каже \"не\", и $S=4$ ако каже \"да\" и промаши. Најзад, ако је $B \\neq 0$, с обзиром на потврдан одговор првог важи $S \\equiv B+2$ или $S \\equiv B+3\\ (\\bmod\\ 4)$, а одговор другог одређује парност $S$, што осталима омогућује да одреде $S$.\n\nОстаје да докажемо да ниједна стратегија не гарантује слободу 2014 затвореника. Претпоставимо да таква стратегија постоји и посматрајмо скуп од пет распореда који се разликују само у боји шешира другог затвореника. У највише једном од ових распореда стратегија предвиђа да први каже \"не\" (у супротном други не би могао да са сигурношћу одреди боју свог шешира). У бар три од преостала четири распореда први би морао да промаши, па други не би могао да одреди боју свог шешира. Овим је доказ завршен.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40869,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nЗа цео број $a, a \\neq 0$, означимо са $v_{2}(a)$ највећи ненегативан цео број $k$ такав да $2^{k} \\mid a$. За дато $n \\in \\mathbb{N}$ одредити највећу могућу кардиналност подскупа $A$ скупа $\\{1,2,3, \\ldots, 2^{n}\\}$ са следећим својством:\n$$\n\\text{за све } x, y \\in A, x \\neq y, \\text{ број } v_{2}(x-y) \\text{ је паран. }\n$$",
    "options": [],
    "answer": "2^{floor((n+1)/2)}",
    "solution": "Solution:\n\nДоказаћемо индукцијом по $k$ да скуп $A$ садржи највише $2^{k}$ различитих елемената по модулу $2^{2k}$. То тривијално важи за $k=0$. Нека је $k>0$. По индуктивној претпоставци, елементи скупа $A$ дају највише $2^{k-1}$ остатака по модулу $2^{2k-2}$. Претпоставимо да елементи $A$ дају више од $2^{k}$ остатака по модулу $2^{2k}$. На основу Дирихлеовог принципа, бар три од ових остатака су једнаки по модулу $2^{2k-2}$. Али међу ова три остатка, два се разликују за $2^{2k-1}(\\bmod\\ 2^{2k})$, противно услову задатка.\nСледи да је $|A| \\leqslant 2^{\\left[\\frac{n+1}{2}\\right]}$. Пример скупа $A$ са $2^{\\left[\\frac{n+1}{2}\\right]}$ елемената добијамо укључивањем бројева облика $\\sum_{i \\in B} 4^{i}$ за све подскупове $B$ скупа $\\{0,1, \\ldots,\\left[\\frac{n-1}{2}\\right]\\}$.\n\n\nДруго решење. Кажемо да је скуп $X$ срећан ако је $v_{2}(x-y)$ парно за све $x, y \\in X(x \\neq y)$, а несрећан ако је $v_{2}(x-y)$ непарно за све $x, y \\in X(x \\neq y)$. Означимо са $a_{n}$ и $b_{n}$ редом максималне кардиналности срећног и несрећног подскупа скупа $T_{n}=\\{1,2, \\ldots, 2^{n}\\}$.\nПосматрајмо срећан скуп $A \\subset T_{n}, n \\geqslant 1$. Како је $v_{2}(2x-2y)=v_{2}(x-y)+1$, скупови $A_{0}=\\left\\{\\frac{x}{2}\\mid x \\in A, 2\\mid x\\right\\}$ и $A_{1}=\\left\\{\\left.\\frac{x-1}{2} \\right| x \\in A, 2 \\nmid x\\right\\}$ су несрећни подскупови скупа $T_{n-1}$ и имају највише по $b_{n-1}$ елемената. С друге стране, ако је $A_{0} \\subset T_{n-1}$ несрећан скуп, скуп $\\{2x, 2x+1 \\mid x \\in A_{0}\\} \\subset T_{n}$ је срећан и има $2|A_{0}|$ елемената. Следи да је $a_{n}=2b_{n-1}$.\nСлично, ако је $B \\subset T_{n}$ непаран скуп, сви његови елементи су исте парности, а скуп $B' = \\left\\{\\left.\\left\\lceil\\frac{x}{2}\\right\\rceil \\right| x \\in B\\right\\} \\subset T_{n-1}$ је срећан. С друге стране, ако је $B' \\subset T_{n-1}$ срећан, скуп $B=\\{2x-1 \\mid x \\in B'\\} \\subset T_{n}$ је несрећан. Одавде је $b_{n}=a_{n-1}$.\nДобијене релације дају $a_{n}=2a_{n-2}$ за $n \\geqslant 2$, па из $a_{0}=1$ и $a_{1}=2$ једноставном индукцијом добијамо $a_{n}=2^{\\left[\\frac{n+1}{2}\\right]}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40870,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nДоказати неједнакост\n$$\n\\frac{x-y}{x y+2 y+1}+\\frac{y-z}{y z+2 z+1}+\\frac{z-x}{z x+2 x+1} \\geqslant 0\n$$\nгде су $x, y$ и $z$ ненегативни реални бројеви.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nОзначимо $a=\\frac{x-y}{x y+2 y+1}$, $b=\\frac{y-z}{y z+2 z+1}$ и $c=\\frac{z-x}{z x+2 x+1}$. Тада је $1+\\frac{1}{a}=\\frac{x y+x+y+1}{x-y}$ и одатле $\\frac{a}{a+1}=\\frac{x-y}{x y+x+y+1}=\\frac{1}{y+1}-\\frac{1}{x+1}$; аналогно је $\\frac{b}{b+1}=\\frac{1}{z+1}-\\frac{1}{y+1}$ и $\\frac{c}{c+1}=\\frac{1}{x+1}-\\frac{1}{z+1}$.\nИз $0<\\frac{1}{x+1}, \\frac{1}{y+1}, \\frac{1}{z+1}<1$ следи $\\frac{a}{a+1}, \\frac{b}{b+1}, \\frac{c}{c+1}<1$, па су $a+1, b+1, c+1$ позитивни. Осим тога, имамо $\\frac{a}{a+1}+\\frac{b}{b+1}+\\frac{c}{c+1}=0$, тј. $\\frac{1}{a+1}+\\frac{1}{b+1}+\\frac{1}{c+1}=3$. Сада је по Коши-Шварцовој неједнакости $(a+1)+(b+1)+(c+1) \\geqslant 3$, тј. $a+b+c \\geqslant 0$.\n\n\nДруго решење. Тражена неједнакост се множењем и груписањем своди на\n$$\n\\begin{aligned}\n& 2(x-1)^{2}(y-z)^{2}+2(y-1)^{2}(z-x)^{2}+2(z-1)^{2}(x-y)^{2} \\\\\n& +9\\left(x y^{2}+y z^{2}+z x^{2}-3 x y z\\right)+3\\left(x^{2} y+y^{2} z+z^{2} x-3 x y z\\right) \\geqslant 0\n\\end{aligned}\n$$\nгде су сви сабирци ненегативни на основу А-Г неједнакости.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40871,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nУ скупу ненегативних целих бројева решити једначину\n$$\n\\left(2^{2015}+1\\right)^{x}+2^{2015}=2^{y}+1.\n$$",
    "options": [],
    "answer": "(x, y) = (0, 2015) and (1, 2016)",
    "solution": "Solution:\n\nЗа $x \\leqslant 1$ једина решења су $(0,2015)$ и $(1,2016)$.\n\nНека је $x>1$. Како је $2^{2015}+1$ дељиво са $3$, имамо\n$$\n\\left(2^{2015}+1\\right)^{x}+2^{2015} \\equiv 2^{2015} \\equiv 5 \\pmod{9},\n$$\nдакле $2^{y} \\equiv 4 \\pmod{9}$, одакле добијамо $y=6k+2$ за неко $k \\in \\mathbb{N}$.\n\nСада по модулу $13$ имамо $2^{y}+1=\\left(2^{6}\\right)^{k} \\cdot 2^{2} \\equiv \\pm 4$ и $2^{2015} \\equiv 7 \\pmod{13}$, па тако добијамо $8^{x}+7 \\equiv \\pm 4 \\pmod{13}$. Међутим, $8^{x}$ даје један од остатака $1,5,8,12$ по модулу $13$, па је последња конгруенција немогућа.\nSolution:\n\nЗа $x>1$ имамо\n$$\n2^{y}=\\left(2^{2015}+1\\right)^{x}+2^{2015}-1=(x+1)2^{2015}+\\sum_{i=2}^{x}\\binom{x}{i}2^{2015i} \\equiv (x+1)2^{2015} \\pmod{2^{2019}},\n$$\nпа због $y>2019$ следи да $16 \\mid x+1$.\n\nСада посматрајмо једначину по модулу $17$. Како је $2^{2015} \\equiv 9 \\pmod{17}$, добијамо $2^{y} \\equiv 10^{x}+8 \\equiv 10^{15}+8 \\equiv 3 \\pmod{17}$, што је немогуће.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40872,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $n$ природан број већи од 1. Доказати да постоји природан број $m$ већи од $n^{n}$ такав да је\n$$\n\\frac{n^{m}-m^{n}}{n+m}\n$$\nприродан број.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nЗа почетак приметимо да за $m>n \\geqslant 3$ важи $n^{m}>m^{n}$, те је $\\frac{n^{m}-m^{n}}{m+n}>0$. Заиста, функција $f(x)=\\frac{\\ln x}{x}$ је опадајућа за $x>e$ јер је $f^{\\prime}(x)=\\frac{1-\\ln x}{x^{2}}<0$, па је $\\frac{\\ln n}{n}>\\frac{\\ln m}{m}$, тј. $m \\ln n>n \\ln m$, и одатле $n^{m}=e^{m \\ln n}>e^{n \\ln m}=m^{n}$.\n\nЗа $n=2$ се може узети $m=10$. Претпоставимо да је $n>2$. Имамо\n$$\nn^{m}-m^{n} \\equiv n^{m}-(-n)^{n}=n^{n}\\left(n^{m-n}-(-1)^{n}\\right) \\quad(\\bmod m+n)\n$$\nПотражићемо $m$ у облику $m=k n^{n}-n\\ (k \\in \\mathbb{N})$. Тада $m+n=k n^{n} \\mid n^{m}-m^{n}$ ако и само ако $k \\mid n^{m-n}-(-1)^{n}$.\n\n$(1^{\\circ})$ Ако је $n$ непарно, онда је $n^{m-n}-(-1)^{n}$ парно, па можемо узети $k=2$, тј. $m=2 n^{n}-n$.\n\n$(2^{\\circ})$ Ако је $n$ парно, онда је $n^{m-n}-(-1)^{n}=n^{m-n}-1$ дељиво са $n-1$, па можемо узети $k=n-1$, тј. $m=(n-1) n^{n}-n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40873,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nДат је природан број $n$. Дефинишимо $f(0, j)=f(i, 0)=0$, $f(1,1)=n$ и\n$$\nf(i, j)=\\left\\lfloor\\frac{f(i-1, j)}{2}\\right\\rfloor+\\left\\lfloor\\frac{f(i, j-1)}{2}\\right\\rfloor\n$$\nза све природне бројеве $i$ и $j$, $(i, j) \\neq (1,1)$. Колико има уређених парова природних бројева $(i, j)$ за које је $f(i, j)$ непаран број? (Душан Ђукић)",
    "options": [],
    "answer": "n",
    "solution": "Solution:\n\nЗа $m \\geqslant 2$ означимо $s_{m}=\\sum_{i+j=m} f(i, j)$. Како је остатак $f(i, j)$ при дељењу са 2 једнак $f(i, j)-2\\left[\\frac{f(i, j)}{2}\\right]$, број непарних међу бројевима $f(i, j)$ за $i, j \\geqslant 0$ и $i+j=m$ једнак је\n$$\n\\begin{aligned}\n\\sum_{i+j=m}\\left(f(i, j)-2\\left[\\frac{f(i, j)}{2}\\right]\\right) & =s_{m}-\\sum_{i+j=m}\\left(\\left[\\frac{f(i-1, j+1)}{2}\\right]+\\left[\\frac{f(i, j)}{2}\\right]\\right) \\\\\n& =s_{m}-s_{m+1}\n\\end{aligned}\n$$\nСледи да је број парова $(i, j)$ за које је $f(i, j)$ непарно и $i+j<m$ једнак $s_{2}-s_{m}=n-s_{m}$.\nОстаје да покажемо да је $s_{m}=0$ за све довољно велике $m$. Јасно је да је низ $s_{m}$ ненегативан и нерастући, па постоје $N$ и $k$ такви да је $s_{m}=k$ за све $m \\geqslant N$. То значи да је $f(i, j)$ парно кад год је $i+j \\geqslant m$. Претпоставимо да је $k>0$ и посматрајмо најмање $i$ такво да је $f(i, m-i)>0$. Једноставном индукцијом добијамо $f(i, m+r-i)=\\left[\\frac{f(i, m-i)}{2^{r}}\\right]$ за $r \\geqslant 1$. Међутим, ако је $2^{r} \\leqslant f(i, m-i)<2^{r+1}$, одавде је $f(i, m+r-i)=1$, противно претпоставци.\nSolution:\n\nДруго решење (У. Динић). Поставимо $n$ жетона у тачку $(1,1)$ у координатној равни. У сваком кораку, из сваке тачке $(i, j)$ ћемо пребацити по цео део половине њених жетона у тачке $(i+1, j)$ и $(i, j+1)$. Приметимо да, ако је нека врста или колона у неком тренутку непразна, она ће увек остати непразна. Тако ниједан жетон не може изаћи из квадрата $[1, n] \\times [1, n]$, па се игра завршава у коначном броју корака.\nНије тешко видети да се након $i+j-2$ корака у тачки $(i, j)$ налази тачно $f(i, j)$ жетона. Шта више, број $f(i, j)$ је непаран ако након следећег корака у тачки $(i, j)$ остане један жетон. Како у завршној позицији има тачно $n$ жетона, следи да међу члановима низа $f(i, j)$ има $n$ непарних бројева.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40874,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $O$ центар кружнице описане око $\\triangle A B C$. Права $t$ додирује кружницу описану око $\\triangle B O C$ и сече странице $A B$ и $A C$ у тачкама $D$ и $E$, редом $(D, E \\not \\equiv A)$. Тачка $A'$ је симетрична тачки $A$ у односу на праву $t$. Доказати да се кружнице описане око $\\triangle A' D E$ и $\\triangle A B C$ додирују. (Душан Ђукић)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nОзначимо са $K$ додирну тачку праве $t$ и круга $B O C$. Нека се кругови описани око троуглова $B D K$ и $C E K$ секу у тачки $X \\neq K$. Како је $\\varangle B X C = \\varangle B X K + \\varangle K X C = \\varangle A D K + \\varangle K E A = 180^\\circ - \\varangle C A B$, тачка $X$ лежи на описаном кругу $k$ троугла $A B C$. Шта више, $\\varangle D X E = \\varangle D X K + \\varangle K X E = \\varangle D B K + \\varangle K C E = \\varangle C K B - \\varangle C A B = \\varangle C A B = \\varangle D X E$, па $X$ такође лежи на описаном кругу $k_1$ троугла $A' D E$. Доказаћемо да се кругови $k$ и $k_1$ додирују у тачки $X$.\n\nАко се праве $C K$ и $X D$ секу у тачки $P$, онда важи $\\varangle X P C = \\varangle X D E - \\varangle C K E = \\varangle X B K - \\varangle C B K = \\varangle X B C$, што значи да $P$ лежи на кругу $k$. Аналогно, праве $B K$ и $X E$ се секу у тачки $Q$ на кругу $k$. Најзад, из $\\varangle X P Q = \\varangle X B Q = \\varangle X D K$ следи да\n\n![](attached_image_1.png)\n\nје $P Q \\parallel D E$. Према томе, троуглови $X D E$ и $X P Q$ су хомотетични са центром хомотетије $X$, па се њихови описани кругови додирују у $X$.\nSolution:\n\nДруго решење. Нека праве $B K$ и $C K$ поново секу описани круг $\\triangle A B C$ редом у тачкама $Q$ и $P$. Из $\\varangle C P Q = \\varangle C B Q = \\varangle C K E$ следи $P Q \\parallel D E$. Нека се праве $D P$ и $E Q$ секу у тачки $X$. Како су тачке $D = P X \\cap A B$, $K = P C \\cap Q B$ и $E = A C \\cap Q X$ колинеарне, по обратној Паскаловој теореми тачка $X$ лежи на истом кругу са тачкама $A, B, C, P, Q$. Према томе, троуглови $X D E$ и $X P Q$ су хомотетични, па се њихови описани кругови додирују у њиховом центру хомотетије $X$. Најзад, тачка $A'$ лежи на описаном кругу $\\triangle D E X$ јер je $\\varangle D X E = \\varangle P X Q = \\varangle P C A + \\varangle A B Q = \\varangle B K C - \\varangle B A C = \\varangle B A C = \\varangle D A' E$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40875,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nУ $\\triangle ABC$ ($AB \\neq AC$) уписана је кружница, чији је центар тачка $I$, и она додирује страницу $BC$ у тачки $D$. Нека је тачка $M$ средиште дужи $BC$. Доказати да се нормале из тачака $M$ и $D$ на праве $AI$ и $MI$, редом, секу на правој која садржи висину $\\triangle ABC$ из темена $A$.\n\n(Душан Ђукић)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nОзначимо са $\\gamma$ уписани круг $\\triangle ABC$, а са $\\gamma_{a}$ и $I_{a}$ приписани круг наспрам $A$ и његов центар. Тачка $E$ симетрична тачки $D$ у односу на $M$ је додирна тачка $\\gamma_{a}$ са страницом $BC$. Нормала $\\ell_{1}$ из $D$ на $MI$ је радикална оса круга $\\gamma$ и круга $\\omega$ над пречником $DE$, док је нормала $\\ell_{2}$ из $AI$ радикална оса кругова $\\gamma$ и $\\gamma_{a}$ (због $MD=ME$). Такође, радикална оса кругова $\\gamma_{a}$ и $\\omega$ је нормала $\\ell_{3}$ из $E$ на $MI_{a}$. Праве $\\ell_{1}, \\ell_{2}$ и $\\ell_{3}$ се секу у радикалном центру $S$ кругова $\\gamma, \\gamma_{a}, \\omega$. С друге стране, познато је (по \"Великом задатку\") да важи $MI \\parallel AE$ и $MI_{a} \\parallel AD$, па праве $\\ell_{1}$ и $\\ell_{3}$ садрже висине из $D$ и $E$ у троуглу $ADE$. Следи да је $S$ ортоцентар $\\triangle ADE$, а он лежи на висини из темена $A$.\n\n![](attached_image_1.png)\n\n\nДруго решење.\n\nНека се нормале из $M$ и $D$ редом на $AI$ и $MI$ секу у тачки $S$ и нека је $J$ тачка у којој права $MI$ сече висину из $A$ у $\\triangle ABC$. Довољно је доказати да је $AJ=ID$. Заиста, тада ће следити да је $AJDI$ паралелограм, па је $MS \\perp DJ$, тако да је $D$ ортоцентар троугла $MSJ$ и одатле $JS \\perp MD$, тј. $AS \\perp BC$.\n\nОво се директно израчунава. Означимо са $H$ и $F$ редом подножја висине и симетрале угла из темена $A$, и $a=BC, b=CA, c=AB$. Тада је $BF=\\frac{ac}{b+c}$, $BD=\\frac{a-b+c}{2}$ и $BH=\\frac{a^{2}-b^{2}+c^{2}}{2a}$, па је $FH=BF-BH=\\frac{|b-c|((b+c)^{2}-a^{2})}{2a(b+c)}$, $FD=BF-BD=\\frac{|b-c|(b+c-a)}{2(b+c)}$ и најзад $\\frac{AJ}{AH}=\\frac{FD}{FH}=\\frac{a}{a+b+c}=\\frac{ID}{AH}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40876,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nДато је $2n-1$ двоелементних подскупова скупа $\\{1,2, \\ldots, n\\}$. Доказати да се може одабрати $n$ од ових подскупова чија унија садржи не више од $\\frac{2}{3}n+1$ елемената.\n\n(Душан Ђукић)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nПоказаћемо индукцијом по $k\\left(k \\leqslant \\frac{2n-1}{3}\\right)$ да се може избацити $3k$ подскупова тако да је кардиналност уније преосталих не већа од $n-k$.\n\nЗа $k=0$ то је тривијално. Претпоставимо да је $k \\geqslant 1$ и да смо избацили $3(k-1)$ подскупова тако да унија преосталих не садржи више од $n-k+1$ елемената. Како је $2n-1-3(k-1)<2(n-k+1)$, постоји елемент $x_{k}$ из уније који се налази у највише три од преосталих подскупова. Дакле, можемо да избацимо још три подскупа тако да унија преосталих $2n-1-3k$ не садржи елемент $x_{k}$, што завршава индуктивни корак.\n\nТврђење задатка следи за $k=\\left[\\frac{n-1}{3}\\right]$ јер је $n-\\left[\\frac{n-1}{3}\\right] \\leqslant n-\\frac{n-3}{3}=\\frac{2}{3}n+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40877,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nДати су природни бројеви $a_{1}, a_{2}, \\ldots, a_{2 \\text{ 2016 }}$ такви да за све $n, 1 \\leqslant n \\leqslant 2^{2016}$, важи\n$$\na_{n} \\leqslant 2016 \\quad \\text{и} \\quad a_{1} a_{2} \\cdots a_{n}+1 \\text{ је потпун квадрат. }\n$$\nДоказати да је неки од бројева $a_{1}, a_{2}, \\ldots, a_{22016}$ једнак 1. (Дуиан Ђукић)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nКључна чињеница је да, ако су $a+1=u^{2}$ и $b=v^{2}$ потпуни квадрати и $a>b$, онда $a b+1$ није квадрат. Заиста, тада је $(u v-1)^{2}<a b+1=u^{2} v^{2}-v^{2}+1<(u v)^{2}$.\n\nНека су $p_{1}, p_{2}, \\ldots, p_{m}$ сви прости бројеви мањи од 2016. За $1 \\leqslant n \\leqslant 2^{2016}$ посматрајмо бинарни низ $\\mathrm{c}_{n}=\\left(r_{1}, r_{2}, \\ldots, r_{m}\\right)$, где је $r_{i}=0$ ако је експонент уз $p_{i}$ у производу $P_{n}=a_{1} a_{2} \\cdots a_{n}$ паран, а $r_{i}=1$ у супротном. Како за низ $c_{n}$ има само $2^{m}$ могућности, за свако $k \\leqslant 2^{2016}-2^{m}$ међу низовима $c_{k+1}, \\ldots, \\mathrm{c}_{k+2^{m}}$ постоје два једнака, рецимо $c_{s}$ и $c_{t}$ ($s<t$), а тада је $P_{t} / P_{s}$ потпун квадрат не већи од $2016^{2^{m}}$.\n\nПретпоставимо да у низу $\\left(a_{n}\\right)$ нема јединица. Узмимо $k=11 \\cdot 2^{m}$; свакако је $k+2^{m}<2^{2016}$. Видели смо да постоје индекси $s, t, k \\leqslant s<t \\leqslant k+2^{m}$, тако да је $b=P_{t} / P_{s}$ потпун квадрат. Међутим, како је $a=P_{s} \\geqslant P_{k} \\geqslant 2^{k}=2048^{2^{m}}>2016^{2^{m}} \\geqslant b$, по чињеници са почетка, $a+1=P_{s}+1$ и $a b+1=P_{t}+1$ не могу истовремено да буду квадрати, што је контрадикција.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40878,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека су $a, b$ и $c$ позитивни реални бројеви за које важи $a+b+c=1$. Доказати:\n$$\na \\sqrt{2 b+1}+b \\sqrt{2 c+1}+c \\sqrt{2 a+1} \\leqslant \\sqrt{2-\\left(a^{2}+b^{2}+c^{2}\\right)}\n$$\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nКвадрирањем обе стране и коришћењем једнакости $1-\\left(a^{2}+b^{2}+c^{2}\\right)=2(ab+bc+ca)$ неједнакост из задатка се своди на\n$$\n\\begin{aligned}\nL= & 2 a^{2} b+2 b^{2} c+2 c^{2} a+ \\\\\n& 2 a b \\sqrt{(2 b+1)(2 c+1)}+2 b c \\sqrt{(2 c+1)(2 a+1)}+2 c a \\sqrt{(2 a+1)(2 b+1)} \\\\\n\\leqslant D= & 4(ab+bc+ca) .\n\\end{aligned}\n$$\nПо АМ-ГМ неједнакости имамо $2ab \\sqrt{(2b+1)(2c+1)} \\leqslant ab(2b+2c+2)$ и аналогно $2bc \\sqrt{(2c+1)(2a+1)} \\leqslant bc(2c+2a+2)$ и $2ca \\sqrt{(2a+1)(2b+1)} \\leqslant ca(2a+2b+2)$, па сабирањем добијамо\n$$\n\\begin{aligned}\nL \\leqslant & 2\\left(a^{2} b+b^{2} c+c^{2} a+ab^{2}+bc^{2}+ca^{2}+3abc\\right)+2(ab+bc+ca)= \\\\\n& 2(a+b+c+1)(ab+bc+ca)=4(ab+bc+ca)=D\n\\end{aligned}\n$$\n\n\nДруго решење. Функција $f(x)=\\sqrt{x}$ је конкавна јер је $f^{\\prime}(x)=2 / \\sqrt{x}$ опадајућа функција. Применом Јенсенове неједнакости са тежинама $a, b$ и $c$ добијамо\n$$\n\\begin{aligned}\na \\sqrt{2 b+1}+b \\sqrt{2 c+1}+c \\sqrt{2 a+1} & \\leqslant \\sqrt{a(2 b+1)+b(2 c+1)+c(2 a+1)} \\\\\n& =\\sqrt{1+2(ab+bc+ca)}=\\sqrt{2-\\left(a^{2}+b^{2}+c^{2}\\right)}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40879,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nУ врсти се налази $2n-1$ сијалица. У почетку је средња ($n$-та) упаљена, а све остале су угашене. У једном кораку је дозвољено одабрати две несуседне угашене сијалице између којих су све сијалице упаљене, и променити стање тим двема сијалицама, као и свим сијалицама између њих (на пример, од конфигурације $\\cdot$ооо $\\cdot$ добија се о $\\cdot$ $\\cdot$ ). Колико највише корака је могуће извршити?\n(Душан Ђукић)",
    "options": [],
    "answer": "floor((2^{n+1} - 5)/3)",
    "solution": "Solution:\n\nОдговор је $\\left[\\frac{2^{n+1}-5}{3}\\right]$.\n\nПридружимо $i$-тој сијалици број $2^{|i-n|}$ и дефинишимо вредност конфигурације као збир бројева на упаљеним сијалицама. Вредност полазне конфигурације је $1$, а при сваком кораку она се повећава за природан умножак броја $3$. Корак повећава вредност за тачно $3$ ако $n$-та сијалица мења стање; овакав корак зовемо добрим.\n\nПошто вредност не може да премаши $2^{n+1}-4$ (јер се не могу упалити све сијалице), није могуће извршити више од $\\left[\\frac{2^{n+1}-5}{3}\\right]$ корака. Овај број се може достићи: довољно је показати да је могуће извршити бар $\\frac{2^{n+1}-7}{3}$ корака.\n\nДоказаћемо индукцијом по $n$ да, почевши од конфигурације вредности највише $3$, можемо да низом добрих корака добијемо конфигурацију вредности бар $2^{n+1}-6$. Ово се директно проверава за $n \\leqslant 2$. Нека је $n \\geqslant 3$. По индуктивној претпоставци за $n-1$, могуће је доћи до конфигурације вредности бар $2^{n}-6$ са првом и последњом сијалицом угашеном. У таквој конфигурацији, осим прве и последње сијалице, могу бити угашене\n\n$(1^{\\circ})$ само $n$-та,\n\n$(2^{\\circ})$ само $n$-та и једна од њој суседних, или\n\n$(3)$ само једна од две суседне.\n\nУ сваком од ова три случаја, у највише три добра корака постижемо да прва и последња сијалица буду упаљене и да вредност остатка конфигурације (без ове две сијалице) буде највише $3$.\n\n![](attached_image_1.png)\n\n$(1^{\\circ})$\n\n![](attached_image_2.png)\n\n$(2^{\\circ})$\n\n![](attached_image_3.png)\n\n$(3^{\\circ})$\n\nПоновна примена индуктивне претпоставке за $n-1$ завршава индукцију.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40880,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $a$ природан број такав да за сваки природан број $n$ број $n^{2} a-1$ има бар један делилац већи од 1 који даје остатак 1 при дељењу са $n$. Доказати да је $a$ потпун квадрат.\n\n(Душан Ђукић)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nНека је $n^{2} a-1 = (n x_{n} + 1) d_{n}$, где $x_{n}, d_{n} \\in \\mathbb{N}$. Тада је $d_{n} \\equiv -1 \\pmod{n}$, па је\n$$\nn^{2} a-1 = (n x_{n} + 1)(n y_{n} - 1) \\quad \\text{за неке } x_{n}, y_{n} \\in \\mathbb{N}\n$$\nшто се своди на $n a - n x_{n} y_{n} = y_{n} - x_{n} > -x_{n} y_{n}$. Одавде добијамо $x_{n} \\leqslant x_{n} y_{n} < \\frac{n}{n-1} a \\leqslant 2a$. Следи да у низу $x_{1}, x_{2}, \\ldots$ постоји члан који се јавља бесконачно много пута. Означимо тај члан са $X$. Тада $n X + 1 \\mid n^{2} a - 1$ и одатле\n$$\nn X + 1 \\mid X^{2}(n^{2} a - 1) - a(n^{2} x^{2} - 1) = a - X^{2}\n$$\nза бесконачно много бројева $n$. Ово је могуће само за $a - X^{2} = 0$, тј. $X^{2} = a$.\n\n\nДруго решење. Као и у првом решењу, нека је $n^{2} a-1 = (n x_{n} + 1)(n y_{n} - 1)$, тј. $y_{n} - x_{n} = n(a - x_{n} y_{n}) = n d_{n}$. Разликујемо три случаја.\n(1) Ако је $d_{n} > 0$, онда је $a = d_{n} + x_{n}(x_{n} + n d_{n}) > n d_{n} x_{n}$, што је немогуће за $n \\geqslant a$.\n(2) Ако је $d_{n} < 0$, онда је $a = d_{n} + y_{n}(y_{n} - n d_{n}) = y_{n}^{2} - d_{n}(n y_{n} - 1) > n y_{n} - 1$, што је немогуће за $n \\geqslant a + 1$.\n(3) Ако је $d_{n} = 0$, онда је $a = x_{n}^{2}$, потпун квадрат.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40881,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nОдредити колико се највише краљица може поставити на таблу $2017 \\times 2017$, при чему свака краљица сме да напада највише једну од преосталих.\n(Бојан Башић и комисија)",
    "options": [],
    "answer": "2689",
    "solution": "Solution:\n\nОзначимо $n=2017$. Претпоставимо да је постављено $m>n$ краљица. Ни у једној врсти нема више од две краљице, па се у бар $m-n$ врста налазе по две краљице, тако да има највише $m-2(m-n)=2 n-m$ краљица које су саме у својој врсти. Слично, највише $2 n-m$ краљица су саме у својој колони. С друге стране, свака краљица је сама у својој врсти или у својој колони, па је $m \\leqslant 2(2 n-m)$, одакле је $m \\leqslant\\left[\\frac{4 n}{3}\\right]=2689$.\n\nНа слици А је приказано постављање 8 краљица на таблу $6 \\times 6$ у складу са захтевом задатка. Пре конструкције примера на табли $2017 \\times 2017$ размотрићемо следећи распоред краљица:\n- На таблу $335 \\times 335$ могуће је поставити 335 краљица које се међусобно не нападају чак ни ако се дијагонале продуже по модулу 335. Заиста, довољно је поставити краљице на сва поља $(x, y), 1 \\leqslant x, y \\leqslant 335$, за која је $y \\equiv 2 x(\\bmod 335)$, као на слици Б. Заиста, тада су сви збирови $x+y$ међусобно различити по модулу 335, све разлике $x-y$ такође, па никоје две краљице нису у истој врсти, колони или дијагонали.\n\nПоделимо таблу $2017 \\times 2017$ на правоугаонике и квадрате страница 335, 6 и 1, као на слици В. Квадрате обележене са Б и А попунићемо редом као на сликама Б и А, а у горње десно поље табле поставићемо још једну краљицу. Овако смо укупно поставили $8 \\cdot 335+8+1=2689$ краљица. Лако се проверава да овакво постављање задовољава услове задатка.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40882,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека је $k$ кружница описана око $\\triangle A B C$, а $k_{a}$ приписана кружница наспрам темена $A$. Две заједничке тангенте кружница $k$ и $k_{a}$ секу праву $B C$ у тачкама $P$ и $Q$. Доказати да важи $\\varangle P A B=\\varangle Q A C$.\n\n(Дуиан Ђукић)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nНека унутрашња и спољна симетрала угла $B A C$ секу праву $B C$ редом у тачкама $D$ и (можда бесконачној) $D_{1}$. Заједничке тангенте се секу у центру $T$ позитивне хомотетије $\\mathscr{H}$ која слика приписани круг $\\omega_{a}$ у описани круг $\\Omega$. Ако је $T$ бесконачна тачка, $\\mathscr{H}$ је транслација, а остатак доказа је исти.\n\nЛема. Нека произвољна права $p$ кроз $D_{1}$ сече круг $\\Omega$ у тачкама $L$ и $K$. Тангенте у $L$ и $K$ на $\\Omega$ секу праву $B C$ редом у тачкама $P$ и $Q$. Тада је $\\varangle P A B=\\varangle C A Q$.\n\nДоказ. Означимо $\\varangle B A C=\\alpha, \\varangle C B A=\\beta, \\varangle A C B=\\gamma, \\varangle P A B=x$ и $\\varangle C A Q=y$.\n\nАко је $D_{1}$ бесконачна тачка, тврђење је тривијално по симетрији. Ако није, из $\\triangle P B L \\sim \\triangle P L C$ следи $\\frac{P B}{P L}=\\frac{P L}{P C}=\\frac{L B}{L C}$ и одатле $\\frac{P B}{P C}=\\left(\\frac{L B}{L C}\\right)^{2}$. Слично je $\\frac{Q B}{Q C}=\\left(\\frac{K B}{K C}\\right)^{2}$. Пошто је $\\frac{L B}{L C} \\cdot \\frac{K B}{K C}=\\frac{|K L B|}{|K L C|}=\\frac{D_{1} B}{D_{1} C}=\\frac{A B}{A C}$, добијамо $\\frac{P B}{P C} \\cdot \\frac{Q B}{Q C}=\\left(\\frac{A B}{A C}\\right)^{2}$. Како је $\\frac{P B}{P C}=\\frac{P B}{P A} \\cdot \\frac{P A}{P C}=\\frac{\\sin x}{\\sin \\beta} \\cdot \\frac{\\sin \\gamma}{\\sin (\\alpha+x)}$ и $\\frac{Q B}{Q C}=\\frac{Q B}{Q A} \\cdot \\frac{Q A}{Q C}=\\frac{\\sin (\\alpha+y)}{\\sin \\beta} \\cdot \\frac{\\sin \\gamma}{\\sin y}$, множење даје $\\left(\\frac{\\sin \\gamma}{\\sin \\beta}\\right)^{2} \\cdot \\frac{\\sin (\\alpha+y) / \\sin y}{\\sin (\\alpha+x) / \\sin x}=\\left(\\frac{A C}{A B}\\right)^{2}=\\left(\\frac{\\sin \\gamma}{\\sin \\beta}\\right)^{2}$, одакле је $\\sin \\alpha \\operatorname{ctg} y+\\cos \\alpha=\\frac{\\sin (\\alpha+y)}{\\sin y}=\\frac{\\sin (\\alpha+x)}{\\sin x}=\\sin \\alpha \\operatorname{ctg} x+\\cos \\alpha$, тј. $x=y$.\n\nАко су $K$ и $L$ додирне тачке заједничких тангенти са $\\Omega$, остаје да се покаже да тачка $D_{1}$ лежи на правој $K L$, тј. на полари тачке $T$ у односу на $\\Omega$. По ставу о полу и полари, довољно је доказати да $T$ лежи на полари $d$ тачке $D_{1}$ у односу на $\\Omega$.\n\nОзначимо са $N$ средиште лука $B A C$ круга $\\Omega$. Слика тачке $D$ при хомотетији $\\mathscr{H}$ је пресек $S$ тангенти на $\\Omega$ у тачкама $A$ и $N$, па тачка $T$ лежи на правој $D S$. С друге стране, тачка $D$ је на полари $d$ јер је четворка $\\left(B, C ; D_{1}, D\\right)$ хармонијска, а тачка $S$ је\n\n![](attached_image_1.png)\n\nтакође на $d$ јер полара тачке $S$ у односу на $\\Omega$, што је права $A N$, садржи тачку $D_{1}$. Према томе, праве $D S$ и $d$ се поклапају, чиме је доказ завршен.\nSolution:\n\nДруго решење. Нека заједничке тангенте додирују круг $\\Omega$ у тачкама $K$ и $L$, при чему је теме $L P$ тангента ближа темену $B$. Означимо са $M$ средиште оног лука $B C$ који не садржи тачку $A$, а са $O$ и $I_{a}$ редом центре описаног и приписаног круга наспрам $A$.\n\nКако је $\\varangle L P I_{a}=90^{\\circ}+\\frac{1}{2} \\varangle L P C$ и $\\varangle L A I_{a}=\\varangle L A M=\\frac{1}{2} \\varangle L O M=\\frac{1}{2} \\varangle L P D_{1}=90^{\\circ}-\\frac{1}{2} \\varangle L P C$, следи да је $\\varangle L P I_{a}+\\varangle L A I_{a}=180^{\\circ}$, па је четвороугао $A L P I_{a}$ тетиван. Слично, и четвороугао $A K Q I_{a}$ је тетиван. Сада имамо $\\varangle P A I_{a}=\\varangle P L I_{a}=\\varangle Q K I_{a}=\\varangle Q A I_{a}$, јер су углови $P L I_{a}$ и $Q K I_{a}$ симетрични у односу на праву $O I_{a}$, а одавде је $\\varangle P A B=\\varangle Q A C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40883,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nКружница уписана у $\\triangle A B C$ има центар у тачки $I$ и додирује страницу $B C$ у тачки $D$. На дужима $B I$ и $C I$ одабране су тачке $P$ и $Q$, редом, такве да важи $\\varangle B A C=2 \\varangle P A Q$. Доказати: $\\varangle P D Q=90^\\circ$.\n\n(Душан Ђукић)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nОзначимо са $E$ и $F$ редом подножја нормала из $P$ и $Q$ на праву $B C$, а са $M$ средиште дужи $P Q$.\n\nПосматрајмо тачку $X$ на страници $B C$ такву да је $\\varangle B A X=2 \\varangle B A P$. Тада је такође $\\varangle C A X=\\varangle B A C-2 \\varangle B A P=2 \\varangle C A Q$, па су $P$ и $Q$ редом центри уписаних кругова троуглова $B A X$ и $C A X$. Следи да су $X P$ и $X Q$ симетрале углова $B X A$ и $C X A$, па је $\\varangle P X Q=90^\\circ$.\n\nУслов $\\varangle P D Q=90^\\circ$ је еквивалентан са $M D=M P=M Q=M X$, а како је $M E=M F$, довољно је доказати да је $D E=X F$. Обе дужине се једноставно рачунају на основу \"великог задатка\":\n\n$D E=B D-B E=\\frac{A B+B C-A C}{2}-\\frac{A B+B X-A X}{2}=\\frac{C X-A C+A X}{2}=X F$.\n\n![](attached_image_1.png)\n\n\nДруго решење. Нека права $B I$ поново сече описани круг $\\triangle A P Q$ у тачки $N$. Имамо $\\varangle A I N=180^\\circ-\\varangle B I A=90^\\circ-\\frac{\\gamma}{2}=\\varangle D I Q$. Такође, како је $\\varangle I N Q=\\varangle P A Q=\\frac{\\alpha}{2}$ и $\\varangle I Q N=\\varangle B I C-\\varangle I N Q=\\left(90^\\circ+\\frac{\\alpha}{2}\\right)-\\frac{\\alpha}{2}=90^\\circ$, имамо $\\frac{I Q}{I N}=\\sin \\frac{\\alpha}{2}=\\frac{I A}{I D}$, одакле је $\\frac{I A}{I N}=\\frac{I D}{I Q}$. Следи да су троуглови $D I Q$ и $A I N$ слични, па је\n\n$\\varangle I D Q=\\varangle I A N=180^\\circ-\\varangle A I N-\\varangle A N I=180^\\circ-\\left(90^\\circ-\\frac{\\gamma}{2}\\right)-\\varangle A N P=90^\\circ+\\frac{\\gamma}{2}-\\varangle A Q P$.\n\nАналогно је $\\varangle I D P=90^\\circ+\\frac{\\beta}{2}-\\varangle A P Q$, па сабирањем добијамо\n\n$\\varangle P D Q=180^\\circ+\\frac{\\beta+\\gamma}{2}-\\left(180^\\circ-\\varangle P A Q\\right)=90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40884,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nДат је природан број $n$, $n>1$. Цео број $x$ зовемо красним ако је остатак броја $x^{2}$ при дељењу са $n$ непаран. Доказати да не постоји више од $1+\\lfloor\\sqrt{3 n}\\rfloor$ узастопних красних природних бројева.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nАко је $n$ паран број, тврђење је тривијално: тада не постоје ни два узастопна красна броја, јер парни бројеви нису красни. Надаље сматрамо да је $n$ непаран.\n\nПошто бројеви дељиви са $n$ нису красни, можемо да се ограничимо на бројеве $x \\in \\{1, \\ldots, n-1\\}$. Претпоставимо да су бројеви $x$ и $x+1$ красни. Остаци бројева $x^{2}$ и $(x+1)^{2}$ при дељењу са $n$, који су непарни, једнаки су $x^{2}-n\\left[\\frac{x^{2}}{n}\\right]$ и $(x+1)^{2}-n\\left[\\frac{(x+1)^{2}}{n}\\right]$ редом, па су $\\left[\\frac{x^{2}}{n}\\right]$ и $\\left[\\frac{(x+1)^{2}}{n}\\right]$ различите парности. Како због $0<(x+1)^{2}-x^{2}<2 n$ важи $0 \\leqslant\\left[\\frac{(x+1)^{2}}{n}\\right]-\\left[\\frac{x^{2}}{n}\\right] \\leqslant 2$, следи да је $\\left[\\frac{(x+1)^{2}}{n}\\right]=\\left[\\frac{x^{2}}{n}\\right]+1$.\n\nПрема томе, ако су бројеви $x, x+1, \\ldots, x+k$ красни, онда је\n$$\nm=y-\\left[\\frac{y^{2}}{n}\\right] \\quad \\text{константно за све} \\quad y=x, x+1, \\ldots, x+k\n$$\nКако је једнакост $y-\\left[\\frac{y^{2}}{n}\\right]=m$ еквивалентна са $y-m \\leqslant \\frac{y^{2}}{n}<y-m+1$, тј. са $n\\left(\\frac{n}{4}-m\\right) \\leqslant\\left(y-\\frac{n}{2}\\right)^{2}<n\\left(\\frac{n}{4}-m+1\\right)$, број узастопних бројева $y$ са овим својством није већи од $\\left[\\sqrt{n\\left(\\frac{n}{4}-m+1\\right)}-\\sqrt{n\\left(\\frac{n}{4}-m\\right)}\\right]+1 \\leqslant[\\sqrt{n}]+1$ ако је $m \\leqslant\\left[\\frac{n}{4}\\right]$, и није већи од $2\\left[\\sqrt{n\\left\\{\\frac{n}{4}\\right\\}}+\\frac{1}{2}\\right] \\leqslant[\\sqrt{3 n}]+1$ за $m=\\left[\\frac{n}{4}\\right]+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40885,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nУ равни је дато $n$ правих међу којима никоје две нису паралелне и никоје три се не секу у једној тачки. Под пресечним тачкама сматрамо све тачке у којима се секу неке две од ових правих.\n\na. Доказати да међу датим правим постоји једна са чије се сваке стране налази бар по\n$$\n\\left\\lfloor\\frac{(n-1)(n-2)}{10}\\right\\rfloor\n$$\nпресечних тачака (тачке на тој правој се не рачунају).\n\nб. За које вредности $n$ се оцена из дела под (а) не може побољшати?",
    "options": [],
    "answer": "n ≤ 7",
    "solution": "Solution:\n\nЗа сваку од датих правих, пресечних тачака ван ње има\n$$\nm=\\frac{(n-1)(n-2)}{2}\n$$\nПретпоставимо да, за сваку праву, с једне њене стране има не више од $k$ пресечних тачака, где је $k \\leqslant \\frac{n}{2}$. Пребројмо на два начина тројке $(p, A, B)$, где су $A$ и $B$ пресечне тачке са различитих страна праве $p$. По претпоставци, за сваку праву $p$ има не више од $k(m-k)$ оваквих тројки, па је њихов укупан број $N \\leqslant n k(m-k)$.\n\nРазмотримо сада два типа оваквих тројки: \nтип 1, када су тачке $A$ и $B$ на истој од датих правих, и \nтип 2, када нису.\nСвака четворка датих правих, заједно са њима одређеним пресечним тачкама, одређује тачно 4 тројке типа 1, а свака тројка је одређена тачно једном четворком правих. Следи да има $4\\binom{n}{4}$ тројки типа 1. Такође, свака петорка датих правих са њима одређеним пресечним тачкама одређује бар 5 тројки типа 2, те тројки типа 2 има бар $5\\binom{n}{5}$. Ово се директно проверава: постоји шест различитих распореда по пет правих у општем положају, као на слици.\n\n![](attached_image_1.png)\n\nПрема томе, $N \\geqslant 4\\binom{n}{4}+5\\binom{n}{5}=n\\binom{n}{4}$.\nСледи да је $k(m-k) \\geqslant\\binom{ n}{4}=\\frac{m(m-1)}{6}$. Одавде се добија\n$$\nk \\geqslant k_{0}=\\frac{1}{2}\\left(m-\\sqrt{\\frac{m^{2}+2 m}{3}}\\right)\n$$\nКако се неједнакост $k_{0}>\\frac{m}{5}=\\frac{(n-1)(n-2)}{10}$ своди на $m>25$, што важи за $n \\geqslant 9$, док за $n=8$ имамо $k_{0}>\\left[\\frac{m}{5}\\right]=4$, остаје да испитамо случајеве $n \\leqslant 7$.\n\nЗа $n \\leqslant 4$ је тврђење тривијално, као и достизање једнакости, јер је тада $k_{0}=\\left[\\frac{m}{5}\\right]=0$. За $n=5,6,7$ је редом $\\left\\lceil k_{0}\\right\\rceil=\\left[\\frac{m}{5}\\right]=1,2,3$, па тврђење задатка и тада важи, а достижу се и једнакости у случајевима на слици.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40886,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nДоказати да постоји тачно један полином $P(x)$ с реалним коефицијентима за који је полином\n$$\n(x+y)^{1000}-P(x)-P(y)\n$$\nдељив полиномом $x y-x-y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nОзначимо $n=1000$. Сменом $x=u+1$ и $y=v+1$ добијамо да полином $u v-1$ дели полином $P(u+1)+P(v+1)-(u+v+2)^{n}$. Еквивалентан услов је да је $P(u+1)+P(v+1)-(u+v+2)^{n}=0$ кад год је $u v-1=0$ (видети напомену). Тако за $u \\neq 0$ и $v=\\frac{1}{u}$ имамо $P(u+1)+P\\left(\\frac{1}{u}+1\\right)=\\left(u+\\frac{1}{u}+2\\right)^{n}=\\frac{(u+1)^{2 n}}{u^{n}}$. Полином $Q(x)=P(x+1)=\\sum_{i=0}^{n} a_{i} x^{i}$ задовољава\n$$\n2 a_{0}+\\sum_{i=1}^{n} a_{i}\\left(u^{i}+u^{-i}\\right)=Q(u)+Q\\left(\\frac{1}{u}\\right)=\\frac{(u+1)^{2 n}}{u^{n}}=\\binom{2 n}{n}+\\sum_{i=1}^{n}\\binom{2 n}{n-i}\\left(u^{i}+u^{-i}\\right)\n$$\nодакле одмах следи да је $a_{0}=\\frac{1}{2}\\binom{2 n}{n}$ и $a_{i}=\\binom{2 n}{n-i}$ за $1 \\leqslant i \\leqslant n$. Дакле,\n$$\nP(x)=\\frac{1}{2}\\binom{2 n}{n}+\\sum_{i=1}^{n}\\binom{2 n}{n-i}(x-1)^{i}\n$$\n\n\nДруго решење. Тражимо полиноме $P(x)=\\sum_{i=0}^{n} p_{i} x^{i}$ и $Q(x, y)=\\sum_{i, j} a_{i, j} x^{i} y^{j}$ такве да је\n$$\nA(x, y)=(x y-x-y) Q(x, y)=(x+y)^{1000}-P(x)-P(y)\n$$\nПриметимо да је $\\operatorname{deg} Q \\leqslant 998$. Заиста, ако је $a_{i, j} x^{i} y^{j}$ моном највећег степена у $Q(x, y)$, онда је коефицијент уз $x^{i+1} y^{j+1}$ у $A(x, y)$ једнак $a_{i, j} \\neq 0$, па је $i+j+2 \\leqslant 1000$. Следи да је $\\operatorname{deg} A \\leqslant 1000$, па је и $\\operatorname{deg} P \\leqslant 1000$.\nИзједначавање коефицијената уз $x^{i} y^{j}$ у (*) даје једнакости $a_{i-1, j-1}=\\binom{1000}{i}$ за $i+j=998(i, j>0), a_{i-1, j-1}=a_{i-1, j}+a_{i, j-1}$ за $i+j<998(i, j>0)$ и $a_{i-1,0}=$ $a_{0, i-1}=p_{i}$, одакле једноставном индукцијом налазимо $a_{i-1, j-1}=\\binom{2000-i-j}{1000-i}$ за $i+j \\leqslant 1000(i, j>0)$ и $p_{i}=\\binom{1999-i}{999}$, тј.\n$$\nP(x)=x^{1000}+\\binom{1000}{999} x^{999}+\\binom{1001}{999} x^{998}+\\cdots+\\binom{1998}{999} x\n$$\n\n\nTреће решење. Не постоје два различита полинома са жељеним својством. Заиста, ако $P_{1}(x) \\not \\equiv P_{2}(x)$ имају то својство, онда $x y-x-y$ дели разлику $P_{1}(x)+P_{1}(y)-P_{2}(x)-P_{2}(y)=(x y-x-y) U(x, y)$. Међутим, ако је $c x^{i} y^{j}$ моном највећег степена у $U(x, y)$, онда је коефицијент уз $x^{i+1} y^{j+1}$ на левој страни ове једнакости једнак $c \\neq 0$, што је немогуће.\nДокажимо сада да за сваки симетричан полином $Q(x, y)$ постоји полином $P(t)$ такав да $x y-x-y \\mid Q(x, y)-P(x)-P(y)$. Довољно је доказати да за полиноме $Q$ облика $x^{i} y^{j}+x^{j} y^{i}(0 \\leqslant i \\leqslant j)$ постоји тражени полином $P_{i, j}(t)$. Тврђење је тривијално за $i=0$. За $i>0$ доказ спроводимо индукцијом по $i+j$. Наиме, $x^{i} y^{j}+x^{j} y^{i} \\equiv(x+y)\\left(x^{i-1} y^{j-1}+x^{j-1} y^{i-1}\\right)=\\left(x^{i} y^{j-1}+x^{j-1} y^{i}\\right)+$ $\\left(x^{i-1} y^{j}+x^{j} y^{i-1}\\right)(\\bmod x y-x-y)$, па можемо узети $P_{i, j}(t)=P_{i, j-1}(t)+P_{i-1, j}(t)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40887,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНека су $a$ и $b$ непарни природни бројеви већи од 1. Посматрајмо таблу $a \\times b$ којој недостају поља $(2,1),(a-2, b)$ и $(a, b)$ (под пољем $(i, j)$ подразумевамо поље у пресеку врсте $i$ и колоне $j$ ). Претпоставимо да је оваква табла поплочана помоћу $2 \\times 1$ домина и $2 \\times 2$ квадрата (домине се могу ротирати).\nДоказати да је употребљено бар $\\frac{3}{2}(a+b)-6$ домина. (Никола Петровић)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nУпишимо у поље $(i, j)$ број $(-1)^{i+j}(i+j)$. Збир уписаних бројева у читавој таблици је $\\sum_{i=1}^{a}(-1)^{i} \\sum_{j=1}^{b}(-1)^{j}(i+j)=\\sum_{i=1}^{a}(-1)^{i+1}\\left(i+\\frac{b+1}{2}\\right)=\\frac{a+b+2}{2}$, а ако се три наведена поља избаце, збир у остатку таблице је\n$$\n\\frac{a+b+2}{2}-(-3)-(a+b-2)-(a+b)=-\\frac{3}{2}(a+b)+6\n$$\nКако је збир бројева у сваком квадрату $2 \\times 2$ једнак нули, а збир у свакој домини је 1 или -1, потребно је бар $\\frac{3}{2}(a+b)-6$ домина.\n\n\nДруго решење. Нека су врсте и колоне нумерисане одоздо нагоре и слева надесно. Можемо да сматрамо да се у поплочавању не појављује ниједна од целина означених са $L$ на слици - заиста, оне се могу заменити одговарајућим целинама означеним са $D$, притом не повећавајући број домина.\nРазмотримо фигуру која покрива поље $(a, 1)$. Ако је то нпр. хоризонтална домина $(a, 1)-(a, 2)$ (аналогно се испитује случај вертикалне домине), онда поље $(a-1,1)$ не може бити покривено ни квадратом ни хоризонталном домином (јер би они са горњом домином градили једну од целина $L$ ), па оно мора бити покривено вертикалном домином $(a-1,1)-(a-2,1)$. Слично, сада поље $(a-1,2)$ мора бити покривено домином $(a-1,2)-(a-1,3)$, итд. до бесконачности, што је немогуће. Према томе, фигура која покрива поље $(a, 1)$ је квадрат.\nНа сличан начин, поља $(a, 3)$ и $(a-2,1)$ су такође покривена квадратима, затим поља $(a, 5),(a-2,3)$ и $(a-4,1)$, итд. Настављајући овај поступак, закључујемо да су сва поља осим оних у врстама $1,2,3$ и оних у колонама $b-2, b-1, b$ покривена квадратима. Преостала поља се могу поплочати на јединствен начин, приказан на слици, што се директно проверава. При томе је употребљено тачно $\\frac{3}{2}(a+b)-6$ домина.\n\n![](attached_image_1.png)\n\n\nTреће решење. Обојићемо поља табле црно и бело попут шаховске табле. Праву која раздваја $i$-ту и $(i+1)$-ву врсту/колону зовемо $i$-том хоризонталом/вертикалом.\nАко је $j$ парно ($2 \\leqslant j \\leqslant b-1$), бар једна (хоризонтална) домина сече $j$-ту вертикалу јер је број поља у првих $j$ колона непаран. С друге стране, ако је $j$ непарно $(1 \\leqslant j \\leqslant b-2)$, бар две домине морају сећи $j$-ту вертикалу јер у првих $j$ колона има за два више црних поља него белих. Овако добијамо бар $\\frac{3b-3}{2}$ хоризонталних домина.\nСлично, ако је $i$ парно ($2 \\leqslant i \\leqslant a-3)$ или $i \\in\\{1, a-2\\}$, бар једна (вертикална) домина сече $i$-ту хоризонталу јер је број поља у првих $i$ врста непаран. С друге стране, ако је $i$ непарно ($3 \\leqslant i \\leqslant a-4$), број црних поља у првих $i$ врста је за два већи од броја белих, па бар две домине морају сећи $i$-ту хоризонталу. Овако добијамо бар $\\frac{3a-9}{2}$ вертикалних домина.\nДакле, укупно има бар $\\frac{3a-9}{2}+\\frac{3b-3}{2}=\\frac{3}{2}(a+b)-6$ домина.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40888,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nЗа задат природан број $k$, нека је $n_{k}$ најмањи природан број такав да постоји коначан скуп $A$ целих бројева са следећим особинама:\n- за свако $a \\in A$ постоје $x, y \\in A$ (не обавезно различити) такви да\n$$\nn_{k} \\mid a-x-y\n$$\n- не постоји подскуп $B$ скупа $A$ за који важи $|B| \\leqslant k$ и $n_{k} \\mid \\sum_{b \\in B} b$.\nДоказати да за све $k, k \\geqslant 3$, важи\n$$\nn_{k}<\\left(\\frac{13}{8}\\right)^{k+2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nСа $F_{i}$ означавамо Фибоначијеве бројеве: $F_{1}=F_{2}=1, F_{i+1}=F_{i}+F_{i-1}$. Посматрајмо скуп $A=\\{a_{1}, a_{2}, \\ldots, a_{k+1}\\}$ са $n=F_{k+2}+3$, где су\n$$\na_{i}=(-1)^{k-1-i} F_{i+1} \\text{ за } i=1,2, \\ldots, k-1, \\quad a_{k}=F_{k}+1 \\quad \\text{и} \\quad a_{k+1}=F_{k}+2\n$$\nУ случају парног $k$ видимо да је $a_{i}=a_{i+1}+a_{i+2}(1 \\leqslant i \\leqslant k-3), a_{k-2} \\equiv a_{k}+a_{k+1}$ $(\\bmod n), a_{k-1}=a_{k+1}+a_{2}, a_{k}=a_{k-1}+a_{1}$ и $a_{k+1}=a_{k}+a_{1}$, па скуп $A$ задовољава први услов задатка. Овај услов се слично проверава и за непарно $k$.\n\nДокажимо индукцијом по $k$ (база $k=3$ се директно проверава) да је задовољен и други услов, тј. да не постоји подскуп $B \\subsetneq A$ такав да $n \\mid \\sum_{b \\in B} b$. Како је збир елемената скупа $A$ једнак нули, можемо да сматрамо без смањења општости (заменом $B$ са $A \\backslash B$ по потреби) да $B$ садржи највише један од елемената $F_{k}, F_{k}+1, F_{k}+2$. Тада је\n$1-F_{k}=-\\left(F_{k-1}+F_{k-3}+\\cdots\\right) \\leqslant \\sum_{b \\in B} b \\leqslant\\left(F_{k}+2\\right)+F_{k-2}+F_{k-4}+\\cdots=F_{k+1}+1$,\nпа мора бити $\\sum_{b \\in B} b=0$. На основу индуктивне претпоставке за $k-1$, скуп $B$ не може бити подскуп скупа $\\{a_{1}, a_{2}, \\ldots, a_{k-2}\\}$, тј. мора бити $a_{k-1}=F_{k} \\in B$. Међутим, тада је $\\sum_{b \\in B} b \\geqslant F_{k}-F_{k-1}-F_{k-3}-\\cdots=1$, што је контрадикција. Доказ је завршен.\n\nПрема томе, $n_{k} \\leqslant F_{k+2}+3<\\phi^{k-3}\\left(F_{5}+3\\right)<\\phi^{k+2}<\\left(\\frac{13}{8}\\right)^{k+2}$, где је $\\phi=\\frac{1+\\sqrt{5}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40889,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nОдредити све природне бројеве $n$ ($n>1$) који имају следеће својство: ако су $a_{1}, a_{2}, a_{3}, \\ldots, a_{k}$ сви природни бројеви мањи од $n$ и узајамно прости са $n$ и важи поредак $a_{1}<a_{2}<a_{3}<\\cdots<a_{k}$, онда ниједан од збирова $a_{i}+a_{i+1}$ за $i=1,2, \\ldots, k-1$ није дељив са 3.",
    "options": [],
    "answer": "2, 4, 10",
    "solution": "Solution:\n\nЗа $n \\leqslant 28$ услов задатка је задовољен само за $n \\in\\{2,4,10\\}$. Нека је $n>28$. Приметимо да је низ $a_{i}$ симетричан у односу на $\\frac{n}{2}$. Дакле, $a_{i}+a_{k+1-i}=n$. Ако $2 \\nmid n$, онда је $a_{1}=1, a_{2}=2$ и $3 \\mid a_{1}+a_{2}$. С друге стране, ако $3 \\mid n$, одаберимо $i$ тако да је $a_{i}<\\frac{n}{2}<a_{i+1}$ : тада $3 \\mid a_{i}+a_{i+1}=n$. Надаље $2 \\mid n$ и $3 \\nmid n$, тако да је $a_{2}=3$. Одавде имамо и $a_{k-1}=n-3$ и $a_{k}=n-1$.\nАко је $n \\equiv 2(\\bmod 3)$, онда $3 \\mid a_{k-1}+a_{k}=2 n-4$. Остаје само случај $n \\equiv 1$ $(\\bmod 3)$.\nДаље, ако је $a_{i}+a_{i+1} \\equiv 2(\\bmod 3)$, онда је $a_{k-i}+a_{k+1-i}=2 n-\\left(a_{i}+a_{i+1}\\right) \\equiv 0(\\bmod 3)$. Зато можемо да сматрамо да је $a_{i}+a_{i+1} \\equiv 1(\\bmod 3)$ за $i=$ $1,2, \\ldots, k-1$. Индукцијом налазимо\n$$\na_{1} \\equiv a_{3} \\equiv \\cdots \\equiv 1 \\quad \\text { и } \\quad a_{2} \\equiv a_{4} \\equiv \\cdots \\equiv 0 \\quad(\\bmod 3)\n$$\nКако је $(n, 9)=1$, следи $a_{4}=9$, а због $a_{3} \\equiv 1(\\bmod 3)$ имамо и $a_{3}=7$. Одавде добијамо $(n, 7)=1$ и $(n, 5) \\neq 1$, тј. $5 \\mid n$.\nСада имамо $(n, 21)=(n, 27)=1$, али бројеви $22,24,25$ и 26 нису узајамно прости са $n$, а на основу $(*)$ ни број 23 се не може појавити у низу $a_{1}, \\ldots, a_{k}$. Дакле, 21 и 27 су суседни у низу $a_{1}, \\ldots, a_{k}$, али $3 \\mid 21+27$, контрадикција.\nПрема томе, једина решења су 2,4 и 10 .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40890,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nЗа низ ненегативних реалних бројева $a_{1}, a_{2}, \\ldots, a_{k}$ кажемо да је уложив у интервал $[b, c]$ ако постоје бројеви $x_{0}, x_{1}, \\ldots, x_{k}$ из интервала $[b, c]$ такви да важи $\\left|x_{i}-x_{i-1}\\right|=a_{i}$ за $i=1,2, \\ldots, k$. Низ је нормиран ако су сви његови чланови не већи од 1. За задат природан број $n$, доказати:\n\na) сваки нормиран низ дужине $2 n+1$ је уложив у интервал $\\left[0,2-\\frac{1}{2^{n}}\\right]$;\n\nб) постоји нормиран низ дужине $4 n+3$ који није уложив у $\\left[0,2-\\frac{1}{2^{n}}\\right]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Довољно је доказати да је сваки нормиран низ $a_{1}, a_{2}, \\ldots, a_{2 n+1}$ уложив у неки интервал дужине $2-\\frac{1}{2^{n}}$. Тврђење доказујемо индукцијом по $n$. Оно је тачно за $n=0$; нека је $n \\geqslant 1$. По индуктивној претпоставци постоји низ $x_{0}, x_{1}, \\ldots, x_{2 n-1} \\in\\left[0,2-\\frac{1}{2^{n-1}}\\right]$ такав да је $\\left|x_{i}-x_{i-1}\\right|=a_{i}$ за $i=1, \\ldots, 2 n-1$. Не умањујући општост, сматраћемо да је $x_{2 n-1} \\leqslant 1-\\frac{1}{2^{n}}$.\n\n(1) Ако је $a_{2 n} \\geqslant \\frac{1}{2^{n}}$, можемо узети $x_{2 n}=x_{2 n-1}+a_{2 n} \\in\\left[1,2-\\frac{1}{2^{n}}\\right]$ и $x_{2 n+1}=x_{2 n}-a_{2 n+1} \\in\\left[0,2-\\frac{1}{2^{n}}\\right]$, чиме је низ уложен у интервал $\\left[0,2-\\frac{1}{2^{n}}\\right]$.\n\n(2) Ако је $a_{2 n}<\\frac{1}{2^{n}}$, узећемо $x_{2 n}=x_{2 n-1}-a_{2 n} \\in\\left[-\\frac{1}{2^{n}}, 1-\\frac{1}{2^{n}}\\right]$ и $x_{2 n+1}=x_{2 n}+a_{2 n+1}$, чиме је низ уложен у један од интервала $\\left[0,2-\\frac{1}{2^{n}}\\right]$ и $\\left[-\\frac{1}{2^{n}}, 2-\\frac{1}{2^{n-1}}\\right]$.\n\nб) Означимо $N=3 \\cdot 2^{n-1}-1$. Доказаћемо да се низ дужине $4 n-1$\n\n$1,1-\\frac{1}{N}, 1,1-\\frac{2}{N}, 1,1-\\frac{2^{2}}{N}, \\ldots, 1,1-\\frac{2^{n-1}}{N}, 1,1-\\frac{2^{n-2}}{N}, 1, \\ldots, 1-\\frac{2}{N}, 1,1-\\frac{1}{N}, 1$\n\nне може уложити у интервал $\\left(-1+\\frac{1}{2 N}, 1-\\frac{1}{2 N}\\right)$, одакле следи тврђење.\n\nПретпоставимо супротно. Једноставном индукцијом се доказује да је:\n\n(i) $\\left|x_{2 i}\\right|<1-\\frac{2^{i+1}-1}{2 N}$ и $\\left|x_{2 i+1}\\right|>\\frac{2^{i+1}-1}{2 N}$ за $i=0, \\ldots, n$;\n\n(ii) $\\left|x_{2 i}\\right|<\\frac{2^{2 n+2-i}-1}{2 N}$ и $\\left|x_{2 i+1}\\right|>1-\\frac{2^{2 n+2-i}-1}{2 N}$ за $i=n+1, \\ldots, 2 n+1$.\n\nТако за $x_{4 n+3}$ добијамо контрадикцију.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40891,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nКонвексан четвороугао $A B C D$ је описан око кружнице $k$. Праве $A D$ и $B C$ се секу у тачки $P$, а кружнице описане око $\\triangle P A B$ и $\\triangle P C D$ се секу у тачки $X$. Доказати да тангенте из тачке $X$ на кружницу $k$ граде једнаке углове са правима $A X$ и $C X$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nПошто је $\\varangle X A D=\\varangle X B C$ и $\\varangle X D P=\\varangle X C P$, важи $\\triangle X A D \\sim \\triangle X B C$.\n\nНека симетрала $s_{X}$ угла $A X C$ сече кругове $P A B$ и $P C D$ у тачкама $K$ и $R$, а симетрала $s_{P}$ угла $A P C$ сече кругове $P A B$ и $P C D$ у тачкама $L$ и $S$. Права $s_{P}$ пролази кроз центар $I$ круга $k$ и важи $L A=L B=L I$ и $S C=S D=S I$.\n\nКако је $\\varangle I L K=\\varangle P X K=\\varangle P X R=\\varangle I S R$, важи $K L \\| R S$. Даље, имамо $\\varangle R X S=\\varangle R X C-\\varangle S P C=\\frac{1}{2}(\\varangle A X C-\\varangle A P C)=\\frac{1}{2} \\varangle B X C$ и, слично, $\\varangle L X K=\\frac{1}{2} \\varangle B X C$. Следи да тетивама $K L$ и $R S$ у круговима $P A B$ и $P C D$, као и тетивама $L B$ и $S D$, одговарају једнаки периферијски углови, па је $\\frac{K L}{R S}=\\frac{L B}{S D}=\\frac{L I}{S I}$. Следи да је $\\triangle I K L \\sim \\triangle I R S$, па тачка $I$ лежи на правој $K R$, што је симетрала угла $A X C$. Тврђење задатка одмах следи.\n\n\nДруго решење. Означимо са $U$ и $V$ редом пресеке симетрала углова $A X D$ и $B X C$ са $A D$ и $B C$. Као и у првом решењу, $\\triangle X A D \\sim \\triangle X B C$, одакле је $\\varangle X U P=\\varangle X V P$, па тачке $X, P, U$ и $V$ леже на истом кругу $\\gamma$. Дока-\n![](attached_image_1.png)\nзаћемо да је и тачка $I$ на овом кругу и да је $I U=I V$. Следиће да $I$ припада симетрали угла $U X V$, што је уједно и симетрала угла $A X C$.\n\nНека су $M$ и $N$ редом тачке додира круга $k$ са страницама $A D$ и $B C$. Означимо $A M=a, B N=b, C N=c$ и $D M=d$. Тада је $A B=a+b, C D=c+d$ и $A U: U D=(a+b):(c+d)$, одакле налазимо $A U=\\frac{a+b}{a+b+c+d} \\cdot A D=\\frac{(a+b)(a+d)}{a+b+c+d}$; слично имамо $B V=\\frac{(b+a)(b+c)}{a+b+c+d}$. Следи да је $A M-A U=B V-B N=\\frac{a c-b d}{a+b+c+d}$, па је $M U=N V$ и троуглови $I M U$ и $I N V$ су подударни и исто оријентисани. Према томе, $I U=I V$ и $\\varangle U I V=\\varangle M I N=180^{\\circ}-\\varangle V P U$, тј. $I$ је средиште лука $U V$ круга $P X U V$.\n\n\nTреће решење. Познато је следеће тврђење из пројективне геометрије.\n- Дезаргова теорема о инволуиији. Коника $\\gamma$ је описана око четвороугла $\\overline{A B C D}$. Права $\\ell$ сече $A B, C D, B C, D A, A C, B D$ редом у тачкама $X_{1}$, $X_{2}, Y_{1}, Y_{2}, Z_{1}, Z_{2}$, и сече конику $\\gamma$ у $W_{1}$ и $W_{2}$. Тада постоји инволуција на правој $\\ell$ која слика $X_{1} \\leftrightarrow X_{2}, Y_{1} \\leftrightarrow Y_{2}, Z_{1} \\leftrightarrow Z_{2}$ и $W_{1} \\leftrightarrow W_{2}$.\n\nДуално тврђење (добијено поларним пресликавањем у односу на $\\gamma$ ) гласи овако:\n- Коника $\\gamma$ је уписана у четвороугао $A B C D$ у коме је $A D \\cap B C=\\{P\\}$ и $A B \\cap C D=\\{Q\\}$. Праве $X U$ и $X V$ су тангенте из произвољне тачке $X$ на $\\gamma$. Тада постоји инволуција на прамену правих кроз $X$ која слика $X A \\leftrightarrow X C, X B \\leftrightarrow X D, X P \\leftrightarrow X Q$ и $X U \\leftrightarrow X V$.\n\nУ нашем случају углови $A X C, B X D$ и $P X Q$ имају заједничку симетралу $s$, па је поменута инволуција управо осна симетрија у односу на $s$. Следи да су две тангенте из $X$ на круг (конику) $k$ симетричне у односу на $s$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40892,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nДат је $\\triangle ABC$. Нека је $A_{1}$ централносиметрична слика пресечне тачке симетрале $\\measuredangle BAC$ и странице $BC$, где је центар симетрије средина странице $BC$. Аналогно дефинишемо тачке $B_{1}$ (на страници $CA$) и $C_{1}$ (на страници $AB$). Пресек кружнице описане око $\\triangle A_{1}B_{1}C_{1}$ с правом $AB$ је скуп $\\{Z, C_{1}\\}$, с правом $BC$ је скуп $\\{X, A_{1}\\}$, а с правом $CA$ је скуп $\\{Y, B_{1}\\}$. Ако се нормале из тачака $X, Y$ и $Z$ на $BC, CA$ и $AB$, редом, секу у једној тачки, доказати да је $\\triangle ABC$ једнакокрак.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nПодсетимо се да тачке $P$ и $Q$ унутар $\\triangle ABC$ зовемо изогонално спрегнутим ако је $\\varangle PAB = \\varangle QAC$ и $\\varangle PBC = \\varangle QBA$. Тада такође важи $\\varangle PCA = \\varangle QCB$.\n\nЛема. Подножја нормала из тачака $P$ и $Q$ на праве $BC, CA$ и $AB$ леже на истом кругу.\n\nДоказ. Нека су $P_{a}$ и $Q_{a}$ редом подножја нормала из $P$ и $Q$ на $BC$; аналогно означавамо $P_{b}, Q_{b}, P_{c}, Q_{c}$. Из $\\varangle AP_{b}P_{c} = \\varangle APP_{c} = \\varangle AQQ_{b} = \\varangle AQ_{c}Q_{b}$ следи $\\triangle AP_{b}P_{c} \\sim \\triangle AQ_{c}Q_{b}$, па су тачке $P_{b}, P_{c}, Q_{b}, Q_{c}$ на истом кругу $k$, а његов центар је пресек симетрала дужи $P_{b}Q_{b}$ и $P_{c}Q_{c}$, што је управо средиште $U$ дужи $PQ$. Аналогно, и тачке $P_{c}, P_{a}, Q_{c}, Q_{a}$ су једнако удаљене од тачке $U$, па и $P_{a}$ и $Q_{a}$ леже на кругу $k$.\n\nПретпоставимо да се нормале из $X, Y$ и $Z$ редом на $BC, CA$ и $AB$ секу у тачки $P$. Ако је тачка $Q$ изогонално спрегнута тачки $P$ у $\\triangle ABC$, подножја нормала из $Q$ на $BC, CA$ и $AB$ су по Леми управо тачке $A_{1}, B_{1}$ и $C_{1}$.\n\nОзначимо са $A_{0}, B_{0}$ и $C_{0}$ редом пресеке унутрашњих симетрала углова код $A, B$ и $C$ с наспрамним страницама. Уобичајено, $BC = a, CA = b$ и $AB = c$. Из односа $BA_{0} : A_{0}C = c : b$ налазимо $BA_{1} = A_{0}C = \\frac{ab}{b+c}$ и, слично, $A_{1}C = \\frac{ac}{b+c}, CB_{1} = \\frac{bc}{c+a}, B_{1}A = \\frac{ba}{c+a}, AC_{1} = \\frac{ca}{a+b}$ и $C_{1}B = \\frac{cb}{a+b}$. Сада имамо\n\n![](attached_image_1.png)\n\n$$\n\\begin{aligned}\n0 & = \\left(BA_{1}^{2} - A_{1}C^{2}\\right) + \\left(CB_{1}^{2} - B_{1}A^{2}\\right) + \\left(AC_{1}^{2} - C_{1}B^{2}\\right) \\\\\n& = \\frac{a^{2}(b-c)}{b+c} + \\frac{b^{2}(c-a)}{c+a} + \\frac{c^{2}(a-b)}{a+b} \\\\\n& = \\frac{a^{4}(b-c) + b^{4}(c-a) + c^{4}(a-b) - (b-c)(c-a)(a-b)(ab+bc+ca)}{(b+c)(c+a)(a+b)} \\\\\n& = -\\frac{(b-c)(c-a)(a-b)(a+b+c)^{2}}{(b+c)(c+a)(a+b)}\n\\end{aligned}\n$$\n\nодакле следи $a = b$ или $a = c$ или $b = c$.\n\n\nДруго решење. Као у првом решењу, $BA_{1} = A_{0}C = \\frac{ab}{b+c}, A_{1}C = \\frac{ac}{b+c}, CB_{1} = \\frac{bc}{c+a}$, $B_{1}A = \\frac{ba}{c+a}, AC_{1} = \\frac{ca}{a+b}$ и $C_{1}B = \\frac{cb}{a+b}$. Означимо $BX = x, CY = y$ и $AZ = z$. Потенција тачке $A$ даје $AB_{1} \\cdot AY = AC_{1} \\cdot AZ$, тј. $\\frac{by}{c+a} + \\frac{cz}{a+b} = \\frac{b^{2}}{c+a}$. Слично добијамо $\\frac{cz}{a+b} + \\frac{ax}{b+c} = \\frac{c^{2}}{a+b}$ и $\\frac{ax}{b+c} + \\frac{by}{c+a} = \\frac{a^{2}}{b+c}$. Одавде следи $\\frac{2ax}{b+c} = \\frac{a^{2}}{b+c} + \\frac{c^{2}}{a+b} - \\frac{b^{2}}{c+a}$ што се своди на $x = \\frac{1}{2}a - \\frac{(b+c)(b-c)(b^{2} + c^{2} + ab + ac + bc)}{2a(a+b)(a+c)}$, итд. Услов да су три нормале конкурентне је\n\n$$\n\\begin{aligned}\n0 & = (a+b)(a+c)(b+c)\\left[x^{2} - (a-x)^{2} + y^{2} - (b-y)^{2} + z^{2} - (c-z)^{2}\\right] \\\\\n& = (b+c)^{2}(c-b)(T - a^{2}) + (c+a)^{2}(a-c)(T - b^{2}) + (a+b)^{2}(b-a)(T - c^{2}) \\\\\n& = a^{2}(b+c)^{2}(b-c) + b^{2}(c+a)^{2}(c-a) + c^{2}(a+b)^{2}(a-b) - (a-b)(b-c)(c-a)T \\\\\n& = -(a-b)(b-c)(c-a)(a+b+c)^{2}\n\\end{aligned}\n$$\n\nгде је $T = a^{2} + b^{2} + c^{2} + ab + bc + ca$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40893,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНа планети $X$ облика лопте се налази $2n$ бензинских пумпи. Притом је свака пумпа упарена с по једном другом пумпом и сваке две упарене пумпе се налазе на дијаметрално супротним тачкама планете. На свакој пумпи се налази одређена количина бензина. Познато је следеће: уколико аутомобил с претходно празним (довољно великим) резервоаром крене с ма које пумпе, увек може стићи до пумпе с њом упарене (уз могуће допуњавање бензина на другим пумпама током пута). Одредити све природне бројеве $n$ такве да, за ма какав распоред $2n$ пумпи који испуњава наведени услов, увек постоји пумпа од које аутомобил може кренути с претходно празним резервоаром и обићи све остале пумпе на планети. (Сматрати да аутомобил троши константну количину бензина по јединици дужине.) (Никола Петровић)",
    "options": [],
    "answer": "n ≤ 3",
    "solution": "Solution:\n\nОдговор је $n \\leqslant 3$.\nПумпу дијаметрално супротну пумпи $X$ означаваћемо са $X'$\n\nЗа $n \\leqslant 1$ тврђење је тривијално. Нека је $n=2$ и нека је $AB=A'B'$ најмање међу свим растојањима између две пумпе. Од пумпе $A$ до $A'$ се може стићи, рецимо путем $AB'A'$ (случај пута $ABA'$ је сличан), али у $B$ има довољно бензина за вожњу до њој најближе пумпе $A$, те је путања $BAB'A'$ могућа.\n\nПокажимо тврђење за $n=3$ и шест пумпи $A, A', B, B', C, C'$. Нека је $AB=A'B'$ најмање међу растојањима између две пумпе и нека је $B$ пумпа најближа пумпи $C$. Означимо $S=\\{A, B, C\\}$ и $S'=\\{A', B', C'\\}$. Полазећи из сваке пумпе једног скупа можемо стићи до другог скупа.\n\n(1) Претпоставимо да се из пумпе $A$ путем $AB$ не може отићи у скуп $S'$. До $S'$ се не може стићи ни путем $AC$ - иначе би могло и путем $ABC$, јер је $BC \\leqslant AC$, а у $B$ има довољно бензина да надокнади утрошак на путу $AB$. Дакле, полазећи из пумпе $A$, до $S'$ можемо стићи само директно. Најближа тачка скупа $S'$ је $C'$, па је цела путања $CBA C'B'A'$ могућа. Случај када се из $A'$ путем $A'B'$ не може стићи до $S$ је аналоган.\n\n(2) Ако не важи случај (1), кренимо из $A$ право у $B$. Како нам је скуп $S'$ у домету, а $BC \\leqslant d(B, S')=BC'$, из $B$ можемо продужити у $C$. Ту ћемо надокнадити бензин потрошен на путу $BC$, а $d(C, S')=CA'<d(B, S')$, па још увек можемо у $S'$, и то у пумпу $A'$. Даље се може ићи у $B'$, и одатле (као малопре) у $C'$. Добијамо путању $ABC A'B'C'$.\n\nОстаје да конструишемо контрапример за $n \\geqslant 4$. Сматраћемо да је полуобим лопте 1. Распоредимо пумпе $A_2, A_3, \\ldots, A_n$ по великом кругу тако да је $A_2A_3=A_3A_4=\\cdots=A_{n-1}A_n=d<\\frac{1}{n-1}$ и пумпу $A_1$ тако да је $A_1A_3=d$ и $A_1A_2=A_1A_4$. Опет означавамо $S=\\{A_1, \\ldots, A_n\\}$ и $S'=\\{A_1', \\ldots, A_n'\\}$. Снабдимо пумпе $A_1, A_1', \\ldots, A_{n-1}, A_{n-1}'$ бензином довољним за прелазак раздаљине $d$, а пумпе $A_n$ и $A_n'$ бензином за прелазак раздаљине $1-(n-1)d$. Из сваке пумпе се може стићи до дијаметрално супротне: заиста, за $2 \\leqslant i \\leqslant n$ могућа је путања $A_iA_{i+1}\\ldots A_nA_2'A_3'\\ldots A_i'$, а могућа је и путања $A_1A_3A_4\\ldots A_nA_2'A_3'A_1'$. С друге стране, у свакој од пумпи $A_1, \\ldots, A_{n-1}$ има тек толико бензина да се дође до најближе пумпе, а у $A_n$ и $A_n'$ таман довољно за прелаз у други скуп. Зато, да би се обишле све пумпе, бар један од скупова, рецимо $S$, морао би се обићи цео без коришћења горива у $A_n$, али за то је потребно прећи пут дужи од $(n-1)d$, а бензина има само за пут дужине $(n-1)d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40894,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНизови $\\left(a_{n}\\right)_{n=0}^{\\infty}$ и $\\left(b_{n}\\right)_{n=0}^{\\infty}$ дефинисани су рекурентним релацијама\n$$\na_{0}=0, \\quad a_{1}=1, \\quad a_{n+1}=\\frac{2018}{n} a_{n}+a_{n-1} \\quad \\text{ за } n \\geqslant 1\n$$\nи\n$$\nb_{0}=0, \\quad b_{1}=1, \\quad b_{n+1}=\\frac{2020}{n} b_{n}+b_{n-1} \\quad \\text{ за } n \\geqslant 1\n$$\nДоказати:\n$$\n\\frac{a_{1010}}{1010}=\\frac{b_{1009}}{1009}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nДефинишимо низ $\\left(c_{m, n}\\right)\\left(m, n \\in \\mathbb{N}_{0}\\right)$ условима\n$$\nc_{m, 0}=0, \\quad c_{m, 1}=1, \\quad c_{m, n+1}=\\frac{2 m}{n} c_{m, n}+c_{m, n-1} \\quad \\text{ за } n \\geqslant 1\n$$\nТада је $a_{n}=c_{1010, n}$ и $b_{n}=c_{1009, n}$.\nВидимо да је нпр. $c_{1, n}=n, c_{2, n}=n^{2}$ и $c_{3, n}=\\frac{2 n^{3}+n}{3}$. Тврдимо да за свако $m \\in \\mathbb{N}$ постоји моничан полином $P_{m}(x)$ такав да је\n$$\nP_{m}(x+1)=\\frac{2 m}{x} P_{m}(x)+P_{m}(x-1)\n$$\nпошто је очигледно $P_{m}(0)=0$, индукцијом ће следити $c_{m, n}=P_{m}(n) / P_{m}(1)$.\n\nЛема. Дефинишимо низ полинома $P_{k}$ условима $P_{0}(x)=0, P_{1}(x)=x$ и\n$$\nP_{k+1}(x)=x P_{k}(x)+\\frac{k(k-1)}{4} \\cdot P_{k-1}(x)\n$$\nТада полиноми $P_{k}$ задовољавају (1).\nШта више, важи $P_{k}(x+1)-2 P_{k}(x)+P_{k}(x-1)=\\frac{k(k-1)}{x} \\cdot P_{k-1}(x)$.\n\nДоказ. Ако означимо\n$$\n\\begin{aligned}\n& A_{k}(x)=P_{k+1}(x)-x P_{k}(x)-\\frac{k(k-1)}{4} P_{k-1}(x) \\equiv 0 \\\\\n& B_{k}(x)=P_{k}(x+1)-2 P_{k}(x)+P_{k}(x-1)-\\frac{k(k-1)}{x} P_{k-1}(x) \\\\\n& C_{k}(x)=P_{k}(x+1)-P_{k}(x-1)-\\frac{2 k}{x} P_{k}(x)\n\\end{aligned}\n$$\nи претпоставимо да је $B_{i}(x) \\equiv C_{i}(x) \\equiv 0$ за све $i \\leqslant k$, тада је\n$$\n\\begin{aligned}\n& B_{k+1}(x)-x B_{k}(x)-\\frac{k(k-1)}{4} B_{k-1}(x)= \\\\\n& C_{k}(x)+A_{k}(x+1)+A_{k}(x-1)-2 A_{k}(x)-\\frac{k(k-1)}{x} A_{k-1}(x)=0\n\\end{aligned}\n$$\nпа је $B_{k+1} \\equiv 0$. С друге стране,\n$$\n\\begin{aligned}\n& C_{k+1}(x)-x C_{k}(x)-\\frac{k(k-1)}{4} C_{k-1}(x)= \\\\\n& B_{k}(x)+A_{k}(x+1)-A_{k}(x-1)-\\frac{2(k+1)}{x} A_{k}(x)=0\n\\end{aligned}\n$$\nпа је и $C_{k+1} \\equiv 0$.\n\nИз (2) следи да полиноми $Q_{0}(x)=0$ и $Q_{k}(x)=\\frac{2^{k-1}}{(k-1)!} P_{k}(x)$ задовољавају везу $Q_{k+1}(x)=\\frac{2 x}{k} Q_{k}(x)+Q_{k-1}(x)$, па индукцијом добијамо $Q_{k}(x)=x c_{x, k}$ за све $x \\in \\mathbb{N}$. Одавде је $P_{k}(x)=\\frac{(k-1)!}{2^{k-1}} \\cdot x c_{x, k}$ и\n$$\n\\frac{c_{m, n}}{n}=\\frac{1}{n} \\cdot \\frac{P_{m}(n)}{P_{m}(1)}=\\frac{c_{n, m}}{c_{1, m}}=\\frac{c_{n, m}}{m}\n$$\nТврђење задатка се добија за $m=1010$ и $n=1009$.\n\n\nДруго решење. За дато $m \\geqslant 0$ посматрајмо генераторску функцију низа $c_{m, n}$ датог условима $(*)$ :\n$$\nf_{m}(x)=\\frac{1}{2 m}+\\sum_{n=1}^{\\infty} \\frac{c_{m, n}}{n} x^{n}\n$$\nИз рекурентне везе $(*)$ следи да функција $f_{m}$ задовољава диференцијалну једначину $\\left(1-x^{2}\\right) f_{m}^{\\prime}(x)=2 m \\cdot f_{m}(x)$. Ова једначина се лако решава: ако је запишемо као $\\frac{f_{m}^{\\prime}(x)}{f_{m}(x)}=\\frac{2 m}{1-x^{2}}$, интеграција по $x$ даје $\\ln \\left|f_{m}(x)\\right|=\\int \\frac{2 m}{1-x^{2}} d x=$\n$m \\ln \\frac{1+x}{1-x}+$ const, тј. $f_{m}(x)=C \\cdot\\left(\\frac{1+x}{1-x}\\right)^{m}$. Услов $f_{m}(0)=\\frac{1}{2 m}$ најзад даје $C=1$, те је\n$$\n\\begin{aligned}\nf_{m}(x) & =\\frac{1}{2 m}\\left(\\frac{1+x}{1-x}\\right)^{m}=\\frac{1}{2 m}(1+x)^{m}(1-x)^{-m}= \\\\\n& =\\frac{1}{2 m} \\sum_{i=0}^{m}\\binom{m}{i} x^{i} \\cdot \\sum_{j=1}^{\\infty}\\binom{m-1+j}{m-1} x^{j}\n\\end{aligned}\n$$\nКоефицијент уз $x^{n}$ је\n$$\n\\frac{c_{m, n}}{n}=\\frac{1}{2 m} \\sum_{i=0}^{m}\\binom{m}{i}\\binom{m+n-1-i}{m-1}=\\frac{1}{2} \\sum_{i} \\frac{(m+n-1-i)!}{i!(m-i)!(n-i)!}\n$$\nОвај израз је симетричан по $m$ и $n$, па је $\\frac{c_{m, n}}{n}=\\frac{c_{n, m}}{m}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40895,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nНаћи све моничне полиноме $P(x)$ такве да је полином $P(x)^2-1$ дељив полиномом $P(x+1)$.",
    "options": [],
    "answer": "P(x) = 1 or P(x) = x - c (where c is a constant)",
    "solution": "Solution:\n\nЈедина решења су полиноми $P(x)=1$ и $P(x)=x-c$, где је $c$ константа.\n\nПретпоставимо да је $P(x)=(x-c)\\left(x-x_{2}\\right) \\cdots\\left(x-x_{n}\\right)$ неконстантан полином, при чему је $c$ његова комплексна нула чији је реални део најмањи.\n\nПо услову задатка, $x+1-c$ дели $P(x)^2-1$, одакле је $P(c-1)= \\pm 1$. С друге стране, за $2 \\leqslant i \\leqslant n$ важи $\\left|c-1-x_{i}\\right| \\geqslant 1$, па је $|P(c-1)|=\\prod_{i=2}^{n}\\left|c-1-x_{i}\\right| \\geqslant 1$, а то је могуће само ако је $\\left|c-1-x_{i}\\right|=1$, тј. $x_{i}=c$ за све $i$.\n\nСледи да је $P(x)=(x-c)^{n}$. Међутим, $P(x+c)^2-1=x^{2 n}-1$ није дељиво са $P(x+c+1)=(x+1)^{n}$ ако је $n \\geqslant 2$, па је једина могућност $n=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40896,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nДат је конвексан полиедар са бар 5 темена у чијем се сваком темену састају тачно по три ивице. Доказати да је могуће доделити сваком темену тог полиедра неки рационалан број тако да буду задовољени следећи услови:\n(i) бар један од додељених бројева је једнак 2020;\n(ii) за сваку страну полиедра, производ бројева у свим теменима те стране је једнак 1 .",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nОзначимо $c = 2020$. Прво размотримо случај када постоји страна $\\mathcal{F}$ са парним бројем темена. Тада је довољно доделити теменима стране $\\mathcal{F}$ наизменично бројеве $c$ и $\\frac{1}{c}$, а свим осталим теменима полиедра број $1$. Заиста, производ бројева на страни $\\mathcal{F}$ је једнак $1$, а свака друга страна или дели ивицу са страном $\\mathcal{F}$ (а на њој је производ $1$) или нема\n\n![](attached_image_1.png)\n\nниједно заједничко теме са $\\mathcal{F}$ (тада су сви бројеви на њој јединице).\n\nНадаље сматрамо да све стране имају непаран број темена. Посматрајмо две дисјунктне стране полиедра $\\mathcal{F}_A$ и $\\mathcal{F}_B$ и темена $A$ и $B$ на њима, редом, тако да је $AB$ ивица полиедра. Теменима ових двеју страна, изузев $A$ и $B$, доделићемо наизменично бројеве $c$ и $\\frac{1}{c}$ као на слици, а свим осталим теменима полиедра број $1$. И у овом случају се лако види да је услов задатка испуњен.\n\nНајзад, уверимо се да се овакве стране $\\mathcal{F}_A$ и $\\mathcal{F}_B$ увек могу наћи. Постоји страна $\\mathcal{F}$ која није троугао (иначе би дати полиедар био тетраедар). Посматрајмо чети-\n\n![](attached_image_2.png)\n\nри узастопне њој суседне стране $\\mathcal{F}_1, \\mathcal{F}_2, \\mathcal{F}_3$ и $\\mathcal{F}_4$. Ако су стране $\\mathcal{F}_1$ и $\\mathcal{F}_3$ дисјунктне, узмимо њих, а ако нису, онда оне имају заједничку целу ивицу, па три стране $\\mathcal{F}_1, \\mathcal{F}_3$ и $\\mathcal{F}$ чине појас који раздваја $\\mathcal{F}_2$ од $\\mathcal{F}_4$, те у том случају можемо узети $\\mathcal{F}_2$ и $\\mathcal{F}_4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40897,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nДат је троугао $ABC$. Тачке $D$ и $E$ на правој $AB$ су такве да је $AD = AC$ и $BE = BC$, уз распоред $D-A-B-E$. Описане кружнице троуглова $DBC$ и $EAC$ секу се у тачки $X \\neq C$, а описане кружнице троуглова $DEC$ и $ABC$ секу се у тачки $Y \\neq C$. Ако важи $DY + EY = 2XY$, одредити $\\varangle ACB$.",
    "options": [],
    "answer": "∠ACB = 60°",
    "solution": "Solution:\n\nНека је $I_{c}$ центар приписаног круга $\\triangle ABC$ наспрам темена $C$. Тада је $AI_{c}$ симетрала угла $CAD$, одакле је $\\triangle I_{c}AD \\cong \\triangle I_{c}AC$. Следи да је $\\varangle I_{c}DB = \\varangle I_{c}DA = \\varangle I_{c}CA = \\varangle I_{c}CB$, па $I_{c}$ лежи на кругу $BCD$. Слично, $I_{c}$ лежи на кругу $ACE$, што значи да је $X \\equiv I_{c}$. Такође, како је $XD = XC$ и аналогно $XE = XC$, тачка $X$ је центар круга $CDEY$.\n\nДаље, $\\varangle DYA = \\varangle CYA - \\varangle CYD = \\varangle CBA - \\varangle CED = \\varangle CED = \\varangle CXA = \\varangle DXA$ и слично $\\varangle EYB = \\varangle EXB$. Зато посматрајмо тачку $X'$ симетричну тачки $X$ у односу на $AB$. Због $\\varangle DX'A = \\varangle DYA$ ова тачка лежи на кругу $ADY$, а аналогно и на кругу $BEY$.\n\n![](attached_image_1.png)\n\nАко је $\\varangle C < 60^{\\circ}$, онда је $\\varangle AX'B = \\varangle AXB = 90^{\\circ} - \\frac{1}{2} \\varangle C > \\varangle C = \\varangle AYB$, па тачка $X'$ лежи унутар $\\triangle ABY$, а самим тим и унутар $\\triangle DEY$. Следи да је $DY + EY > DX' + EX' = DX + EX = 2XY$. Слично, ако је $\\varangle C > 60^{\\circ}$, тачка $Y$ лежи унутар $\\triangle DEX$, те је тада $DY + EY < 2XY$. Према томе, ако је $DY + EY = 2XY$, мора бити $\\varangle C = 60^{\\circ}$, а тада је $X' \\equiv Y$ па заиста важи једнакост.\n\n\nДруго решење. Углове троугла $ABC$ означавамо уобичајено са $\\alpha, \\beta, \\gamma$. Као у првом решењу, тачке $C, D, E, Y$ леже на кругу са центром $X$. Такође, због $\\varangle CYD = \\varangle CED = \\frac{\\beta}{2}$, права $DY$ садржи средиште $N$ лука $AC$ круга $ABC$.\n\nОзначимо $\\varangle YDE = x$ и $\\varangle YED = y$. Услов $DY + EY = 2XY$ даје $\\sin x + \\sin y = \\sin x + \\sin \\left(90^{\\circ} - \\frac{\\gamma}{2} - x\\right) = 1$.\n\n$$\n\\left(1 - \\sin \\frac{\\gamma}{2}\\right) \\sin x + \\cos \\frac{\\gamma}{2} \\cos x = 1\n$$\n\nС друге стране, синусна теорема у $\\triangle ADN$ даје $2 \\sin x \\sin \\frac{\\alpha + \\gamma}{2} = \\cos \\left(\\frac{\\alpha - \\gamma}{2} - x\\right)$, одакле следи $\\operatorname{tg} x = \\frac{\\cos \\frac{\\alpha - \\gamma}{2}}{2 \\sin \\frac{\\alpha + \\gamma}{2} - \\sin \\frac{\\alpha - \\gamma}{2}}$. Одавде налазимо\n$$\n\\sin x = \\frac{\\cos \\frac{\\alpha - \\gamma}{2}}{\\sqrt{3 - 2 \\cos \\gamma + 4 \\sin \\frac{\\gamma}{2} \\cos \\varphi}}, \\quad \\cos x = \\frac{2 \\sin \\frac{\\alpha + \\gamma}{2} - \\sin \\frac{\\alpha - \\gamma}{2}}{\\sqrt{3 - 2 \\cos \\gamma + 4 \\sin \\frac{\\gamma}{2} \\cos \\varphi}}\n$$\nгде је $\\varphi = \\alpha + \\frac{\\gamma}{2} - 90^{\\circ} = \\frac{\\alpha - \\beta}{2}$. Заменом у једначину (*) добијамо\n$$\n2 \\cos \\frac{3\\gamma - 180^{\\circ}}{4} \\cos \\frac{\\varphi}{2} = \\sqrt{3 - 2 \\cos \\gamma + 4 \\sin \\frac{\\gamma}{2} \\cos \\varphi}\n$$\nшто се након квадрирања своди на\n$$\n\\cos \\varphi = \\frac{2 - 2 \\cos \\gamma - \\sin \\frac{3\\gamma}{2}}{1 - 4 \\sin \\frac{\\gamma}{2} + \\sin \\frac{3\\gamma}{2}} = \\frac{-3t + 4t^{2} + 4t^{3}}{1 - t - 4t^{3}}, \\quad \\text{где је} \\quad t = \\sin \\frac{\\gamma}{2}\n$$\nАко је $\\gamma \\neq 60^{\\circ}$, скраћивање $2t - 1 \\neq 0$ даје $\\cos \\varphi = -\\frac{2t^{2} + 3t}{2t^{2} + t + 1} < 0$, што је немогуће. Према томе, мора бити $\\gamma = 60^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40898,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nУ трапезу $ABCD$ чији унутрашњи углови нису прави, дијагонале $AC$ и $BD$ секу се у тачки $E$. Нека су $P$ и $Q$ редом подножја нормала из темена $A$ и $B$ на праве $BC$ и $AD$. Описане кружнице троуглова $CEQ$ и $DEP$ секу се у тачки $F \\neq E$. Доказати да се праве $AP$, $BQ$ и $EF$ секу у једној тачки или су паралелне.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nСлучај када је $AD \\parallel BC$ је једноставан. Наиме, тада је $E$ средиште дијагонале $AC$, те је $EA = EC = EP$, а слично важи и $EB = ED = EQ$. Одатле следи да су кругови $CEQ$ и $DEP$ симетрични у односу на симетралу дужи $CP$ и $DQ$, те је $EF \\perp CP$, тј. $EF \\parallel AP \\parallel BQ$.\n\nОзначимо са $O$ тачку пресека правих $AD$ и $BC$. Троуглови $OAP$ и $OBQ$ су слични, па је $\\frac{OP}{OQ} = \\frac{OA}{OB} = \\frac{OD}{OC}$, тј. $OC \\cdot OP = OD \\cdot OQ$, одакле следи да тачке $C, D, P$ и $Q$ леже на неком кругу $\\gamma$.\n\nНека се праве $AP$ и $BQ$ секу у тачки $H$, а праве $DP$ и $CQ$ у тачки $G$. Како је $GC \\cdot GQ = GD \\cdot GP$, тачка $G$ има једнаку потенцију у односу на кругове $CEQ$ и $DEP$, па она лежи на њиховој радикалној оси $EF$. С друге стране, тачке $E, G$ и $H$ су колинеарне на\n\n![](attached_image_1.png)\n\nоснову Папосове теореме за тројке тачака $B, C, P$ и $A, D, Q$. Следи да су све четири тачке $H, E, G$ и $F$ колинеарне.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40899,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nЗа природан број $n$, са $v_{2}(n)$ означавамо највећи цео број $k \\geqslant 0$ такав да $2^{k} \\mid n$. Претпоставимо да функција $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ задовољава услове:\n(i) $f(x) \\leqslant 3 x$ за све $x \\in \\mathbb{N}$;\n(ii) $v_{2}(f(x)+f(y))=v_{2}(x+y)$ за све $x, y \\in \\mathbb{N}$.\nДоказати да за сваки природан број $a$ постоји тачно један природан број $x$ такав да је $f(x)=3 a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nЗаменом $x=y$ добијамо $v_{2}(f(x))=v_{2}(x)$. Ако је $v_{2}(a)=k>0$, посматрањем функције $g(x)=f\\left(2^{k} x\\right) / 2^{k}$ тврђење сводимо на случај непарног $a$. Зато надаље сматрамо да $2 \\nmid a$.\n\nПриметимо да, ако је $x \\not \\equiv y\\left(\\bmod 2^{k}\\right)$, онда је $f(x) \\not \\equiv f(y)\\left(\\bmod 2^{k}\\right)$. Заиста, ако је $z \\equiv-x\\left(\\bmod 2^{k}\\right)$, онда $2^{k} \\nmid z+y$, па је $f(y) \\not \\equiv-f(z) \\equiv f(x)\\left(\\bmod 2^{k}\\right)$. Одавде такође следи да је функција $f$ инјективна.\n\nНека је $2^{k-1}<3 a<2^{k}$, где је $k \\in \\mathbb{N}$. Како су $f(1), f(3), \\ldots, f\\left(2^{k}-1\\right)$ међусобно различити по модулу $2^{k}$, постоји непаран број $x<2^{k}$ такав да је $f(x) \\equiv 3 a$ $\\left(\\bmod 2^{k}\\right)$. Претпоставимо да је $f(x) \\neq 3 a$. Тада је $f(x)>2^{k}$, па из $f(x)+$ $f\\left(2^{k}-x\\right) \\equiv 2^{k}\\left(\\bmod 2^{k+1}\\right)$ следи да је $f(x)+f\\left(2^{k}-x\\right) \\geqslant 3 \\cdot 2^{k}$. Међутим, с друге стране је $f(x)+f\\left(2^{k}-x\\right) \\leqslant 3\\left(x+\\left(2^{k}-x\\right)\\right)=3 \\cdot 2^{k}$, па је то могуће једино ако је $f(x)=3 x$. Одавде је $x \\equiv a\\left(\\bmod 2^{k}\\right)$, па је $x=a$, тј. опет је $f(x)=3 a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40900,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nДат је природан број $k$. Посматрајмо следећу игру на бесконачној једнодимензионалној табли. На почетку игре, на поља постављамо укупно $n$ жетона, при чему може бити више жетона на истом пољу. Након тога, у сваком потезу извршавамо једну од следећих операција:\n$(1^{\\circ})$ бирамо два суседна поља која су оба непразна и са једног од њих преносимо све жетоне на друго;\n$(2^{\\circ})$ бирамо поље са бар два жетона и са њега премештамо по један жетон $k$ поља улево и $k$ поља удесно.\n\na. Ако је $n \\leqslant k+1$, доказати да ће се извршити само коначно много потеза.\n\nб. За које вредности $k$ се може одабрати $n$ и поставити $n$ жетона тако да буде могућ бесконачан низ потеза?",
    "options": [],
    "answer": "For all k, there exists a choice of n and an initial placement that allows an infinite sequence of moves.",
    "solution": "Solution:\n\n(a) Потезе врсте $(1^{\\circ})$ зваћемо скупљањем, а потезе врсте $(2^{\\circ})$ сејањем. Поља нумеришемо редом бројевима $\\ldots, -1, 0, 1, 2, \\ldots$. Поља се могу поделити на $k$ класа по модулу $k$, тако да при сејању жетони не мењају своју класу.\n\nПретпоставимо прво да у једној од класа нема ниједног жетона. Поља ове класе деле таблу на блокове дужине $k-1$, које опет нумеришемо целим бројевима. Како приликом скупљања ниједан жетон не излази из свог блока, унутар сваког блока може се извршити само коначно много узастопних скупљања. С друге стране, при сваком сејању се из неког блока $i$ пребацује по један жетон у блокове $i-1$ и $i+1$. Овако се задатак своди на следећи.\n\nЛема. Нека се у сваком кораку бира $i$ и из блока $i$ премешта по један жетон у блокове $i-1$ и $i+1$. Тада се игра завршава у коначно много корака.\n\nДоказ. Кад год нека два суседна блока $x$ и $x+1$ по први пут размене жетон, доделимо овом жетону број $x$. Приликом сваке следеће размене између ових двају блокова можемо сматрати да се размењује тај исти жетон. Овако сваки жетон само циркулише између два суседна блока. Следи да ће највише $n$ парова блокова икада разменити жетоне. Међутим, ако неки блок сеје жетоне бесконачно много пута, онда то мора да важи и за суседне блокове (иначе би се у њима гомилало бесконачно много жетона), а индукцијом и за све блокове, што је немогуће.\n\nОстаје случај када се ниједна класа поља никад не испразни. Тада се увек у једној класи налазе два жетона, а у осталима по један. Сејања и скупљања се врше из класе са два жетона. Нека се у неком тренутку сеје са поља $p$ у $i$-тој класи (на коме су два жетона) на поља $p-k$ и $p+k$. Следи скупљање, и то без смањења општости удесно ка ($i+1$)-тој класи - са поља $p \\pm k$ на поље $p \\pm k+1$. Затим се из овог поља сеје на поља $p+1$ и $p \\pm 2k+1$. Следи скупљање из ($i+1$)-те класе, али не у $i$-ту, јер је у њој једини жетон остао на пољу $p \\mp k$. Дакле, сада се скупља у $(i+2)$-ту класу. Закључујемо да се наизменично сеје и скупља, при чему се класа из које се сеје и скупља циклично помера за један удесно. Може се сматрати без умањења општости да је жетон који се скупља увек исти - назовимо га црним. Црни жетон периодично учествује у сејању из сваке класе, дакле са свим осталим жетонима.\n\nПри сваком сејању црни жетон баца други жетон $k$ поља на једну страну, а себе $k$ поља на другу. Тако се у сваком циклусу од $k$ сејања позиција најлевљег жетона помера бар $k$ поља улево, а најдешњег бар $k$ поља удесно, те се опсег игре увећава за бар $2k$. Међутим, у току једног циклуса црни жетон не може прећи пут дужи од $k(k+1)$, па након $\\frac{k+1}{2}$ циклуса он више не може стићи од најлевљег до најдешњег, што је коначна контрадикција.\n\n\n(б) За свако $k$ даћемо пример периодичне (и самим тим бесконачне) игре. Посматрајмо $5k$ узастопних поља означених редом бројевима $1,2, \\ldots, 5k$, свако са по једним жетоном. Делимо их на групе од по $k$ поља:\n$$\n1,1, \\cdots, 1,1 \\quad 1,1, \\cdots, 1,1 \\quad 1,1, \\cdots, 1,1 \\quad 1,1, \\cdots, 1,1 \\quad 1,1, \\cdots, 1,1\n$$\nПочињемо скупљањем жетона са поља $1, \\ldots, k$ на поље $k+1$. Затим, за $i=1,2, \\ldots, k$, са поља $k+i$ једном сејемо, а преостале жетоне с тог поља скупљамо на поље $k+i+1$ (осим за $i=k$). Након $i$-тог циклуса стање је овакво:\n$$\n\\underbrace{1,1, \\ldots, 1}_i, 0,0, \\ldots, 0 \\quad \\underbrace{0,0, \\ldots, 0}_i, k+1-i, 1,1, \\ldots, 1 \\quad \\underbrace{2,2, \\ldots, 2}_i, 1,1, \\ldots, 1 \\quad 1,1, \\cdots, 1,1 \\quad 1,1, \\cdots, 1,1\n$$\nКада аналогне потезе одиграмо и здесна, добићемо стање\n$$\n1,1, \\cdots, 1,1 \\quad 0,0, \\cdots, 0,0 \\quad 3,3, \\cdots, 3,3 \\quad 0,0, \\cdots, 0,0 \\quad 1,1, \\cdots, 1,1\n$$\nНајзад, ако сада посејемо по једном са сваког од поља $2k+1, \\ldots, 3k$, добијамо конфигурацију идентичну полазној.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40901,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$ with incenter $I$, let $P$ be the intersection point of the bisector of the angle $A$ with the circumcircle other than $A$, $D$ the point of tangency of the incircle to the side $BC$, and $Q$ the intersection point of $PD$ with the circumcircle other than $P$. Show that $PI = QI$ if $PD$ is equal to the inradius.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nIn the triangle $BIA$, $\\angle BIP = \\angle IBA + \\angle IAB = \\angle B/2 + \\angle A/2$. On the other hand, $\\angle PBI = \\angle PBC + \\angle CBI = \\angle A/2 + \\angle B/2$, as $\\angle PBC = \\angle PAC = \\angle A/2$. Hence, in the triangle $BPI$, $BP = IP$.\n\nSince $\\angle QPB = \\angle BPD$ and $\\angle BQP = \\angle BAP = \\angle A/2 = \\angle PAC = \\angle PBC = \\angle PBD$, the triangles $QPB$ and $BPD$ are similar. In particular, $\\frac{PD}{PB} = \\frac{BP}{QP}$.\n\nCombining this with the fact $BP = IP$, gives\n$$ \\frac{PD}{PI} = \\frac{PI}{PQ}. $$\nTherefore, the triangles $DPI$ and $IPQ$ are similar, and $\\frac{QI}{PI} = \\frac{ID}{PD} = 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40902,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer with $n > 2$ and $a_1, a_2, \\dots, a_n \\in \\mathbb{R}^+$ positive real numbers. Given any positive integers $t, k, p$ with $1 < t < n$, set $m = k+p$. Prove the following inequalities:\n\n$$\n1) \\quad \\frac{a_1^p}{a_2^k + a_3^k + \\cdots + a_t^k} + \\frac{a_2^p}{a_3^k + a_4^k + \\cdots + a_{t+1}^k} + \\cdots + \\frac{a_{n-1}^p}{a_n^k + a_1^k + \\cdots + a_{t-2}^k} + \\\\\n\\qquad + \\frac{a_n^p}{a_1^k + a_2^k + \\cdots + a_{t-1}^k} \\ge \\frac{(a_1^p + a_2^p + \\cdots + a_n^p)^2}{(t-1)(a_1^m + a_2^m + \\cdots + a_n^m)}\n$$\n\n$$\n2) \\quad \\frac{a_2^k + a_3^k + \\cdots + a_t^k}{a_1^p} + \\frac{a_3^k + a_4^k + \\cdots + a_{t+1}^k}{a_2^p} + \\cdots + \\frac{a_n^k + a_1^k + \\cdots + a_{t-2}^k}{a_{n-1}^p} + \\\\\n\\qquad + \\frac{a_1^k + a_2^k + \\cdots + a_{t-1}^k}{a_n^p} \\ge \\frac{(t-1)(a_1^k + a_2^k + \\cdots + a_n^k)^2}{a_1^m + a_2^m + \\cdots + a_n^m}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1) By the rearrangement inequality we have:\n$$\n\\begin{align*} \na_1^m + a_2^m + \\dots + a_n^m &\\ge a_1^p a_2^k + a_2^p a_3^k + \\dots + a_{n-1}^p a_n^k + a_n^p a_1^k \\\\ \na_1^m + a_2^m + \\dots + a_n^m &\\ge a_1^p a_3^k + a_2^p a_4^k + \\dots + a_{n-1}^p a_1^k + a_n^p a_2^k \\\\ \n\\multicolumn{2}{c}{\\text{...................}} &\\quad \\text{...................} \\\\ \na_1^m + a_2^m + \\dots + a_n^m &\\ge a_1^p a_t^k + a_2^p a_{t+1}^k + \\dots + a_{n-1}^p a_{t-2}^k + a_n^p a_{t-1}^k \n\\end{align*}\n$$\nAdding above $(t-1)$ inequalities we get\n$$\n(1) \\quad (t-1)(a_1^m + a_2^m + \\dots + a_n^m) \\ge a_1^p (a_2^k + a_3^k + \\dots + a_t^k) + a_2^p (a_3^k + a_4^k + \\dots + a_{t+1}^k) + \\dots + a_n^p (a_1^k + a_2^k + \\dots + a_{t-1}^k)\n$$\n\nand using Cauchy-Schwartz inequality we have\n$$\n(t-1)(a_1^m + a_2^m + \\dots + a_n^m) \\left( \\frac{a_1^p}{a_2^k + a_3^k + \\dots + a_t^k} + \\frac{a_2^p}{a_3^k + a_4^k + \\dots + a_{t+1}^k} + \\dots + \\frac{a_n^p}{a_1^k + a_2^k + \\dots + a_{t-1}^k} \\right) \\ge (a_1^p + a_2^p + \\dots + a_n^p)^2\n$$\nNow, our conclusion is obtained by dividing both sides of above equation by $(t-1)(a_1^m + a_2^m + \\dots + a_n^m)$.\n\n2) Multiplying both sides of (1) by\n$$\n\\frac{a_2^k + a_3^k + \\dots + a_t^k}{a_1^p} + \\frac{a_3^k + a_4^k + \\dots + a_{t+1}^k}{a_2^p} + \\dots + \\frac{a_1^k + a_2^k + \\dots + a_{t-1}^k}{a_n^p}\n$$\nand using Cauchy-Schwartz inequality too we have\n$$\n(t-1)(a_1^m + a_2^m + \\dots + a_n^m) \\left( \\frac{a_2^k + a_3^k + \\dots + a_t^k}{a_1^p} + \\frac{a_3^k + a_4^k + \\dots + a_{t+1}^k}{a_2^p} + \\dots + \\frac{a_1^k + a_2^k + \\dots + a_{t-1}^k}{a_n^p} \\right) \\ge [(t-1)(a_1^k + a_2^k + \\dots + a_n^k)]^2\n$$\nAgain, our conclusion is obtained by dividing both sides of above equation by $(t-1)(a_1^m + a_2^m + \\dots + a_n^m)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40903,
    "subject": "Mathematics (Multi-modal)",
    "question": "An integer is assigned to each unit square of a finite set of unit squares on an infinite chessboard such that the sum of numbers lying on each row as well as each column is divisible by $2002$. Prove that every number $a$ can be replaced by some number $a'$ divisible by $2002$ such that $|a - a'| < 2002$ and the sums of rows and columns remain unchanged.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider any number $a_1$ which is not divisible by $2002$. Then there exists a number $a_2$ not divisible by $2002$ lying on the same row with $a_1$. In turn, there exists a number $a_3$ not divisible by $2002$ lying on the same column with $a_2$. We continue this process till we first reach a number $a_k$ which lies on the same row or column with some number $a_l$, $1 \\le l \\le k-2$.\n\nTake $l$ as a maximal number with this property. One easily sees that the collection $C$ of numbers $a_l, a_{l+1}, \\dots, a_k$ includes an even number of elements.\n\nFor each number $a_i$, $l \\le i \\le k$ set $d(i) = \\min_{k \\in \\mathbb{Z}} |a_i - 2002k|$.\n\nTake $m \\in \\{l, \\dots, k\\}$ with $d(m) \\le d(i)$, $l \\le i \\le k$ and replace $a_m$ with the closest integer number $a'_m$ which is divisible by $2002$. By definition, $d(m) = |a_m - a'_m| < 2002$. Set $a_m - a'_m = d$ and replace the remaining elements in $C$ as follows: if $|i - m|$ is even, replace $a_i$ by $a_i - d$; if $|i - m|$ is odd, replace $a_i$ by $a_i + d$.\n\nIt can be easily seen that after these replacements the number of integers not divisible by $2002$ is decreased at least by one, the sums of rows and columns are unchanged, and if $2002k < a_j < 2002(k+1)$ then $2002k < a'_j < 2002(k+1)$. Thus, we can reach a desired configuration repeating this procedure finitely many times.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40904,
    "subject": "Mathematics (Multi-modal)",
    "question": "Observe that the fraction $1/7 = 0.\\overline{142857}$ is a pure periodical decimal with period $6 = 7 - 1$, and in one period one has $142 + 857 = 999$. For $n = 1, 2, \\dots$, find a sufficient and necessary condition that the fraction $1/(2n+1)$ has the same properties as above and find two such fractions other than $1/7$.",
    "options": [],
    "answer": "Necessary and sufficient condition: 2n+1 divides 10^n + 1 and 2n+1 does not divide 10^i + 1 for any i with 1 ≤ i < n. Examples: 1/17 and 1/19.",
    "solution": "Suppose $1/(2n+1) = \\dot{a}_1 a_2 \\cdots a_n b_1 b_2 \\cdots \\dot{b}_n$, $a_i + b_i = 9$, $i = 1, 2, \\dots, n$, and $2n$ is its period. Since\n$$\n1 + \\frac{1}{10^{2n}} + \\frac{1}{10^{4n}} + \\dots = \\frac{1}{1 - \\frac{1}{10^{2n}}} = \\frac{10^{2n}}{10^{2n} - 1}\n$$\nwe have\n$$\n\\begin{align*}\n(1) \\quad \\frac{1}{2n+1} &= \\frac{10^{2n}}{10^{2n}-1} \\left( \\sum_{i=1}^{n} \\frac{a_i}{10^i} + \\frac{1}{10^n} \\sum_{i=1}^{n} \\frac{b_i}{10^i} \\right) \\\\\n&= \\frac{10^{2n}}{10^{2n}-1} \\left( \\sum_{i=1}^{n} \\frac{a_i}{10^i} + \\frac{1}{10^n} \\sum_{i=1}^{n} \\frac{9-a_i}{10^i} \\right) \\\\\n&= \\frac{10^{2n}}{10^{2n}-1} \\left[ \\left(1-\\frac{1}{10^n}\\right) \\sum_{i=1}^{n} \\frac{a_i}{10^i} + \\frac{9}{10^n} \\sum_{i=1}^{n} \\frac{1}{10^i} \\right] \\\\\n&= \\frac{10^{2n}}{10^{2n}-1} \\left[ \\left(1-\\frac{1}{10^n}\\right) \\sum_{i=1}^{n} \\frac{a_i}{10^i} + \\frac{9}{10^n} \\cdot \\frac{1}{10} \\cdot \\frac{1-\\frac{1}{10^n}}{1-\\frac{1}{10}} \\right] \\\\\n&= \\frac{10^{2n}}{10^n+1} \\left( \\frac{1}{10^n} \\sum_{i=1}^{n} \\frac{a_i}{10^i} + \\frac{1}{10^{2n}} \\right) = \\frac{\\sum_{i=1}^{n} 10^{n-i} a_i + 1}{10^n+1}\n\\end{align*}\n$$\nTherefore,\n$2n+1 \\mid 10^n + 1.$\nBy (1) and the knowledge of numbers, we guess the sufficient and necessary condition is\n$$\n(2) \\quad 2n+1 \\mid 10^n+1 \\text{ and } 2n+1 \\nmid 10^i+1 \\text{ for } i=1,2,\\dots,n-1.\n$$\nIn fact, since $2n$ is the period of $1/(2n+1)$, we have\n$$\n2n+1 \\mid 10^{2n}-1 \\text{ and } 2n+1 \\nmid 10^i-1 \\text{ for } i=1,2,\\dots,n-1.\n$$\nSo $2n+1 \\nmid (10^n+1) + (10^{n+i}-1) = 10^n(10^i+1)$ for $i=1,2,\\dots,n-1$. Namely, $2n+1 \\nmid 10^i+1$ for $i=1,2,\\dots,n-1$. The condition (2) is necessary.\n\nNow we assume that $1/(2n + 1)$ satisfies the condition (2). Let\n$$\n\\frac{10^n + 1}{2n + 1} - 1 = \\sum_{i=1}^{n} 10^{n-i}a_i,\n$$\nwhere $0 \\le a_i \\le 9$, $i = 1, \\dots, n$ are integers. Then\n$$\n\\frac{1}{2n + 1} = \\frac{\\sum_{i=1}^{n} 10^{n-i}a_i + 1}{10^n + 1}.\n$$\nPut $b_i = 9 - a_i$. Using (1) we have\n$$\n\\frac{1}{2n + 1} = \\dot{a}_1 a_2 \\cdots a_n b_1 b_2 \\cdots \\dot{b}_n.\n$$\nFor $1 \\le i \\le n$, because\n$$\n(10^n + 1) + (10^i - 1) = 10^i(10^{n-i} + 1),\n$$\nso $2n + 1 \\nmid 10^i - 1$.\nFor $n < i < 2n$, because\n$$\n(10^n + 1) + (10^i - 1) = 10^n(10^{i-n} + 1),\n$$\nwe also have $2n + 1 \\nmid 10^i - 1$.\nTherefore the period of $1/(2n + 1)$ is $2n$. That is, the condition (2) is sufficient.\nBy the condition (2), if $1/(2n + 1)$ is a fraction we look for, then $2n + 1$ does not have factors 3 and 5. So the possible values of $1/(2n + 1)$ are 11, 13, 17, 19, 23, .... Since $11, 13 \\nmid 10^3 + 1 = 1001$, 11 and 13 do not satisfy the condition. Consider $17 = 2 \\times 8 + 1$. We have\n$$\n10^8 + 1 = (17 \\times 4 - 2)^4 = 2^4 + 1 = 0 \\pmod{17}.\n$$\nFor $i = 1, 2, 3, 4$ it is obvious that $17 \\nmid 10^i + 1$. For $i = 5, 6, 7$, since $(10^8 + 1) - (10^i + 1) = 10^i(10^{8-i} - 1)$, it is obvious too that $17 \\nmid 10^i + 1$. Therefore, $1/17$ is a fraction we need.\nSimilarly, consider $19 = 2 \\times 9 + 1$.\n$$\n10^9 + 1 = (53 \\times 19 - 7)^3 + 1 = -342 = 0 \\pmod{19}.\n$$\nIt is easy to check that $19 \\nmid 10^i + 1$ for $1 \\le i < 9$. So $1/19$ is another fraction we want to seek.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40905,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_{2003}$ be a sequence of real numbers. A term $a_k$, $1 \\le k \\le 2003$, is said to be a *leading term*, if at least one of the expressions $a_k, a_k+a_{k+1}, \\dots, a_k+a_{k+1}+\\dots+a_{2003}$ is positive. Prove that the sum of all leading terms is positive provided that the sequence has at least one leading term.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We solve this problem for any sequence having $n$ terms applying induction with respect to $n$.\n\nThe case $n = 1$ is clear.\n\nSuppose that the statement is true for all sequences of length less than $n$.\n\nNow consider a sequence $a_1, a_2, \\dots, a_n$.\n\n*Case 1. $a_1$ is not a leading term.*\n\nThen the set of all leading terms of the sequence $a_1, a_2, \\dots, a_n$ coincides with the set of all leading terms of the sequence $a_2, a_3, \\dots, a_n$. And by inductive hypothesis we are done.\n\n*Case 2. $a_1$ is a leading term.*\n\nConsider the smallest nonnegative integer $m$, with positive $a_1 + a_2 + \\dots + a_m$. Then the terms $a_2, a_3, \\dots, a_m$ are also leading terms and their sum is positive. The sum of all remaining leading terms also is nonnegative by induction hypothesis.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40906,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $s = (AB + BC + AC)/2$ be the semiperimeter of triangle $ABC$. We choose two points $L$ and $N$ lying on the rays $AB$ and $CB$ satisfying $AL = CN = s$. Let $K$ be the point symmetric to $B$ with respect to the center of the circumcircle of triangle $ABC$. Prove that the perpendicular drawn from the point $K$ to the line $NL$ passes through the incenter of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $BC = a$, $AC = b$, $AB = c$ and $I$ be the incenter of the triangle $ABC$.\n\n![](attached_image_1.png)\n\nFrom the point $I$ let us draw parallel lines to $AB$ and $BC$, intersecting $AK$ and $KC$ at $P$ and $Q$, respectively. In the triangle $IPQ$ we have $\\angle PIQ = \\angle ABC$, $IQ = s - c$, $IP = s - a$. Since $\\angle NBL = \\angle ABC$, $NB = s - a$, $LB = s - c$, it follows that $\\angle PIQ = \\angle NBL$.\n\nOn the ray $IP$ let's choose a point $L_1$ satisfying $IL_1 = IQ$, and on the ray $IQ$ choose a point $N_1$, such that $IN_1 = IP$. Then $\\Delta IN_1L_1 = \\Delta NBL$, $BL||IL_1$, $BN||IN_1$ and it follows that $L_1N_1||LN$.\n\n$IPKQ$ is cyclic. Therefore, $\\angle IKP = \\angle IQP$,\n$$\n\\angle KIP + \\angle IL_1N_1 = \\angle KIP + \\angle IQP = \\angle KIP + \\angle IKP = 90^\\circ\n$$\nand it follows that $IK \\perp L_1N_1$ and $IK \\perp LN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40907,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $0 < a < b < 1$ be real numbers and\n$$\ng(x) = \\begin{cases} x + 1 - a, & \\text{if } 0 < x < a, \\\\ b - a, & \\text{if } x = a, \\\\ x - a, & \\text{if } a < x < b, \\\\ 1 - a, & \\text{if } x = b, \\\\ x - a, & \\text{if } b < x < 1. \\end{cases}\n$$\nAssume that for some positive integer $n$ there are $n+1$ real numbers $0 < x_0 < x_1 < \\dots < x_n < 1$ such that $g^n(x_i) = x_i$ for $0 \\le i \\le n$. Prove that there is a positive integer $N$ such that $g^N(x) = x$ for all $0 < x < 1$. (Notation: $g^k(x) = \\underbrace{g(g(\\dots(g(x))\\dots))}_{k \\text{ times}}$).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For real numbers $r$ and $s$ we call them equivalent (and write $r \\equiv s$), if $r - s \\in \\mathbb{Z}$. Then,\n$$\ng(x) \\equiv \\begin{cases} x - a, & \\text{if } x \\not\\equiv a, b, \\\\ b - a, & \\text{if } x \\equiv a, \\\\ -a, & \\text{if } x \\equiv b. \\end{cases}\n$$\nNote that $g$ is a bijection on $(0,1)$. Let\n$$\nS = \\{g^{-i}(a) : 0 \\le i \\le n-1\\} \\cup \\{g^{-i}(b) : 0 \\le i \\le n-1\\}.\n$$\nThen $g^n(x) = x - na$ for $x \\notin S$.\n**Case 1.** $g^n$ has a fixed point $x'$ outside $S$.\nThen $x' \\equiv g^n(x') \\equiv x' - na$, and therefore, $na \\equiv 0$ and $g^n(x) = x$ for all $x \\notin S$. On the other hand, $g^n$ is a permutation of $S$, and for some positive integer $m$, $(g^n)^m$ be the identity in $S$. So we can take $N = mn$ in this case.\n**Case 2.** All fixed points of $g^n$ are in $S$.\nIf $g(a) \\notin S$, then none of $g^{-i}(a)$ for $0 \\le i \\le n-1$ can be a fixed point of $g^n$, and $g^n$ can not have more than $n$ fixed points. Similarly, $g(b) \\notin S$ is impossible. Therefore, $g(S) = S$. In particular, as in **Case 1**, there is a positive integer $m$ such that $g^{nm}(x) = x$ for all $x \\in S$.\nSince $g(S) = S$, $g^n$ is a bijection on $(0,1)\\setminus S$. As $g^n$ is defined by $x \\to x - na$ on $(0,1)\\setminus S$, it means that $x \\to x - na$ is also bijection on $S$. It follows that $lna \\equiv 0$ for some positive integer $l$, and therefore $g^{nl}(x) = x$ for all $x \\notin S$.\nHence we can choose $N = nml$ in this case.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40908,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the sum\n$$\n\\sum_{k \\in A} \\frac{1}{k-1}\n$$\nif $A = \\{m^n : m, n \\in \\mathbb{Z}, m, n \\ge 2\\}$.",
    "options": [],
    "answer": "1",
    "solution": "Answer: the sum is equal $1$.\nThis can be seen as follows:\n$$\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 1} \\frac{1}{k^i}\n$$\n**Proposition:** For $x \\in \\mathbb{Z}$ the sets $\\{(m,n) : x = m^n; m, n \\in \\mathbb{Z}, m, n \\ge 2\\}$ and $\\{(k,i) : x = k^i; k \\in A, i \\in \\mathbb{N}\\}$ have the same number of elements.\n**Proof of the proposition:** Let $x = p_1^{a_1} \\dots p_t^{a_t}$ be the prime factorization of $x$. $a = \\text{GCD}(a_1, \\dots, a_t)$, $a_j = ab_j$ for $0 \\le j \\le t$ and $y = p_1^{b_1} \\dots p_t^{b_t}$. Then $x = m^n$ iff $n \\mid a$ and $m = y^{a/n}$. So both of the sets above have $\\tau(a) - 1$ elements, where $\\tau(a)$ stands for the number of positive divisors of $a$. $\\square$\n\nTherefore:\n$$\n\\begin{aligned}\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} \\\\\n&= \\sum_{m \\ge 2} \\frac{1/m^2}{1 - 1/m} = \\sum_{m \\ge 2} \\frac{1}{m(m-1)}\n\\end{aligned}\n$$\nAnd finally,\n$$\n\\sum_{m=2}^{l} \\frac{1}{m(m-1)} = \\sum_{m=2}^{l} \\left( \\frac{1}{m-1} - \\frac{1}{m} \\right) = 1 - \\frac{1}{l} \\to 1 \\text{ for } l \\to \\infty.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40909,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all prime numbers $p$ for which there are integers $m$ and $n$ such that $p = m^2 + n^2$ and $p \\mid m^3 + n^3 - 4$.",
    "options": [],
    "answer": "2, 5, 13",
    "solution": "Considering all integers $|m|, |n| \\le 3$ we can get the solutions $2 = 1^2 + 1^2$, $5 = 1^2 + 2^2$ and $13 = (-3)^2 + (-2)^2$. Now, let's prove that there is no other prime numbers satisfying the statement. We have\n$$\np = m^2 + n^2 \\Rightarrow p = (m + n)^2 - 2mn \\Rightarrow mn = \\frac{(m + n)^2 - p}{2},\n$$\n$$\nm^3 + n^3 - 4 = (m + n)^3 - 3mn(m + n) - 4 = \\frac{-(m + n)^3 + 3p(m + n) - 8}{2},\n$$\n$$\np \\mid m^3 + n^3 - 4 \\Rightarrow p \\mid (m + n)^3 + 8.\n$$\nHence, $p \\mid m+n+2$ or $p \\mid (m+n)^2 - 2(m+n) + 4$, and from last we conclude that $p \\mid mn - m - n + 2$, if we set $p > 2$.\nIn a first case we have:\n$$\np \\mid m + n + 2 \\Rightarrow m^2 + n^2 \\le |m + n + 2|,\n$$\ni.e.\n$$\nm^2 + n^2 \\le m + n + 2 \\Rightarrow (2m - 1)^2 + (2n - 1)^2 \\le 10 \\Rightarrow -1 \\le m, n \\le 2.\n$$\n$$\nm^2 + n^2 \\le -(m+n+2) \\Rightarrow (2m+1)^2 + (2n+1)^2 \\le -6,\n$$\nwhich is a contradiction.\nIn a second case:\n$$\np \\mid mn - m - n + 2 \\Rightarrow m^2 + n^2 \\le |mn - m - n + 2|,\n$$\ni.e.\n$$\nm^2 + n^2 \\le mn - m - n + 2 \\Rightarrow (2m - n + 1)^2 + 3(n + 1)^2 \\le 12 \\Rightarrow -3 \\le m, n \\le 1.\n$$\n$$\nm^2 + n^2 \\le -(mn - m - n + 2) \\Rightarrow (2m + n - 1)^2 + 3(n - 1/3)^2 \\le -20/3,\n$$\nwhich is a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40910,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of $\\triangle ABC$ with center $I$ touches the sides $AB$ and $AC$ at points $P$ and $Q$, respectively. $BI$ and $CI$ intersect $PQ$ at $K$ and $L$, respectively. Prove that circumcircle of $\\triangle ILK$ touches the incircle of $\\triangle ABC$ if and only if $AB + AC = 3BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $BC = a$, $AC = b$, $AB = c$ and $\\angle CAB = \\alpha$, $\\angle ABC = \\beta$, $\\angle BCA = \\gamma$. Let $D$ be the intersection point of $BL$ and $CK$. Note that $\\triangle PAQ$ is isosceles.\n$$\n\\angle BKL = \\angle APK - \\angle ABK = \\frac{\\pi - \\alpha}{2} - \\frac{\\beta}{2} = \\frac{\\alpha + \\beta + \\gamma - \\alpha - \\gamma}{2} = \\frac{\\gamma}{2} = \\frac{\\angle ACB}{2}.\n$$\nSince $\\angle IKL = \\angle BKL = \\angle ABC/2 = \\angle ACI$, the points $I, K, Q, C$ are concyclic, and hence, $\\angle IKC = \\angle IQC = \\pi/2$. Similarly, $\\angle ILB = \\pi/2$. Therefore, the points $B, L, K, C$ are concyclic, and the points $I, L, D, K$ are also. Particularly, $BC$ is a diameter of the circumscribed circle of $\\triangle CLK$ and $ID$ is a diameter of the circumscribed circle of $\\triangle ILK$.\n\nUsing the sines law in $\\triangle ILK$ we have\n$$\nID = \\frac{IK}{\\sin \\angle LDK} = \\frac{LK}{\\cos \\angle LCK},\n$$\nand in $\\triangle CLK$ we have $a = \\frac{LK}{\\sin \\angle LCK}$.\nTherefore, $ID = a \\tan \\angle LCK$.\nSince $\\sin \\angle LCK = \\sin \\angle IQK = \\sin(\\alpha/2)$, we also can write that $ID = \\tan(\\alpha/2)$. On the other hand, $r = AQ \\tan(\\alpha/2)$ and $AQ = (b+c-a)/2$, where $r$ is a radius of incircle of $\\triangle ABC$.\n\nCircumcircle of $\\triangle ILK$ touches the incircle of $\\triangle ABC \\iff$ diameter of the circumcircle of $\\triangle ILK$ is equal to the radius of incircle of $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40911,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer. Prove that\n$$\n(1^{n-1} + 2^{n-1} + \\dots + (n-1)^{n-1}) + 1 \\text{ is divisible by } n\n$$\nif and only if, for each prime factor $p$ of $n$\n$$\n\\frac{n}{p}-1 \\text{ is divisible by } p \\text{ and } \\frac{n}{p}-1 \\text{ is divisible by } p-1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For simplicity let's enumerate the statements.\n(1) $(1^{n-1} + 2^{n-1} + \\dots + (n-1)^{n-1}) + 1$ is divisible by $n$\n(2) $\\frac{n}{p}-1$ divisible by $p$ and $\\frac{n}{p}-1$ divisible by $p-1$\nLet $n = Ap$. Firstly, observe that\n$$\n\\sum_{k=1}^{n-1} k^{n-1} \\equiv \\begin{cases} -A & (\\text{mod } p), & \\text{if } n-1 \\text{ is divisible by } p-1, \\\\ 0 & (\\text{mod } p), & \\text{if } n-1 \\text{ is not divisible by } p-1 \\end{cases}\n$$\nThe first case is clear, it follows from Fermat's little theorem. And the second can be shown, for instance, as follows: let $g$ be a primitive root by modulo $p$, then\n$$\n\\sum_{k=1}^{p-1} k^{n-1} \\equiv \\sum_{k=1}^{p-1} (gk)^{n-1} \\equiv g^{n-1} \\sum_{k=1}^{p-1} k^{n-1} \\pmod{p}\n$$\nand $g^{n-1} \\not\\equiv 1 \\pmod{p}$ implies that $\\sum_{k=1}^{n-1} k^{n-1} \\equiv 0 \\pmod{p}$.\nSuppose that (1) holds. Then either $n-1$ divisible by $p-1$ and $A-1$ divisible by $p$, or $n-1$ divisible by $p-1$ and $1$ divisible by $p$. Therefore, $n-1$ divisible by $p-1$ and $A-1$ divisible by $p$. Hence, $n-1 = (p-1)A + A-1$ divisible by $p-1 \\Rightarrow A-1$ divisible by $p-1$, we obtain (2).\nNow suppose that (2) holds. $A-1$ divisible by $p \\Rightarrow n$ is not divisible by $p^2$, so $n$ is squarefree. $A-1$ divisible by $p-1 \\Rightarrow p(A-1) = n-1-(p-1)$ divisible by $p-1 \\Rightarrow n-1$ divisible by $p-1$. Therefore,\n$$\n1 + \\sum_{k=1}^{n-1} k^{n-1} \\equiv 1 - A \\equiv 0 \\pmod{p},\n$$\nsince $A-1$ divisible by $p$. Because this holds for every prime divisor $p$ of $n$, and $n$ is squarefree, using the Chinese Remainder Theorem, we obtain (1).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40912,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $(m, n)$ for which it is possible to paint each unit square of an $m \\times n$ chessboard either white or black in such a way that, for any unit square of the board, the number of unit squares which are painted the same color as that square and which have at least one common vertex with it (including the square itself) is even.",
    "options": [],
    "answer": "All pairs with mn even (i.e., at least one of m or n is even).",
    "solution": "We shall call a painting satisfying the condition of the problem a *good* painting. We shall also call squares sharing a vertex *neighbors*. Note that every square is a neighbor of itself.\n\nIf $mn$ is even, then there is a good painting: Suppose that the number of rows is even. Then we paint 1st and 2nd rows white, 3rd and 4th rows black, 5th and 6th rows white and so on. This is a good painting.\n\nNow assume that $mn$ is odd and there is a good painting. Either the number of white squares or the number of black squares is odd. Let us assume the first. Consider the set $W$ of ordered pairs of white squares which are not neighbors.\n\nSince each square is a neighbor of itself, $W$ has an even number of elements.\n\nOn the other hand, there is an odd number of white squares and, since the painting is good, for each white square there is an odd number of white squares which are not neighbors of it. This implies that $W$ has an odd number of elements, a contradiction.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 40913,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$, $C$ be three collinear points such that the point $B$ lies between $A$ and $C$. Let $AA'$ and $BB'$ be parallel lines such that the points $A'$ and $B'$ lie on the same side of the line $AB$, and $A'$, $B'$, $C$ are not collinear. Let $O_1$ be the center of the circle passing through the points $A$, $A'$, $C$, and $O_2$ be the center of the circle passing through the points $B$, $B'$, $C$. Find all possible values of the angle $CAA'$, if triangles $A'CB'$ and $O_1CO_2$ have the same area.",
    "options": [],
    "answer": "all values strictly between 0 and 180 degrees except 90 degrees",
    "solution": "If $\\angle CAA'$ is acute, then $\\angle CO_1A' = 2\\angle CAA' = 2\\angle CBB' = \\angle CO_2B'$. And if $\\angle CAA'$ is obtuse, then $\\angle CO_1A' = 2(180^\\circ - \\angle CAA') = 2(180^\\circ - \\angle CBB') = \\angle CO_2B'$. In particular, $\\angle CO_1A' = \\angle CO_2B'$. (It can be easily seen that $S_{CA'B'} \\neq S_{CO_1O_2}$ is not possible if $\\angle CAA' = 90^\\circ$.) Since triangles $CO_1A'$ and $CO_2B$ are both isosceles, it follows that they are similar.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40914,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a(1), a(2), \\dots$ be the infinite sequence defined by:\n$a(1) = a(2) = 1$ and\n$$\na(n) = a(a(n-1)) + a(n - a(n-1)) \\text{ for } n \\ge 3.\n$$\nProve that $a(2n) \\le 2a(n)$ for all $n \\ge 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first prove that:\n$$\n(1) \\qquad a(n+1) - a(n) = 0 \\text{ or } 1\n$$\nfor all positive integers $n$. It is true if $n = 1, 2$, since $a(3) = 2$. Assume that $n \\ge 3$ and (1) is true for all $k \\le n$. In particular, $1 \\le a(k) \\le k$ for all $k \\le n$\n\n$$\na(n+1) - a(n) = a(a(n)) - a(a(n-1)) + a(n+1 - a(n)) - a(n - a(n-1)) = \\\\\n= \\begin{cases} a(n - a(n) + 1) - a(n - a(n)), & \\text{if } a(n) = a(n-1), \\\\ a(a(n-1) + 1) - a(a(n-1)), & \\text{if } a(n) = a(n-1) + 1, \\end{cases}\n$$\nand (1) follows by induction.\n\nNow assume that $a(2n) \\le 2a(n)$ not for all $n \\ge 1$, and let $m$ be the smallest integer for which this inequality is false. Hence, $a(2m) > 2a(m)$ and $a(2(m-1)) \\le 2a(m-1)$. Moreover, if $a(m) = a(m-1)+1$, then $a(2m) > 2a(m) = 2(a(m-1)+1) \\ge a(2(m-1))+2$ and we get a contradiction to (1). Therefore, $a(m) = a(m-1)$.\n\nLet $M = a(m) = a(m-1)$. Since $a(2m) > 2M$ and $a(2m-2) \\le 2M$, we have $a(2m-1) = 2M$ or $2M+1$.\n\nIf $a(2m-1) = 2M$, then $a(2m) = a(2M) + a(2m-2M) \\le 2a(M) + 2a(m-M) = 2a(m)$, which contradicts with our assumption $a(2m) > 2a(m)$.\n\nIf $a(2m-1) = 2M+1$, then $a(2m-2) = 2M$ and $a(2m-1) = a(2M) + a(2m-1-2M) \\le a(2M) + a(2m-2M) \\le 2a(M) + 2a(m-M) = 2a(m)$, which means $2M+1 \\le 2M$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40915,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$, satisfying the identity\n$$\nf(x^2 + xy + f(y)) = (f(x))^2 + x f(y) + y\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = x for all real x",
    "solution": "Using the identity $x^2 + x y = (-x - y)^2 + (-x - y) y$ we obtain the following identity for the function $f$:\n$$\n(f(x))^2 + x f(y) = (f(-x - y))^2 - (x + y) f(y).\n$$\nLet's write down some particular cases of the last identity:\n$$\n(1) \\quad (f(x))^2 + x f(-x) = (f(0))^2 \\text{ (it is obtained by putting } y = -x)\n$$\n$$\n(2) \\quad (f(x))^2 + x f(0) = (f(-x))^2 - x f(0) \\text{ (it is obtained by putting } y = 0).\n$$\n(1) implies $(f(x))^2 + x f(-x) = (f(-x))^2 - x f(x)$ which is equivalent to\n$$\n(f(x) + f(-x))(f(x) - f(-x) + x) = 0\n$$\nSuppose that for some $a \\neq 0$ we have $f(a) + f(-a) + a = 0$. Then by (2) $(f(a))^2 + 2 a f(0) = (f(-a))^2 = (f(a) + a)^2$, it follows that\n$$\n(3) \\qquad 2 f(0) = 2 f(a) + a.\n$$\nEliminating $f(0)$ and $f(-a)$ in (1) by $a, f(a)$, we obtain $a^2 = a^2/4$, hence, $a = 0$.\nThe contradiction says that $f(x) + f(-x) = 0$ for all $x \\neq 0$. Now from (3) we get $f(0) = 0$. The identity (1) is transformed to $f(x)(f(x) - x) = 0$. If we put $x = 0$ in the original identity we obtain $f(f(y)) = y$, which means that $f$ is injective. Hence, for $x \\neq 0$ we have $f(x) \\neq 0$. So, $f(x) = x$, and it obviously satisfies the given identity.\nFor $x = 0$ we get $f(f(y)) = y + (f(0))^2$, which implies that $f$ is injective and subjective. Let $f(a) = 0$, then the substitution $x = y = a$ gives that $f(2 a^2) = a$, which implies that $2 a^2 + (f(0))^2 = 0$ or $a = f(0) = 0$. Particularly, $f(f(y)) = y$.\nSubstitution $y = 0$ now gives $f(x^2) = (f(x))^2$, from which using that $f$ is injective we obtain $f(x) = -f(-x)$.\nSubstitution $y = -x$ gives $f(x)(f(x) - x) = 0$, and also using that $f$ is injective and applying $x \\neq 0 \\Rightarrow f(x) \\neq 0$, we have $f(x) = x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40916,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove the inequality\n$$\n4 \\left( \\sqrt[3]{\\frac{a}{b}} + \\sqrt[3]{\\frac{b}{c}} + \\sqrt[3]{\\frac{c}{a}} \\right) \\le 3 \\left( 2 + a + b + c + \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right)^{2/3}\n$$\nfor positive real numbers $a$, $b$ and $c$ with $abc = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $x$, $y$, $z$ be positive real numbers such that $a = x/y$, $b = y/z$, $c = z/x$. After substitution and simplifying the given inequality is transformed to the following one:\n$$\n4(x + y + z)^{\\sqrt{3}xyz} \\le 3(x + y)^{2/3}(y + z)^{2/3}(z + x)^{2/3}\n$$\nFor triangle with the sides lengths $u = x + y$, $v = y + z$, $w = z + x$ the last inequality gives\n$$\n4p\\sqrt[3]{(p-u)(p-v)(p-w)} \\le 3(uvw)^{2/3}\n$$\nwhere $p$ is a semiperimeter of the triangle. By Heron's formula we obtain that the inequality is equivalent to:\n$$\n(2p)^{2/3} \\le (3\\sqrt{3}\\frac{uvw}{4s})^{2/3} = (3\\sqrt{3}R)^{2/3}\n$$\nwhere $S$ is area and $R$ is a radius of the circumcircle of the triangle. But the inequality\n$$\nu + v + w = 2p \\le 3\\sqrt{3}R$$\nis well-known; there are several ways to prove that.\nBy raising to the third power the given inequality and taking into account the condition $abc = 1$ we eliminate radicals. After simplifying we get the following inequality:\n$$\n114 + 30 \\sum_{cyc} \\left(a + \\frac{1}{a}\\right) + 10 \\sum_{cyc} \\frac{a}{b} \\le 54 \\sum_{cyc} \\frac{b}{a} + 27 \\sum_{cyc} \\left(a^2 + \\frac{1}{a^2}\\right)\n$$\nThe last inequality can be obtained from the following elementary inequalities:\n$$\n2 \\sum_{cyc} \\frac{a}{b} \\le \\sum_{cyc} \\left(a^2 + \\frac{1}{a^2}\\right), \\quad 3 \\le \\sum_{cyc} \\frac{b}{a},\n$$\n$$\na + \\frac{1}{a} \\le \\frac{1}{4} \\left(a^2 + \\frac{1}{a^2} + 6\\right), \\quad 2 \\le a^2 + \\frac{1}{a^2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40917,
    "subject": "Mathematics (Multi-modal)",
    "question": "A subset $S$ of the set $M = \\{1, 2, \\dots, p-1\\}$, where $p$ is a prime number of the kind $12n + 11$ is *essential*, if the product $\\Pi_s$ of all elements of the subset is not less than the product $\\overline{\\Pi}_s$ of all other elements of the set. The difference $\\Delta_s = \\Pi_s - \\overline{\\Pi}_s$ is called the *deviation* of the subset $S$. Determine the least possible remainder of division by $p$ of the deviation of an essential subset, containing $\\frac{p-1}{2}$ elements.",
    "options": [],
    "answer": "2",
    "solution": "Answer: 2.\nLet's consider the subset $S = \\{(p+1)/2, \\dots, p-2, p-1\\}$. Then\n$$\n\\Pi_s = (-1)^{\\frac{p-1}{2}} \\left(\\frac{p-1}{2}\\right)! = - \\left(\\frac{p-1}{2}\\right)! \\equiv a \\pmod{p},\n$$\n$$\n\\overline{\\Pi}_s = \\left(\\frac{p-1}{2}\\right)! \\equiv -a \\pmod{p}.\n$$\nBy the Wilson's theorem $a^2 \\equiv 1 \\pmod{p}$. Hence, two cases are possible:\n**Case 1.** $a \\equiv 1 \\pmod{p}$. Then the deviation of the essential subset $S$ is equal to $\\Delta_s = 2 \\pmod{p}$.\n**Case 2.** $a \\equiv -1 \\pmod{p}$. Then the deviation of the essential subset $T$, where $T$ is obtained from $S$ by substituting $\\frac{p+1}{2}$ to $\\frac{p-1}{2} \\equiv -\\frac{p+1}{2} \\pmod{p}$, is equal to $\\Delta_T = \\Pi_T - \\overline{\\Pi}_T \\equiv 1 - (-1) \\equiv 2 \\pmod{p}$.\nNow we prove that $2$ is the least possible remainder.\nNote that $\\Pi_s \\cdot \\overline{\\Pi}_s = -1$ by the Wilson's theorem again. Since $-1$ is not a quadratic residue by modulo a prime number of the kind $4k+3$, we have certainly $\\Pi_s \\neq \\overline{\\Pi}_s \\pmod{p}$. In other words, the deviation is not divisible by $p$. Suppose to the contrary, let $S$ be a subset with deviation $\\Pi_s - \\overline{\\Pi}_s = 1$. Then $\\Pi_s \\equiv m+1 \\pmod{p}$, $\\overline{\\Pi}_s \\equiv m \\pmod{p}$ for some $m \\in \\{2, 3, \\dots, p-3\\}$ and $m^2 + m \\equiv -1 \\pmod{p}$. This implies $m^3 \\equiv 1 \\pmod{p}$ and by the Fermat's (little) theorem $p-1$ is divisible by $3$, which is a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40918,
    "subject": "Mathematics (Multi-modal)",
    "question": "A family $L$ of $2006$ lines on the plane is given in such a way that it doesn't contain parallel lines and it doesn't contain three concurrent lines. We say that the line $l_1 \\in L$ is *bounding* the line $l_2 \\in L$, if all intersection points of the line $l_2$ with other lines from $L$ lie on the one side of the line $l_1$. Prove that in the family $L$ there are two lines $l$ and $l'$ such that the following two conditions are hold simultaneously:\n1) the line $l$ is bounding the line $l'$;\n2) the line $l'$ is not bounding the line $l$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume the contrary, i.e. there aren't lines $l$ and $l'$, satisfying the conditions 1) and 2).\nLet's choose an arbitrary $l \\in L$. There are two points $a, b \\in l$ such that $a, b$ are intersection points of $l$ with lines from $L$, and all intersection points of the line $l$ with lines from $L$ lie on the segment $[a, b]$. This $[a, b]$ is called a *big segment* (of the line $l$).\nThere is another line, say $l'$, that passes through the endpoint $a$ of the big segment $[a, b]$. Then, by the assumption, this $a$ will be an endpoint of the big segment of the line $l'$.\nDenote by $A$ the set of all endpoints of big segments; and if $[a, b]$ is a big segment then we write $aRb$ and say that $a$ and $b$ are in the relation $\\mathcal{R}$.\n**Lemma 1.** Let $\\mathcal{R}$ be an irreflexive, symmetric relation on the finite set $A$, i.e. $x\\mathcal{R}y \\Rightarrow x \\neq y \\& y\\mathcal{R}x$. Suppose that for any point $a \\in A$ there are exactly two points $b \\in A$ such that $a\\mathcal{R}b$. Then $A$ is partitioned into a finite $\\mathcal{R}$-cycles of the length $\\ge 3$.\n**Proof.** is well-known.\nBut in our situation there is an unique $\\mathcal{R}$-cycle of the odd length.\n**Lemma 2.** $\\mathcal{R}$-cycles have odd lengths.\n\nSOLUTIONS\n**Proof.** Let $a_0\\mathcal{R}a_1\\mathcal{R}a_2\\mathcal{R}\\dots a_n\\mathcal{R}a_0$ be a cycle of the length $n+1$ and $a=a_0, b=a_1$. The points $a_i$ and $a_{i+1}$ lie on different sides of the line $l_{ab}$, $2 \\le i \\le n-1$, since any two big segments must intersect. Also, $a_n$ and $a_2$ should lie on the same side, otherwise the big segments $[a_n, a_0]$ and $[a_1, a_2]$ would not intersect. So, $n$ is even. $\\square$\n**Lemma 3.** There is an unique $\\mathcal{R}$-cycle.\n**Proof.** Let $a_0\\mathcal{R}a_1\\mathcal{R}a_2\\mathcal{R}\\dots a_n\\mathcal{R}a_0$ be a cycle. Assume that there exists a big segment $[b, c]$ outside this cycle. The line of segment $[b, c]$ divides the plane into 2 semiplanes. Since big segments should intersect, adjacent (by $\\mathcal{R}$) points of that cycle should lie on a different semiplanes. But it is impossible, because cycle contains an odd number of points. $\\square$\nLemmas give a contradiction with assumptions, because $2006$ is even number.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40919,
    "subject": "Mathematics (Multi-modal)",
    "question": "The numbers $2$, $3$, $5$, $\\ldots$, $2003$ are written on the board (here listed all prime numbers from the segment $[2; 2007]$). By operation of *simplification* we call a changing of two numbers $a$, $b$ by maximal prime number not exceeding $\\sqrt{a^2 - ab + b^2}$. At the beginning student removes the number $q$, $2 < q < 2003$, and then repeats the simplification operation until the board contains at least two numbers. Find the maximal and minimal possible values of the number which he obtain at the end. How these values depend on $q$?",
    "options": [],
    "answer": "Maximum: 1999, independent of the initial removal. Minimum: 2 if the removed prime q ≥ 17; otherwise 3 for q ∈ {3, 5, 7, 11, 13}.",
    "solution": "Note that $a < \\sqrt{a^2 - ab + b^2} < b$ if $a < b$. That's why simplification of two successive prime numbers is equivalent to removing the maximal (from that two) number. If between prime numbers $a < b$ there is exactly one prime number $c$, then operation of simplification to pair $(a, b)$ gives $a$ or $c$. Hence, the \"hole\" between $a$ and $b$ can vanish or move down to one \"step\".\n\nThe maximal possible value is $1999$, and this value doesn't depend on first removed number $2 < q < 2003$. The algorithm of getting this number can be described as a repeating the operation of simplification to the maximal numbers of the set, except the number $2003$. Before the last operation board will have a pair $2$, $2003$ or $3$, $2003$. And last simplification gives $1999$.\n\nAlgorithm of getting the minimal number can be described as a repeating the operation of simplification to the maximal numbers of the set. If the \"hole\" obtained by removing the number $q$ disappear at some step, then we get the minimal number $2$, else $3$.\n\nIt is easy to see that the \"hole\" vanishes if $q \\ge 17$. In that case the minimal number is equal to $2$. On the other hand, if $3 \\le q \\le 13$ then easy analysis shows that it vanishes for all steps of student; in this case, the minimal number is equal to $3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40920,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle $\\omega$ of the triangle $ABC$ touches the side $BC$ at point $K$. Let's draw a circle passing through $B$, $C$ and touching $\\omega$ at $S$. Prove that the line $SK$ passes through the center of the excircle touching the side $BC$ of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The lengths of the sides of $ABC$ denote by $a, b, c$. Let $I$ be the center of the incircle, $r$ its radius, $M$ the midpoint of $BC$. Denote by $r_a$ the radius of the excircle touching $BC$. Let $L$ be the point of tangency of excircle with $BC$. $\\omega_0$ be the circle with center $O$, passing through $B$ and $C$, touching $\\omega$. W.l.o.g. assume that $b \\ge c$.\n\nFirstly, $SK$ is a bisector of the angle $CSB$, since the homothety with center $S$, which transforms the incircle to $\\omega_0$, transforms $BC$ to some line touching $\\omega_0$ at $E$ the midpoint of $BC$.\n\nIt is well-known fact that $K$ and $L$ are symmetric with respect to $M$. So, if we prove that $ME = I_aL/2$ then we can immediately conclude that $K, E, I_a$ are collinear.\n\nLet $OT \\perp IK$, $OM = |x|$, where $x \\ge 0$, if $O$ lies below $BC$, and $x < 0$ otherwise. Let $R$ be the radius of $\\omega_0$. Then $ME = R - x$. Consider the right triangle $OTI$, $OI = R - r$, $OT = (b - c)/2$, $IT = r - x$. By Pythagorean theorem\n\n$$\n(1) \\qquad (R-r)^2 = (r-x)^2 + (b-c)^2/4\n$$\nFrom the right triangle $OMB$ we have $R^2 - x^2 = a^2/4$. Then using (1) we get $ME = R-x = (p-b)(p-c)/(2r)$, where $p$ is a semiperimeter of $ABC$. Then it is sufficient to prove that\n$$\n(2) \\qquad r_a = \\frac{(p-b)(p-c)}{r}\n$$\nWe have $r = S/p$. On the other hand, $S = r_a(p-a)$. Then (2) true iff $S^2 = p(p-a)(p-b)(p-c)$, which is Heron's formula.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40921,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the maximal value of the real number $M$ such that for all positive real numbers $a$, $b$, $c$ the following inequality holds:\n$$\na^3 + b^3 + c^3 - 3abc \\ge M(|a-b|^3 + |b-c|^3 + |c-a|^3)\n$$",
    "options": [],
    "answer": "M = 1/2",
    "solution": "Answer: $M = \\frac{1}{2}$.\nThe inequality $a^3 + b^3 + c^3 - 3abc \\ge M(|a-b|^3 + |b-c|^3 + |c-a|^3)$ for $M = 1/2$ implies from\n$$\n2(a^3 + b^3 + c^3 - 3abc) = (a + b + c)((a - b)^2 + (b - c)^2 + (c - a)^2)\n$$\nand from obvious inequalities\n$$\n(x + y + z)(x - y)^2 \\ge |x - y|^3\n$$\n\nThe value $M = 1/2$ is maximal, because for very large values of $a$ and very small values of $b$, $c$, the value of $a^3 + b^3 + c^3 - 3abc$ can be very close to $a^3$, and the value of $|a-b|^3 + |b-c|^3 + |c-a|^3$ can be very close to $2a^3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40922,
    "subject": "Mathematics (Multi-modal)",
    "question": "The set $f_1, f_2, \\dots, f_n$ of polynomials with real coefficients is called *special*, if for all distinct $i, j, k \\in \\{1, 2, \\dots, n\\}$ the polynomial $\\frac{2}{3}f_i + f_j + f_k$ has no real roots, but for all distinct $p, q, r, s \\in \\{1, 2, \\dots, n\\}$ the polynomial $f_p + f_q + f_r + f_s$ has a real root.\n1) Give an example of the special set of four polynomials with nonzero sum.\n2) Is there exists a special set of five polynomials?",
    "options": [],
    "answer": "1) x^2 + x + 100, x^2 + x + 100, -x^2 + x - 200, 1. 2) No; a special set of five polynomials does not exist.",
    "solution": "1) The polynomials $x^2 + x + 100$, $x^2 + x + 100$, $-x^2 + x - 200$, $1$ give a needed example.\n\n2) To the contrary, assume that there exists a special set of polynomials $f_1, f_2, f_3, f_4, f_5$ with real coefficients. Consider a complete graph with a set of vertices $\\{1, 2, 3, 4\\}$. An edge $(i, j)$ is called *positive* (*negative*), if the polynomial $\\frac{2}{3}f_5 + f_i + f_j$ is positive (*negative*). By the assumption in this graph there is no any positive (*negative*) triangles. Indeed, if for example $\\frac{2}{3}f_5 + f + g > 0$, $\\frac{2}{3}f_5 + g + h > 0$ and $\\frac{2}{3}f_5 + f + h > 0$, then $f_5 + f + g + h > 0$ which is a contradiction with the definition of a special set. The next lemma can be easily proved.\n\n**Lemma.** In a two-colored complete graph with four vertices without one-colored triangles there is a pair of skew (crossed) edges of every color.\n\nUsing the lemma we have conflicting inequalities:\n$$\nf_1 + f_2 + f_3 + f_4 + \\frac{4}{3}f_5 > 0,\n$$\n$$\nf_1 + f_2 + f_3 + f_4 + \\frac{4}{3}f_5 < 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40923,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be positive integers such that $d$ divides $a^{2b} + c$ and $d \\ge a + c$. Prove that $d \\ge a + \\sqrt[2b]{a}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have $a^{2b} + c \\equiv (d - a)^{2b} + c \\pmod d$, since\n$$\na^{2b} - (d - a)^{2b} = (a^2 - (d - a)^2) \\times \\\\\n\\times (a^{2(b-1)} + a^{2(b-2)}(d - a)^2 + \\dots + a^2(d - a)^{2(b-2)} + (d - a)^{2(b-1)})\n$$\n$$\na^{2b} - (d - a)^{2b} \\vdots (a + (d - a)) = d.\n$$\nWe deduce consequently $(d - a)^{2b} + c \\ge d$\n$$\n\\Rightarrow (d - a)^{2b} \\ge (d - c) \\ge a \\Rightarrow d - a \\ge \\sqrt[2b]{a} \\Rightarrow d \\ge a + \\sqrt[2b]{a}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40924,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$ let $A_0, B_0, C_0$ are the midpoints of the sides $BC, CA, AB$, respectively, and $A_1, B_1, C_1$ are the midpoints (by length) of the broken lines $BAC, CBA, BCA$, respectively. Prove that the lines $A_0A_1, B_0B_1, C_0C_1$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nW.l.o.g. assume that $AC \\le AB$. Then $A_1$ belongs to the segment $AB$, moreover, it lies between $A$ and $C_0$. Then notice that $AC + AA_1 = BA_1 \\iff AC + AC_0 - A_1C_0 = C_0B + A_1C_0 \\iff A_1C_0 = \\frac{1}{2}AC \\Rightarrow A_1C_0 = A_0C_0$ i.e. $\\angle C_0A_1A_0 = \\angle A_1A_0C_0 \\Rightarrow \\angle A_1A_0C_0 = \\angle A_1A_0B_0$ since $AB \\parallel A_0B_0$. So, $A_0A_1$ is bisector of the angle $\\angle C_0A_0B_0$.\n\nSimilarly, $B_0B_1$ is bisector of $\\angle C_0B_0A_0$ and $C_0C_1$ is bisector of $\\angle A_0C_0B_0$. So the lines $A_0A_1, B_0B_1$ and $C_0C_1$ intersect in the incenter of the triangle $A_0B_0C_0$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40925,
    "subject": "Mathematics (Multi-modal)",
    "question": "A (non-oriented) graph (without loops) with $2n$ vertices and $2n(n-1)$ edges is given, $n > 1$. Prove that some vertices and edges of the graph can be coloured in red in such way that each red edge connects red vertices and each red vertex belongs to exactly $n$ red edges.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider a regular $2n$-gon (inscribed in a unit circle), which we label $n$ segments from the set of all sides and all diagonals.\nThe problem is equivalent to the following one: prove that some of unlabelled segments (including endpoints) can be coloured in red in such way that each red vertex belongs to exactly $n$ red segments.\nWe define *k-segment* as a segment spanning an arc of the length $\\pi k/n$.\nThe following cases are possible:\n*Case 1.* Let some two labelled segments have a common vertex. Then we choose this vertex and other $n-2$ vertices in such way that at least a vertex is chosen in each labelled segment (so, totally we choose $1+(n-2)=n-1$ vertices). The other vertices will form a complete graph (i.e. a clique) with $n+1$ vertices: each two vertices are connected by an *unlabeled* segment. Colour these segments in red and we are done.\n*Case 2.* The labelled segments have no common points.\n\n*Subcase 2.1.* $n$ is even.\nW.l.o.g. (using an appropriate numeration of vertices) we can assume that all longest diagonals (i.e. $n$-segments) are labelled. Colour all $k$-segments for $k = 1, 2, \\dots, n/2$ in red and we are done.\n*Subcase 2.2.* $n$ is odd.\nW.l.o.g. (using an appropriate numeration of vertices again) we can assume that $n$ 1-segments are labelled (i.e. $n$ sides of the polygon, in alternating manner). Colour all unlabeled $k$-segments for $k = 1, 2, \\dots, (n+1)/2$ in red and we are done.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 40926,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all polynomials $P(x)$ with real coefficients such that for any rational (number) $r$ the equation $P(x) = r$ has a rational solution.",
    "options": [],
    "answer": "P(x) = a x + b, where a and b are rational and a ≠ 0",
    "solution": "Answer: $P(x) = a x + b$, where $a \\neq 0$, $b$ are rational numbers.\n\nObviously, $P(x)$ is not constant, so $\\deg P > 0$.\n\nFirst of all we prove that the coefficients of $P(x)$ are rational numbers. Let $\\deg P = n$, consider $n+1$ distinct rational numbers $r_1, r_2, \\dots, r_{n+1}$ and rationals $x_1, x_2, \\dots, x_{n+1}$ such that $P(x_i) = r_i, \\forall 1 \\le i \\le n+1$. Then by the Lagrange's interpolation theorem we have\n$$\nP(x) = \\sum_{i=1}^{n+1} r_i \\frac{\\prod_{j \\neq i} (x - x_j)}{\\prod_{j \\neq i} (x_i - x_j)}\n$$\nand it easily follows all coefficients of the polynomial are rational numbers.\n\nMultiplying by appropriate integer we can assume that $P(x)$ has integer coefficients (and, of course, it satisfies the condition of the problem).\n\nLet $a_n$ be the leading coefficient of $P(x)$, w.l.o.g. $a_n > 0$ and let $a_0$ be the last coefficient of $P(x)$.\n\nConsider numbers $r$ of the kind $r = p_1 + a_0$, where $p_1$ is a prime number. There exist an integer $p_r$ and a positive integer $q_r$ with $(p_r, q_r) = 1$ such that $P(p_r/q_r) = r$. Then it is easy to see that $p_r \\mid r - a_0 = p_1$ and $q_r \\mid a_n$, hence, $p_r \\in \\{-1, 1, -p_1, p_1\\}$.\n\nThere are infinitely many different $r$, so taking them into consideration we obtain infinitely many different corresponding pairs $(p_r, q_r)$.\n\nWe state that among these pairs there are infinitely many $(p_r, q_r)$, for which $p_r \\in \\{-p_1, p_1\\}$, since otherwise $|p_r/q_r| \\le 1$, but $r$ can take very large values; then at some moment the equation $P(p_r/q_r) = r$ is impossible.\n\nW.l.o.g. assume that there are infinitely many pairs $(p_r, q_r)$ with $p_r = p_1$ (if sign is negative solution is similar). Since the number of divisors of $a_n$ is finite, there are infinitely many prime numbers $p_1$ and some constant $d$ (the divisor of $a_n$) with $P(p_1/d) = p_1 + a_0$.\n\nThen consider another polynomial $f(x) = P(x/d)$. So, there are infinitely many numbers $a \\in \\mathbb{Z}$ for which $f(a) = a + a_0$ and, therefore, $f(x) \\equiv x + c$. Finally, we have that $P(x) = d x + c$ is a linear polynomial.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40927,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$ the angle bisectors of $A$ and $C$ intersect the sides $BC$ and $AB$ at the points $A_1$ and $C_1$, respectively, and the circumcircle of the triangle $ABC$ at the points $A_2$ and $C_2$, respectively. Let $K$ be the point of intersection of $A_1C_2$ and $C_1A_2$, and $I$ be the incenter of triangle $ABC$. Prove that $KI$ passes through the midpoint of $AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's define notations of the points and angles as in the figure. Then\n$$\n\\angle C_2AB = \\angle C_2CB = \\angle C_2CA = \\angle C_2A_2A = \\gamma,\n$$\n$$\n\\angle A_2CB = \\angle A_2AB = \\angle A_2AC = \\angle A_2C_2C = \\alpha\n$$\nand\n$$\n\\angle C_2AC = \\angle A_2CA = \\angle ABC = 2\\beta.\n$$\n\nBy the law of sines in $\\Delta C_2AC_1$ and $\\Delta C_2AI$ we easily obtain:\n$$\n\\frac{C_2C_1}{C_1I} = \\frac{\\sin \\gamma}{\\sin \\alpha} \\cdot \\frac{\\sin \\angle C_2IA}{\\sin \\angle AC_2I} = \\frac{\\sin \\gamma}{\\sin \\alpha} \\cdot \\frac{\\sin (\\alpha + \\gamma)}{\\sin 2\\beta}.\n$$\nSimilarly, in $\\Delta IA_2C$ we get:\n$$\n\\frac{IA_1}{A_1A_2} = \\frac{\\sin \\gamma}{\\sin \\alpha} \\cdot \\frac{\\sin \\angle IA_2C}{\\sin \\angle A_2IC} = \\frac{\\sin \\gamma}{\\sin \\alpha} \\cdot \\frac{\\sin 2\\beta}{\\sin (\\alpha + \\gamma)}.\n$$\nThen\n$$\n(1) \\qquad \\frac{C_2C_1}{C_1I} \\cdot \\frac{IA_1}{A_1A_2} = \\left( \\frac{\\sin \\gamma}{\\sin \\alpha} \\right)^2\n$$\nBy Ceva's theorem in $\\Delta C_2A_2I$ we have\n$$\n(2) \\qquad \\frac{C_2N}{NA_2} = \\frac{C_2C_1}{C_1I} \\cdot \\frac{IA_1}{A_1A_2} = \\left( \\frac{\\sin \\gamma}{\\sin \\alpha} \\right)^2\n$$\nIn $\\Delta C_2IN$ and $\\Delta A_2IN$, using (2) we obtain\n$$\n(3) \\qquad \\frac{C_2IN}{NIA_2} = \\frac{C_2N \\sin \\alpha}{N_2A \\sin \\gamma} = \\left( \\frac{\\sin \\gamma}{\\sin \\alpha} \\right)^2 \\cdot \\frac{\\sin \\alpha}{\\sin \\gamma} = \\frac{\\sin \\gamma}{\\sin \\alpha}\n$$\nIn $\\Delta AIM$ and $\\Delta CIM$, using (3) we get\n$$\nAM = \\frac{MI \\sin \\alpha}{\\sin \\angle AIM} = \\frac{MI \\sin \\alpha}{\\sin \\angle NIA_2} = \\frac{MI \\sin \\gamma}{\\sin \\angle C_2IN} = \\frac{MI \\sin \\gamma}{\\sin \\angle MIC} = MC\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40928,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) In this country there are $1024$ cities, numbered with integers from $0$ to $1023$;\nb) Two cities with numbers $m$ and $n$ are connected by a single road if and only if the binary notations of $m$ and $n$ differ in exactly one digit;\nc) During the tourist's trip in that country, $8$ roads will be closed for repairing.\n\nProve that the tourist can organize a closed path by working roads of the Compland which passes through every city exactly once.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Remind that $n$-dimensional *binary cube* (boolean) is a graph with vertices labeled by binary sequences of length $n$, and there is an edge between any two vertices if and only if their sequences differ only in one corresponding coordinate (digit).\n\nThen our problem is equivalent to the following: prove that if from a $10$-dimensional binary cube we remove $8$ edges, then we always obtain a *hamiltonian* graph (i.e., a graph having a hamiltonian cycle – a closed path passing through all vertices of the graph exactly once).\n\nBy induction on $n$ let's prove a more general statement:\nIf from an $n$-dimensional binary cube, $n > 1$, we remove arbitrary $(n-2)$ edges, then the obtained graph is hamiltonian.\n\n*Proof:* For $n=2$ the statement is obvious, there is nothing to remove.\n\nAssume that for $n = k > 1$ the statement is true, and prove it for $n = k + 1$.\n\nLet $V = \\{(a_0, a_1, \\dots, a_k) : a_i \\in \\{0, 1\\}, 0 \\le i \\le k\\}$ be the set of vertices of the $(k+1)$-dimensional cube $C = \\langle V, E \\rangle$, where $E$ is a set of edges. Let's take an arbitrary removed edge $e \\in E$. W.l.o.g. we may assume that it connects vertices $u = (0, b_1, \\dots, b_k)$ and $v = (1, b_1, \\dots, b_k)$.\n\nThen consider the sets of vertices\n$$\nV^0 = \\{(0, a_1, \\dots, a_k) : a_i \\in \\{0, 1\\}, 1 \\le i \\le k\\}\n$$\nand\n$$\nV^1 = \\{(1, a_1, \\dots, a_k) : a_i \\in \\{0, 1\\}, 1 \\le i \\le k\\},\n$$\nwhich form a partition of $V$; and corresponding subgraphs $C^0 = \\langle V^0, E \\setminus V^0 \\rangle$ and $C^1 = \\langle V^1, E \\setminus V^1 \\rangle$.\n\nNote that:\n1) Subgraphs $C^0$ and $C^1$ are identical to $k$-dimensional binary cubes;\n2) A mapping $(0, a_1, \\dots, a_k) \\to^f (1, a_1, \\dots, a_k)$ is an isomorphism from $C^0$ to $C^1$.\n\nLet $D \\subset E$ be the set of all removed edges, $|D| = n - 2 = k - 1$. Then we have a partition $D = D_0 \\cup D_1 \\cup D_2$, where $D_i$ is the set of edges removed from subgraph $C^i$, $i = 0, 1$, and $D_2$ is the set of removed edges which \"connect\" $C^0$ and $C^1$. In particular, $e \\in D_2$.\n\nSince $|f(D_0) \\cup D_1| \\le k - 2$, then by the induction hypothesis in $C^1$ there is a hamiltonian cycle $v = v_1, v_2, \\dots, v_{2^k}$, which does not use the edges from $f(D_0) \\cup D_1$. Then $u = u_1, u_2, \\dots, u_{2^k}$ is a hamiltonian cycle in $C^0$, where $f(u_i) = v_i$ for $1 \\le i \\le 2^k$, not passing through the edges of $D_0$.\n\n$$\nu = u_1, u_2, \\dots, u_i, v_i, v_{i-1}, \\dots, v_1, v_{2^k}, v_{2^k-1}, \\dots, v_{i+1}, u_{i+1}, \\dots, u_{2^k}$$\nis a hamiltonian cycle in $C$, not passing through the edges of $D$. The statement is proved. $\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40929,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every prime number $p$ there exists infinitely many 4-tuples $(x, y, z, t)$ of pairwisely distinct positive integers such that the number\n$$\n(x^2 + pt^2)(y^2 + pt^2)(z^2 + pt^2)\n$$\nis a square of an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Firstly, note that the equation $x^2 - py^2 = 1$ has infinitely many solutions in positive integers. (Pell's equation)\nThen for every prime number $p$ there exist infinitely many positive integers $s$ and $t$ such that $s^2 - 1 = pt^2$.\nPutting $x = s^2 - 1$, $y = s + 1$, $z = s - 1$ we have\n$$\n(x^2 + s^2 - 1)(y^2 + s^2 - 1)(z^2 + s^2 - 1) = \\\\\n= (s^2 - 1)s^2(s + 1)(2s)(s - 1)(2s) = ((s^2 - 1)2s^2)^2.\n$$\nIt is remained to check that $x, y, z \\ne t$.\nNote that $x = s^2 - 1 = pt^2 \\ne t$.\nIf $y = s + 1 = t$ then\n$$\n(s-1)(s+1) = pt^2 = p(s+1)^2 \\Rightarrow s-1 = p(s+1) > s+1 > s-1.\n$$\nIf $z = s - 1 = t$, then\n$$\n(s-1)(s+1) = pt^2 = p(s-1)^2 \\Rightarrow s+1 = p(s-1) \\ge 2(s-1) > s+1,\n$$\nfor $s > 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40930,
    "subject": "Mathematics (Multi-modal)",
    "question": "In convex quadrilateral $ABCD$\n$$\n\\angle ADB + \\angle ACB = \\angle CAB + \\angle DBA = 30^\\circ \\text{ and } AD = BC.\n$$\nProve that segments $DB$, $CA$ and $DC$ form the sides of the right triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40931,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N = 2010! + 1$. Prove that\n\na) $N$ is not divisible by $4021$;\n\nb) $N$ is not divisible by $2027$, $2029$, $2039$;\n\nc) $N$ has a prime divisor greater than $2050$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40932,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive real numbers $a$, $b$, $c$, $d$ satisfy:\n$$\na(c^2 - 1) = b(b^2 + c^2) \\text{ and } d \\le 1.\n$$\nProve that\n$$\nd(a\\sqrt{1-d^2} + b^2\\sqrt{1+d^2}) \\le \\frac{(a+b)c}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40933,
    "subject": "Mathematics (Multi-modal)",
    "question": "In countries of the Silk Road there are finite number of cities and some of them are connected by directed roads (one pair of cities can have several roads, moreover they can have an opposite direction). It is known that every two paths by these roads form the city $A$ to the city $B$ use at least one same road. Prove that some of the roads is common part for all the paths from $A$ to $B$ ($A$ and $B$ are two given cities).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40934,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be real numbers satisfying\n$$\n|(a - b)(b - c)(c - a)| = 1.\n$$\nFind minimum value of $|a| + |b| + |c|$.",
    "options": [],
    "answer": "2^{2/3}",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40935,
    "subject": "Mathematics (Multi-modal)",
    "question": "Acute angled triangle $ABC$ ($AC > CB$) has circumscribed circle with point $N$ in the middle of the arc $ACB$. Let points $A_1$ and $B_1$ be feets of perpendiculars to the line $NC$, drawn from the points $A$ and $B$ respectively (the segment $NC$ lies inside $A_1B_1$). The height $A_1A_2$ of triangle $A_1AC$ and the height $B_1B_2$ of triangle $B_1BC$ intersect at point $K$. Prove that $\\angle A_1KN = \\angle B_1KM$, where $M$ is the middle of the segment $A_2B_2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40936,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ be positive integers and function $f: \\mathbb{N} \\to \\mathbb{N}$ so that for each positive integer $n$, the number $f(n+a)$ is divisible by $f([\\sqrt{n}] + b)$. Prove that for each positive integer $n$ there exist $n$ pairwise distinct and pairwise relatively prime positive integers $a_1, a_2, \\dots, a_n$ so that the number $f(a_i+1)$ is divisible by $f(a_i)$ for all $i = 1, 2, \\dots, n-1$. (Here $[x]$ is an integer part of $x$, i.e. greatest integer not exceeding $x$; $\\mathbb{N}$ the set of positive integers.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40937,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(n)$ be the number of ways of partitioning a positive integer number $n$ as a sum of powers of $2$, where the order of summands is not important. For example, $P(5) = 4$, since $5 = 4+1 = 2+2+1 = 2+1+1+1 = 1+1+1+1+1$. Prove that for each positive integer $n$ the following identity holds:\n$$\nP(n) + (-1)^{a_1}P(n-1) + (-1)^{a_2}P(n-2) + \\dots + (-1)^{a_{n-1}}P(1) + (-1)^{a_n} = 0,\n$$\nwhere $a_k$ is the number of ones in the binary notation of $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40938,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $Q$ be a square of size $12 \\times 12$ given on an infinite cell grid. Student wants to paint some (not necessary all!) cells with seven colors (each cell with one color), such that no two three-cell rectangles (out of 288) with centers in $Q$ are colored identically. Is this possible?\n(Two three-cell rectangles are colored identically, if it is possible to shift and/or rotate so that each cell of a rectangle matches the same color cell of the other rectangle.)",
    "options": [],
    "answer": "impossible",
    "solution": "Answer: impossible.\nSuppose, that there is a coloring, that satisfy above conditions. Let's count the number of different colorings of $1 \\times 3$ rectangle using 8 colors.\nLet's fix the center cell of rectangle, it can be colored in 8 ways. Other 2 cells can be colored in 8 ways, if they have the same color. If not, then its $C_8^2 = \\frac{7 \\cdot 8}{2} = 28$. So the number of ways to color edge cells is $8 + 28 = 36$. Therefore, total number of different colorings of $1 \\times 3$ rectangle is $8 \\cdot 36 = 288$. It means that every coloring appears exactly once.\nSuppose that there are $k$ white cells in $Q$. Then number of different rectangles with white center cell equals $2k$, hence $2k = 36$ or $k = 18$.\nOn the one hand, let's count the number of rectangles with at least one non-center cell. If both of them are white, then the number of such rectangles equal to 8, if exactly one - number is $7 \\cdot 8 = 56$ (7 ways to color other non-center cell, 8 ways to color center). Means that there are $8 + 56 = 64$ such rectangles in total. On the other hand, we have 48 rectangles with white non-center cell, that lie outside $Q$. Also, for every white cell inside of $Q$, at least one among two vertical and at least one among two horizontal rectangles with white non-center cell is fully lie in $Q$, moreover every such rectangle can be counted no more than twice. Hence, there are at least 18 rectangles in $Q$ with white non-center cell. Then total number of such rectangles at least $48 + 18 = 66 > 64$ - contradiction.\nAs in first solution, let's notice that there are 288 colorings of 3-cell rectangle with 8 colors. Therefore, every coloring appears exactly once (if it is possible).\nIn particular, each of 7 colorings of type «red - red - not red» should appear exactly once. On the other hand, notice that each row and each column contain even number of such rectangles (every continuous block of exactly two red cells gives exactly two rectangles). Therefore, total number of such rectangles should be even, and cannot be equal to 7.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40939,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral inscribed in a circle $\\omega$. Its diagonals $AC$ and $BD$ intersect at the point $O$. Let $E$ and $F$ be points on the segments $AO$ and $DO$ respectively. The line $EF$ intersects $\\omega$ at the points $E_1$ and $F_1$. Circumcircles of triangles $ADE$ and $BCF$ intersect the line $EF$ at the points $E_2$, $F_2$ respectively. Prove that $E_1E_2 = F_1F_2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is enough to show that points $E_2$ and $F_2$ are symmetrical with respect to the middle of the chord $E_1F_1$. In order to ignore cases of mutual arrangement of points, lines and circles, we use oriented angles. Let $M$ be the center of circle $\\omega$. Denote the intersection point of lines $DE_2$ and $CF_2$ as $K$, and let $\\angle(AD, AC) = \\alpha$. Then lines $E_2F_2$, $KE_2$ and $KF_2$ form an isosceles triangle, since $\\angle(E_2K, E_2F_2) = \\angle(F_2E_2, F_2K) = \\alpha$. Then, the exterior angle at vertex $K$ in this triangle equals $2\\alpha$. Since points $K$ and $M$ lie on one side of the line $CD$ and $\\angle CMD = 2\\alpha$, then points $C$, $M$, $D$ and $K$ lie on one circle. Since triangle $CMD$ is isosceles, we have: $\\angle(KM, KD) = \\angle(CM, CD) = 90^\\circ - \\alpha$ or $\\angle(KM, F_2E_2) = \\angle(KM, KD) + \\angle(KD, F_2E_2) = 90^\\circ$. Then, we have that $KM$ passes through the center of circle $\\omega$ and is perpendicular to the chord $E_1F_1$, as well as splits segment $E_2F_2$ in half. Hence, pairs of points $E_2, F_2$ and $E_1, F_1$ are symmetrical with respect to the middle of the chord $E_1F_1$. Proved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40940,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that among every 42 numbers from the interval $[1, 10^6]$ it is possible to choose four numbers in such a way that for any permutation $(a, b, c, d)$ of these numbers, the following inequality holds:\n$$\n25(ab + cd)(ad + bc) \\geq 16(ac + bd)^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $1 \\leq x_1 \\leq x_2 \\leq \\dots \\leq x_{42} \\leq 10^6$ be the given numbers. Suppose that $x_{2k+1}x_{2k+2} > 4x_{2k}x_{2k-1}$ for every $k = 1, 2, \\dots, 20$. Multiplying all these inequalities for every $k = 1, 2, \\dots, 20$ we obtain $x_{41}x_{42} > 4^{20}x_1x_2 \\geq 4^{20}$, hence $x_{42} > 2^{20} > 10^6$, contradiction. Therefore, there is $k$ ($1 \\leq k \\leq 20$) such that $x_{2k+2}x_{2k+1} \\leq 4x_{2k}x_{2k-1}$ (*). Let's prove that $(x_{2k-1}, x_{2k}, x_{2k+1}, x_{2k+2})$ satisfies the constraints. In fact, let $(a, b, c, d)$ be some permutation of these numbers. From (*) it follows that $4ac \\geq bd$, $4bd \\geq ac$. Then\n$$\n\\begin{aligned}\n& 25(ab + cd)(ad + bc) - 16(ac + bd)^2 = \\\\\n& \\quad 25(ac(b^2 + d^2) + bd(a^2 + c^2)) - 100abcd + (68abcd - 16a^2c^2 - 16b^2d^2) = \\\\\n& \\quad 25ac(b-d)^2 + 25bd(a-c)^2 + 4(4ac-bd)(4bd-ac) \\geq 0,\n\\end{aligned}\n$$\nproved.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 40941,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$, points $H, L, K$ are chosen on the sides $AB, BC, AC$, respectively, so that $CH \\perp AB$, $HL \\parallel AC$, $HK \\parallel BC$. In the triangle $HBL$, let $P$ and $Q$ be feets of altitudes from $H$ and $B$, respectively. Prove that in the triangle $AKH$, the feets of altitudes from $A$ and $H$ lie on the line $PQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose $PQ$ intersects $AK$ and $HK$ at $R$ and $S$, respectively. From the fact that $P$ and $Q$ lie on a circle with diameter $HB$ and from $HL \\parallel AC$, it follows that $\\angle LPQ = \\angle BHQ = \\angle BAR$, i.e. the points $A, R, P$ and $B$ lie on the same circle. Therefore, $CP \\cdot CB = CR \\cdot CA$. Also from the right triangle $BCH$, we have $CH^2 = CP \\cdot CB$, implying $CH^2 = CR \\cdot CA$ and hence $HR \\perp AC$.\n\n![](attached_image_1.png)\n\nIt remains to prove that $AS \\perp HK$. From the fact that $A, R, P$ and $B$ lie on one circle and $HK \\parallel BC$ it follows that $\\angle CRP = \\angle CBA = \\angle KHA$, i.e. the points $A, R, S$ and $H$ lie on the same circle. Since $\\angle ARH = 90^\\circ$, we have $\\angle ASH = 90^\\circ$ and $AS \\perp HK$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40942,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that both equations\n$$\nf(x+1) = 1 + f(x) \\text{ and } f(x^4 - x^2) = f(x)^4 - f(x)^2\n$$\nsimultaneously hold for all real $x$. ($\\mathbb{R}$ is the set of real numbers.)",
    "options": [],
    "answer": "f(x) = x for all real x",
    "solution": "Answer: $f(x) = x$ for all real $x$.\nFrom the first equation we have by induction that\n$$\nf(x+n) = f(x) + n \\quad (1)\n$$\nfor every real $x$ and integer $n$. Since $x^4 - x^2 = (x^2 - \\frac{1}{2})^2 - \\frac{1}{4}$, for every $y \\ge -\\frac{1}{4}$ there is $x$ such that $y = x^4 - x^2$, implying\n$$\nf(y) = f(x^4 - x^2) = f(x)^4 - f(x)^2 = \\left(f(x)^2 - \\frac{1}{2}\\right)^2 - \\frac{1}{4} \\ge -\\frac{1}{4}.\n$$\n\nFor every $x$ we have $x = \\{x\\} + [x]$ and $0 \\le \\{x\\} < 1$ (in particular, $\\{x\\} > -\\frac{1}{4}$ and $[x] > x - 1$),\nimplying\n$$\nf(x) = f(\\{x\\}) + [x] > -\\frac{1}{4} + (x-1) > x - 2. \\quad (2)\n$$\nSuppose there is $x$ such that $t = f(x) - x \\ne 0$. Then for every integer $n$ we have\n$$\nf(x+n)^4 - f(x+n)^2 = f((x+n)^4 - (x+n)^2),\n$$\ngiving\n$$\n(x+n+t)^4 - (x+n+t)^2 > (x+n)^4 - (x+n)^2 - 2,\n$$\nand thus\n$$\n4t(x+n)^3 + 6t^2(x+n)^2 + 2t(2t^2-1)(x+n) + t^4 - t^2 + 2 > 0 \\quad (3)\n$$\nfor every integer $n$, which is impossible as if $t > 0$, then for $n \\to -\\infty$ we get a contradiction. If $t < 0$, then for $n \\to \\infty$ we again get a contradiction. Therefore, $f(x) = x$ for all real $x$ which is easy to check works for both given equations.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40943,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$. Let us call by a *word* any sequence of $n$ letters of an alphabet. Define the distance $\\rho(A, B)$ between the words $A = a_1a_2...a_n$ and $B = b_1b_2...b_n$ as the number of positions in which they differ (i.e. the number of indices $i$ for which $a_i \\ne b_i$). We say that the word $C$ lies in between the words $A$ and $B$ if $\\rho(A, B) = \\rho(A, C) + \\rho(C, B)$. What is the largest number of words can be chosen so that among every three of them there is a word that lies in between the other two?",
    "options": [],
    "answer": "n+1 for n ≠ 2; 4 for n = 2",
    "solution": "Answer: $n+1$ for $n \\ne 2$ and $4$ for $n = 2$.\nFor $n \\ne 2$, take all words consisting only of letters $a$ and $b$, in which each letter $a$ to the left of each letter $b$, and for $n = 2$ the words $aa, ab, ba, bb$.\n\nLet us prove that more words can not be chosen. Consider two selected words $A$ and $B$, located at the largest distance $m$. Suppose that the selected word $C$ is from $A$ at a distance $k > 0$. Then $\\rho(B, C)$ is equal to the sum or difference of $\\rho(A, B) = m$ and $\\rho(A, C) = k$ and does not exceed $m$, hence $\\rho(B, C) = m - k$. Now, if some selected word $C' \\ne C$ is also from $A$ at distance $k$, we can assume without loss of generality $k \\ge m/2$ (otherwise, we can swap $A$ and $B$). The distance between $C$ and $C'$ can not be equal to $k - k = 0$, therefore, it is equal to $k + k \\ge m$, and hence $k = m/2$.\n\nSo, we have that for each $k$, $0 \\le k \\le m$, except, perhaps, for $k = m/2$, among the selected words no more than one, located from $A$ at distance $k$. The selected words, which are at a distance of $m/2$ from $A$, are no more than 2 (we showed that the distance between any two such words is $m$, therefore none of the three words lies between the two others).\n\nIt remains to check that if $m > 2$ and two words are chosen at a distance $k = m/2$ from $A$, then there are at most 4 words. Indeed, let $\\rho(A, C) = \\rho(A, C') = k = m/2$, and we have chosen some other word $D$. As in the previous argument, we can assume without loss of generality that $\\rho(A, D) = s < k$. Then $\\rho(B, D) = 2k - s$ and $\\rho(D, C) = \\rho(D, C') = k - s$ (the distance between $D$ and $C$ can not be equal to the sum $\\rho(B, D) + \\rho(B, C) = 2k - s + k > 2k$, so it is equal to\n\nthe difference $\\rho(B, D) - \\rho(B, C)$). It turns out that among the three distances $k-s, k-s, 2k$, the words $D, C, C'$ are not equal to the sum of the other two, a contradiction.\nThis proves the estimate: if $m > 2$, then for each $k \\le n$ there is at most one selected word at a distance of $k$ from $A$, and there are no more than $n+1$ words in total, and if $m = 1$ or $2$, then there are no more than 4 words.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40944,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is there a positive integer sequence $a_1, a_2, \\dots$ that contains every positive integer exactly once and such that\n$$\n\\tau(na_{n+1}^n + (n+1)a_n^{n+1})\n$$\nis divisible by $n$ for every positive integer $n$? (Here $\\tau(n)$ is the number of positive divisors of $n$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the statement we have\n$$\n\\tau(na_{n+1}^n + (n+1)a_n^{n+1}) \\vdash n. \\qquad (1)\n$$\nLet $a_1 = 1$ and $a_2 = 2$. Suppose that we have constructed the numbers $a_1, a_2, \\dots, a_m$ so that they are different natural numbers and (1) satisfied for each $n = 1, 2, \\dots, m-1$. Let $L$ be the smallest positive integer that does not occur among the numbers $a_1, a_2, \\dots, a_m$. We set $a_{m+2} = L$ and choose $a_{m+1}$ so that $a_{m+1} > a_1 + a_2 + \\dots + a_m + L$ and (1) is true for $n = m$ and $n = m+1$. Repeating this process we obtain a permutation of all positive integers satisfying (1) for any positive integer $n$.\n\nFor a prime $p$ and a positive integer $k$, we denote by $\\psi_p(k)$ the largest integer $l$ such that $k$ is divisible by $p^l$.\n\n**Lemma.** Given positive integers $a, b, k, l, M$. Then there is a prime number $p > M$ and a positive integer $x$ such that $\\psi_p(ax^k + b) = l$.\n\n**Proof.** Let $d = M + k + b$ and $p$ be a prime divisor of $ab^{k-1}(d!)^k + 1$. Then $p > d > M$. By induction on $N$, we prove that for any positive integer $N$ there exists a positive integer $x$ such that the number $ax^k + b$ is divisible by $p^N$.\n\nBase: for $N = 1$ the number $x = d!b$.\n\nSuppose that for $N$ there exist positive integers $x$ and $t$ such that $ax^k + b = p^N t$. Let $z$ be a positive integer such that the number $ax^{n-1} z k + t$ is divisible by $p$ (such a number exists, since the number $p$ is coprime with $d!a$, and hence with the number $akx$). Then\n$$\n\\begin{aligned}\na(x + z p^N)^k + b &\\equiv a x^k + a x^{k-1} z p^N k + b \\equiv p^N t + a x^{k-1} z p^N k \\\\\n&\\equiv p^N(a x^{k-1} z k + t) \\equiv 0 \\pmod{p^{N+1}},\n\\end{aligned}\n$$\nand the induction step is proved.\n\nTherefore, there are positive integers $x, t$ with $a x^k + b = p^{l+1} t$. Then\n$$\na(x + p^l)^k + b \\equiv a x^k + a x^{k-1} p k + b \\equiv a x^{k-1} k p^l \\pmod{p^{l+1}},\n$$\ni.e. $\\psi_p(a(x + p^l)^k + b) = l$, and the lemma is proved.\n\nFrom the lemma for $a = m, b = (m+1)a_m^{m+1}, M = 1, k = m, l = m-1$ there is a prime $p > 1$ and a positive integer $x$ such that $\\psi_p(m x^m + (m+1)a_m^{m+1}) = m-1$ and from the lemma for $a = m+2, b = (m+1)k^{m+1}, M = p, k = m+2, l = m$ there is a prime $q > p$ and a positive integer $y$ such that $\\psi_q((m+1)k^{m+1} + (m+2)y^{m+2}) = m$. By the Chinese remainder theorem, there is a positive integer $z$ such that $z > a_1 + a_2 + \\dots + a_m + L, z \\equiv x \\pmod{p^m}$ and $z \\equiv y \\pmod{q^{m+1}}$. Then it is easy to see that $a_{m+1} = z$ works.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40945,
    "subject": "Mathematics (Multi-modal)",
    "question": "An infinite strictly increasing sequence of positive integers $\\{a_n\\}_{n\\ge 1}$ is given. It is also given that $a_n \\le n + 2020$ and $n^3 a_n - 1$ is divisible by $a_{n+1}$ for any positive integer $n$. Prove that $a_n = n$ for any positive integer $n$. (Kanat Satylkhanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By induction on $n$ it is easy to show that $a_n \\ge n$ for any $n$. Suppose that there exists a positive integer $k$ such that $a_k > k$. Let's choose such positive integer $m$ that $m : 2021!$ and $m > k$. Then for any $i = 2, 3, \\dots, 2021$, $\\text{GCD}(m, m + i) > 1$. It follows from the problem statement that $\\text{GCD}(m, a_{m+1}) = 1$. Since $\\{a_n\\}$ is strictly increasing and $a_k > k$, then $a_{m+1} > m + 1$. Therefore, $m + 2 \\le a_{m+1} \\le m + 2021$, but then $\\text{GCD}(m, a_{m+1}) > 1$ — a contradiction.\nLet $b_n = a_n - n$ for each $n$. By induction on $n$ it is easy to show that $b_n \\ge 0$ for any $n$. If $b_k > b_{k+1}$ for some $k$, then\n$$\na_k - k > a_{k+1} - k - 1 \\implies a_k + 1 > a_{k+1} \\implies a_k \\ge a_{k+1}\n$$\n— a contradiction. Thus, the sequence $\\{b_n\\}$ is non-decreasing. On the other hand, it has an upper bound: $b_n = a_n - n \\le 2020$. Hence, there exists such non-negative integer $k$ and a positive integer $t$ that $b_n = k$ for each $n \\ge t$. So, for any $n \\ge t$\n$$\n\\begin{aligned}\na_{n+1} \\mid n^3 a_n - 1 &\\implies n + k + 1 \\mid n^3(n + k) - 1 \\\\\n&\\implies n + k + 1 \\mid n^3(n + k) - 1 - (n + k + 1)(n^3 - n^2 + n(k + 1) - (k + 1)^2) = (k + 1)^3 - 1 \\\\\n&\\implies n + k + 1 \\mid (k + 1)^3 - 1.\n\\end{aligned}\n$$\nBut this is only possible when $k = 0$. Therefore, $b_n = 0$ for each sufficiently large $n$, and thus for all $n$, i.e. $a_n = n$ for all $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40946,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle $ABC$ is inscribed into circle $\\omega$. On sides $AB$, $BC$, $CA$ there are points $K$, $L$, $M$, respectively, such that $CM \\cdot CL = AM \\cdot BL$. Ray $LK$ intersects line $AC$ at point $P$. The common chord of $\\omega$ and the circumscribed circle of $KMP$ intersects segment $AM$ at point $S$. Prove that $SK \\parallel BC$. (Medeubek Kungozhin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nSuppose that $AC \\le BC$. Let $D$ be a point on the side $AB$ such that $DM \\parallel BC$. Then\n$$\n\\frac{DB}{DA} = \\frac{CM}{AM} = \\frac{BL}{CL},\n$$\ni. e. $DL \\parallel AC$. On the tangent line to $\\omega$ at point $C$ let's choose a point $T$, such that $KT \\parallel BC$. Then $\\angle TKA = \\angle CBA = \\angle TCA$. Therefore, $AKCT$ is cyclic. Let segments $KT$ and $AC$ intersect at point $S_1$. Then\n$$\n\\frac{S_1P}{S_1C} = \\frac{KP}{KL} = \\frac{KA}{KD} = \\frac{S_1A}{S_1M} \\implies S_1P \\cdot S_1M = S_1C \\cdot S_1A = S_1K \\cdot S_1T.\n$$\nThus, $TMKP$ is inscribed into the circumcircle of triangle $KMP$. It is known that the common chords of three pairs of circles, centers of which are not collinear, are concurrent. It means that the common chord of $\\omega$ and the circumcircle of $KMP$ passes through point $S_1$. Therefore, point $S_1$ coincides with $S$, and $S_1K \\parallel BC$ follows from the definition of $T$.\n\n\nSolution 2:\n\nLet's introduce some notation:\n$$\nb = AC, \\ m = AM, \\ s = AS, \\ p = AP.\n$$\nSince $S$ lies on the radical axis of the circumcircles of $ABC$ and $PKM$, then\n$$\nSM \\cdot SP = SA \\cdot SC \\implies (m-s)(s+p) = s(b-s) \\implies s = \\frac{pm}{b+p-m}.\n$$\nSo,\n$$\n\\frac{AS}{SC} = \\frac{s}{b-s} = \\frac{pm}{(b-m)(b+p)}.\n$$\nAccording to Menelaus' theorem for triangle $ABC$ and transversal line $PKL$:\n$$\n\\frac{AK}{KB} \\cdot \\frac{BL}{LC} \\cdot \\frac{CP}{PA} = 1 \\implies \\frac{AK}{KB} = \\frac{CL}{LB} \\cdot \\frac{AP}{PC} = \\frac{AM}{MC} \\cdot \\frac{AP}{PC} = \\frac{m}{b-m} \\cdot \\frac{p}{b+p} = \\frac{AS}{SC} \\implies SK \\parallel BC,\n$$\nQ. E. D.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40947,
    "subject": "Mathematics (Multi-modal)",
    "question": "Polynomial $Q(x) = k_n x^n + k_{n-1} x^{n-1} + \\dots + k_1 x + k_0$ with real coefficients is called *mighty* if $|k_0| = |k_1| + |k_2| + \\dots + |k_{n-1}| + |k_n|$, and *non-increasing* if $k_0 \\ge k_1 \\ge \\dots \\ge k_{n-1} \\ge k_n$.\nLet $P(x) = a_d x^d + a_{d-1} x^{d-1} + \\dots + a_1 x + a_0$ be a polynomial with real non-zero coefficients, such that $a_d > 0$ and $P(x)(x-1)^t (x+1)^s$ is *mighty* for some non-negative integers $s$ and $t$ ($s+t > 0$). Prove that at least one of the polynomials $P(x)$ and $(-1)^d P(-x)$ is *non-increasing*. (Navid Safaei, Iran)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that if for some real numbers $x_1, x_2, \\dots, x_m$ the following equality holds:\n$$\n|x_1| + |x_2| + \\dots + |x_m| = |x_1 + x_2 + \\dots + x_m|,\n$$\nthen they are of the same sign.\nLet\n$$\nQ(x) = P(x)(x-1)^t (x+1)^s = b_n x^n + b_{n-1} x^{n-1} + \\dots + b_0,\n$$\nwhere $b_n = a_d > 0$. From the problem statement it follows that\n$$\n|b_0| = |b_1| + |b_2| + \\dots + |b_n|.\n$$\n**Lemma:** If $t \\ge 1$, then $b_1, b_2, \\dots, b_n \\ge 0$.\n**Proof:**\n$$\n\\begin{align*} Q(1) = 0 &\\implies b_0 + b_1 + \\dots + b_n = 0 \\\\ &\\implies |b_1 + b_2 + \\dots + b_n| = |b_0| = |b_1| + |b_2| + \\dots + |b_n|, \\end{align*}\n$$\nhence, $b_1, b_2, \\dots, b_n$ are of the same sign. Since $b_n > 0$, then $b_1, b_2, \\dots, b_{n-1} \\ge 0$. We proved the lemma.\nAssume that $t \\ge 2$. According to the lemma,\n$$\nb_1, b_2, \\dots, b_{n-1} \\ge 0 \\implies b_1 + 2b_2 + \\dots + nb_n > 0.\n$$\nOn the other hand, let $R(x) = \\frac{Q(x)}{(x-1)^2}$. Then\n$$\nQ'(x) = 2(x-1)R(x) + (x-1)^2 R'(x) \\implies Q'(1) = 0 \\implies b_1 + 2b_2 + \\dots + nb_n = 0\n$$\n— a contradiction. Thus, $t \\le 1$.\n\nSimilarly, we can show that $s \\le 1$. Let's consider three cases.\n\n**I) $t = 1, s = 1$**\n$$\n\\begin{align*} Q(x) = P(x)(x^2 - 1) &\\implies Q(1) = Q(-1) = 0 \\\\ &\\implies b_0 + b_1 + \\dots + b_n = b_0 - b_1 + b_2 + \\dots + (-1)^n b_n = 0 \\\\ &\\implies b_1 + b_2 + b_3 + \\dots = 0. \\end{align*}\n$$\nAccording to the lemma,\n$$\nb_1, b_2, \\dots, b_n \\ge 0 \\implies b_1 = b_2 = b_3 = \\dots = 0,\n$$\n— a contradiction, since $b_1 = -a_1$ and by the problem statement $a_1 \\ne 0$.\n\n**II) $t = 1, s = 0$**\n$$\nQ(x) = P(x)(x-1) = -a_0 + (a_0 - a_1)x + \\dots + (a_{d-1} - a_d)x^d + a_d x^{d+1}.\n$$\nBy the lemma,\n$$\na_0 - a_1 = b_1 \\ge 0, \\dots, a_{d-1} - a_d = b_d \\ge 0 \\implies a_0 \\ge a_1 \\ge \\dots \\ge a_d.\n$$\n\nTherefore, $P(x)$ is non-increasing.\n\n**III) $t=0, s=1$**.\n$$\nQ(-1) = 0 \\implies b_0 - b_1 + \\dots + (-1)^n b_n = 0 \\implies \\\\\n\\implies |b_1 - b_2 + \\dots + (-1)^n b_n| = |b_0| = |b_1| + |-b_2| + \\dots + |(-1)^n b_n|.\n$$\nThus, $b_1, -b_2, \\dots, (-1)^n b_n$ are of the same sign, and since $b_n > 0$, then $(-1)^{n-i}b_i \\ge 0$ for each $1 \\le i \\le n$.\n$$\nQ(x) = P(x)(x+1) = a_0 + (a_0 + a_1)x + \\dots + (a_{d-1} + a_d)x^d + a_d x^{d+1} \\implies \\\\\n\\implies (-1)^{d+1-i}(a_{i-1} + a_i) \\ge 0 \\text{ (for all } 1 \\le i \\le d) \\implies \\\\\n\\implies a_d \\le -a_{d-1} \\le a_{d-2} \\le \\dots \\le (-1)^d a_0.\n$$\nIt follows that $(-1)^d P(-x) = a_d x^d - a_{d-1} x^{d-1} + \\dots + (-1)^d a_0$ is non-increasing.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40948,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any positive integer $m$ there exists a positive integer $n$, such that any $n$ different points on a plane can be partitioned into $m$ non-empty sets, *convex hulls* of which would share a common point.\n\n*Convex hull* of a finite set $X$ of points on a plane is a set of points that lie inside or on the border of at least one convex polygon with vertices in $X$, including degenerate ones, i.e. a segment and a point are considered to be convex polygons. No three vertices of a convex polygon are collinear. A polygon contains its border.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's remind **Helly's theorem**: if in a finite set of convex sets of points on a plane each three intersect, then all of them intersect.\n\nLet's prove that $n = 9m$ satisfies the problem statement. Let $X$ be an arbitrary set of $9m$ different points on a plane, and $Y$ — the set of subsets of $X$ of size $6m + 1$.\n\nSuppose that there exist such $A, B, C \\in Y$ that their intersection is empty. Let's enumerate all points in $X$ by numbers from $1$ to $9m$. Let's write down on a sheet of paper the numbers of points in $A$, then the numbers of points in $B$, and after that the numbers of points in $C$. We wrote $|A| + |B| + |C| = 18m + 3$ numbers in total. Since these sets do not intersect, then we couldn't write any number more than twice. Thus, we wrote no more than $2 \\cdot 9m = 18m$ numbers — a contradiction. Therefore, any three elements of $Y$ intersect.\n\nSince the convex hull of a set of points contains the set itself, then the convex hulls of any three elements of $Y$ intersect. According to Helly's theorem, the convex hulls of all elements of $Y$ share some common point $O$.\n\nLet's prove the following **lemma**: if the convex hull of a finite set of points $Z$ contains some point $P$, then there exists such $W \\subseteq Z$ that $|W| \\le 3$ and the convex hull of $W$ contains $P$. By definition of convex hull, there exists a convex polygon with the set of vertices $V \\subseteq Z$ (possibly, degenerate) that contains $P$. If $|V| \\le 3$, then $V$ works as $W$. Otherwise, let's perform an arbitrary triangulation of the polygon with vertices in $V$. Point $P$ has to lie in at least one of the obtained triangles. The set of vertices of such triangle works as $W$.\n\nSuppose we have a bag into which we can put non-empty subsets of $X$. Let's denote the following operation, which modifies $X$ and $Y$: take any $A \\in Y$. Since the convex hull of $A$ contains $O$, then, according to the lemma, there exists such $B \\subseteq A$ that $|B| \\le 3$ and the convex hull of $B$ contains $O$. Let's put $B$ into our bag (obviously, $B$ is non-empty), delete elements of $B$ from $X$, and delete sets from $Y$ that contain element of $B$.\n\nAfter one such operation the size of $X$ decreases by at most three, and $Y$ remains non-empty as long as $|X| \\ge 6m + 1$. Therefore, we can perform the operation at least $m$ times. Let's perform it exactly $m$ times. Distribute the remaining elements of $X$ randomly among the sets in the bag.\n\nSo, the sets in the bag constitute a partition of the initial set of points into $m$ non-empty sets and the convex hull of each of them contains point $O$, which is what we wanted.\nLet's prove by induction on $m$ that any finite set of at least $4m^2$ different points on a plane can be partitioned into $m$ non-empty sets, convex hulls of which intersect. Obviously, the claim holds for $m=1$. Assume that it holds for $m=k-1$, where $k \\ge 2$. Let's prove that it also holds for $m=k$. Let's consider an arbitrary finite set $X$ consisting of at least $4k^2$ different points on a plane. Let $Y$ be the subset of points of $X$ that lie on the border of the convex hull of $X$.\n\nIf $|Y| < 4k$, then $|X \\setminus Y| > 4k^2 - 4k > 4(k-1)^2$. By the induction hypothesis, $X \\setminus Y$ can be partitioned into $k-1$ non-empty sets, convex hulls of which intersect. If we add $Y$ to these $k-1$ sets, then we would get $k$ sets, convex hulls of which intersect (since the convex hull of $Y$ contains all points from $X \\setminus Y$).\n\nIf $|Y| \\ge 4k$, then there are two cases.\n\nIf all points from $Y$ lie on the same line, then all points from $X$ lie on the same line. Let's draw a coordinate axis along this line and denote the points from $X$ by $A_1, A_2, \\dots, A_{|X|}$ in the order of increasing coordinates. Since $4k^2 > 2k$, then the following partition works:\n$$\nX = \\{A_1, A_{|X|}\\} \\cup \\{A_2, A_{|X|-1}\\} \\cup \\dots \\cup \\{A_{k-1}, A_{|X|-k+2}\\} \\cup \\{A_k, A_{k+1}, \\dots, A_{|X|-k+1}\\}\n$$\n\nOtherwise points of $Y$ lie on the border of some non-degenerate convex polygon. Denote the points from $Y$ in clockwise order by\n$$\nA_1, A_2, \\dots, A_k, B_k, B_{k-1}, \\dots, B_1, C_1, C_2, \\dots, C_k, D_k, D_{k-1}, \\dots, D_1, E_1, E_2, \\dots, E_{|Y|-4k}.\n$$\nLet\n$$\nZ = \\{A_k, B_k, C_k, D_k, E_1, \\dots, E_{|Y|-4k}\\} \\cup (X \\setminus Y).\n$$\nLet's prove that the following partition works:\n$$\nX = \\left( \\bigcup_{i=1}^{k-1} \\{A_i, B_i, C_i, D_i\\} \\right) \\cup Z.\n$$\nIt is enough to prove the **key assertion**: convex hulls of\n$$\n\\{A_1, B_1, C_1, D_1\\}, \\{A_2, B_2, C_2, D_2\\}, \\dots, \\{A_k, B_k, C_k, D_k\\}\n$$\nintersect. Denote the convex hull of a set of points $M$ by $f(M)$. Let\n$$\nT \\in A_1B_1 \\cap A_kD_k,\n$$\n$$\nH_i = f(\\{A_1, A_2, \\dots, A_i, B_i, B_{i-1}, \\dots, B_1, C_1, C_2, \\dots, C_i, D_i, D_{i-1}, \\dots, D_1\\})\n$$\nand\n$$\nV_i = f(\\{A_i, A_{i+1}, \\dots, A_k, B_k, B_{k-1}, \\dots, B_i, C_i, C_{i+1}, \\dots, C_k, D_k, D_{k-1}, \\dots, D_i\\}).\n$$\nThen for each $1 \\le i \\le k$ holds\n$$\nT \\in A_1 B_1 \\subseteq H_i \\text{ and } T \\in A_k D_k \\subseteq V_i.\n$$\nTherefore,\n$$\nT \\in \\bigcap_{i=1}^{k} (H_i \\cap V_i) = \\bigcap_{i=1}^{k} f(\\{A_i, B_i, C_i, D_i\\}),\n$$\nQ.E.D.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40949,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence $s$ consisting of zeroes and ones is given. For each positive integer $k$ we define $v_k$ as the maximum number of ways to find consecutive digits forming the sequence $s$ in a sequence of $k$ digits. (For example, if $s = 0110$, then $v_7 = v_8 = 2$, since in the sequences 0110110 and 01101100 consecutive digits 0110 are found in two places, and three pairs of ones surrounded by zeroes can not occur in a sequence of 7 or 8 digits.) It is known that $v_n < v_{n+1} < v_{n+2}$ for some positive integer $n$. Prove that all the digits in the sequence $s$ are identical.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the sequence $s = c_1c_2 \\dots c_m$ have length $m$, and some sequence $X$ of length $n$ be $a_1a_2 \\dots a_n$. If two occurrences of $s$ in $X$ begin with $a_p$ and $a_{p+k}$, then $a_{p+i} = a_{p+i+k} = c_{i+1}$ for $0 \\le i < m$. When $k \\le m$ this means that $c_i = c_{i+k}$ for $1 \\le i \\le m-k$, that is, $s$ is the beginning of the (infinite) periodic sequence $a_1a_2 \\dots a_k a_1a_2 \\dots a_k a_1 \\dots$ with period $k$. We choose the minimum such $k$; if no such $k < m$ exists, we let $k = m$ (since $s$ is obviously the beginning of the periodic sequence $a_1a_2 \\dots a_m a_1a_2 \\dots a_m a_1 \\dots$).\n\nWe have seen that the number of places in $X$ where the occurrences of $s$ begin must differ at least by $k$. If there are $t$ such occurrences, the last one should begin at the place with number not less than $1+k(t-1)$, so $m+k(t-1) \\le n$. It follows that the number of occurrences of $s$ in any sequence of length $n$ does not exceed $\\lfloor \\frac{n-m}{k} \\rfloor + 1$.\n\nOn the other hand, an example of a sequence where $s$ can be found in this number of ways, is given by the periodic sequence $a_1a_2 \\dots a_k a_1a_2 \\dots a_k a_1 \\dots$.\n\nWe have proved that $v_n = \\lfloor \\frac{n-m}{k} \\rfloor + 1$, and the problem states that\n$$\n\\lfloor \\frac{n-m}{k} \\rfloor < \\lfloor \\frac{n-m+1}{k} \\rfloor < \\lfloor \\frac{n-m+2}{k} \\rfloor .\n$$\nIf the inequality $\\lfloor \\frac{a}{k} \\rfloor < \\lfloor \\frac{a+1}{k} \\rfloor$ holds for an integer $a$ and a positive integer $k$, then $k$ divides $a+1$ (since an integer which is greater than $\\frac{a}{k}$ and not greater than $\\frac{a+1}{k}$, can be presented as a fraction with denominator $k$ and therefore equals $\\frac{a+1}{k}$). Therefore, the problem implies that $k$ divides both $n-m+1$ and $n-m+2$, which is possible for $k=1$ only. By the definition of $k$ we have $c_i = c_{i+1}$ for $1 \\le i \\le m-1$, that is, every two consecutive digits in $s$ are equal. Thus all the digits of $s$ are equal.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40950,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every positive integer $m$ prove the inequality $|\\{\\sqrt{m}\\} - \\frac{1}{2}| > \\frac{1}{8(\\sqrt{m}+1)}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $[\\sqrt{m}] = k$. Then\n$$\n|\\{\\sqrt{m}\\} - \\frac{1}{2}| = \\frac{1}{2}|2k + 1 - 2\\sqrt{m}| = \\frac{1}{2} \\cdot \\frac{|4k^2 + 4k + 1 - 4m|}{2k + 1 + 2\\sqrt{m}}.\n$$\nThe numerator of the last fraction is an odd positive integer, and therefore at least $1$. Thus,\n$$\n|\\{\\sqrt{m}\\} - \\frac{1}{2}| \\ge \\frac{1}{2} \\cdot \\frac{1}{2k + 1 + 2\\sqrt{m}} \\ge \\frac{1}{2(4\\sqrt{m} + 1)} > \\frac{1}{8(\\sqrt{m} + 1)},\n$$\n\nQ.E.D.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 40951,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the midpoint of side $AB$ in triangle $ABC$. $B_1$ is a point on segment $AC$ such that $CB = CB_1$. The circumcircles of triangles $ABC$ and $BMB_1$, $\\omega$ and $\\omega_1$, intersect for the second time at point $K$. Let $Q$ be the midpoint of arc $ACB$ of $\\omega$. Lines $B_1Q$ and $BC$ intersect at point $E$. Prove that line $KC$ passes through the midpoint of segment $B_1E$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the bisector of $\\angle ACB$ intersect $\\omega$ for the second time at point $N$. Note that $N$ is the midpoint of arc $AB$ (that does not contain $C$) of $\\omega$. Also, it is easy to see that line $CN$ is the perpendicular bisector of segment $BB_1$. Thus, $NA = NB = NB_1$, i.e. points $A, B$ and $B_1$ lie on a circle centered at $N$ with radius $NA$. Since $\\angle NAQ = \\angle NBQ = 90^\\circ$, then lines $QA$ and $QB$ are tangent to this circle. Therefore, line $QB_1$ contains the symmedian of triangle $ABB_1$ corresponding to vertex $B_1$.\n![](attached_image_1.png)\nLet line $KC$ intersect $\\omega_1$ for the second time at point $P$. From the properties of symmedian it follows that\n$$\n\\angle(B_1C, B_1E) = \\angle(B_1C, B_1Q) = \\angle(B_1M, B_1B) = \\angle(PM, PB) = \\alpha. \\qquad (1)\n$$\nAlso, the following equalities hold:\n$$\n\\angle(CK, CA) = \\angle(BK, BA) = \\angle(BK, BM) = \\angle(PK, PM) = \\beta. \\qquad (2)\n$$\nFrom (1) and (2) it follows that $\\angle(PK, B_1E) = \\angle(CK, B_1E) = \\alpha + \\beta = \\angle(PK, PB)$. The latter equality gives\n$$\nB_1E \\parallel PB. \\qquad (3)\n$$\nLet line $BP$ intersect line $AC$ at point $R$. Then, from (2) it follows that $CA \\parallel PM$, or that $MP$ is a midsegment of triangle $ARB$, i.e. $P$ is the midpoint of segment $BR$. Now it is enough to note that if line $KC$ bisects segment $BR$, then it also bisects segment $B_1E$, since (3) holds.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40952,
    "subject": "Mathematics (Multi-modal)",
    "question": "Integers $x$, $y$, $z$, $t$ satisfy $x^2 + y^2 = z^2 + t^2$ and $xy = 2zt$. Prove that $xyzt = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that $xyzt \\neq 0$.\nIf three of the numbers $x$, $y$, $z$, $t$ have a common divisor $d$, the remaining number is also a multiple of $d$. The division of all the four numbers by $d$ preserves the validity of our equations. Since $x$, $y$, $z$, $t$ are not zero, we can divide them by their greatest common divisor and henceforth assume that they are coprime.\n\nThe numbers $x$ and $y$ can not be both odd (since $xy$ is even) or both even (since in this case $zt$ is also even and at least three of the numbers $x$, $y$, $z$, $t$ are even). Thus $x$ and $y$ and therefore $z$ and $t$ have different parity.\n\nDenote $s = x^2 + y^2$, $n = zt$. Without loss of generality, $n > 0$.\nThe condition on $x$, $y$, $z$, $t$ means that $s - 4n$, $s - 2n$, $s + 2n$, $s + 4n$ are perfect squares. We will prove that this is not possible for any odd $s$ and positive $n$. First, we find a statement equivalent to these four numbers being squares. Their product $(s^2 - 4n^2)(s^2 - 16n^2) = (s^2 - 10n^2)^2 - 36n^4$ is a square of some odd $T$. The numbers $T$, $6n^2$, $s^2 - 10n^2$ form a primitive Pythagorean triple, hence there exist coprime $u$, $v$ of different parity such that $2uv = 6n^2$ and $u^2 + v^2 = s^2 - 10n^2$. Since $u$ and $v$ are coprime, it follows from $uv = 3n^2$ that one of these numbers is a square and another is a triple square. Without loss of generality, assume that $u = 3k^2$ and $v = l^2$; $k$ and $l$ are coprime. Multiplying $u^2 + v^2 = s^2 - 10n^2$ by $3$ we get $3s^2 - 30n^2 = 3s^2 - 10uv = 3u^2 + 3v^2$, that is, $3s^2 = 3u^2 + 10uv + 3v^2 = (3u + v)(3v + u)$, where $3u + v$ and $3v + u$ are coprime (they have different parity, and their common divisor must also divide $8u = 3(3u + v) - (3v + u)$ and, similarly, $8v$). It follows that one of the numbers $3u + v$ and $3v + u$ is a square and another is a triple square. Since $3$ does not divide $3u + v$, the square is $3u + v = 9k^2 + l^2$, and the triple square is $u + 3v = 3(k^2 + l^2)$. We have proved that the original equations imply the existence of coprime $k$ and $l$ such that $9k^2 + l^2$ and $k^2 + l^2$ are perfect squares.\n\nIt remains to prove that there are no such $k$, $l$. If $k^2 + l^2$ and $9k^2 + l^2$ are perfect squares, the pairs $(k, l)$ and $(3k, l)$ are pairs of legs in primitive Pythagorean triples. Unfortunately, we do not know which of the numbers $k$ and $l$ is even, so two cases are possible. But let's prove the following lemma first.\n\n**Lemma.** Let $a$, $b$, $c$, $d$ be pairwise coprime positive integers such that $c^2(a^2 + d^2) = b^2(9a^2 + d^2)$. Then, $a^2 + d^2$ and $9a^2 + d^2$ are perfect squares.\n\n**Proof.** Since $\\text{GCD}(b, c) = 1$, then there exists such positive integer $f$ that\n$$\na^2 + d^2 = b^2 f, \\quad 9a^2 + d^2 = c^2 f.\n$$\n$$\nf = \\text{GCD}(a^2 + d^2, 9a^2 + d^2) \\implies f \\mid 8a^2, \\quad f \\mid 8d^2.\n$$\n$$\n\\text{GCD}(a, d) = 1 \\implies f \\mid 8.\n$$\nIf $a$, $d$ are of different parity, then $a^2 + d^2$ is odd, and therefore $f = 1$. Let both $a$ and $d$ be odd (they can't be simultaneously even). Then\n$$\na^2 + d^2 \\equiv 2 \\pmod{8} \\implies f \\mid 2.\n$$\nLet $f = 2$.\n$$\n9a^2 + d^2 = c^2 f = 2c^2 \\implies d^2 \\equiv 2c^2 \\pmod{3} \\implies 3 \\mid c, \\quad 3 \\mid d\n$$\n— a contradiction, since $\\text{GCD}(c, d) = 1$. Thus, $f = 1$, $a^2 + d^2 = b^2$, $9a^2 + d^2 = c^2$. The lemma is proven.\n\n**Case 1.** $k$ is even. Then $k = 2u_1v_1$, $3k = 2u_2v_2$, $l = u_1^2 - v_1^2 = u_2^2 - v_2^2$, where $(u_1, v_1)$ and $(u_2, v_2)$ are two pairs of coprime numbers with different parity. Since $u_2v_2 = 3u_1v_1$, then there exist pairwise coprime positive integers $a$, $b$, $c$, $d$ such that $u_2 = 3ab$, $v_2 = cd$, $u_1 = ac$, $v_1 = bd$ (the case when $u_2$ is not divisible by $3$ can be considered similarly). Substituting this in the expression for $l$ we obtain $a^2c^2 + c^2d^2 = 9a^2d^2 + d^2c^2$, or $c^2(a^2 + d^2) = d^2(9a^2 + c^2)$. According to the lemma, $a^2 + d^2$ and $9a^2 + c^2$ are squares, and, obviously, $a < k$, $d < l$.\n\n**Case 2.** $l$ is even. Then $k = u_1^2 - v_1^2$, $3k = u_2^2 - v_2^2$, $l = 2u_1v_1 = 2u_2v_2$. Since $(u_2 - v_2)(u_2 + v_2) = 3(u_1 - v_1)(u_1 + v_1)$, then we have $u_1 - v_1 = ab$, $u_1 + v_1 = cd$, $u_2 - v_2 = ac$, $u_2 + v_2 = 3bd$ for some pairwise coprime positive integers $a$, $b$, $c$, $d$ (the case when $u_2 + v_2$ is not divisible by $3$ can be considered similarly). Expressing $u_1$, $v_1$, $u_2$, $v_2$ in terms of $a$, $b$, $c$, $d$ and putting the result in the formula for $l$, we get $(ab+cd)(cd-ab) = (3bd+ac)(3bd-ac)$, that is, $c^2(a^2+d^2) = b^2(a^2+9d^2)$. Hence, by the lemma, we obtain that $a^2+d^2$ and $a^2+9d^2$ are perfect squares, and $a<l$, $d<k$.\n\nIn both cases, for each pair $(k, l)$ such that $k^2 + l^2$ and $9k^2 + l^2$ are perfect squares we constructed a pair of smaller numbers with the same property. It follows that such numbers do not exist. Thus, the original system does not admit a solution in non-zero integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40953,
    "subject": "Mathematics (Multi-modal)",
    "question": "Convex quadrilateral *ABCD* is inscribed in circle *ω*. Rays *AB* and *DC* intersect at *K*. *L* is chosen on the diagonal *BD* so that $\\angle BAC = \\angle DAL$. *M* is chosen on the segment *KL* so that $CM \\parallel BD$. Prove that the line $BM$ touches $\\omega$. (Kungozhin M.)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let *N* be a point on the line $AK$ so that $MN \\parallel AL$. Since $\\frac{CM}{DL} = \\frac{NM}{AL} = \\frac{KM}{KL}$ and $\\angle CMN = \\angle DLA$, it follows that $K$ is a center of homothety which sends $\\triangle CMN$ into similar $\\triangle DLA$. On the other hand, $\\triangle DLA$ is similar to $\\triangle CBA$ because $\\angle DAL = \\angle BAC$ and $\\angle ADL = \\angle ACB$. Consequently, $\\angle (BN, BC) = \\angle (MN, MC)$, and thus points $N$, $B$, $M$, $C$ are cyclic. Therefore, $\\angle CBM = \\angle CNM = \\angle CAB$ from which it follows that $BM$ touches $\\omega$.\nFor this solution we need the following theorem.\n\n**Pascal's theorem.** Points $A$, $B$, $C$, $D$, $E$, $F$ (not necessarily in this order) lie on some circle. Then the intersections of the lines $AB$ and $DE$, $BC$ and $EF$, $CD$ and $FA$ lie on a straight line.\n\nBack to the problem. Let $E$ be the second intersection of the line $AL$ and $\\omega$. Define $l_b$ as a tangent line to $\\omega$ at $B$. Let's apply Pascal's theorem on points $B$, $B_1$, $A$, $E$, $C$, $D$ (here $B_1$ coincides with $B$) and pairs of lines ($BB_1$, $EC$), ($B_1A$, $CD$), ($AE$, $DB$). These pairs of lines coincide with pairs of lines ($l_b$, $EC$), ($BA$, $CD$), ($AE$, $DB$). Then from Pascal's theorem, the straight line connecting $K = BA \\cap CD$ and $L = AE \\cap DB$ will also contain $l_b \\cap EC$. Since $M = EC \\cap KL$, it follows that $BM$ touches $\\omega$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40954,
    "subject": "Mathematics (Multi-modal)",
    "question": "Distinct positive integers $A$ and $B$ are given. Prove that there exist infinitely many positive integers that can be represented both as $x_1^2 + Ay_1^2$ for some positive coprime integers $x_1$ and $y_1$, and as $x_2^2 + By_2^2$ for some positive coprime integers $x_2$ and $y_2$. (Golovanov A.S.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality $A > B$.\nChoose an arbitrary prime $p > 2$ and let's find $x_1$ and $x_2$ so that\n$$\nx_1^2 + A(2p)^2 = x_2^2 + B(2p)^2,\n$$\nHence, $x_2^2 - x_1^2 = 4Cp^2$ where $C = A - B$. Set $x_1 = Cp^2 - 1$ and $x_2 = Cp^2 + 1$. If $x_1$ and $x_2$ are both odd, then they are both coprime with $y = 2p$, and we have $x_1^2 + Ay^2 = x_2^2 + By^2$. If they are both even, then $\\frac{x_1}{2}$ and $\\frac{x_2}{2}$ are both coprime with $y = p$, and we have $(\\frac{x_1}{2})^2 + Ay^2 = (\\frac{x_2}{2})^2 + By^2$.\nThe number to which we have found two such forms will be not less than $p^2$, thus proving there are infinitely many such numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40955,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an infinite sequence $\\{\\alpha\\}$, $\\{\\alpha^2\\}$, $\\{\\alpha^3\\}$, ... there are only finitely many distinct values. Show that $\\alpha$ is an integer. ($\\{x\\}$ denotes the fractional part of $x$, i.e. $\\{x\\} = x - [x]$, where $[x]$ is the greatest integer not greater than $x$.) (Golovanov A.S.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Step 1. We show that there is a positive integer $l$ such that $\\alpha^l$ is rational. Say the sequence is of length $k-1$. For any positive integer $n$ the sequence $\\{\\alpha^{nk}\\}$, $\\{\\alpha^{nk+1}\\}$, ..., $\\{\\alpha^{nk+k-1}\\}$ contains two equal elements. Hence, there are infinitely many pairs $i, j$, $0 < i - j < k$, such that $\\{\\alpha^i\\} = \\{\\alpha^j\\}$, i.e. $\\alpha^j(\\alpha^{i-j} - 1)$ is an integer. Since there are finitely many possible values of $i-j$, at least one of them occurs infinitely often. So we can find $m$ such that $\\alpha^j(\\alpha^m - 1)$ is an integer for infinitely many $j$. We can divide two such numbers to get $\\alpha^l$ is rational for some positive integer $l$.\n\nStep 2. Conclusion. Now if $\\alpha^l$ is not an integer, say, is equal to $\\frac{a}{b}$ for $b > 1$, $\\text{gcd}(a, b) = 1$, then $\\{\\alpha^{ln}\\}$ is an irreducible fraction with denominator $b^n$. This is true for any $n$ so we get infinitely many values, contradiction. So $\\alpha^l$ is an integer. If $\\alpha$ is irrational, then $\\alpha^{nl+1} = \\alpha^{nl} \\cdot \\alpha$ is irrational for any natural $n$ and have distinct fractional parts (Indeed, $\\alpha^{il+1} - \\alpha^{jl+1} = \\alpha(\\alpha^{il} - \\alpha^{jl})$ cannot be an integer). This contradicts finiteness as well. Thus $\\alpha$ is rational, and similar to above we can conclude it is an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40956,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a language, an alphabet with 25 letters is used; *words* are exactly all sequences of (not necessarily different) letters of length $17$. Two ends of a paper strip are glued so that the strip forms a ring; the strip bears a sequence of $5^{18}$ letters. Say that a word is *singular* if one can cut out a piece bearing exactly that word from the strip, but one cannot cut out two such non-overlapping pieces. It is known that one can cut out $5^{16}$ non-overlapping pieces each containing the same word. Determine the largest possible number of singular words. (Bogdanov I.)",
    "options": [],
    "answer": "2*5^17",
    "solution": "Let the alphabet consist of letters $a_1, a_2, \\dots, a_{25}$. By a *piece* we always mean a piece of the strip containing exactly $17$ consecutive letters; different pieces may contain the same word. Say that a piece is *singular* if the word it contains is such.\n\nWe start with constructing an example containing $N = 2 \\cdot 5^{17}$ singular words. Define a word $W = a_1a_2 \\dots a_{17}$; this will be the word having $k = 5^{16}$ non-overlapping copies on the strip. There exist exactly $25^8 = k$ possible $8$-letter sequences, consisting of letters $a_{18}, a_{19}, \\dots, a_{25}$; put them onto the strip in an arbitrary order, separating each two sequences by an instance of $W$. Each segment of the strip containing one $8$-sequence mentioned above (and no other letters) will be referred to as a *part*. Notice that the strip contains exactly $(8+17)k = 5^{18}$ letters.\n\nClearly, the obtained strip contains $k$ non-overlapping copies of $W$. Now we show that any piece containing a whole part is singular — moreover, that the word it contains is met on no other piece. Since a part can be situated in a piece at $10$ different positions (starting from the $1$-st, from the $2$-nd, ..., or from the $10$-th letter of a piece), we will get that there are at least $10 \\cdot 5^{16} = N$ singular words.\n\nConsider an arbitrary piece $p$ containing a word $P$. Either this piece contains a unique nonempty prefix which coincides with some suffix of $W$, or there is no such prefix — only in this case we will say that such prefix is empty. Let $b$ be the length of the defined prefix. Define similarly a suffix of $P$ which coincides with a prefix of $W$, and denote its length by $e$. Notice that the defined prefix and suffix do not overlap whenever $P \\neq W$ (if $P = W$, we have $b = e = 17$).\n\nIf the piece contains no whole part, then $\\max\\{b, e\\} > 9$. If the piece contains a part, then $b + e = 9$ and $0 \\le b, e \\le 9$. Thus, piece $p$ contains a part if and only if $\\max\\{b, e\\} \\le 9$, and in this case the position of the part at $P$ (and hence the position of $p$ at the strip) is uniquely determined. Therefore, in this case $P$ is met only on piece $p$, so this piece is singular. We have proven that the constructed example works.\n\nIt remains to prove that the number of singular words cannot exceed $N$. Enumerate the positions in the strip successively by $1, 2, \\dots, 5^{18}$ (the numeration is cyclic modulo $5^{18}$). Let $p_i$ denote the piece starting at position $i$, and let $P_i$ be the word on that piece. Let $n_1, \\dots, n_k$ be positions such that the pieces $p_{n_1}, p_{n_2}, \\dots, p_{n_k}$ are pairwise disjoint and contain the same word $W$ (from the problem statement). Clearly, those pieces are not singular.\n\nFor $i = 1, 2, \\dots, 8$ and $1 \\le s \\le k$, we say that a piece $p_{n_s+i}$ is a *rank $i$ follower*, while $p_{n_s-i}$ is a *rank $i$ predecessor*. All these pieces (followers and predecessors) are distinct; moreover, followers of a fixed rank are pairwise disjoint, and the same holds for predecessors. We will show that *among* $8 \\cdot 5^{16}$ *followers of all ranks*, *at most* $5^{16}$ pieces are singular (we will call this statement a *quoted claim* in the future); by symmetry, the same bound holds for predecessors. This will yield that there are at least $5^{16} + 7 \\cdot 5^{16} + 7 \\cdot 5^{16} = 3 \\cdot 5^{17}$ non-singular pieces, which implies the desired bound.\n\n**Remark.** We present a shorter (yet more ideological) proof of the quoted claim on the number of singular followers. Say that a singular follower's tail $T$ is *minimal* if none of its proper prefixes is a singular follower's tail. In particular, no minimal tail can be a proper prefix of other minimal tail.\n\nFor every minimal tail $T$ let us write down all $8$-letter sequences starting with $T$; if the length of $T$ is $d$, then the number of such sequences is $25^{8-d}$. No sequence could be written down twice; therefore, if there are $M$ minimal tails of lengths $d_1, \\dots, d_M$, then\n$$\n\\sum_{i=1}^{M} 25^{8-d_i} \\le 25^8.\n$$\nOn the other hand, each singular follower's tail has a prefix which is a minimal tail. For a minimal tail $T$ of length $d$, there are at most $9-d$ singular followers whose tails start with $T$ — at most one per tail's length. Therefore, the number of singular followers does not exceed\n$$\n\\sum_{i=1}^{M} (9 - d_i) \\le \\sum_{i=1}^{M} 25^{8-d_i} \\le 25^8,\n$$\nsince $9 - d \\le 25^{8-d}$ for all $d = 1, 2, \\dots, 8$.\n\nFinally, if there is a singular follower $P_{n_s+8-m}$ whose tail is $U$, then such follower is unique. Therefore, all followers of larger ranks whose tails start with $U$ correspond to the same copy $p_{n_s}$ of $W$. Then the number of such followers (including $P_{n_s+8-m}$ itself) is at most $m+1 \\le 25^m$, as desired again. The claim, and the bound, are proven.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40957,
    "subject": "Mathematics (Multi-modal)",
    "question": "A set $S$ of real numbers is such that $1 + \\frac{1}{x} \\in S$ for each $x \\in S$. Is it possible that $S$ contains exactly 2025 elements?",
    "options": [],
    "answer": "No",
    "solution": "**Answer.** The answer is no.\n\nLet $f(x) = 1 + \\frac{1}{x}$ and $f^{(n)}(x) = f(f(\\dots f(x)))$ (where $f$ is applied $n$ times). For each $x \\in S$ the set $S$ contains all numbers of the form $f^{(n)}(x)$. If $S$ is finite, then $f^{(k)}(x) = f^{(\\ell)}(x)$ for some $k > \\ell$. Since $f(a) = f(b)$ implies $a = b$, $f^{(k-\\ell)}(x) = x$. Thus each $x \\in S$ is a root of equation $f^{(n)}(x) = x$ for some positive integer $n$.\n\nThe functions $f^{(n)}(x)$ obviously are of the form $\\frac{px+q}{rx+s}$ with some $p, q, r, s$ (it is immediate for $n=1$, and $f$ transforms every such function into a function of this form). Therefore, the equation $f^{(n)}(x) = \\frac{px+q}{rx+s} = x$ is reduced to an equation of the form $rx^2 + (s-p)x - q = 0$ which is an equation of degree at most 2. This equation is not identical (since, for example, $f(-2) = \\frac{1}{2}$ and therefore $f^{(n)}(-2)$ is positive for all positive integers $n$), hence it cannot have more than two real roots. However, the roots $\\frac{1+\\sqrt{5}}{2}$ and $\\frac{1-\\sqrt{5}}{2}$ of $f(x) = x$ are also roots of $f^{(n)}(x) = x$ and therefore its only roots.\n\nWe see now that if $S$ is finite, it can contain at most two elements.\n\n\nSecond solution. There is an alternative way to prove that for every $x$ except $\\varphi = \\frac{1+\\sqrt{5}}{2}$ and $\\bar{\\varphi} = \\frac{1-\\sqrt{5}}{2}$, $f^{(n)}(x)$ and $x$ would not coincide.\n\nIf $x < -1$, then $f(x) > 0$ and therefore all $f^{(n)}(x)$ are positive.\n\nIf $0 < x \\le 1$, then $f(x) > 1$, and all of $f^{(n)}(x)$ are greater than 1.\n\nIf $x > 1$, then $|f(x) - \\varphi| = |1 + \\frac{1}{x} - 1 - \\frac{1}{\\varphi}| = \\frac{|x-\\varphi|}{x\\varphi} < |x - \\varphi|$, that is, the distance from $f^{(n)}(x)$ to $\\varphi$ decreases as $n$ increases.\n\nFinally, if $-1 < x < 0$, while the terms of the sequence $x, f(x), f(f(x)), \\dots$ remain in the interval $(-1, 0)$, their distances between each term of the sequence and $\\bar{\\varphi}$ would increase (indeed, $|f(t) - \\bar{\\varphi}| = |1 + \\frac{1}{t} - 1 - \\frac{1}{\\bar{\\varphi}}| = \\frac{|t-\\bar{\\varphi}|}{t\\bar{\\varphi}} > |t - \\bar{\\varphi}|$), and when a term appears outside this interval, all the proceeding terms would also be outside of this interval.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40958,
    "subject": "Mathematics (Multi-modal)",
    "question": "Altitudes of an acute scalene triangle $ABC$ meet at point $H$. Points $M$ and $N$ are the midpoints of the segments $AB$ and $CH$, respectively. $R$ is the foot of the perpendicular from $H$ to line $CM$. $T$ is the second intersection of the line $MN$ with the circumcircle of triangle $CNR$. Let $P$ be the circumcentre of the triangle formed by the lines $CM$, $AH$ and $BH$. Prove that the lines $HP$ and $CT$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will make several uses of the fact that the lines connecting a vertex of a triangle with the circumcentre and the orthocentre are symmetrical with respect to the angle bisector at this vertex.\nLet $AC > BC$, $AA_1$ and $BB_1$ be the altitudes of the triangle $ABC$, $O$ its circumcentre. The points $A_1$, $B_1$ and $R$ lie on the circle with diameter $CH$. The triangles $A_1B_1C$ and $ABC$ are similar, and their similarity ratio is $\\frac{A_1C}{AC} = \\cos C$. Then the ratio of the circumradii of these triangles is also $\\cos C$, that is, $\\frac{CN}{AO} = \\cos C$, hence $CN = AO \\cos C = OM$. Since $OM \\parallel CN$, $COMN$ is a parallelogram. Let $\\angle ACO = \\angle BCH = x$, $\\angle OCM = \\angle NMC = y$, $\\angle RCT = \\angle RNT = z$. Then, since $NC = NR$, we have $z + \\angle TCN = \\angle RCN = \\angle CRN = y + z$, whence, $\\angle TCN = y$ and the lines $CT$ and $CM$ are symmetrical with respect to the bisector of $ACB$. Since $\\angle AHB_1 = \\angle ACB$ and $\\angle AHP = RHB_1 = \\angle B_1CR = \\angle BCT$, we have $\\angle RCT = \\angle RHP$, that is, $C$, $R$, $Q$ and $H$ belong to the same circle with diameter $CH$ (here $Q = CT \\cap PH$). The desired perpendicularity follows immediately.\n\n**Note.** We have proved in this solution that $CT$ lies on the symmedian of the triangle $ABC$.\n\n![](attached_image_1.png)\n![](attached_image_2.png)\nLet $AA_1$, $BB_1$, and $CC_1$ be the altitudes of the triangle $ABC$, and $T_1$ the midpoint of $A_1B_1$. Then $MA_1 = MB_1 = AB/2$, $NA_1 = NB_1 = CH/2$, therefore $MN$ is the perpendicular bisector of $A_1B_1$. We will prove that $T_1 = T$.\nSince $CM$ and $CT_1$ are respective medians in similar triangles $ABC$ and $A_1B_1C$, while $AB$ and $A_1B_1$ are anti-parallel, $CT_1$ lies on the symmedian $CS$ of the triangle $ABC$. Let $\\angle NRC = \\angle RCN = \\delta$. Then $90^\\circ - \\delta = \\angle CMB = \\angle CT_1B_1 = \\angle ST_1A_1$, hence $\\delta = \\angle MT_1S = \\angle CT_1N$, that is, $\\angle NRC = \\angle CT_1N$, and it follows that $T_1$ belongs to the circumcircle of the triangle $CNR$.\nNext, as in the first solution, we can prove $HP \\perp CT$. The proof is indeed complete.\n\n![](attached_image_1.png)\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40959,
    "subject": "Mathematics (Multi-modal)",
    "question": "An odd integer $a > 1$ is given. Initially, Basil chooses an even positive integer $b$ such that $b < a$ and tells it to Pete. Basil then writes down three integers on a blackboard. After that, Pete makes a sequence of moves. By a move, Pete can either add $a$ to one of the numbers on the blackboard, add $b$ to the second number, and subtract $a+b+1$ from the third number, or, conversely, subtract $a$ from one number on the blackboard, subtract $b$ from the second number, and add $a+b+1$ to the third one. At each move, Pete can independently choose which number on the blackboard is the first, the second, and the third. Pete wins if, after some moves, all three numbers on the blackboard are zero. For which $a$ Basil cannot prevent Pete's win?",
    "options": [],
    "answer": "All odd a > 1 for which Basil cannot prevent Pete's win are a = (4^n − 1)/3 for an integer n > 1, and a = (2p − 1)/3 where p is an odd prime with p ≡ 2 (mod 3).",
    "solution": "**Answer.** $a = \\frac{4^n - 1}{3}$ for an integer $n > 1$ and $a = \\frac{2p-1}{3}$, where $p$ is an odd prime number (necessarily having remainder 2 when divided by 3).\n\nWe start with describing the set of pairs $(a, b)$ for which Pete can win.\n\n**Lemma.**\n(a) If $\\gcd(2a + b + 1, a - b) > 1$, then Pete cannot win.\n(b) If $\\gcd(2a + b + 1, a - b) = 1$, then Pete can win.\n\n**Proof.**\n(a) Set $d = \\gcd(2a+b+1, a-b) > 1$. Then we have $b = a - (a-b) \\equiv a \\pmod d$ and $-(a+b+1) = a - (2a+b+1) \\equiv a \\pmod d$. This means that, on each move, all three numbers on the blackboard increased by $a$ modulo $d$ (or, conversely, decreased by $a$ modulo $d$). Hence, if the three numbers written by Basil are not all congruent modulo $d$, then this property will be preserved during Pete's moves, so that they will never be all equal (so they will never become all zero).\n\n(b) Let $x, y, z$ be three numbers on the blackboard at some moment. Then Pete can perform, by several moves, any of the following operations:\n\n1. To yield numbers $x + (a-b)$, $y - (a-b)$, and $z$. For this purpose, he can add $a$ to the first number and $b$ to the second number, and then subtract $a$ from the second number and $b$ from the first one.\n\n2. To yield numbers $x - (2a + b + 1)$, $y + (2a + b + 1)$, and $z$. For that, he can subtract $a$ from the first number and add $a+b+1$ to the second one, and then subtract $a+b+1$ from the first number and add $a$ to the second one.\n\n3. To yield numbers $x + 1$, $y - 1$, $z$. Indeed, since $2a + b + 1$ and $a - b$ are coprime, there exist positive integers $u$ and $v$ such that $u(a - b) - v(2a + b + 1) = 1$. Applying operation 1 $u$ times and operation 2 $v$ times, Pete gets the desired situation.\n\n4. Pete can, by one move, either decrease the sum of all numbers by 1 (using the first described move) or increase that sum by 1 (using the second move).\n\nPerforming operations 3 and 4, Pete can reach his goal: first, using operation 4, he gets three numbers summing up to zero; then, using operation 3, he makes one number on the blackboard zero, and then another number becomes zero. Since the sum of numbers on the blackboard does not change during these operations, the third number also vanishes at the end. $\\square$\n\nDue to the lemma, an odd number $a > 1$ satisfies the requirements if and only if the following condition holds:\n\n(*) For every even $b < a$, the numbers $a-b$ and $2a+b+1$ are coprime.\n\nNotice that\n$$\n\\gcd(a - b, 2a + b + 1) = \\gcd(a - b, 2a + b + 1 + (a - b)) = \\gcd(a - b, 3a + 1).\n$$\nNotice also that the number $a-b$ runs over all odd positive integers smaller than $a$. Therefore, property (*) is equivalent to the fact that the number $3a+1$ is coprime to any odd positive integer smaller than $a$; or, in other words, that $3a+1$ has no odd divisors smaller than $a$ and greater than $1$ — let us call such divisors *bad*.\n\nNotice that the number $3a+1$ is even; then $3a+1 = 2^k\\ell$, where $k$ is a positive integer, and $\\ell$ is an odd positive integer. If $k \\ge 2$ and $\\ell > 1$, then $\\ell \\le \\frac{3a+1}{4} < a$ is a bad divisor. If $\\ell$ is composite, i.e., $\\ell = \\ell_1\\ell_2$ with $\\ell_1, \\ell_2 \\ge 3$, then $\\ell_1 = \\frac{3a+1}{2^k\\ell_2} \\le \\frac{3a+1}{6} < a$ is a bad divisor.\n\nThe remaining cases are (1) $\\ell = 1$, and (2) $k = 1$ and $\\ell$ is a prime number. It is easy to see that, in both cases, $3a+1$ has no bad divisors. Therefore, the desired cases are precisely $3a+1 = 2^k$ and $3a+1 = 2p$, where $p$ is a prime. Checking modulo 3 it follows that $k$ is even in the former case, and $\\ell$ has a remainder 2 when divided by 3 in the latter. The result follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40960,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p > 200$ be a prime number. We call a positive integer $n$ *good* if $p$ divides the numerator of the irreducible fraction $\\frac{a_n}{b_n} = 1 + \\frac{1}{2} + \\cdots + \\frac{1}{n}$. Prove that for all large enough $N$ the number of good numbers not exceeding $N$ is not greater than $C N^{\\frac{3}{4}}$, where $C$ is a constant; possibly depends on $p$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will use congruences modulo $p$ for fractions, writing $\\frac{a}{b} \\equiv x \\pmod{p}$ for $b \\not\\equiv 0 \\pmod{p}$ if $b x \\equiv a \\pmod{p}$. A sum of such fractions is congruent to $0 \\pmod{p}$ if and only if $p$ divides the numerator of the (reduced) sum of respective usual fractions. We shall prove the statement through the following claims;\n\n**Claim 1.** For a given integer $r$, the number of positive integers $n < p$ such that $1 + \\frac{1}{2} + \\cdots + \\frac{1}{n} \\equiv r \\pmod{p}$ is not greater than $\\frac{3}{2} p^{\\frac{2}{3}}$.\nLet $n_1 < n_2 < \\cdots < n_{s+1}$ be all such $n$. Every integer $k$ appears at most $k-1$ times among the differences $n_{i+1} - n_i$ (since $n_i$ should be a root of the congruence\n$$\n\\frac{1}{x+1} + \\cdots + \\frac{1}{x+k} \\equiv 0 \\pmod{p}\n$$\nhaving at most $k-1$ roots: its left-hand side becomes a polynomial of degree $k-1$ when multiplied by the denominator).\nConsider the largest $m$ such that $1+2+\\cdots+(m-1) = \\frac{m(m-1)}{2} \\le s$. Then $s < \\frac{m(m+1)}{2}$. On the other hand, $n_{s+1}-n_1$ is the sum of $s$ differences $n_{i+1}-n_i$, and\n$$\np > n_{s+1} - n_1 \\ge 1 \\cdot 2 + 2 \\cdot 3 + \\cdots + (m-1)m = \\frac{m^3 - m}{3}.\n$$\nIf $m \\le 3$, the number of $n_i$, that is, $s+1$, is not greater than $6 = \\frac{3}{2} \\cdot 8^{2/3} < \\frac{3}{2} p^{2/3}$. And, for $m > 3$ we have\n$$\ns + 1 \\le \\frac{m(m+1)}{2} < \\frac{3}{2} \\left( \\frac{m^3 - m}{3} \\right)^{2/3} < \\frac{3}{2} p^{2/3}.\n$$\n(The middle inequality $m(m+1) < 3 \\left(\\frac{m^3-m}{3}\\right)^{2/3}$ follows from $m^3(m+1)^3 < 3(m^3-m)^2$, i.e., $m^2+m < 3(m-1)^2$.)\n\n**Claim 2.** The number of good $n$ less than $p^k$ is not greater than $\\left(\\frac{3}{2} p^{\\frac{2}{3}}\\right)^k$.\nThis can be proved by induction on $k$. The base case $k=1$ is given by Claim 1 for $r=0$. Note that for a good $n = c + n' p$, $0 \\le c < p$, the number $n' = c_1 + \\cdots + c_{k-1} p^{k-2}$ is also good. Indeed, in the sum $1 + \\frac{1}{2} + \\cdots + \\frac{1}{n}$ all the terms with denominators divisible by $p$ make together $\\frac{1}{p}(1 + \\frac{1}{2} + \\cdots + \\frac{1}{n'})$, and since the remaining sum has no $p$ in the denominator, the irreducible form of $\\frac{1}{p}(1 + \\frac{1}{2} + \\cdots + \\frac{1}{n'})$, too, should not have $p$ in the denominator).\nThe number of good $n'$ does not exceed $\\left(\\frac{3}{2} p^{\\frac{2}{3}}\\right)^{k-1}$ by the induction hypothesis. For each good $n'$ we set $-r \\equiv 1 + \\frac{1}{2} + \\cdots + \\frac{1}{n-c_0}$ (mod $p$); the sum of $c$ last terms in the sum $1 + \\frac{1}{2} + \\cdots + \\frac{1}{n}$ should be $r$ (mod $p$). By Claim 1, there are at most $\\frac{3}{2} p^{\\frac{2}{3}}$ such $c$, which gives the desired bound.\nTurning to the main claim, we choose an integer $k$ such that $p^{k-1} < N \\le p^k$. The number of good numbers not exceeding $N$ is not greater than the number of good numbers not exceeding $p^k$, and this in its turn does not exceed\n$$\n\\left(\\frac{3}{2} p^{\\frac{2}{3}}\\right)^k < \\left(200^{\\frac{1}{12}} p^{\\frac{2}{3}}\\right)^k = \\left(p^{\\frac{1}{12}} p^{\\frac{2}{3}}\\right)^k = p^{\\frac{3}{4}k} = p^{\\frac{3}{4}} (p^{k-1})^{\\frac{3}{4}} < p^{\\frac{3}{4}} N^{\\frac{3}{4}},\n$$\nwhich proves the desired claim for $C = p^{\\frac{3}{4}}$ (here we use the inequality $\\left(\\frac{3}{2}\\right)^{12} < 200$).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40961,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the diagonals of the square $ABCD$ intersect at $S$ and let $P$ be the midpoint of $AB$. Let $M$ be the intersection of $AC$ and $PD$ and $N$ the intersection of $BD$ and $PC$. A circle is inscribed in the quadrilateral $PMSN$. Prove that the radius of the circle is $MP - MS$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the centre and $r$ the radius of the circle. Let $X, Y$ be its points of contact with the sides $PM, MS$, respectively.\n\nSince $OY \\perp MS$ and $\\angle YSO = \\angle ASP = 45^\\circ$, $SY = YO = r$. Also $\\angle OPX = \\angle PDA$\n(since $OP \\parallel DA$) and $\\angle OXP = \\angle PAD = 90^\\circ$. Therefore $\\triangle OXP \\sim \\triangle PAD$. Hence\n$OX/XP = PA/AD = 1/2$. Hence $PX = 2r$. Therefore $PM - MS = 2r + MX -\nMY - r = r$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40962,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $CD$ be a chord of a circle $\\Gamma_1$ and $AB$ a diameter of $\\Gamma_1$ perpendicular to $CD$ at $N$ with $AN > NB$. A circle $\\Gamma_2$ centred at $C$ with radius $CN$ intersects $\\Gamma_1$ at points $P$ and $Q$. The line $PQ$ intersects $CD$ at $M$ and $AC$ at $K$; and the extension of $NK$ meets $\\Gamma_2$ at $L$. Prove that $PQ$ is perpendicular to $AL$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1. Extend $DC$ meeting $\\Gamma_2$ at $H$. Let the radius of $\\Gamma_2$ be $r$. Note that $DN = NC = CH = r$. Since $M$ is of equal power with respect to $\\Gamma_1$ and $\\Gamma_2$. Thus $MN \\cdot NH = MC \\cdot MD$. That is $MN(MC + r) = MC(MN + r)$ giving $MN = MC$. Thus $M$ is the midpoint of $NC$.\n\n![](attached_image_1.png)\n\nAs $K$ lies on the radical axis of $\\Gamma_1$ and $\\Gamma_2$, the points $C$, $N$, $A$, $L$ are concyclic. Thus $\\angle ALC = \\angle ANC = 90^\\circ$ so that $AL$ is tangent to $\\Gamma_2$. It follows that $AC$ is perpendicular to $NL$ at $K$, and hence $MN = MC = MK$.\n\nNow let $PQ$ intersect $AL$ at $T$. We have $\\angle TAK = \\angle KNM = \\angle NKM = \\angle LKT$ and similarly $\\angle TLK = \\angle AKT$. Consequently, $2\\angle KTL = 2(\\angle TAK + \\angle AKT) = \\angle TAK + \\angle AKT + \\angle LKT + \\angle TLK = 180^\\circ$, which means $\\angle KTL = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40963,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$ with $AC > AB$, $D$ is the foot of the perpendicular from $A$ onto $BC$ and $E$ is the foot of the perpendicular from $D$ onto $AC$. Let $F$ be the point on the line $DE$ such that $EF \\cdot DC = BD \\cdot DE$. Prove that $AF$ is perpendicular to $BF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since we are supposed to prove $\\angle AFB = 90^\\circ$, it means that the 4 points $A$, $B$, $D$, $F$ are concyclic. Note that $AC > AB$ implies that $\\angle B > \\angle C$. If $TD$ is the tangent to the circumcircle of the triangle $ABD$ with $B$ and $T$ lying on opposite sides of the line $AD$, then $\\angle ADT = \\angle B > \\angle C = \\angle ADE$ so that $\\omega$ intersects the interior of $DE$ at $F$. Therefore $F$ can only be in the interior of $DE$.\n\nNow observe that the triangles $ADE$ and $DCE$ are similar so that $AD/AE = DC/DE$. By the given condition, this can be written as $AD/AE = BD/EF$. This means the triangles $ABD$ and $AFE$ are similar. Thus $\\angle ABD = \\angle AFE$. This shows that $A$, $B$, $D$, $F$ are concyclic. Therefore $\\angle AFB = \\angle ADB = 90^\\circ$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40964,
    "subject": "Mathematics (Multi-modal)",
    "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$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nNote 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": 40965,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a set of positive integers and let $k$ be a positive integer. Set $A$ is said to be $k$-thin if there are $k$ positive integers $n_1, \\dots, n_k$ so that for all $1 \\le i < j \\le k$ and all $a, a' \\in A$, $n_i + a \\ne n_j + a'$. Suppose that $A_r$ is a $k_r$-thin set for $1 \\le r \\le m$ and that $\\cup_{r=1}^m A_r$ is the set of all positive integers. Show that\n$$\n\\frac{1}{k_1} + \\dots + \\frac{1}{k_m} \\ge 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any positive integer $n$, let $[n] = \\{1, 2, \\dots, n\\}$. Consider a $k$-thin set $A$ with the associated integers $n_1, \\dots, n_k$. Let $N$ be the maximum of these integers. We let\n$$\nA + n = \\{a + n : a \\in A\\}.\n$$\nFor any fixed integer $L$, let $B = A \\cap [L]$. Now $B + n_i$, $i = 1, \\dots, k$, are mutually disjoint subsets of $[N+L]$. Also $|B + n_i| = |B|$. Thus $k|B| = \\sum |B + n_i| \\le L + N$. (Note: this yields $\\frac{|A \\cap [L]|}{L} \\le \\frac{1}{k} + \\frac{N}{kL}$. This shows that the probability that any positive integer belongs to $A$ is $\\le 1/k$.)\nNow applying this result to the sets $A_1, \\dots, A_m$, with $N$ being the maximum of all the associated integers, we get $|A_i \\cap [L]| \\le (L+N)/k_i$. Since $\\cup_i A_i \\cap [L] = [L]$ and letting $p = \\sum \\frac{1}{k_i}$, we have, by summing over all $i$,\n$$\nL \\le (L + N)p, \\quad \\text{or} \\quad (1-p)L \\le Np.\n$$\nThe above must hold for all positive integer $L$. Since $Np$ is a constant, this is impossible unless $1-p \\le 0$ or $p \\ge 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40966,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the sum of all the 5-digit integers which are not multiples of 11 and whose digits are 1, 3, 4, 7, 9.",
    "options": [],
    "answer": "5842368",
    "solution": "First note that an integer is divisible by 11 if and only if the alternating sum of the digits is divisible by 11. In our case, these are the integers where $1$, $4$ and $7$ are at the odd positions. Let $S$ be the sum of all the 5-digit integers formed by $1$, $3$, $4$, $7$, $9$ and let $T$ be the sum of those which are multiples of $11$. Then\n$$\n\\begin{align*}\nS &= 4!(1 + 3 + 4 + 7 + 9)(1 + 10 + 100 + 1000 + 10000) \\\\\n&= 6399936 \\\\\nT &= 2!2!(1 + 4 + 7)(1 + 100 + 10000) + 3!(3 + 9)(10 + 1000) \\\\\n&= 557568.\n\\end{align*}\n$$\nThus the sum is $6399936 - 557568 = 5842368$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40967,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_n, b_n$, $n = 1, 2, \\ldots$ be two sequences of integers defined by $a_1 = 1$, $b_1 = 0$ and for $n \\ge 1$,\n$$\na_{n+1} = 7a_n + 12b_n + 6\n$$\n$$\nb_{n+1} = 4a_n + 7b_n + 3.\n$$\n\nProve that $a_n^2$ is the difference of two consecutive cubes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First we shall prove that $a_n^2$ is the difference of two consecutive cubes. To do so, we shall prove by induction that $a_n^2 = (b_n + 1)^3 - b_n^3$.\n\nWhen $n = 1$, this is true.\n\nSuppose for $n \\ge 1$, this is true. We have\n$$\n\\begin{aligned}\n(b_{n+1} + 1)^3 - b_{n+1}^3 &= 3b_{n+1}^2 + 3b_{n+1} + 1 \\\\\n&= 3(4a_n + 7b_n + 3)^2 + 3(4a_n + 7b_n + 3) + 1 \\\\\n&= 48a_n^2 + 147b_n^2 + 168a_n b_n + 84a_n + 147b_n + 37 \\\\\n&= (7a_n + 12b_n + 6)^2 + (3b_n^2 + 3b_n + 1) - a_n^2 \\\\\n&= a_{n+1}^2 + (b_n + 1)^3 - b_n^3 - a_n^2 = a_{n+1}^2\n\\end{aligned}\n$$\nwhere the last equality is by induction hypothesis.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40968,
    "subject": "Mathematics (Multi-modal)",
    "question": "A subset $T$ of $\\{7, 8, ..., 26\\}$ does not contain three elements whose product is a perfect square. Determine the maximum number of elements in $T$.",
    "options": [],
    "answer": "15",
    "solution": "By checking the parity of the prime factorization of all the integers in $T$, we partition them into disjoint triples, whose products are all squares:\n$$\n\\{8, 13, 26\\}, \\{11, 18, 22\\}, \\{14, 21, 24\\}, \\{12, 15, 20\\}, \\{9, 16, 25\\}.\n$$\nIf $|T| \\ge 16$, at least one of these triples will be in $T$, a contradiction.\n\nWe can easily check that the subset (obtained by strategically removing one element $8$, $18$, $24$, $12$, $9$ from each of the triples above):\n$$\n\\{7, 10, 11, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 25, 26\\}\n$$\nsatisfies the condition given. Thus the maximum number of elements in $T$ is $15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40969,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, ..., a_n$ be positive integers, not necessarily distinct but with at least five distinct values. Suppose that for any $1 \\le i < j \\le n$, there exist $k, \\ell$, both different from $i$ and $j$ such that $a_i + a_j = a_k + a_\\ell$. What is the smallest possible value of $n$?",
    "options": [],
    "answer": "13",
    "solution": "$a_1 \\le a_2 \\le \\dots \\le a_n$. Suppose $x < y$ are the two smallest values. Then $a_1 = x$ and let $s$ be the smallest index such that $a_s = y$. Now there are two other terms whose sum is $x+y$. Thus we have $a_2 = x$ and $a_{s+1} = y$. Since $a_1 + a_2 = 2x$, we must have $a_3 = a_4 = x$. Similarly, by considering the largest two values $w < z$, we have $a_n = a_{n-1} = a_{n-2} = a_{n-3} = z$ and another two terms equal to $w$. Since there is one other value, there are at least $4+2+4+2+1=13$ terms. The following 13 numbers\n\nsatisfy the required property: 1, 1, 1, 1, 2, 2, 3, 4, 4, 5, 5, 5, 5. Thus the smallest possible value of $n$ is 13.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40970,
    "subject": "Mathematics (Multi-modal)",
    "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": "Detailed 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) = n(n+1)/2$. Let $T_2 = T_1 \\cup \\{n(n+1)/2\\}$. Then $\\sigma(T_2) = n(n+1)$.\n\nFinally we add another $n-2$ integers\n$$\n(n-j)(n-j+2)(n-j+3)(n-j+4)\\cdots n(n+1), \\quad j=2,3,\\dots,n-1\n$$\nto obtain $T_3$. Again, it is clear that the integers are pairwise distinct and greater than $n(n+1)/2$. Now\n$$\n\\begin{aligned}\n\\sigma(T_3) &= n(n+1) + \\sum_{j=2}^{n-1} (n-j)(n-j+2)(n-j+3)(n-j+4)\\cdots n(n+1) \\\\\n&= n(n+1) + \\sum_{j=2}^{n-1} [(n-j+1)-1](n-j+2)(n-j+3)(n-j+4)\\cdots n(n+1) \\\\\n&= n(n+1) + (n+1)! - n(n+1) = (n+1)!\n\\end{aligned}\n$$\nHence by our construction, $T_3$ is a $T$-set, and since $S \\subseteq T_3$, we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40971,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $a_1 \\ge 1$ and $a_{k+1} \\ge a_k + 1$ for all $k = 1, 2, \\dots, n$, show that\n$$\na_1^3 + a_2^3 + \\dots + a_n^3 \\ge (a_1 + a_2 + \\dots + a_n)^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove it by induction. First, it is clear that $a_1^3 \\ge a_1^2$ since $a_1 \\ge 1$.\n\nNext, suppose it is true for $n$ terms. Then\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n+1} a_k^3 &\\ge a_{n+1}^3 + \\sum_{k=1}^n a_k^3 \\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$$\n\nTo complete the induction, we'll now show that $a_{n+1}^3 - a_{n+1}^2 - 2a_{n+1} \\sum_{k=1}^n a_k \\ge 0$.\n\nSince $a_{k+1} - a_k \\ge 1$, we have $a_{k+1}^2 - a_k^2 \\ge a_{k+1} + a_k$. Summing up over $k=1, \\dots, n$, and using $a_1^2 - a_1 \\ge 0$, we have\n$$\na_{n+1}^2 - a_1^2 \\ge a_{n+1} + 2 \\sum_{k=1}^n a_k - a_1 \\Rightarrow a_{n+1}^3 - a_{n+1}^2 - 2a_{n+1} \\sum_{k=1}^n a_k \\ge 0\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40972,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $a_1, \\dots, a_{15}$ are prime numbers forming an arithmetic progression with common difference $d > 0$. If $a_1 > 15$, prove that $d > 30,000$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Lemma: Suppose $p$ is prime and $a_1, \\dots, a_p$ are primes forming an A.P. with common difference $d$. If $a_1 > p$, we claim that $p \\mid d$.\n\n*Proof:* Since $p$ is prime and every $a_i$ is a prime $> p$, $p$ does not divide $a_i$ for any $i$. By the pigeonhole principle, there exist $1 \\le i < j \\le p$ so that $a_i \\equiv a_j \\pmod{p}$. Now $a_j - a_i = (j-i)d$, and $p$ does not divide $j-i$. So $p$ must divide $d$.\n\nApply the Lemma to the sequences $a_1, \\dots, a_k$ for $k = 2, 3, 5, 7, 11$ and $13$. Then all such $k$'s are factors of $d$. So $d > 2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 > 30,000$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40973,
    "subject": "Mathematics (Multi-modal)",
    "question": "An infinite sequence of integers, $a_0, a_1, a_2, \\dots$, with $a_0 > 0$, has the property that for any $n \\ge 0$, $a_{n+1} = a_n - b_n$, where $b_n$ is the number having the same sign as $a_n$, but having the digits written in the reverse order. For example if $a_0 = 1210$, $a_1 = 1089$ and $a_2 = -8712$, etc. Find the smallest value of $a_0$ so that $a_n \\ne 0$ for all $n \\ge 1$.",
    "options": [],
    "answer": "1012",
    "solution": "If $a_0$ has a single digit, then $a_1 = 0$. Thus $a_0$ has at least 2 digits. If $a_0 = \\overline{ab} = 10a + b$, then $a_1 = 9(a-b)$ which is divisible by 9. It follows that all subsequent terms are divisible by 9. Checking all 2-digit multiples of 9 shows that eventually 9 appears (Note that $\\overline{ab}$ and $\\overline{ba}$ give rise to the same sequence, but with opposite signs):\n$$\n81 \\rightarrow 63 \\rightarrow 27 \\rightarrow 45 \\rightarrow 9.\n$$\nIf $a_0 = \\overline{abc}$, then $a_1 = 99(a-c)$. Thus it suffices to investigate 3-digit multiples of 99, i.e., 198..., 990. Here we find that 99 will eventually appear:\n$$\n990 \\rightarrow 891 \\rightarrow 693 \\rightarrow 297 \\rightarrow -495 \\rightarrow 99.\n$$\nIf $a_0 = \\overline{abcd}$, then $a_1 = 999(a-d) + 90(b-c)$. If $b, c$ are both 0, then $a_1$ and all subsequent terms are multiples of 999. However, if such numbers appear in the sequence, eventually 999 will appear:\n$$\n9990 \\rightarrow 8991 \\rightarrow 6993 \\rightarrow 2997 \\rightarrow -4995 \\rightarrow 999.\n$$\nFor 1010, we get 909 and for 1011 we get -90. For 1012, we get\n$$\n1012 \\rightarrow -1089 \\rightarrow -8712 \\rightarrow 6534 \\rightarrow 2178 \\rightarrow -6534\n$$\nand the sequence becomes periodic thereafter. Thus the smallest $a_0 = 1012$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40974,
    "subject": "Mathematics (Multi-modal)",
    "question": "A student divides an integer $m$ by a positive integer $n$, where $n \\le 100$, and claims that\n$$ \\frac{m}{n} = 0.167a_1a_2\\dots $$\nShow the student must be wrong.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have\n$$\n0.167 \\le \\frac{m}{n} < 0.168 \\Rightarrow 167n \\le 1000m < 168n.\n$$\nMultiply by $6$, we get\n$$\n1002n \\le 6000m < 1008n \\Rightarrow 6000m - 1000n < 8n \\le 800.\n$$\nBut $6000m - 1000n \\ge 2n > 0$. Thus $6000m - 1000n \\ge 1000$ since it is a multiple of $1000$. We thus get a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40975,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all couples of coprime numbers $(p, q)$ satisfying:\n$$\np^3 - q^5 = (p+q)^2.\n$$",
    "options": [],
    "answer": "(7, 3)",
    "solution": "Consider the equation modulo $q$, which reduces to $p^3 \\equiv p^2 \\pmod{q}$. Since $p$ and $q$ are coprime ($p \\neq q$ clearly), we must have $p \\equiv 1 \\pmod{q}$.\n\nConsider now the equation modulo $q^2$. Then the equation becomes $p^3 \\equiv p^2 + 2pq \\pmod{q^2}$, and since $\\gcd(p, q) = 1$, $p^2 \\equiv p + 2q \\pmod{q^2}$. Let $p = 1 + aq$ for some integer $a$. Then the congruence becomes\n$$\n1 + 2aq \\equiv 1 + aq + 2q \\pmod{q^2}\n$$\nand we deduce that $a \\equiv 2 \\pmod{q}$ and hence $p \\equiv 2q + 1 \\pmod{q^2}$.\n\nWe will now that $p < (q+1)^2 = q^2+2q+1$, which will mean that $p = 2q+1$. From the equation we get $p^3 \\ge q^5$ and hence $p \\ge q$, and hence $(p+q)^2 \\le 4p^2$. Hence $p^3 - 4p^2 \\le q^5$, or\n$$\np^2(p - 4) \\le q^5.\n$$\nBut if $p \\ge (q+1)^2$, then $p^2 > q^4$ and $p-4 \\ge q^2$ (since $q \\ge 2$). We obtain the desired contradiction.\n\nNow substituting $p = 2q + 1$ into the original equation we obtain\n$$\nq^3 - 8q - 3 = 0.\n$$\nThe only integer solution is $q = 3$. Hence the only possible solution is $q = 3$, which gives $p = 7$. It is easy to check that this gives a solution to the equation.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40976,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Find the smallest positive integer $k$ with the property that for any colouring of the squares of a $2n \\times k$ chessboard with $n$ colours, there are 2 columns and 2 rows such that the 4 squares in their intersections have the same colour.",
    "options": [],
    "answer": "2n^2 - n + 1",
    "solution": "The answer is $2n^2 - n + 1$.\n\nConsider an $n$-colouring of the $2n \\times k$ chessboard. A vertical-pair is a pair of squares in the same column that are coloured the same. In every column there are at least $n$ vertical-pairs. Let $P$ be the total number of vertical-pairs and $P_i$ be the number of vertical-pairs with colour $i$. Then $P = P_1 + \\dots + P_n \\ge nk$. Thus there is colour $i$ with $P_i \\ge k$. There are $\\binom{2n}{2} = 2n^2 - n$ pairs of rows. Thus if $k \\ge 2n^2 - n + 1$, there is a pair of rows that contains two vertical-pairs with colour $i$.\n\nNext for $k = 2n^2 - n$, exhibit an $n$-colouring where no such sets of 4 squares exists. Note that it suffices to find such an $n$-colouring for the $2n \\times (2n-1)$ board. We can then rotate the colours to obtain $n$ of these boards which can then be put together to obtain the requiring $n$-colouring of the $2n \\times (2n^2 - n)$ board. For each $i = 1, 2, \\dots, 2n-1$, let $A_i = \\{(i, 2n-1+i), (i+1, 2n-2+i), \\dots, (n-1+i, n+i)\\}$, where $2n+k, k > 0$, is taken to be $k$. Note that the pairs in each $A_i$ give a partition of $\\{1, 2, \\dots, 2n\\}$. Moreover, each pair of elements appears in exactly one $A_i$. Now colour the squares of column $i$ using $n$ colours so that the two squares in each pair of $A_i$ receive the same colour and the colours the $2n$ pairs are mutually distinct. This gives an $n$-colouring of the $2n \\times 2n - 1$ board with the required property and we are done.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 40977,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $OT$ be a diameter of a circle. Let $A$ and $B$ be two distinct points on the circle both on the same side of $OT$, and let $C$ be the intersection of the tangents to the circle at $A$ and $B$. The tangent to the circle at $T$ meet the lines $OA$, $OB$ and $OC$ at $A'$, $B'$ and $C'$ respectively. Prove that $C'$ is the midpoint of $A'B'$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nConsider the inversion in the circle $\\omega$ centred at $O$ with radius $OT$. The circle $\\alpha$ with diameter $OT$ is inverted into the tangent line $\\alpha'$ to $\\alpha$ at $T$. Thus $A'$ and $B'$ are the inverses of $A$ and $B$ respectively. The circle $\\beta$ centred at $C$ with radius $CA$ or $CB$ is orthogonal to $\\alpha$ so that its inverse $\\beta'$ remains orthogonal to $\\alpha'$. This implies that the segment $A'B'$ is the diameter of $\\beta'$ and its midpoint is the centre of $\\beta'$. As the points $O$ and the centres of $\\beta$ and $\\beta'$ are collinear under the inversion, the point $C'$ is the midpoint of $A'B$'.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40978,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number and let $a_1, a_2, \\dots, a_k$ be distinct integers chosen from $1, 2, \\dots, p-1$. For $1 \\le i \\le k$, let $r_i^{(n)}$ denote the remainder of the integer $n a_i$ upon division by $p$, so $0 \\le r_i^{(n)} < p$. Define\n$$\nS = \\{ n : 1 \\le n \\le p-1,\\ r_1^{(n)} < \\dots < r_k^{(n)} \\}.\n$$\nShow that $S$ has less than $\\frac{2p}{k+1}$ elements.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $r_0^{(n)} = 0$ and $r_{k+1}^{(n)} = p$. Set\n$$\nS' = \\{ n : 1 \\le n \\le p-1,\\ \\sum_{i=0}^k | r_{i+1}^{(n)} - r_i^{(n)} | = p \\}.\n$$\nNote that\n$$\n\\sum_{i=0}^k | r_{i+1}^{(n)} - r_i^{(n)} | = p \\quad \\text{iff} \\quad r_0^{(n)} \\le r_1^{(n)} \\le \\dots \\le r_{k+1}^{(n)}\n$$\nThus $|S| = |S'|$. Since for $n \\in S'$, $| r_{i+1}^{(n)} - r_i^{(n)} | = r_{i+1}^{(n)} - r_i^{(n)} \\equiv n(a_{i+1} - a_i) \\pmod{p}$ and $p \\nmid (a_{i+1} - a_i)$, the numbers $r_{i+1}^{(n)} - r_i^{(n)}$, $1 \\le n \\le p-1$, are all distinct. Therefore\n$$\np|S'| = \\sum_{n \\in S'} \\sum_{i=0}^k | r_{i+1}^{(n)} - r_i^{(n)} | \\ge \\sum_{i=0}^k \\sum_{n \\in S'} | r_{i+1}^{(n)} - r_i^{(n)} | \\ge \\sum_{i=0}^k \\sum_{j=1}^{|S'|} j \\ge \\frac{(k+1)|S'|(|S'|+1)}{2}\n$$\nTherefore $|S| < \\frac{2p}{k+1}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40979,
    "subject": "Mathematics (Multi-modal)",
    "question": "The numbers $\\frac{1}{2}, \\frac{1}{3}, \\dots, \\frac{1}{2010}$ are written on a blackboard. A student chooses any two of the numbers, say $x, y$, erases them and then writes down $x + y + xy$. He continues to do this until only one number is left on the blackboard. What is this number?",
    "options": [],
    "answer": "2009/2",
    "solution": "We shall prove by induction that if the original numbers are $a_1, \\dots, a_n$, $n \\ge 2$, then the last number is $(1+a_1) \\cdots (1+a_n) - 1$.\nThe assertion is certainly true for $n=2$, the base case. Now suppose it is true for $n=k \\ge 2$. Consider $k+1$ numbers $a_1, \\dots, a_{k+1}$ written on the board. After one operation, we are left with $k$ numbers. Without loss of generality, we can assume that the student erases $a_k$ and writes $b_k = a_k+a_{k+1}+a_k a_{k+1} = (1+a_k)(1+a_{k+1})-1$. After a further $k$ operations, we are left with the number\n$$\n(1+a_1) \\cdots (1+a_{k-1})(1+b_k) - 1 = (1+a_1) \\cdots (1+a_{k-1})(1+a_k)(1+a_{k+1}) - 1.\n$$\nThis completes the proof of the inductive step. Thus the last number is\n$$\n(1 + \\frac{1}{2})(1 + \\frac{1}{3}) \\cdots (1 + \\frac{1}{2010}) - 1 = 2010\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40980,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all finite sets $A$ of non-negative real numbers, containing at least four elements, and such that for all distinct $a, b, c, d \\in A$, $ab + cd \\in A$.",
    "options": [],
    "answer": "All such sets are exactly A = {x, 1, 1/x, 0} for some positive real x ≠ 1 (equivalently x > 1 up to relabeling).",
    "solution": "Let $a_1 > a_2 > a_3 > a_4$ be the four largest elements in $A$. Let $a = a_1a_2 + a_3a_4$, $b = a_1a_3 + a_2a_4$ and $c = a_1a_4 + a_2a_3$. Then $a > b > c$ by the rearrangement inequality. We also know that $a, b, c \\in A$.\n\nIf there is a fifth term $a_5$, then as before, the 4 terms $a_2, a_3, a_4, a_5$ generate 3 numbers $a' > b' > c'$ and we have $c = a_1a_4 + a_2a_3 > a' = a_2a_3 + a_4a_5$. Thus we have 6 distinct numbers. In general, if we have $n + 4$ terms, $n \\ge 1$, then every 4 consecutive terms generate 3 distinct elements, giving a total of $\\ge 3(n+1) > n + 4 = |A|$. Thus $A = \\{a_1, a_2, a_3, a_4\\}$.\n\nSuppose $a_4 > 0$. If $c = a_4$, then\n$$\n(1 - a_1)a_4 = a_2a_3 \\Rightarrow 1 - a_1 > a_2 \\Rightarrow a_2 < \\frac{1}{2} \\Rightarrow a_4 < \\frac{1}{2} \\Rightarrow a < \\frac{a_1+a_3}{2} < a_1.\n$$\nHence $a = a_2$. Therefore\n$$\n1 = a_1 + \\frac{a_3a_4}{a_2} = a_1 + \\frac{a_2a_3}{a_4} \\Rightarrow a_2 = a_4, \\text{ contradiction.}\n$$\nIf $c > a_4$, then $a_1 = a = a_1a_2 + a_3a_4$, $a_2 = b = a_1a_3 + a_2a_4$, $a_3 = c = a_1a_4 + a_2a_3$. Therefore\n$$\n1 = a_2 + \\frac{a_3a_4}{a_1} = a_2 + \\frac{a_1a_4}{a_3} \\Rightarrow a_4 = 0, \\text{ contradiction.}\n$$\nHence $a_4 = 0$ and since $c > 0$, $c = a_3$, $b = a_2$ and $a = a_1$. Thus\n$$\na_3 = a_4a_1 + a_2a_3 = a_2a_3 \\Rightarrow a_2 = 1; \\quad a_2 = a_1a_3 + a_2a_4 \\Rightarrow a_1a_3 = 1.\n$$\nIt is easy to check that $A = \\{x, 1, \\frac{1}{x}, 0\\}$, satisfies the conditions given.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40981,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer $\\ge 2$. Let $S$ be a set of triples of nonnegative integers $(a, b, c)$ such that $a + b + c = n$ and if $(a, b, c)$ and $(a', b', c')$ are distinct triples in $S$, then $a \\ne a'$, $b \\ne b'$ and $c \\ne c'$. Find the maximum value of $|S|$.",
    "options": [],
    "answer": "⌊2n/3⌋ + 1",
    "solution": "Suppose $|S| = s$. Then $\\sum_{(a, b, c) \\in S} a \\ge \\sum_{i=0}^{s-1} i = \\frac{s(s-1)}{2}$. Likewise $\\sum_S b \\ge \\frac{s(s-1)}{2}$ and $\\sum_S c \\ge \\frac{s(s-1)}{2}$. Therefore\n$$\n\\frac{3s(s-1)}{2} \\le \\sum_{S} a + b + c = ns \\Rightarrow s \\le \\left\\lfloor \\frac{2n}{3} \\right\\rfloor + 1.\n$$\nThe following shows the bound can be attained.\n\n|   |   |   |   |   |   |   |   |   |\n|---|---|---|---|---|---|---|---|---|\n| **n = 3k - 1** |   |   | **n = 3k** |   |   | **n = 3k + 1** |   |   |\n| $a$ | $b$ | $c$ | $a$ | $b$ | $c$ | $a$ | $b$ | $c$ |\n| 0 | $k + 1$ | $2k - 2$ | 0 | $k$ | $2k$ | 0 | $k$ | $2k + 1$ |\n| 1 | $k + 2$ | $2k - 4$ | 1 | $k + 1$ | $2k - 2$ | 1 | $k + 1$ | $2k - 1$ |\n| ... | ... | ... | ... | ... | ... | ... | ... | ... |\n| $k - 1$ | $2k$ | 0 | $k$ | $2k$ | 0 | $k$ | $2k$ | 1 |\n| $k$ | 0 | $2k - 1$ | $k + 1$ | 0 | $2k - 1$ | $k + 1$ | 0 | $2k$ |\n| $k + 1$ | 1 | $2k - 3$ | $k + 2$ | 1 | $2k - 3$ | $k + 2$ | 1 | $2k - 2$ |\n| ... | ... | ... | ... | ... | ... | ... | ... | ... |\n| $2k - 1$ | $k - 1$ | 1 | $2k$ | $k - 1$ | 1 | $2k$ | $k - 1$ | 2 |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40982,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots$ be a non-constant sequence of positive integers such that $m-n \\mid a_m - a_n$ for all distinct pairs of positive integers $m, n$. Prove that there is an infinite set of primes such that each divides $a_n$ for some $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the union of the sets of the prime divisors of $a_n$, $n \\ge 1$. Denote by $v_p(n)$ the highest power of $p$ that divides $n$. Suppose on the contrary, that $|P|$ is finite. Let $A = \\{\\prod_{p \\in P} p^k : k \\in \\mathbb{Z}, k > v_p(a_1) \\ \\forall p \\in P\\}$. Then $|A| = \\infty$. Let $t \\in A$. Assume that $a_{t+1} \\ne a_1$. Then there exists $q \\in P$ such that $v_q(a_{t+1}) \\ne v_q(a_1)$. Thus $v_q(a_{t+1} - a_1) = \\min(v_q(a_{t+1}), v_q(a_1)) \\le v_q(a_1) < v_q(t)$. But this contradicts the fact that $n \\mid (a_{t+1} - a_1)$. Thus we must have $a_{t+1} = a_1$. Now for any positive integer $m$, we have $(t+1-m) \\mid (a_{t+1} - a_m) = (a_1 - a_m)$. Since this is true for all $t \\in A$ and $|A| = \\infty$, $a_1 = a_m$ for all $m$, a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40983,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the acute-angled non-isosceles triangle $ABC$, $O$ is its circumcentre, $H$ is its orthocentre and $AB > AC$. Let $Q$ be a point on $AC$ such that the extension of $HQ$ meets the extension of $BC$ at the point $P$. Suppose $BD = DP$, where $D$ is the foot of the perpendicular from $A$ onto $BC$. Prove that $\\angle ODQ = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Drop perpendiculars $OM$ and $QX$ onto $BC$, and $QY$ from $Q$ onto $AD$. First $2DM = DM + BD - BM = BD - (BM - DM) = PD - (CM - DM) = PD - CD = PC$. It is a well-known fact that $2OM = AH$.\n\n![](attached_image_1.png)\n\nNext $\\angle CPQ = \\angle DBH = \\angle HAQ$ so that the triangles $CPQ$ and $HAQ$ are similar. Thus the triangles $XPQ$ and $YAQ$ are similar. Therefore\n$$\n\\frac{QX}{DX} = \\frac{QX}{QY} = \\frac{PC}{AH} = \\frac{DM}{OM}.\n$$\nHence the triangles $DXQ$ and $OMD$ are similar. It follows that $\\angle ODQ = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40984,
    "subject": "Mathematics (Multi-modal)",
    "question": "If 46 squares are colored red in a $9 \\times 9$ board, show that there is a $2 \\times 2$ block on the board in which at least 3 of the squares are colored red.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that at most 2 squares are colored red in any $2 \\times 2$ square. Then in any $9 \\times 2$ block, there are at most 10 red squares. Moreover, if there are 10 red squares, then there must be 5 in each row. This can be seen as follows. There are $8 \\times 2 \\times 2$ blocks. Counting multiplicity, there are altogether 16 red squares. Each red square in the interior is counted twice while each red square at the edge is counted once. If there are 11 red squares, then there are at least 7 red squares in the interior. Thus the total count is at least $4+7 \\times 2 = 18 > 16$, a contradiction. If there are exactly 10 red squares, then 4 of them must be at the edge and the red squares in each row are not next to each other and hence there are 5 in each row.\n\nNow let the number of red squares in row $i$ be $r_i$. Then $r_i + r_{i+1} \\le 10$, $1 \\le i \\le 8$. Suppose that some $r_i \\le 5$ with $i$ odd. Then\n$$\n(r_1 + r_2) + \\dots + (r_{i-2} + r_{i-1}) + r_i + \\dots + (r_8 + r_9) \\le 4 \\times 10 + 5 = 45\n$$\nwhich leads to a contradiction. On the other hand, suppose that $r_1, r_3, r_5, r_7, r_9 \\ge 6$.\nThen the sum of any 2 consecutive $r_i$'s is $\\le 9$. Again we get a contradiction as\n$$\n(r_1 + r_2) + \\cdots + (r_7 + r_8) + r_9 \\le 4 \\times 9 + 9 = 45.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40985,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x, y, z > 0$ such that $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} < \\frac{1}{xyz}$. Show that\n$$\n\\frac{2x}{\\sqrt{1+x^2}} + \\frac{2y}{\\sqrt{1+y^2}} + \\frac{2z}{\\sqrt{1+z^2}} < 3.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $r = 1/x$, $s = 1/y$, $t = 1/z$. There exists $\\alpha < 1$ such that $r+s+t = \\alpha^2 rst$ or $\\alpha(r+s+t) = \\alpha^3 rst$. Let $a = \\alpha r$, $b = \\alpha s$, $c = \\alpha t$. Write $a = \\tan A$, $b = \\tan B$, $c = \\tan C$, then $A+B+C = \\pi$. It is clear that\n$$\n\\begin{aligned}\n\\frac{1}{2} \\times \\text{LHS} &= \\frac{1}{\\sqrt{1+r^2}} + \\frac{1}{\\sqrt{1+s^2}} + \\frac{1}{\\sqrt{1+t^2}} \\\\\n&< \\frac{1}{\\sqrt{1+a^2}} + \\frac{1}{\\sqrt{1+b^2}} + \\frac{1}{\\sqrt{1+c^2}} \\\\\n&= \\cos A + \\cos B + \\cos C \\\\\n&\\le 3 \\cos \\left( \\frac{A+B+C}{3} \\right) = \\frac{3}{2} = \\frac{1}{2} \\times \\text{RHS.}\n\\end{aligned}\n$$\n\n\n2nd soln:\nNote that\n$$\n\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} < \\frac{1}{xyz} \\Rightarrow xy + yz + xz < 1.\n$$\nHence\n$$\n\\frac{2x}{\\sqrt{1+x^2}} < \\frac{2x}{\\sqrt{x^2+xy+xz+yz}} = \\frac{2x}{\\sqrt{(x+y)(x+z)}}.\n$$\nBy AM-GM we have\n$$\n\\frac{2x}{\\sqrt{(x+y)(x+z)}} \\le \\frac{x}{x+y} + \\frac{x}{x+z}.\n$$\nSimilarly,\n$$\n\\frac{2y}{\\sqrt{(y+z)(y+x)}} \\le \\frac{y}{y+z} + \\frac{y}{y+x}, \\quad \\frac{2z}{\\sqrt{(z+x)(z+y)}} \\le \\frac{z}{z+x} + \\frac{z}{z+y}.\n$$\nThe desired inequality then follows by adding up the three inequalities.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40986,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $P(x)$ with real coefficients such that\n$$\nP(a) \\in \\mathbb{Z} \\text{ implies that } a \\in \\mathbb{Z}.\n$$",
    "options": [],
    "answer": "All constant polynomials with non-integer value, and all linear polynomials of the form P(x) = (x + q)/p where p and q are integers with p nonzero.",
    "solution": "Let $P(x) = a_n x^n + \\cdots + a_1 x + a_0$. Define $Q(x) = P(x+1) - P(x)$. Then $Q(x)$ is of degree $n-1$. We'll prove by contradiction that $|Q(x)| \\le 3$ for all $x$. This will imply that $n \\le 1$.\n\nAssume that $|Q(a)| > 3$ for some $a \\in \\mathbb{R}$. Then $|P(a+1) - P(a)| > 3$. Thus there are 3 integers between $P(a)$ and $P(a+1)$. Hence there exists three values of $x \\in [a, a+1]$ such that $P(x)$ is an integer. Thus there are three integers in $[a, a+1]$, a contradiction.\n\nThere are two cases:\nCase (i) $n = 0$. Here we have $P(x) = c$ where $c \\notin \\mathbb{Z}$.\n\nCase (ii) $n = 1$. Here $P(x) = s x + t$. There are two integers $m, n$ such that $P(m) = s m + t = 0$ and $P(n) = s n + t = 1$. Thus $s(n - m) = 1$ implying that $1/s \\in \\mathbb{Z}$ and $s m + t = 0$ implying that $t/s \\in \\mathbb{Z}$. Letting $1/s = p$ and $t/s = q$, $P(x) = \\frac{x}{p} + \\frac{q}{p}$ where $p, q \\in \\mathbb{Z}$ and $p \\neq 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 40987,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $(m, n)$ such that\n$$\nm+n - \\frac{3mn}{m+n} = \\frac{2011}{3}.\n$$",
    "options": [],
    "answer": "(1144, 377) and (377, 1144)",
    "solution": "Answer: $(m, n) = (1144, 377)$ or $(377, 1144)$.\n\nLet $m$ and $n$ be positive integers satisfying the given equation. That is $2011(m + n) = 3(m^2 - mn + n^2)$. Since the equation is symmetric in $m$ and $n$, we may assume $m \\ge n$. If $m = n$, then $m = n = 4022/3$ which is not an integer. So we may further assume $m > n$. Let $p = m+n$ and $q = m-n > 0$. Then $m = (p+q)/2$ and $n = (p-q)/2$, and the equation becomes $8044p = 3(p^2 + 3q^2)$. Since 3 does not divide 8044, it must divide $p$. By letting $p = 3r$, the above equation reduces to $8044r = 3(3r^2 + q^2)$. From this, 3 must divide $r$. By letting $r = 3s$, we get $8044s = 27s^2 + q^2$, or equivalently\n$$\ns(8044 - 27s) = q^2. \\quad (*)\n$$\nFor $s$ between 1 and $\\lfloor 8044/27 \\rfloor = 297$, the number $s(8044 - 27s)$ is a square only when $s = 169$. To narrow down the values of $s$, we proceed as follows.\nLet $s = 2^\\alpha u$, where $\\alpha$ is a nonnegative integer and $u$ is an odd positive integer. Suppose $\\alpha$ is odd and $\\alpha \\ge 3$. Then (*) becomes $2^{\\alpha+2}u(2011 - 27 \\times 2^{\\alpha-2}u) = q^2$ which is a square. Since $\\alpha+2$ is odd, 2 must divide $2011 - 27 \\times 2^{\\alpha-2}u$ implying 2 divides 2011 which is a contradiction. Next suppose $\\alpha = 1$. Then we have $u(2 \\times 2011 - 27u) = (q/2)^2$. If $u$ is not a square, then there exists an odd prime factor $t$ of $u$ such that $t$ divides $2 \\times 2011 - 27u$. Thus $t$ divides $2 \\times 2011$ so that $t$ must be 2011 since 2011 is a prime. But then $u \\ge t = 2011$ contradicting $2 \\times 2011 - 27u > 0$. Therefore $u$ must be a square. This implies that $2 \\times 2011 - 27u$ is also a square. Taking mod 4, we have $u \\equiv 0$ or $1$ (mod 4) so that $2 \\times 2011 - 27u \\equiv 2$ or $3$ (mod 4) which contradicts the fact that $2 \\times 2011 - 27u$ is a square. Thus $\\alpha \\ne 1$ too. Consequently $\\alpha$ must be even. Then dividing both sides of (*) by $2^\\alpha$, we obtain $u(8044 - 27 \\times 2^\\alpha u) = q^2/2^\\alpha$ which is a square. Now suppose $u$ is not a square. Then there exists an odd prime factor $v$ of $u$ such that $v$ divides $8044 - 27 \\times 2^\\alpha u$. Then $v$ must divide 8044 so that $v = 2011$. Thus $u \\ge v = 2011$. This again contradicts the fact that $8044 - 27 \\times 2^\\alpha u > 0$. Therefore $u$ is a square. Consequently $s$ is also a square. Write $s = w^2$. Then (*) becomes $w^2(8044 - 27w^2) = q^2 \\ge 0$. From this $w \\le \\lfloor (8044/27)^{1/2} \\rfloor = 17$. A direct verification shows that $8044 - 27w^2$ is a square only when $w = 13$. Thus $s = w^2 = 169$. Then $p = 3r = 9s = 1521$, and by (*) $q = (169 \\times (8044 - 27 \\times 169))^{1/2} = 767$. Lastly, $m = (p+q)/2 = 1144$ and $n = (p-q)/2 = 377$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40988,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ so that\n$$\n(x+y)(f(x)-f(y)) = (x-y)f(x+y)\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = a x + b x^2 for arbitrary real constants a and b",
    "solution": "Suppose that $f$ is a solution. Let\n$$\na = \\frac{1}{2}(f(1) - f(-1)), \\quad b = \\frac{1}{2}(f(1) + f(-1))\n$$\nand $g(x) = f(x) - a x - b x^2$. Then\n$$\n(x+y)(g(x)-g(y)) = (x-y)g(x+y)\n$$\nand $g(1) = g(-1) = 0$. Letting $y = 1$ and $y = -1$ above give\n$$\n\\begin{aligned}\n(x+1)g(x) &= (x-1)g(x+1) \\\\\nxg(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$$\nfor all $x$. So $g(x) = 0$ for all $x$. Hence $f(x) = a x + b x^2$. We can check directly that any function of this form (for some $a, b \\in \\mathbb{R}$) satisfies the given equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40989,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each $i = 1, 2, \\dots, N$, let $a_i, b_i, c_i$ be integers such that at least one of them is odd. Show that one can find integers $x, y, z$ such that $xa_i + yb_i + zc_i$ is odd for at least $4N/7$ different values of $i$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider all the 7 triples $(x, y, z)$, where $x, y, z$ are either $0$ or $1$ but not all $0$. For each $i$, at least one of the numbers $a_i, b_i, c_i$ is odd. Thus among the 7 sums $xa_i + yb_i + zc_i$, 3 are even and 4 are odd. Thus there are altogether $4N$ odd sums. Thus there is choice of $(x, y, z)$ for which at least $4N/7$ of the corresponding sums are odd. (You can think of a table where the rows are numbered $1, 2, \\ldots, N$ and the columns correspond to the 7 choices of the triples $(x, y, z)$. The 7 entries in row $i$ are the 7 sums $xa_i + yb_i + zc_i$. Thus there are 4 odd numbers in each row, making a total of $4N$ odd sums in the table. Since there are 7 columns, one of the columns must contain at least $4N/7$ odd sums.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40990,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be an odd prime. Prove that\n$$\n1^{p-2} + 2^{p-2} + 3^{p-2} + \\dots + \\left(\\frac{p-1}{2}\\right)^{p-2} \\equiv \\frac{2-2^p}{p} \\pmod{p}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, for each $i = 1, 2, \\dots, \\frac{p-1}{2}$,\n$$\n\\frac{2i}{p} \\binom{p}{2i} = \\frac{(p-1)(p-2)\\cdots(p-(2i-1))}{(2i-1)!} \\equiv \\frac{(-1)(-2)\\cdots(-(2i-1))}{(2i-1)!} \\equiv -1 \\pmod{p}.\n$$\nHence\n$$\n\\begin{aligned}\n\\sum_{i=1}^{(p-1)/2} i^{p-2} &\\equiv - \\sum_{i=1}^{(p-1)/2} i^{p-2} \\frac{2i}{p} \\binom{p}{2i} \\\\\n&\\equiv -\\frac{2}{p} \\sum_{i=1}^{(p-1)/2} i^{p-1} \\binom{p}{2i} \\\\\n&\\equiv -\\frac{2}{p} \\sum_{i=1}^{(p-1)/2} \\binom{p}{2i} \\pmod{p} \\quad \\text{(by Fermat's Little Theorem.)}\n\\end{aligned}\n$$\nThe last summation counts the even-sized nonempty subsets of a $p$-element set, of which there are $2^{p-1} - 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40991,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $2012$ distinct points in the plane each of which is to be coloured using one of $n$ colours so that the number of points of each colour are distinct. A set of $n$ points is said to be multi-coloured if their colours are distinct. Determine $n$ that maximizes the number of multi-coloured sets.",
    "options": [],
    "answer": "61",
    "solution": "Let $m_1 < m_2 < \\dots < m_n$ be the number of points of each colour. We call $m_1, m_2, \\dots, m_n$ the colour distribution. Then $m_1 + \\dots + m_n = 2012$ and the number of multi-coloured sets is $M = m_1 m_2 \\dots m_n$. We have the following observations.\n\n(i) $m_1 > 1$. For if $m_1 = 1$, then $m_1 m_2 \\dots m_n < m_2 m_3 \\dots m_{n-1} (1 + m_n)$. This means if we use $n-1$ colours with colour distribution $m_2, m_3, \\dots, m_{n-1}, (1+m_n)$, we obtain a larger $M$.\n\n(ii) $m_{i+1} - m_i \\le 2$ for all $i$. For if there exists $k$ with $m_{k+1} - m_k \\ge 3$, then the colour distribution with $m_k, m_{k+1}$ replaced by $m_k + 1, m_{k+1} - 1$ yields a larger $M$.\n\n(iii) $m_{i+1} - m_i = 2$ for at most one $i$. For if there exist $i < j$ with $m_{i+1} - m_i = m_{j+1} - m_j = 2$, the colour distribution with $m_i, m_{j+1}$ replaced by $m_i + 1, m_{j+1} - 1$ yields a larger $M$.\n\n(iv) $m_{i+1} - m_i = 2$ for exactly one $i$. For if $m_{i+1} - m_i = 1$ for all $i$, then $m_1 + \\dots + m_n = n m_1 + \\frac{n(n-1)}{2} = 2012 = 4 \\cdot 503$. Thus $n(2m_1 - 1 + n) = 8 \\cdot 503$. Since $503$ is prime, the parity of $n$ and $2m_1 - 1 + n$ are opposite and $2m_1 - 1 + n > n$, we have $n=8$ and $m_1 = 248$. The colour distribution with $m_1$ replaced by two numbers $2, 246$ (using $n+1$ colours) yields a larger $M$.\n\n(v) $m_1 = 2$. If $m_n - m_{n-1} = 2$, then from (iv), we have $m_1 + \\dots + m_n = n m_1 + \\frac{n(n-1)}{2} + 1 = 2012$. Thus $n(2m_1 - 1 + n) = 2 \\cdot 2011$. Since $2011$ is prime, we get $n=2$ and $m_1 = 1005$ which will lead to a contradiction as in (iv). Thus $m_n - m_{n-1} = 1$. $m_{i+1} - m_i = 2$ for some $1 \\le i \\le n-2$. Suppose $m_1 \\ge 3$. Let $m' = m_{i+2} - 2$. Then $m_i < m' < m_{i+1}$ with replacing $m_{i+2}$ by $2, m'$ yields a larger $M$. Thus $m_1 = 2$.\n\nFrom the above analysis, with $n$ colours, we see that the colour distribution $2, 3, \\dots, i-1, i+1, i+2, \\dots, n+1, n+2$, with $3 \\le i \\le n$, yields the maximum $M$. Now we have $\\sum m_i = \\frac{1}{2}(n+1)(n+4) - i = 2012$. Thus $n^2 + 5n - 4020 = 2i$, $3 \\le i \\le n$, i.e., $n^2 + 5n \\ge 4026$ and $n^2 + 3n \\le 4020$. Thus $n=61$ and $i=3$. Thus the maximum is achieved when $n=61$ with the colour distribution $2, 4, 5, 6, \\dots, 63$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40992,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots$ be a sequence of integers defined recursively by $a_1 = 2013$ and for $n \\ge 1$, $a_{n+1}$ is the sum of the $2013$th power of the digits of $a_n$. Do there exist distinct positive integers $i, j$ such that $a_i = a_j$?",
    "options": [],
    "answer": "Yes",
    "solution": "The answer is yes. For any positive integer $n$, let $f(n)$ be the sum of $2013$th power of the digits of $n$. Let $S = \\{1, 2, \\dots, 10^{2017} - 1\\}$, and $n = \\overline{a_1a_2\\dots a_{2017}} \\in S$. Then\n$$\nf(n) = \\sum a_i^{2013} \\le 2017 \\cdot 9^{2013} < 10^4 \\cdot 10^{2013} = 10^{2017} \\in S.\n$$\nSince $a_i = f^{(i)}(2013) \\in S$, there exist distinct positive integers $i, j$ such that $a_i = a_j$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40993,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle and let $D$, $E$ and $F$ be the midpoints of $BC$, $CA$ and $AB$ respectively. Construct a circle, centred at the orthocentre of triangle $ABC$, such that triangle $ABC$ lies in the interior of the circle. Extend $EF$ to intersect the circle at $P$, $FD$ to intersect the circle at $Q$ and $DE$ to intersect the circle at $R$. Show that $AP = BQ = CR$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the radius of the circle be $r$. Let $X$, $Y$ and $Z$ be the feet of the altitudes from $A$, $B$ and $C$ respectively. Let $PE$ intersect the altitude from $A$ at $U$. We have\n$$\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$$\nSimilarly, $BQ = r^2 + BH \\cdot HY$, and $CR = r^2 + CH \\cdot HZ$. Since $AH \\cdot HX = BH \\cdot HY = CH \\cdot HZ$, we have $AP = BQ = CR$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40994,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N$ be a positive integer. Prove that there exists a positive integer $n$ such that $n^{2013} - n^{20} + n^{13} - 2013$ has at least $N$ distinct prime factors.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The result is true for any nonconstant polynomial $f(n) = a_m n^m + a_{m-1} n^{m-1} + \\dots + a_0$ with integer coefficients. We may assume that $a_m > 0$. Thus there exists a positive integer $n_0$ such that $f(n)$ is positive and increasing on $(n_0, \\infty)$.\n\nIt suffices to show that if for some $n_1 > n_0$, $f(n_1) = p_1^{r_1} \\cdots p_k^{r_k}$ has exactly $k$ distinct prime factors, then for some $n_2 > n_1$, $f(n_2)$ has more than $k$ prime factors. Given such an $n_1$, let $n_2 = n_1 + p_1^{r_1+1} \\cdots p_k^{r_k+1}$. Then\n$$\nf(n_2) \\equiv p_1^{r_1} \\cdots p_k^{r_k} \\pmod{p_1^{r_1+1} \\cdots p_k^{r_k+1}}\n$$\nHence, for each $j$, $1 \\le j \\le k$, we have that $p_j^{r_j}$ divides $f(n_2)$ but $p_j^{r_j+1}$ does not divide $f(n_2)$. As $f(n_2) > f(n_1) = p_1^{r_1} \\cdots p_k^{r_k}$, it follows that $f(n_2)$ must have at least $k+1$ prime factors.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 40995,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $F$ be a finite nonempty set of integers and let $n$ be a positive integer. Suppose that\n* Any $x \\in F$ may be written as $x = y + z$ for some $y, z \\in F$;\n* If $1 \\le k \\le n$ and $x_1, \\dots, x_k \\in F$, then $x_1 + \\dots + x_k \\ne 0$.\nShow that $F$ has at least $2n + 2$ distinct elements.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because of the second condition above, $0 \\notin F$. If $F$ contains only positive elements, let $x$ be the smallest element in $F$. But then $x = y+z$, and $y, z > 0$ imply that $y, z < x$, a contradiction. Hence $F$ contains negative elements. A similar argument shows that $F$ contains positive elements.\n\nPick any positive element of $F$ and label it as $x_1$. Assume that positive elements of $F$, $x_1, \\dots, x_k$, have been chosen. We can write $x_k = y + z$, where $y, z \\in F$. We may assume that $y > 0$. Label $y$ as $x_{k+1}$. Carry on in this manner to choose positive elements $x_1, x_2, \\dots$ of $F$, not necessarily distinct. Since $F$ is a finite set, there exist positive integers $i < j$ such that $x_i, \\dots, x_{j-1}$ are distinct and $x_j = x_i$. There are $x_i, \\dots, x_{j-1} \\in F$ such that\n$$\n\\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$$\nSince $x_i = x_j$, we see that $x_i + x_{i+1} + \\dots + x_{j-1} = 0$. By the assumption, $j-i > n$. Since the elements $x_i, \\dots, x_{j-1}$ are distinct, $F$ contains at least $j-i \\ge n+1$ positive elements. Similarly, $F$ contains at least $n+1$ negative elements. The result follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40996,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with integral side lengths such that $\\angle A = 3\\angle B$. Find the minimum value of its perimeter.",
    "options": [],
    "answer": "21",
    "solution": "Let the sides be $a, b, c$. From the sine rule, we have\n$$\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$$\nThus\n$$\n2 \\cos B = \\frac{a^2 + c^2 - b^2}{ac} \\in \\mathbb{Q}.\n$$\n\nHence there exist coprime positive integers $p, q$ such that $2 \\cos B = \\frac{2}{q}$. Hence\n$$\n\\frac{a}{b} = \\frac{p^2}{q^2} - 1 \\quad \\Leftrightarrow \\quad \\frac{a}{p^2 - q^2} = \\frac{b}{q^2};\n$$\n$$\n\\frac{c}{b} = \\frac{p^3}{q^3} - \\frac{2p}{q} \\quad \\Leftrightarrow \\quad \\frac{c}{p^3 - 2pq^2} = \\frac{b}{q^3}.\n$$\nThus\n$$\n\\frac{a}{(p^2 - q^2)q} = \\frac{b}{q^3} = \\frac{c}{p^3 - 2pq^2} = \\frac{e}{f}, \\quad \\text{gcd}(e, f) = 1.\n$$\nSince perimeter is minimum, $\\text{gcd}(a, b, c) = 1$. From $\\text{gcd}(e, f) = 1$, we have $f \\mid q^3$ and $f \\mid p^3 - 2pq^2$. We'll prove that $f = 1$.\nIf $f > 1$, then it has a prime divisor $f' > 1$ such that $f' \\mid q^3$ and $f' \\mid p^3 - 2pq^2$. Thus $f' \\mid q$ and $f' \\mid p$, contradicting $\\text{gcd}(p, q) = 1$. Thus $f = 1$. From $\\text{gcd}(a, b, c) = 1$, we conclude that $e = 1$. Thus\n$$\na = (p^2 - q^2)q, \\quad b = q^3, \\quad c = p^3 - 2pq^2.\n$$\nFrom $0^\\circ < \\angle A + \\angle B = 4\\angle B < 180^\\circ$, we get $0^\\circ < \\angle B < 45^\\circ$ and hence $\\sqrt{2} < 2 \\cos B < 2$ implying that $\\sqrt{q} < p < 2q$. The smallest positive integers satisfying this inequality is $p = 3, q = 2$. Since $a + b + c = p^2q + p(p^2 - 2q^2)$ and $p^2 - 2q^2 = 1$, we see that the minimum perimeter is achieved when $p = 3, q = 2$ and the value is $21$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40997,
    "subject": "Mathematics (Multi-modal)",
    "question": "The quadrilateral $ABCD$ is inscribed in a circle which has diameter $BD$. Points $A'$ and $B'$ are symmetric to $A$ and $B$ with respect to the line $BD$ and $AC$ respectively. If the lines $A'C$, $BD$ intersect at $P$ and $AC$, $B'D$ intersect at $Q$, prove that $PQ$ is perpendicular to $AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AC$ intersect $BD$ at $R$. Then $\\angle BAR = \\angle BAC = \\angle BA'P = \\angle BAP$. That is $AB$ bisects $\\angle PAR$. As $\\angle BAD = 90^\\circ$, we also have $AD$ is the external bisector of $\\angle PAR$. By the angle bisector theorem, we have\n$$\n\\frac{BR}{BP} = \\frac{DR}{DP} = \\frac{AR}{AP} \\qquad (1)\n$$\nAs $B$ and $B'$ are symmetric with respect to the line $AC$, we have $\\angle BQR = \\angle B'QR = \\angle DQR$. Thus $QR$ bisects $\\angle BQD$. By the angle bisector theorem and (1), we have\n$$\n\\frac{QD}{QB} = \\frac{RD}{RB} = \\frac{PD}{PB}.\n$$\nThus $QP$ is the external bisector of $\\angle BQD$. Hence $\\angle RQP = 90^\\circ$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40998,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{R}$ be the set of real numbers. Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for all $x, y \\in \\mathbb{R}$,\n$$\nf(xf(y) + x) = xy + f(x).\n$$",
    "options": [],
    "answer": "f(x) = x and f(x) = -x",
    "solution": "Let $x = 1$, we have $f(f(y) + 1) = y + f(1)$. From this we conclude that $f$ is a bijection.\nLet $y = 0$, we have $f(xf(0) + x) = f(x)$. Since $f$ is injective, we have $xf(0) + x = x$, i.e., $xf(0) = 0$ for all $x$. Hence $f(0) = 0$.\nFor $x \\neq 0$, take $y = -f(x)/x$, then $f(xf(y) + x) = 0$. Thus $xf(y) + x = 0$ or $f(y) = -1$. Thus $f(-f(x)/x) = -1$. Hence $-f(x)/x = c$, where $c$ is the constant satisfying $f(c) = -1$. Thus $f(x) = -cx$ for $x \\neq 0$. Since $f(0) = 0$, $f(x) = -cx$ for all $x \\in \\mathbb{R}$.\n\nSubstitute this into the original equation, we get $f(-xcy+x) = -c(-xcy+x) = xy-cx$.\nThus $c^2xy = xy$. Hence $c^2 = 1$ or $c = \\pm 1$. It is easy to check that both $f(x) = x$ and\n$f(x) = -x$ satisfy the functional equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 40999,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $0 < a_1 < a_2 < \\dots < a_n$ be real numbers. Prove that\n$$\n\\left(\\frac{1}{1+a_1} + \\frac{1}{1+a_2} + \\dots + \\frac{1}{1+a_n}\\right)^2 \\le \\frac{1}{a_1} + \\frac{1}{a_2-a_1} + \\frac{1}{a_3-a_2} + \\dots + \\frac{1}{a_n-a_{n-1}}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By Cauchy-Schwarz inequality,\n$$\n\\text{LHS} \\le \\left( \\frac{1}{a_1} + \\frac{1}{a_2 - a_1} + \\dots + \\frac{1}{a_n - a_{n-1}} \\right) \\left( \\frac{a_1}{(1+a_1)^2} + \\frac{a_2 - a_1}{(1+a_2)^2} + \\dots + \\frac{a_n - a_{n-1}}{(1+a_n)^2} \\right).\n$$\nNote that $\\frac{a_1}{(1+a_1)^2} \\le \\frac{a_1}{1+a_1}$ and for $i=2, \\dots, n$,\n$$\n\\frac{a_i - a_{i-1}}{(1+a_i)^2} \\le \\frac{a_i - a_{i-1}}{(1+a_{i-1})(1+a_i)} = \\frac{1}{1+a_{i-1}} - \\frac{1}{1+a_i}.\n$$\nThus, after telescoping,\n$$\n\\frac{a_1}{(1+a_1)^2} + \\frac{a_2 - a_1}{(1+a_2)^2} + \\dots + \\frac{a_n - a_{n-1}}{(1+a_n)^2} \\le \\frac{a_1}{1+a_1} + \\frac{1}{1+a_1} - \\frac{1}{1+a_n} < 1\n$$\nand we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41000,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a $50 \\times 50$ grid, an integer is written in each of the $2500$ cells. Let $G$ be the configuration of $8$ cells formed by removing the central cell of a $3 \\times 3$ grid. It is given that for any group of $8$ cells in the $50 \\times 50$ grid forming the configuration $G$, the sum of the numbers written in the $8$ cells is positive. Prove that there is a $2 \\times 2$ grid so that the sum of the numbers in the $4$ cells is positive.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the $4 \\times 4$ grid. Place $4$ overlapping copies of $G$ as shown (left figure), where the number $i = 1, 2, 3, 4$ indicates the cells of the $i^{\\text{th}}$ copy of $G$. The same grid is also covered by $8$ overlapping copies of $2 \\times 2$ grid (right), with each cell covered the same number of times. (Note that in the first figure the top left cells of the $4$ copies of $G$ form a $2 \\times 2$ grid while in the second figure, the top left cells of the $8 \\times 2 \\times 2$ grid form the configuration $G$. This is important for the general case.)\n\n| 1 | 12 | 12 | 2 |\n|---|----|----|---|\n| 13 | 234 | 134 | 24 |\n| 13 | 124 | 123 | 24 |\n| 3 | 34 | 34 | 4 |\n\n| 1 | 12 | 23 | 3 |\n|---|----|----|---|\n| 14 | 124 | 235 | 35 |\n| 46 | 467 | 578 | 58 |\n| 6 | 67 | 78 | 8 |\n\nThus the sum of the sums of the numbers in the cells of the $4$ copies of $G$ is equal to that of the $8$ copies of the $2 \\times 2$ grid. Since the former is positive, one of the $2 \\times 2$ grid must be positive as well.\n\nIt is easy to see that the result holds for any two configurations $G_1, G_2$, provided the grid is large enough. Choose a pair of coordinate axes. Let the number of cells in $G_1$ and $G_2$ be $m$ and $n$ respectively. Place $G_1$ and $G_2$ in some position. Let $a_1, \\dots, a_m$ be the centres of the cells covered by $G_1$ and let $b_1, \\dots, b_n$ be the centres of the cells covered by $G_2$. Let $G_1(b_i)$ be the position of $G_1$ after it has been translated by $b_i$ and let $P_1(b_i)$ be the corresponding sum. The sum $P_2(a_i)$ is similarly defined. It is clear that the cells, counting multiplicity, covered by $G_1$ when it is translated by the vectors $b_1, b_2, \\dots, b_n$ are the same as the cells, counting multiplicity, covered by $G_2$ when it is translated by the vectors $a_1, a_2, \\dots, a_m$. Thus\n$$\n\\sum P_1(b_i) = \\sum P_2(a_j).\n$$\nSince the LHS is positive, at least one of the terms on the RHS is positive.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 41001,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the largest odd positive integer $N$ such that every odd integer $k$ with $1 < k < N$ and $(k, N) = 1$ is a prime.",
    "options": [],
    "answer": "105",
    "solution": "The largest such integer $N$ is $105$. Let $p_i$ denote the $i$th prime, i.e. $p_1 = 2$, $p_2 = 3$, etc. We call a positive integer $N$ *admissible* if it is odd and has the property stated in the problem. Any admissible number exceeding $p_i^2$ must clearly contain the factor $p_i$ for $i \\ge 2$. By Bertrand's postulate, $p_{n+1} < 2p_n$. Thus for any $n \\ge 5$, we have $p_{n+1}^2 < 4p_n^2 < 8p_{n-1}p_n$. Since $p_2p_3 = 15 > 8$,\n$$\np_{n+1}^2 < p_2 p_3 \\cdots p_n \\quad (n \\ge 5) \\qquad (1)\n$$\nIf there were an admissible number $N$ such that $p_{n+1}^2 \\ge N > p_n^2$ for some $n \\ge 5$, then $N$ would be divisible by $p_2p_3 \\cdots p_n$ and so $N \\ge p_2p_3 \\cdots p_n > p_{n+1}^2$ by (1) which gives a contradiction. Therefore no admissible number can exceed $p_5^2 = 121$.\nThe number $105$ can be checked to be admissible. Also since any admissible number exceeding $49$ must be divisible by $3 \\times 5 \\times 7 = 105$, there can be no admissible number $N$ between $105$ and $121$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41002,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n = \\overline{30x070y03}$ be a 9-digit integer. Find all possible values of the pair $(x, y)$, so that $n$ is a multiple of 37.",
    "options": [],
    "answer": "(x, y) = (8, 1), (4, 4), (0, 7)",
    "solution": "We have\n$$\nn = 300070003 + 10^6x + 10^2y = 37(8110000 + 27027x + 3y) + (3 + x - 11y).\n$$\nSince $0 \\le x, y \\le 9$, we have $-96 \\le 3 + x - 11y \\le 12$. Also $37 \\mid 3 + x - 11y$. Thus $3 + x - 11y = 0, -37$ or $-74$ and we get $(x, y) = (8, 1), (4, 4), (0, 7)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41003,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute-angled triangle $ABC$, $M$ is a point on the side $BC$, the line $AM$ meets the circumcircle $\\omega$ of $ABC$ at the point $Q$ distinct from $A$. The tangent to $\\omega$ at $Q$ intersects the line through $M$ perpendicular to the diameter $AK$ of $\\omega$ at the point $P$. Let $L$ be the point on $\\omega$ distinct from $Q$ such that $PL$ is tangent to $\\omega$ at $L$. Prove that $L$, $M$ and $K$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet $D$ be the foot of the perpendicular from $M$ onto the diameter $AK$. Since $AK$ is a diameter of $\\omega$, we have $\\angle MQK = 90^\\circ = \\angle MDK$ so that $M$, $Q$, $K$, $D$ are concyclic. Join $QK$. Then $\\angle PMQ = \\angle QKD = \\angle QKA = \\angle PQA = \\angle PQM$. Thus the triangle $PQM$ is isosceles with $PQ = PM$. Therefore, $PQ = PM = PL$ and $P$ is the circumcentre of the triangle $QML$. It follows that $\\angle QLM = \\frac{1}{2}\\angle QPM = 90^\\circ - \\angle PQM = 90^\\circ - \\angle QKA = \\angle QAK = \\angle QLK$. This shows that $L$, $M$ and $K$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41004,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute-angled triangle $ABC$, $D$ is the point on $BC$ such that $AD$ bisects $\\angle BAC$, $E$ and $F$ are the feet of the perpendiculars from $D$ onto $AB$ and $AC$ respectively. The segments $BF$ and $CE$ intersect at $K$. Prove that $AK$ is perpendicular to $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet the extension of $AK$ intersect $BC$ at $N$. Note that $AE = AF$ and $BD : DC = c : b$, where $b = AC$ and $c = AB$. The cevians $AN$, $BF$, $CE$ concur at $K$. By Ceva's theorem, we have $(BN/NC)(CF/FA)(AE/EB) = 1$. Thus $BN/NC = EB/CF$. On the other hand, $EB = BD \\cos B$ and $CF = DC \\cos C$ so that $EB/CF = (BD \\cos B)/(DC \\cos C) = (c \\cos B)/(b \\cos C)$. Therefore, $BN/NC = (c \\cos B)/(b \\cos C)$. This shows that $N$ is the foot of the perpendicular from $A$ onto $BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41005,
    "subject": "Mathematics (Multi-modal)",
    "question": "A boy lives in a small island in which there are three roads at every junction. He starts from his home and walks along the roads. At each junction he would choose to turn to the road on his right or left alternately, i.e., his choices would be ..., left, right, left, .... Prove that he will eventually return to his home.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "After walking for a long time, he must walk along a certain road 6 times and in the direction at least 3 times. When he reaches the junction along this direction, he must choose the same direction twice. Both times, tracing backwards, we conclude that both times the trails are the same. Since first one traces back to his home, we conclude that the second time, the trails also go back to his home. Thus the boy must return to his home.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41006,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a convex hexagon $ABCDEF$, $AB$ is parallel to $DE$, $BC$ is parallel to $EF$ and $CD$ is parallel to $FA$. Prove that the triangles $ACE$ and $BDF$ have the same area.",
    "options": [],
    "answer": "null",
    "solution": "Let the coordinates of $A$ be $(0,0)$, $B$ be $(1,0)$, $D$ be $(c,b)$, $E$ be $(a,b)$, $F$ be $(e,f)$ and $C$ be $(x,y)$. Then $AB \\parallel ED$. For $AF \\parallel CD$, we need $(b-y)/(c-x) = f/e$. For $EF \\parallel CB$ we need $(x-1)/y = (a-e)/(b-f)$. From these equations, we get\n$$\nxb - ay = b - f + cf - be. \\quad (*)\n$$\nThus the proof is now complete since the following is true by $(*)$.\n$$\n[ACE] = [BDF] \\Leftrightarrow \\begin{vmatrix} 0 & 0 & 1 \\\\ x & y & 1 \\\\ a & b & 1 \\end{vmatrix} = \\begin{vmatrix} 1 & 0 & 1 \\\\ c & b & 1 \\\\ e & f & 1 \\end{vmatrix} \\Leftrightarrow xb - ay = b - f + cf - be.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41007,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 30 children, $a_1, a_2, \\dots, a_{30}$, seated clockwise in a circle on the floor. The teacher walks behind the children in the clockwise direction with a box of 1000 candies. She drops a candy behind the first child $a_1$. She then skips one child and drops a candy behind the third child, $a_3$. Now she skips two children and drops a candy behind the next child, $a_6$. She continues this way, at each stage skipping one child more than at the preceding stage before dropping a candy behind the next child. How many children will never receive a candy? Justify your answer.",
    "options": [],
    "answer": "18",
    "solution": "When the $k^{\\text{th}}$ candy is dropped, the teacher has skipped $1+2+\\dots+(k-1) = \\dfrac{k(k-1)}{2}$ children. Then it is dropped behind $a_i$, where $i = k + \\dfrac{k(k-1)}{2} = \\dfrac{k(k+1)}{2}$ (mod 30). Thus $2i = k(k+1)$ (mod 60). From here it is clear that the $i^{\\text{th}}$ and $j^{\\text{th}}$ candies are given to the same child if $i \\equiv j$ (mod 60). Thus the sequence of children receiving candies is periodic with period 60. Note that $k(k+1) \\equiv (60-k-1)(60-k) \\pmod{60}$. Thus the $k^{\\text{th}}$ and $(60-k-1)^{\\text{th}}$ are given to the same child. Thus we only need to compute $\\dfrac{k(k+1)}{2}$ (mod 30) for $k = 1, \\dots, 29$. This yields the following sequence of children receiving the first 29 candies. Thus 18 children never receive any candy.\n\n1, 3, 6, 10, 15, 21, 28, 6, 15, 25, 6, 18, 1, 15, 30\n16, 3, 21, 10, 30, 21, 13, 6, 30, 25, 21, 18, 16, 15",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41008,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer. Prove that there exist positive integers $\\ge 2$, $a_1, a_2, \\dots, a_n$, such that $a_1 a_2 \\cdots \\hat{a}_i \\cdots a_n \\equiv 1 \\pmod{a_i}$, for $i = 1, \\dots, n$. Here $\\hat{a}_i$ means the term $a_i$ is omitted.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_1 = 2$, $a_2 = 3$. For $i = 3, \\dots, n-1$, let $a_i = a_1 a_2 \\cdots a_{i-1} + 1$. Let $a_n = a_1 a_2 \\cdots a_{n-1} - 1$. Clearly, $a_1 a_2 \\cdots a_{n-1} \\equiv 1 \\pmod{a_n}$. Also $a_{i+1} \\equiv a_{i+2} \\equiv \\cdots \\equiv a_{n-1} \\equiv 1 \\pmod{a_i}$. For $i = 1, \\dots, n-1$, we have\n$$\na_1 a_2 \\cdots \\hat{a}_i \\cdots a_n = (a_1 \\cdots a_{i-1})(a_{i+1} \\cdots a_{n-1})a_n \\equiv (-1)(-1)(-1) \\equiv 1 \\pmod{a_i}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41009,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$, where $\\mathbb{R}$ is the set of real numbers, such that\n$$\nf(x)f(yf(x)-1) = x^2f(y) - f(x) \\quad \\text{for all } x, y \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "f(x) = 0 for all real x; f(x) = x for all real x",
    "solution": "The constant function $f(x) = 0$ is a solution.\nLet $f$ be a solution that is not identically $0$. We shall show that $f(x) = x$ for all $x$. Letting $x = 0$ in the given equation, we get\n$$\nf(0)[f(yf(0) - 1) + 1] = 0.\n$$\nSuppose $f(0) \\neq 0$. Let $x = yf(0) - 1$. As $y$ ranges over all real numbers, so does $x$. Thus we get $f(x) = -1$ for all $x$. But this does not satisfy the given equation. So $f(0) = 0$.\nNow suppose that $f(a) = 0$ for some $a \\neq 0$. Then the original equation becomes $0 = a^2f(y)$ for all $y$, implying that $f(y) = 0$ for all $y$. This contradicts our assumption that $f$ is not identically $0$. Thus $f(x) = 0$ iff $x = 0$.\nLetting $x = y = 1$ in the given equation, we have $f(f(1) - 1) = 0$. Thus $f(1) = 1$. When $x = 1$, the original equation becomes\n\n$$\nf(x)f(f(x)) = x^2 \\qquad (2)\n$$\nNow replace $x$ by $x-1$ in (2) and apply (1) 3 times, and finally apply (2)\n$$\n\\begin{aligned}\n(x-1)^2 &= f(x-1)f(f(x-1)) = (f(x)-1)[f(f(x)-1)] \\\\\nx^2 - 2x + 1 &= (f(x)-1)[f(f(x))-1] = f(x)f(f(x)) - f(x) - f(f(x)) + 1 \\\\\n&= x^2 - f(x) - f(f(x)) + 1\n\\end{aligned}\n$$\nTherefore\n$$\nf(x) + f(f(x)) = 2x. \\qquad (3)\n$$\nEliminating $f(f(x))$ from (2) and (3) gives\n$$\n[x - f(x)]^2 = 0,\n$$\nso that $f(x) = x$, as claimed. It is clear that this is also a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41010,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to color each square on a $9 \\times 9$ board so that each $2 \\times 3$ or $3 \\times 2$ block contains exactly $2$ black squares? If so, what is/are the possible total number(s) of black squares?",
    "options": [],
    "answer": "Yes; 27",
    "solution": "We assume each nonblack square is white. Repeat the $3 \\times 3$ block with black squares down the main diagonal $9$ times. This is a possible construction with $3 \\times 9 = 27$ black squares. We shall prove that this is the only answer.\n\nSuppose there are $26$ black squares. We name the bottom left square as $(1, 1)$. Remove the squares in the block $\\{(1, 1), (1, 2), (1, 3)\\}$. The remaining squares can be partitioned into $3 \\times 2$ or $2 \\times 3$ blocks which contain $26$ black squares. Thus the removed squares are white. Repeat the argument with the blocks $\\{(3, 1), (3, 2), (3, 3)\\}$, $\\{(1, 1), (2, 1), (3, 1)\\}$, $\\{(1, 3), (2, 3), (3, 3)\\}$, we conclude that all these squares are white. Thus we have a $3 \\times 3$ block with at most $1$ black square (see figure below). This is a contradiction.\n\n![](attached_image_1.png)\n\nSuppose there are $28$ black squares. Using the same argument as above, we conclude that there are $2$ black squares in each of the blocks $\\{(1, 1), (1, 2), (1, 3)\\}$, $\\{(3, 1), (3, 2), (3, 3)\\}$, $\\{(1, 1), (2, 1), (3, 1)\\}$ and $\\{(1, 3), (2, 3), (3, 3)\\}$. Thus all the squares in the three blocks $\\{(2, 1), (2, 2), (2, 3)\\}$, $\\{(1, 2), (2, 2), (3, 2)\\}$, $\\{(1, 4), (2, 4), (3, 4)\\}$ are white as shown in the following diagram, again leading to a contradiction.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41011,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f_0, f_1, \\dots$ be the Fibonacci sequence: $f_0 = f_1 = 1$, $f_n = f_{n-1} + f_{n-2}$ if $n \\ge 2$. Determine all possible positive integers $n$ so that there is a positive integer $a$ such that $f_n \\le a \\le f_{n+1}$ and that\n$$\na\\left(\\frac{1}{f_1} + \\frac{1}{f_1f_2} + \\frac{1}{f_1f_2f_3} + \\dots + \\frac{1}{f_1f_2\\dots f_n}\\right)\n$$\nis an integer.",
    "options": [],
    "answer": "n = 1, 2",
    "solution": "The number may be rewritten as\n$$\n\\frac{a}{f_1 \\cdots f_n} (f_2 \\cdots f_n + f_3 \\cdots f_n + \\cdots + f_{n-2}f_{n-1}f_n + f_{n-1}f_n + 1).\n$$\nIf this is an integer, then\n$$\nf_n|a(f_2 \\cdots f_n + f_3 \\cdots f_n + \\cdots + f_{n-2}f_{n-1}f_n + f_{n-1}f_n + 1).\n$$\nThus $f_n|a$. First consider the case where $n \\ge 2$. Then\n$$\nf_n \\le a \\le f_{n+1} < 2f_n.\n$$\nHence $a = f_n$. Then\n$$\n\\frac{1}{f_1 \\cdots f_{n-1}} (f_2 \\cdots f_n + f_3 \\cdots f_n + \\cdots + f_{n-2}f_{n-1}f_n + f_{n-1}f_n + 1)\n$$\nis an integer. Thus $f_{n-1}|1$. Hence $n-1=0$ or $1$. This shows that the only possible values of $n$ are 1 and 2.\nFor $n=1$, $1=f_1 \\le 1 \\le f_2$ and $a/f_1 = 1/1 = 1$ is an integer.\nFor $n=2$, $f_2 \\le 2 \\le f_3$ and\n$$\n2\\left(\\frac{1}{f_1} + \\frac{1}{f_1 f_2}\\right) = 2\\left(1 + \\frac{1}{2}\\right)\n$$\nis an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41012,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a set of numbers chosen from $1, 2, \\ldots, 2015$ with the property that any two distinct numbers, say $x$ and $y$, in $A$ determine a unique isosceles triangle (which is non-equilateral) whose sides are of length $x$ or $y$. What is the largest possible size of $A$?",
    "options": [],
    "answer": "11",
    "solution": "Let $x < y$ be two numbers in $A$. For them to determine a unique isosceles triangle, we must have $2x \\le y$. If $A = \\{2^0, 2^1, 2^2, \\ldots, 2^{10}\\}$, then any two of the numbers $x < y$ satisfy $2x \\le y$. So the maximum size is $\\ge 11$.\n\nNow suppose that there is a set $A$ with $|A| = 12$ that has the property. Let $a_1, a_2, \\ldots, a_{12}$ be the elements of $A$ in increasing order. Then $a_2 \\ge 2a_1$, $a_3 \\ge 2a_2 \\ge 2^2a_1$, $\\ldots$, $a_{12} \\ge 2a_{11} \\ge 2^{11}a_1 \\ge 2^{11} = 2048$, a contradiction (since $2015 < 2048$). Thus the maximum size is $11$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41013,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $k$ such that $k^k + 1$ is divisible by $30$. Justify your answer.",
    "options": [],
    "answer": "All positive integers k with k ≡ 29 mod 30, i.e., k = 30n + 29 for n = 0, 1, 2, …",
    "solution": "An integer is divisible by $30$ iff it is divisible by $2$, $3$ and $5$.\n\nNote that $2 \\mid k^k + 1$ iff $k$ is odd. Thus we may assume that $k$ is odd. Write $k = 2t + 1$.\n\nIf $k \\equiv 0$ or $1 \\pmod{3}$, then $3 \\nmid k^k + 1$. If $k \\equiv 2 \\equiv -1 \\pmod{3}$, then $3 \\mid k^k + 1$ iff $k$ is odd, i.e. iff $k = 6t + 5$.\n\nIf $k \\equiv 0$ or $1 \\pmod{5}$, $5 \\nmid k^k + 1$.\nIf $k \\equiv 2$ or $3 \\pmod{5}$, then $k \\equiv \\pm 2 \\pmod{5}$. Therefore\n$$\nk^k + 1 \\equiv (\\pm 2)^k + 1 \\equiv (\\pm 2)^{2t+1} + 1 \\equiv (\\pm 2)4^t + 1 \\\\\n\\equiv (\\pm 2)(-1)^t + 1 \\not\\equiv 0 \\pmod{5}.\n$$\nIf $k \\equiv 4 \\equiv -1 \\pmod{5}$, then $5 \\mid k^k + 1$ iff $k$ is odd.\n\nThus $30 \\mid k^k + 1$ iff $k$ is odd and $k \\equiv 5 \\pmod{6}$ and $k \\equiv 9 \\pmod{10}$.\n\nThus $k = 30n + 29$, $n = 0, 1, 2, \\ldots$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41014,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 3$ be a given integer. Find the largest integer $d$ (in terms of $n$) such that for any set $S$ of $n$ integers, there are four distinct (but not necessarily disjoint) nonempty subsets, the sum of the elements of each of which is divisible by $d$.",
    "options": [],
    "answer": "n-2",
    "solution": "$d \\ge n$ is not possible. To see this, take a set $S$ of $n$ integers so that each element of $S$ is equal to $1 \\pmod{n}$. The sum of any nonempty subset $T$ of $S$ is equal to $\\#T \\pmod{d}$. Since $d \\ge n$, the only possibility for this to hold is if $d=n$ and $\\#T=n$, i.e., $T=S$. This proves that $d \\ge n$ fails the given condition.\n\n$d=n-1$ is not possible. In this case, take a set $S = \\{a_1, \\dots, a_n\\}$ of $n$ integers so that $a_1 = 0$ and $a_i = 1 \\pmod{d}$ if $1 < i \\le n$. The only subsets of $S$ whose sums are divisible by $d$ are:\n$$\n\\{a_1\\}, \\{a_2, \\dots, a_n\\} \\quad \\text{and} \\quad S.\n$$\nWe will show that $d=n-2$ satisfies the given condition. So the largest $d$ is $n-2$.\n\n**Lemma.** (Erdös) Any set of $n$ integers has nonempty subset whose sum is divisible by $n$.\n\nConsider a set of $n$ integers $\\{a_1, \\dots, a_n\\}$. Consider the set of $n$ numbers $\\{a_1, a_1 + a_2, \\dots, a_1 + \\dots + a_n\\}$. If one of these numbers is equal to $0 \\pmod{n}$, then we are done. Otherwise, there are two of them that are equal (mod $n$). Say\n$$\na_1 + \\dots + a_i = a_1 + \\dots + a_j \\pmod{n}, i < j.\n$$\nThen $a_{i+1} + \\dots + a_j = 0 \\pmod{n}$.\n\nWe return to the proof that $n-2$ satisfies the given condition. Let $S = \\{a_1, \\dots, a_n\\}$ be a set of $n$ integers. By the lemma, there is a subset $T_1$ of $\\{a_1, \\dots, a_{n-2}\\}$ whose sum is divisible by $n-2$. We may also assume that $T_1$ is minimal in the sense that removal of any element from $T_1$ would either render it empty, or make the sum of its elements indivisible by $n-2$. Take $a_i \\in T_1$. Choose nonempty subset $T_2$ of $S \\setminus \\{a_i, a_{n-1}\\}$ whose sum is divisible by $n-2$. Clearly, $T_1 \\neq T_2$.\n\nIf $T_1 \\cap T_2 = \\emptyset$, pick $a_i \\in T_1, a_j \\in T_2$ and let $I = S \\setminus \\{a_i, a_j\\}$. By the lemma, there exists a subset $T_3$ of $I$ whose sum is divisible by $n-2$. Clearly, $T_1, T_2, T_1 \\cup T_2$ and $T_3$ are four distinct sets, the sum of the elements of each of which is divisible by $n-2$.\n\nThe previous paragraph shows that if there are two disjoint nonempty subsets of $S$ whose sums are both divisible by $n-2$, then we are done. Assume that this does not occur. Let $T_1$ and $T_2$ be two distinct subsets of $S$ whose sums are divisible by $n-2$. Choose $a_i \\neq a_j$ so that $a_i \\in T_1$ and $a_j \\in T_2$. By the Lemma, there exists a subset $T_3$ of $S \\setminus \\{a_i, a_j\\}$ whose sum is divisible by $n-2$. In particular, $T_3$ is different from both $T_1$ and $T_2$. Moreover, by the previous paragraph, we may assume that $T_i \\cap T_j \\neq \\emptyset$ if $1 \\le i < j \\le 3$. Then we can choose two distinct points $a_r$ and $a_s$ so that $\\{a_r, a_s\\} \\cap T_i \\neq \\emptyset$ for $i=1, 2, 3$. Use the lemma again to find $T_4 \\subseteq S \\setminus \\{a_r, a_s\\}$ so that the sum of the elements in $T_4$ is divisible by $n-2$. By construction $T_4$ is different from $T_1, T_2$ and $T_3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41015,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n$ such that $n^2 + 24n + 35$ is a square.",
    "options": [],
    "answer": "43 and -67",
    "solution": "Suppose $n^2 + 24n + 35$ is a square. Let $n^2 + 24n + 35 = m^2$. This equation can be written as $(2n + 2m + 24)(2n - 2m + 24) = 436$. Since $436 = 2^2 \\times 109$, we have $\\{2n + 2m + 24, 2n - 2m + 24\\} = \\{1, 436\\}, \\{-1, -436\\}, \\{2, 218\\}, \\{-2, -218\\}, \\{4, 109\\}, \\{-4, -109\\}$.\n\nFor instance, we have $2n + 2m + 24 = 1$, $2n - 2m + 24 = 436$ giving $n = 389/4$, $m = -435/4$, and $2n + 2m + 24 = 436$, $2n - 2m + 24 = 1$ giving $n = 389/4$, $m = 435/4$. Since the solutions are not integers, they are rejected.\n\nSolving $2n + 2m + 24 = 2$, $2n - 2m + 24 = 218$ gives $n = 43$, $m = -54$, and solving $2n + 2m + 24 = -2$, $2n - 2m + 24 = -218$ gives $n = -67$, $m = 54$. The other systems do not give integer solutions.\n\nThus the only solutions are $n = 43$ and $-67$. In both cases, the square is $54^2 = 2916$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41016,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$, $M$ is the midpoint of $BC$ and $D$ is the point on $BC$ such that $AD$ bisects $\\angle BAC$. The line through $B$ perpendicular to $AD$ intersects $AD$ at $E$ and $AM$ at $G$. Prove that $GD$ is parallel to $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet $BE$ intersect $AC$ at $H$. Thus $AE$ is the perpendicular bisector of $BH$ and so $ME$ is parallel to $CA$ and $\\frac{HC}{ME} = 2$. Hence the triangles $MEG$ and $AHE$ are similar. Also the triangles $DME$ and $DCA$ are similar. It follows that $\\frac{AG}{GM} = \\frac{AH}{ME} = \\frac{AC - HC}{ME} = \\frac{AC}{ME} - \\frac{HC}{ME} = \\frac{CD}{DM} - 2 = \\frac{MC + DM}{DM} - 2 = \\frac{BM + DM}{DM} - 2 = \\frac{BM}{DM} - 1 = \\frac{BM - DM}{DM} = \\frac{BD}{DM}$. Therefore, $GD$ is parallel to $AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41017,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be the point in the interior of a triangle $ABC$ such that $AB = ab$, $AC = ac$, $AD = ad$, $BC = bc$, $BD = bd$ and $CD = cd$. Prove that $\\angle ABD + \\angle ACD = 60^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_1$, $B_1$, $C_1$ be the feet of the perpendiculars from $D$ onto the sides $BC$, $CA$, $AB$, respectively.\n![](attached_image_1.png)\nThen, $B_1C_1 = DA \\sin A$, $C_1A_1 = DB \\sin B$, $A_1B_1 = DC \\sin C$. Thus, $B_1C_1 : C_1A_1 : A_1B_1 = (ad)(bc) : (bd)(ac) : (cd)(ab) = 1 : 1 : 1$. Therefore, $A_1B_1C_1$ is an equilateral triangle. Consequently, $\\angle ABD + \\angle ACD = \\angle C_1A_1D + \\angle B_1A_1D = \\angle B_1A_1C_1 = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41018,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ be real numbers such that $0 < a, b, c < 1/2$ and $a + b + c = 1$. Prove that for all real numbers $x, y, z$,\n$$\nabc(x + y + z)^2 \\geq ayz(1 - 2a) + bxz(1 - 2b) + cxy(1 - 2c).\n$$\nWhen does equality hold?",
    "options": [],
    "answer": "Equality holds if and only if x/a = y/b = z/c.",
    "solution": "By symmetry, we may assume that $\\frac{x}{a} \\ge \\frac{y}{b} \\ge \\frac{z}{c}$. Let $\\frac{y}{b} = q$, $\\frac{x}{a} = q + \\alpha$ and $\\frac{z}{c} = q - \\beta$ where $\\alpha, \\beta \\ge 0$. Thus $x = a(q + \\alpha)$, $y = bq$, $z = c(q - \\beta)$. We have\n$$\n\\begin{align*}\n& abc(x + y + z)^2 \\ge ayz(1 - 2a) + bxz(1 - 2b) + cxy(1 - 2c) \\\\\n\\Leftrightarrow\\ & (aq + a\\alpha + bq + cq - c\\beta)^2 \\ge q(q + \\alpha)(1 - 2c) + q(q - \\beta)(1 - 2a) \\\\\n& \\phantom{q(q + \\alpha)(1 - 2c) +} + (q - \\beta)(q + \\alpha)(1 - 2b) \\\\\n\\Leftrightarrow\\ & [q + (a\\alpha - c\\beta)]^2 \\ge q^2 + 2q(a\\alpha - c\\beta) - \\alpha\\beta(1 - 2b) \\\\\n\\Leftrightarrow\\ & (a\\alpha - c\\beta)^2 \\ge -\\alpha\\beta(1 - 2b)\n\\end{align*}\n$$\nSince $\\alpha, \\beta \\ge 0$ and $b \\le 0.5$, the last inequality holds. Also equality holds iff $x/a = y/b = z/c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41019,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_9$ be a sequence of numbers satisfying $0 < p \\le a_i \\le q$ for each $i = 1, 2, \\dots, 9$. Prove that\n$$\n\\frac{a_1}{a_9} + \\frac{a_2}{a_8} + \\dots + \\frac{a_9}{a_1} \\le 1 + \\frac{4(p^2 + q^2)}{pq}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that for any positive numbers $a, b$ between $p$ and $q$, we may assume (due to symmetry) that $0 < p \\le a \\le b \\le q$. We have\n$$\n\\frac{a}{b} + \\frac{b}{a} \\le \\frac{p}{q} + \\frac{q}{p} \\iff \\frac{a^2 + b^2}{ab} \\le \\frac{p^2 + q^2}{pq} \\\\\n\\iff p^2 ab + q^2 ab - a^2 pq - b^2 pq = (bq - ap)(aq - bp) \\ge 0\n$$\nSince the last inequality is true, the first is true as well. Thus\n$$\n\\text{LHS} = \\left(\\frac{a_1}{a_9} + \\frac{a_9}{a_1}\\right) + \\dots + \\left(\\frac{a_4}{a_6} + \\frac{a_6}{a_4}\\right) + \\frac{a_5}{a_5} \\le 1 + \\frac{4(p^2 + q^2)}{pq}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41020,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Determine the minimum number of lines that can be drawn on the plane so that they intersect in exactly $n$ distinct points.",
    "options": [],
    "answer": "the smallest integer k such that k(k−1)/2 ≥ n",
    "solution": "Let $m$ be the integer so that $\\left(\\frac{m}{2}\\right) < n \\le \\left(\\frac{m+1}{2}\\right)$. Then since $m$ lines intersect in at most $\\left(\\frac{m}{2}\\right)$ points, we have $n > m$. We shall show that there exist $m + 1$ lines that intersect in exactly $n$ points. Let $p = n - \\left(\\frac{m}{2}\\right)$ and $q = \\lfloor (m-p)/2 \\rfloor$.\n\nCase (i): $m-p$ is even. Draw $q$ pairs of lines so that they intersect in $q$ distinct points on the $x$-axis. Draw another $p$ lines, not parallel to the $x$-axis, so that the $2q+p=m$ lines are pairwise nonparallel. Then draw another line on the $x$-axis. These $m+1$ lines intersect in exactly $n$ points.\n\nCase (ii): $m-p$ is odd. Draw $q$ pairs of lines so that they intersect in $q$ distinct points on the $x$-axis. Draw another $p+1$ lines so that exactly one of them is parallel but not on the $x$-axis, and that the $2q+p+1=m$ lines are pairwise nonparallel. Then draw another line on the $x$-axis. These $m+1$ lines intersect in exactly $n$ points.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41021,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any integer $n \\ge 1$, show that\n$$\n\\sum_{k=1}^{n} \\frac{2^k}{\\sqrt{k+0.5}} \\le 2^{n+1}\\sqrt{n+1} - \\frac{4n^{3/2}}{3}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, note that\n$$\n\\frac{1}{\\sqrt{k+0.5}} \\le 2(\\sqrt{k+1}-\\sqrt{k})\n$$\nwhich follows from the fact that\n$$\n\\sqrt{k+1}-\\sqrt{k} = \\frac{1}{\\sqrt{k+1}+\\sqrt{k}} \\ge \\frac{1}{2\\sqrt{k}+0.5}\n$$\nsince $\\sqrt{x}$ is concave. Hence\n$$\n\\frac{2^k}{\\sqrt{k+0.5}} \\le 2^{k+1}(\\sqrt{k+1}-\\sqrt{k})\n$$\n\nSumming up, we have\n$$\n\\sum_{k=1}^{n} \\frac{2^k}{\\sqrt{k+0.5}} \\le 2^{n+1}\\sqrt{n+1} - 2\\sum_{k=1}^{n} \\sqrt{k}.\n$$\nHowever,\n$$\n\\sqrt{k} \\ge \\frac{2}{3}((k)^{3/2} - (k-1)^{3/2}).\n$$\nThe inequality is now clear.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41022,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$, $\\angle A = 90^\\circ$, the bisector of $\\angle B$ meets the altitude $AD$ at the point $E$, and the bisector of $\\angle CAD$ meets the side $CD$ at $F$. The line through $F$ perpendicular to $BC$ intersects $AC$ at $G$. Prove that $B, E, G$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nFirst $\\triangle ABC \\sim \\triangle DAC$ so that $AC/BA = DC/AD$. Also $\\triangle DAC \\sim \\triangle DBA$. It follows that $\\triangle AFC \\sim \\triangle BEA$ so that $FC/AC = EA/BA$. Thus $FC/EA = AC/BA = DC/AD$. This shows that $EF$ is parallel to $AC$. Hence $AEFG$ is a parallelogram. Since $\\angle EFA = \\angle FAG = \\angle FAE$, we have $AEFG$ is a rhombus so that $AG = FG$. Thus $\\triangle BFG \\cong \\triangle BAG$. Consequently, $\\angle FBG = \\angle ABG = \\frac{1}{2}\\angle ABC = \\angle FBE$. This means $B, E, G$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41023,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a prime number. Show that there is a permutation $a_1, a_2, \\dots, a_n$ of $1, 2, \\dots, n$ so that $a_1, a_1a_2, \\dots, a_1a_2 \\dots a_n$ leave distinct remainders when divided by $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the Chinese remainder Theorem, for every $k=2,3,\\dots,n$ there exists $b_k$ so that\n$$\nb_k \\equiv 0 \\pmod{(k-1)}, \\quad b_k \\equiv k \\pmod{n}.\n$$\nLet $a_1 = 1$ and for $k = 2, \\dots, n$, $a_k$ is the remainder when $b_k/(k-1)$ is divided by $n$. Since $b_n \\equiv 0 \\pmod{n}$, we have $a_n = 0$. Also if $a_i = a_j$, then $b_i/(i-1) \\equiv b_j/(j-1) \\pmod{n}$, i.e., $i(j-1) \\equiv j(i-1) \\pmod{n}$. Since $n$ is prime, $i = j$. Thus $a_1, \\dots, a_n$ are distinct. Now\n$$\na_1 a_2 \\dots a_k \\equiv \\frac{b_2 \\dots b_k}{(k-1)!} \\equiv k \\pmod{n}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41024,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $b$ be a number with $-2 < b < 0$. Prove that there exists a positive integer $n$ such that all the coefficients of the polynomial $(x+1)^n(x^2+bx+1)$ are positive.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Expanding $(x+1)^n(x^2+bx+1)$, we obtain\n$$\nx^{n+2} + (n+b)x^{n+1} + \\sum_{k=0}^{n-2} \\left[ \\binom{n}{k+2} + b\\binom{n}{k+1} + \\binom{n}{k} \\right] x^{n-k} + (n+b)x + 1.\n$$\nWe require $n > 2$ so that $n+b > 0$. We also require $\\binom{n}{k+2} + b\\binom{n}{k+1} + \\binom{n}{k} > 0$ for all $k=0, \\dots, n-2$ which is equivalent to require $H \\equiv \\frac{\\binom{n}{k+2} + b\\binom{n}{k+1} + \\binom{n}{k}}{\\binom{n}{k}} > 0$ for all $k=0, \\dots, n-2$. Direct simplification gives\n$$\nH = \\frac{(2-b)k^2 + (b-2)(n-2)k + 2bn + n^2 - n + 2}{(k+2)(k+1)}.\n$$\nThe numerator of $H$ as a quadratic expression in $k$ has leading coefficient $(2-b) > 0$ and discriminant $[(b-2)(n-2)]^2 - 4(2-b)(2bn+n^2-n+2) = (n+2)(b-2)(bn+2b+2n)$. Since $b-2 < 0$, $H$ is positive if and only if $bn+2b+2n > 0$, which is equivalent to $n > -\\frac{2b}{b+2}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41025,
    "subject": "Mathematics (Multi-modal)",
    "question": "A group of tourists get on 10 buses in the outgoing trip. The same group of tourists get on 8 buses in the return trip. Assuming each bus carries at least 1 tourist, prove that there are at least 3 tourists such that each of them has taken a bus in the return trip that has more people than the bus he has taken in the outgoing trip.",
    "options": [],
    "answer": "at least 3",
    "solution": "We prove the general case where there are $p$ outgoing buses and $q$ returning buses with $p > q$. The number of such tourists is $p - q + 1$. Let $T$ be the set of all tourists. For each $t \\in T$, let $p_t$ be the number of tourists in the bus that $t$ takes in the outgoing trip and let $q_t$ be the number of tourists in the bus that $t$ takes in the return trip. Then\n$$\n\\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$$\nSince $|\\frac{1}{p_t} - \\frac{1}{q_t}| < 1$, at least $p-q+1$ terms in the above sum is positive. That is, there are at least $p-q+1$ tourists such that $\\frac{1}{p_t} - \\frac{1}{q_t} > 0$, or equivalently $p_t < q_t$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 41026,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a $2016$-sided polygon with all its adjacent sides perpendicular to each other, i.e., all its internal angles are either $90^\\circ$ or $270^\\circ$. If the lengths of its sides are odd integers, prove that its area is an even integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We may assume the sides are either parallel to the $x$- or $y$-axes and that the lowest horizontal sides are at the $x$-axis. Start with a horizontal side that is on the $x$-axis, go around the polygon in clockwise direction, label the horizontal sides $1, 2, \\ldots, 1008$ and put an arrow on the side along the direction that you proceed. Each horizontal side $s_i$ forms a rectangle with the $x$-axis whose height is the height of $s_i$ above the $x$-axis and whose width is $s_i$. If the arrow on $s_i$ points in the direction positive $y$-axis, its area $A_i$ is positive. Otherwise it is negative. The area of $P$ is the absolute value of\n$$\nA_1 + A_2 + \\dots + A_{1008}.\n$$\nSince the side lengths are all odd, $A_i$ and $A_{i+1}$ for $i = 1, \\dots, 1007$, have opposite parities. Hence there are $504$ terms which are odd. Therefore the area is even.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41027,
    "subject": "Mathematics (Multi-modal)",
    "question": "A total of $731$ objects are put into $n$ nonempty bags where $n$ is a positive integer. These bags can be distributed into $17$ red boxes and also into $43$ blue boxes so that each red and each blue box contain $43$ and $17$ objects, respectively. Find the minimum value of $n$.",
    "options": [],
    "answer": "59",
    "solution": "We shall prove this general with $p$ red boxes and $q$ blue boxes and $pq$ objects, with $p, q$ coprime. Suppose the bags have been distributed as described, with each red box containing $q$ objects and each blue box containing $q$ objects. Suppose we have a collection of $r$ red and $b$ blue boxes. We say that a bag is internal (with respect to the collection) if it is contained in a red box and a blue box in the collection. It is external if it is contained in only one of the boxes in the collection. If the collection does not have any external bags, then $rq = bp$. Since $p, q$ are coprime, we have $r = p, b = q$. This means any collection of $x$ boxes, with $x < p+q$, must have an external bag.\n\nWe shall prove that for every $x < p+q$, there is a collection of $r$ red and $b$ blue boxes with $r+b = x$ that has at least $x-1$ internal bags and at least 1 external bag. This is certainly true for $x = 2$ since every bag is contained in a red box and a blue box and that these 2 boxes have an external bag. Inductively, suppose $x < p+q-1$ and there is a collection of $x$ boxes with $x-1$ internal bags and 1 external bag. By putting the box that contains the external bag into the collection, we have a collection of $x+1$ boxes with $x$ internal bags. Since $x+1 < p+q$, the collection also has an external bag. This can continue until $x = p+q-1$. For this collection, we have $x-1$ internal and 1 external bag giving a total $x = p+q-1$ bags. Thus $n \\ge p+q-1$.\n\nNext we shall show that $n = p+q-1$ is sufficient. Apply the division algorithm, assuming that $p > q$, we have $p = kq+r$ where $0 \\le r < q-1$. Suppose that we have solved the problem for $(q, r)$, i.e., we have $q+r-1$ bags containing a total of $qr$ objects and these bags can be distributed into $q$ red boxes, $R_1, \\dots, R_q$ with $r$ objects in each box and into $r$ blue boxes, $B_1, \\dots, B_r$, with $q$ objects in each box. Now add another $kq$ bags with $q$ objects in each bag. We have a total $kq + q + r - 1 = p + q - 1$ bags with a total of $kq^2 + qr = pq$ objects. Put each of these new bags into a new blue box, then, together with $B_1, \\dots, B_r$, we have $kq + r = p$ blue boxes each with $q$ objects. Next into each of the red boxes $R_1, \\dots, R_q$, put $k$ of these new bags so that each box has $kq + r = p$ objects. Thus we have a solution for $(p, q)$. Since $p, q$ are coprime, applying the division algorithm repeatedly, we eventually get to the situation where $r=1$. For the solution to the problem with $q \\times 1$ objects, we simply put 1 object in each bag.\n\nFor $p=43, q=17$, we have $43 = 2 \\times 17 + 9$, $17 = 1 \\times 9 + 8$, $9 = 1 \\times 8 + 1$. Thus our solution is 8 bags with 1 object, 8 bags with 8 objects, 9 bags with 9 objects, 34 bags with 17 objects.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41028,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each integer $n > 1$, find a set of $n$ integers $\\{a_1, a_2, \\dots, a_n\\}$ such that the set of numbers $\\{a_i + a_j \\mid 1 \\le i \\le j \\le n\\}$ leave distinct remainders when divided by $n(n+1)/2$. If such a set of integers does not exist, give a proof.",
    "options": [],
    "answer": "Such a set exists only when the set size is two (for example, {1, 2}); for any larger size, no such set exists.",
    "solution": "For $n = 2$, $n(n+1)/2 = 3$. Thus $a_1 = 1$, $a_2 = 2$ work.\n\nNow suppose that $n \\ge 3$ and that the set $\\{a_1, \\dots, a_n\\}$ has the desired properties. We may assume without loss of generality, that $1 \\le a_1 < a_2 < \\dots < a_n \\le \\frac{n(n+1)}{2}$. Also let $a_{n+1} = a_1 + \\frac{n(n+1)}{2}$.\n\nFirst we note that for any two distinct pairs of integers $\\{i, j\\}$ and $\\{k, \\ell\\}$, $i < j, k < \\ell$, chosen from $\\{1, 2, \\dots, n+1\\}$, $a_j - a_i \\ne a_\\ell - a_k$ for if they are equal, we would have $a_j + a_k = a_i + a_\\ell$. In particular, the integers $a_{i+1} - a_i$, $i = 1, \\dots, n$ are distinct. Since $a_{n+1} - a_1 = \\frac{n(n+1)}{2}$, these differences are $1, 2, \\dots, n$. This also means that $a_j - a_i \\ge n+1$ if $j-i \\ge 2$.\n\nLet $t$ be the integer such that $a_{t+1} - a_t = 1$. Then since both $a_{t+2} - a_t$ and $a_{t+1} - a_{t-1}$ are $\\ge n+1$, $a_{t+2} - a_{t+1} = a_t - a_{t-2} = n$ which is impossible unless $t=1$ or $t=n$. By symmetry, we can take $t=1$. Then, $a_2 - a_1 = 1$ and $a_3 - a_2 = n$. For $n=3$, this means $a_3 = a_2 + 3$. Also we have $n(n+1)/2 = 6$. Thus $a_2 + a_2 \\equiv a_3 + a_3 \\pmod 6$. So such a set does not exist for $n=3$.\n\nNow let $n \\ge 4$ and $s$ be such that $a_{s+1} - a_s = 2$. Then $s \\ge 3$. By the same argument as before, we have $a_{s+2} - a_{s+1}$ and $a_s - a_{s-1}$ are both either $n$ or $n-1$. This is possible only when $s = 3$ or $n$. Thus either $a_5 - a_3 = n+1$ or $a_{n+1} - a_{n-1} = n+1$. Therefore one of these equals $a_3 - a_1 = n+1$ which is not possible. Thus the desired set of integers does not exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41029,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the minimum number of lines that can be drawn on the plane so that they intersect in exactly $200$ distinct points.\n(Note that for $3$ distinct points, the minimum number of lines is $3$ and for $4$ distinct points, the minimum is $4$.)",
    "options": [],
    "answer": "21",
    "solution": "Let $m$ be the integer so that $\\left(\\frac{m}{2}\\right) < n \\le \\left(\\frac{m+1}{2}\\right)$. Then since $m$ lines intersect in at most $\\left(\\frac{m}{2}\\right)$ points, we have $n > m$. We shall show that there exist $m + 1$ lines that intersect in exactly $n$ points. Let $p = n - \\left(\\frac{m}{2}\\right)$ and $q = \\lfloor (m-p)/2 \\rfloor$.\n\nCase (i): $m-p$ is even. Draw $q$ pairs of lines so that they intersect in $q$ distinct points on the $x$-axis. Draw another $p$ lines, not parallel to the $x$-axis, so that the $2q+p=m$ lines are pairwise nonparallel. Then draw another line on the $x$-axis. These $m+1$ lines intersect in exactly $n$ points.\n\nCase (ii): $m-p$ is odd. Draw $q$ pairs of lines so that they intersect in $q$ distinct points on the $x$-axis. Draw another $p+1$ lines so that exactly one of them is parallel but not on the $x$-axis, and that the $2q+p+1=m$ lines are pairwise nonparallel. Then draw another line on the $x$-axis. These $m+1$ lines intersect in exactly $n$ points.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41030,
    "subject": "Mathematics (Multi-modal)",
    "question": "A square is cut into several rectangles, none of which is a square, so that the sides of each rectangle are parallel to the sides of the square. For each rectangle with sides $a$, $b$, $a < b$, compute the ratio $a/b$. Prove that sum of these ratios is at least $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, we may assume the square has area $1$. Let the sides of the rectangles be $a_i$, $b_i$, $a_i \\leq b_i$ and $S_i = a_i b_i$. We have $\\sum S_i = 1$ and\n$$\n\\sum \\frac{a_i}{b_i} = \\sum \\frac{S_i}{b_i^2} \\geq \\sum S_i = 1.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41031,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of $\\triangle ABC$ touches the sides $BC$, $CA$, $AB$, at $D$, $E$, $F$ respectively. A circle through $A$ and $B$ encloses $\\triangle ABC$ and intersects the line $DE$ at points $P$ and $Q$. Prove that the midpoint of $AB$ lies on the circuncircle of $\\triangle PQF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ be the midpoint of $AB$. If $DE \\parallel AB$, then $\\triangle ABC$ is isosceles with $CA = CB$, and $F$ coincides with $M$.\n\n![](attached_image_1.png)\n\nConsider the case where $DE \\parallel AB$. Let the lines $DE$ and $AB$ intersect at $X$. By Menelaus theorem, $\\frac{AX}{XB} = \\frac{BD}{CD} = \\frac{CE}{EA} = 1$.\n\n$\\therefore\\ XA \\cdot BF = XB \\cdot AF$ (since $CD = CE$, $BD = BF$, $AE = AF$.)\n\n$\\therefore\\ XA \\cdot (XB - XF) = XB \\cdot (XF - XA)$ (since $BF = XB - XF$, $AF = XF - XA$.)\n\n$\\therefore\\ 2XA \\cdot XB = (XA + XB) \\cdot XF = 2XM \\cdot XF$\n\n$\\therefore\\ XM \\cdot XF = XA \\cdot XB = X'P \\cdot XQ \\Rightarrow P, Q, F, M$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41032,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n, p$ be real numbers with $p > -1$. Prove that\n$$\n\\sum_{i=1}^{n} (a_i - b_i)(a_i(a_1^2 + a_2^2 + \\dots + a_n^2)^{p/2} - b_i(b_1^2 + b_2^2 + \\dots + b_n^2)^{p/2}) \\ge 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Write $A = (a_1^2 + a_2^2 + \\dots + a_n^2)^{1/2}$ and $B = (b_1^2 + b_2^2 + \\dots + b_n^2)^{1/2}$. Then the LHS of the inequality becomes\n$$\nA^{p+2} + B^{p+2} - \\sum_{i=1}^{n} a_i b_i (A^p + B^p).\n$$\nSince $\\sum a_i b_i \\le AB$ it suffices to prove that\n$$\nA^{p+2} + B^{p+2} \\ge BA^{p+1} + AB^{p+1}.\n$$\nThis follows readily from rearrangement inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41033,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and $a_1, a_2, \\dots, a_{2n}$ be $2n$ distinct integers. Given that the equation\n$$\n|x - a_1| |x - a_2| \\dots |x - a_{2n}| = (n!)^2\n$$\nhas an integer solution $x = m$, find $m$ in terms of $a_1, \\dots, a_{2n}$.",
    "options": [],
    "answer": "(a_1 + a_2 + \\cdots + a_{2n})/(2n)",
    "solution": "We have\n$$\n|m - a_1| |m - a_2| \\cdots |m - a_{2n}| = (n!)^2.\n$$\nFirst we show that we cannot have distinct $i$, $j$, $k$ so that $|m - a_i| = |m - a_j| = |m - a_k|$. If so, then two of $(m - a_i)$, $(m - a_j)$, $(m - a_k)$ must be of the same sign, say $(m - a_i)$, $(m - a_j)$. Then $a_i = a_j$, a contradiction. Thus the values of $|m - a_1|$, $|m - a_2|$, ..., $|m - a_{2n}|$ are $1$, $1$, $2$, $2$, ..., $n$, $n$. Also if $|m - a_i| = |m - a_j|$, then we must have $m - a_i = -(m - a_j)$ otherwise $a_i = a_j$. Therefore $\\sum (m - a_i) = 0$ and $m = \\frac{a_1 + \\cdots + a_{2n}}{2n}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41034,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the cyclic quadrilateral $ABCD$, the sides $AB$, $DC$ meet at $Q$, the sides $AD$, $BC$ meet at $P$, $M$ is the midpoint of $BD$. If $\\angle APQ = 90^\\circ$, prove that $PM$ is perpendicular to $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Drop perpendicular $DE$ from $D$ onto $AB$. Join $PE$. In the cyclic quadrilaterals $DEQP$ and $ABCD$, we have $\\angle PEB = \\angle PDQ = \\angle PBE$. Thus $\\triangle PBE$ is isosceles. Let $F$ be the midpoint of $BE$. Then both $PF$ and $MF$ are perpendicular to $AB$. Thus $P$, $M$, $E$ are collinear and $PM$ is perpendicular to $AB$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41035,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $n$ so that $\\sqrt{\\frac{1^2+2^2+\\dots+n^2}{n}}$ is an integer.",
    "options": [],
    "answer": "337",
    "solution": "Let $\\frac{1^2+2^2+\\dots+n^2}{n} = m^2$, $m \\in \\mathbb{Z}^+$. Then $(n+1)(2n+1) = 6m^2$. Thus $n$ is odd and $n = 6p \\pm 1$ or $n = 6p+3$ for some integer $p$.\n\nIf $n = 6p+3$, then $6m^2 = (6p+4)(12p+7)$ which gives no solution since $3$ does not divide the RHS.\n\nIf $n = 6p-1$, $m^2 = p(12p-1)$. Since $p$ and $12p-1$ are coprime, both are squares. Let $p = t^2$, $12p-1 = s^2$. Then $s^2 = 12t^2-1$, i.e., $s^2 \\equiv -1 \\pmod 3$ (mod 4), a contradiction.\n\nIf $n = 6p+1$, $m^2 = (3p+1)(4p+1)$. Since $3p+1$ and $4p+1$ are coprime, both are squares. Let $3p+1 = u^2$ and $4p+1 = v^2$. Thus $4p+1 \\equiv 1 \\pmod 8$ implying that $p$ is even and therefore $u$ is odd. Therefore $3p+1 \\equiv 1 \\pmod 8$ implying that $p \\equiv 0 \\pmod 8$. By checking small even multiples of $8$, we find that $p=56$ is the smallest so that $3p+1$ and $4p+1$ are squares. Thus $n=337$ is the smallest value that works.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41036,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $2017$ distinct points in the plane. For each pair of these points, construct the midpoint of the segment joining the pair of points. What is the minimum number of distinct midpoints among all possible ways of placing the points?",
    "options": [],
    "answer": "4031",
    "solution": "Suppose the points are placed on the $x$-axis with coordinates $(i, 0)$, $i = 0, \\dots, 2016$. Then midpoints are $(i/2, 0)$, $i = 1, 2, \\dots, 4031$. Thus there are $4031$ distinct midpoints.\n\nNext we shall prove that there are at least $4031$ distinct midpoints. Let $A_1, \\dots, A_{2017}$ be the points and assume that $A_1, A_2$ are the pair that are furthest apart. Consider the $4030$ segments from $A_1$ and $A_2$ to $A_3, \\dots, A_{2015}$. The midpoints are distinct. For if $X, Y$ are two points so that the midpoints of $A_1X$ and $A_2Y$ coincide, then we have two cases. If the four points are not collinear, then they are vertices of a parallelogram with $A_1X, A_2Y$ as diagonals and $A_1A_2$ as a side. This is not possible as the longer diagonal is longer than a side. Otherwise $A_1, A_2, X, Y$ are collinear. Then it is easy to verify that if $X$ is in the segment $A_1A_2$, then $Y$ must be outside making $A_1A_2 < A_2Y$, a contradiction. Also none of these midpoints is the midpoint of $A_1A_2$. Thus we have at least $4031$ distinct midpoints.\n\nIn conclusion, the minimum number of midpoints is $4031$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41037,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $AB = AC$, $D$ is a point on the side $BC$ and $E$ is a point on the segment $AD$. Given that $\\angle BED = \\angle BAC = 2\\angle CED$, prove that $BD = 2CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\angle CED = x$ and $\\angle ABE = y$. Then\n$$\n\\angle BAC = \\angle BED = 2x \\Rightarrow \\angle BAE = 2x - y \\text{ and } \\angle EAC = y.\n$$\nLet $F$ be the point on $BE$ so that $\\angle AFE = \\pi$. (Note that $x = \\angle CED = y + \\angle ECA$ implying $x > y$. Thus $F$ is in fact in the interior of the segment $BE$.) Since $\\angle AFE = x$ and $\\angle FED = 2x$, we have $AE = EF$. Next we have\n$$\nAB = AC \\Rightarrow \\triangle BAF \\cong \\triangle ACE \\ (\\text{ASA}) \\Rightarrow BF = AE = EF\n$$\n∴ $[AEC] = [BFA] = [AFE] \\Rightarrow [BEA] = 2[AEC] \\Rightarrow BD = 2CD$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41038,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a polygon with $m+n$ sides where $m, n$ are positive integers. Colour $m$ of its vertices red and the remaining $n$ vertices blue. A side is given the number $2$ if both its end vertices are red, the number $\\frac{1}{2}$ if both its end vertices are blue and the number $1$ otherwise. Let the product of these numbers be $P$. Find the largest possible value of $P$.",
    "options": [],
    "answer": "2^(m-n)",
    "solution": "We first show that if two adjacent vertices have different colours, then swapping the colours of these vertices leaves $P$ unchanged. To see this, we only need to consider the four possible cases:\n\n$RRBR$, $BRBB$, $RRBB$, $BRBR$\n\nwhich change to $RBRR$, $BBRB$, $RBRB$, $BBRP$.\n\nThe value of $P$ remains unchanged in each case. Thus by swapping the colours of adjacent vertices repeatedly, we get to the case where vertices of the same colour form one block. We now have $m-1$ edges whose end vertices are red and $n-1$ edges whose end vertices are blue. The value of $P$ is $2^{m-n}$, a constant.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41039,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ such that $f(k+1) > f(f(k))$ for all $k \\ge 1$, where $\\mathbb{Z}^+$ is the set of positive integers.",
    "options": [],
    "answer": "f(n) = n",
    "solution": "We shall show that $f(k) = k$ for all $k$. It suffices to show that $f$ is a strictly increasing function for if $f$ is a strictly increasing function, then for an arbitrary $k$,\n$$\nf(k+1) > f(f(k)) \\Rightarrow k+1 > f(k) \\Rightarrow f(k) = k.\n$$\nConsider the set of integers $\\{f(1), f(2), \\dots\\}$. By the given condition, each member $f(k+1), k \\ge 1$ is bigger than another member $f(f(k))$, with $f(1)$ as the only exception. Therefore $f(1)$ is the smallest element of the set.\nThis argument can be extended. Let $g(k) = f(k) - 1$, for $k \\ge 2$. Then $\\{g(2), g(3), \\dots\\}$ is a set of positive integers. For $k \\ge 2$, $g(k+1) = f(k+1) - 1 > f(f(k)) - 1 = g(f(k)) = g(g(k) + 1)$. This means $g(2)$ is the smallest element of the set. Therefore $f(1) < f(2) < f(k)$ for all $k \\ge 3$. Continuing this way, we see that $f(1) < f(2) < f(3) < \\dots$, i.e., the function $f$ is strictly increasing.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41040,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer. Prove that there exist positive integers $x_1, \\dots, x_n$ in geometric progression and positive integers $y_1, \\dots, y_n$ in arithmetic progression such that $x_1 < y_1 < x_2 < y_2 < \\dots < x_n < y_n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $x_k = n^{2n} \\left( 1 + \\frac{1}{n^2} \\right)^k$ and $y_k = n^{2n} + (k+1)n^{2n-2}$, $k = 1, \\dots, n$. Then\n$x_1 < y_1 < x_2 < y_2 < \\dots < x_n < y_n$.\n\nBy the binomial theorem, we have for $k \\ge 2$ and $a \\le 1/k^2$,\n$$\n\\begin{align*}\n(1+a)^k &= 1+ka+a \\left( \\frac{ak(k-1)}{2!} + \\frac{a^2k(k-1)(k-2)}{3!} + \\dots + \\frac{a^{k-1}k!}{k!} \\right) \\\\\n&\\le 1+ka+a \\left( \\frac{k(k-1)}{k^2} + \\frac{1}{2!} + \\frac{k(k-1)(k-2)}{k^4} \\frac{1}{3!} + \\dots + \\frac{k!}{k^2k^{k-2}k!} \\right) \\\\\n&\\le 1+ka+a \\left( \\frac{1}{2!} + \\dots + \\frac{1}{k!} \\right) < 1+ka+a \\left( \\frac{1}{1 \\cdot 2} + \\dots + \\frac{1}{(k-1)!} \\right) \\\\\n&\\le 1+ka+a \\left( 1-\\frac{1}{k} \\right) < 1+(k+1)a.\n\\end{align*}\n$$\nLet $X_k = \\left( 1 + \\frac{k}{n^2} \\right)^k$, $k = 1, \\dots, n$. Then, for $2 \\le k \\le n$, $\\frac{1}{n^2} \\le \\frac{1}{k^2}$. Therefore\n$$\n1 + \\frac{k}{n^2} < X_k < 1 + \\frac{k+1}{n^2}.\n$$\nMultiplying throughout by $n^{2n}$, we have\n$$\nn^{2n} + k n^{2n-2} < n^{2n} \\left( 1 + \\frac{1}{n^2} \\right)^k < n^{2n} + (k+1)n^{2n-2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41041,
    "subject": "Mathematics (Multi-modal)",
    "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": "Detailed 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$, $i = 1, \\dots, 5$. For example, we can take $a = d = 1$. Choose such an AP with maximal $d$. Then not 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]$.\n\nWe have two cases.\n\n(i) At least 3 of $k_1, \\dots, k_5$ are odd: use $[a+d, 2d]$.\n\n(ii) 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$. We are now left with the case where $[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": 41042,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be nonzero integers, with $1$ as their only positive common divisor, such that $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = 0$. Find the number of such triples $(a, b, c)$ with $50 \\geq |a| \\geq |b| \\geq |c| \\geq 1$.",
    "options": [],
    "answer": "20",
    "solution": "Since the sum is $0$, there are positive and negative integers among the three. Without loss of generality, we assume that there is $1$ negative and two positive integers. (The other case is obtained by changing the signs of all the integers.) Then $c < 0$ since it has the smallest absolute value. Let $\\gcd(a, b) = p$, $\\gcd(b, c) = q$ and $\\gcd(a, c) = r$. Then $\\gcd(p, q) = \\gcd(q, r) = \\gcd(r, p) = 1$. Therefore $a = pr s$, $b = pq t$ and $c = -qr$ for some integers $s$, $t$, $u$ with $\\gcd(s, t) = \\gcd(s, u) = \\gcd(t, u) = 1$. Substitute into the original equation and simplify, we have\n$$\nqt + rs = \\frac{pst}{u}.\n$$\nNow, $u$ has no common factor with $s$ and $t$. Thus $u \\mid p$. This means $u$ is a common divisor of $a$, $b$, $c$. By our assumption, $u = 1$. By symmetry, we also have $s = t = 1$. Therefore $q + r = p$ and $a = pr$, $b = pq$, $c = -qr$. Thus every triple $p$, $q$, $r$, pairwise coprime with $q + r = p$ uniquely determines the triple $a$, $b$, $c$. We have the following triples, with the product of the larger $2 \\leq 50$:\n$$\n(1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (1, 6, 7), (2, 3, 5), (2, 5, 7), (3, 4, 7), (3, 5, 8), (4, 5, 9).\n$$\nTherefore, we have $20$ such triples.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41043,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ and $B$ be two $n \\times n$ square arrays. The cells of $A$ are labelled by the numbers from $1$ to $n^2$ from left to right starting from the top row; whereas the cells of $B$ are labelled by the numbers from $1$ to $n^2$ along rising north-easterly diagonals starting with the upper left-hand corner. Stack the array $B$ on top of the array $A$. If two overlapping cells have the same number, they are coloured red. Determine those $n$ for which there is at least one red cell other than the cells at top left corner, bottom right corner and the centre (when $n$ is odd). Below shows the arrays for $n=4$.\n$$\nA = \\begin{bmatrix} 1 & 2 & 3 & 4 \\\\ 5 & 6 & 7 & 8 \\\\ 9 & 10 & 11 & 12 \\\\ 13 & 14 & 15 & 16 \\end{bmatrix} \\qquad B = \\begin{bmatrix} 1 & 3 & 6 & 10 \\\\ 2 & 5 & 9 & 13 \\\\ 4 & 8 & 12 & 15 \\\\ 7 & 11 & 14 & 16 \\end{bmatrix}\n$$",
    "options": [],
    "answer": "There is at least one nontrivial red cell if and only if n has at least two distinct prime factors (i.e., n is not a prime power).",
    "solution": "For each $i = 1, 2, \\dots, n^2$, define $f(i) = j$ if the cells containing the number $i$ in $A$ and the cell containing $j$ in $B$ overlap. The two cells are coloured red iff $i = j$. We claim that $f$ has a fixed point not in the set $\\{1, n^2, (n^2+1)/2\\}$ if and only if $n$ has at least two different prime factors. We may assume $n > 1$.\n\nLet the rising diagonals be numbered $1$ through $2n-1$, beginning at the upper left-hand corner and proceeding to the lower right-hand corner. By symmetry a number $x$ in diagonal $k$ with $k \\le n$ is fixed if and only if $n^2+1-x$ in diagonal $2n-k$ is fixed. It thus suffices to consider numbers in the first $n$ diagonals. The number of the card originally in diagonal $k$ and row $i$, where $1 \\le i \\le k \\le n$, is $x = (i-1)n + k - i + 1$. Further\n$$\nf(x) = \\left( \\sum_{j=1}^{k-1} j \\right) + k - i + 1.\n$$\nHence $x = f(x)$ if and only if\n$$\n(i-1)n = k(k-1)/2. \\quad (*)\n$$\nIt follows from $(*)$ that each diagonal contains at most one fixed point. Thus $f$ has a fixed point not in the set $\\{1, n^2, (n^2+1)/2\\}$ if and only if $(*)$ has a solution $(i, k)$ such that $1 < k < n$ and $1 \\le i \\le n$. Since $k$ and $k-1$ are coprime, no such solution exists if $n$ is a prime or a prime-power.\n\nIf $n$ is not a prime or a prime power, then $n = rs$, where $r$ and $s$ are coprime and both greater $1$. Further, we may assume that $s$ is odd. Since $2r$ and $s$ are coprime, there are integers $\\alpha$ and $\\beta$ such that\n$$\n2r\\alpha + s\\beta = 1 \\quad (***)\n$$\nNow the pair $(\\alpha, \\beta)$ satisfies $(**)$, so does the pair $(\\alpha + st, \\beta - 2rt)$ for any integer $t$. Thus we may assume that $-r < \\beta \\le r$. In view of $(**)$, we must have $\\beta \\ne 0, r$, and thus there are two cases to consider.\n\nCase I, $-r < \\beta < 0$, $\\alpha > 0$. In this case, we take $k = 2r\\alpha$, $i = 1 - \\alpha\\beta$.\n\nCase II, $0 < \\beta < r$, $\\alpha < 0$. In this case, we take $k = s\\beta$, $i = 1 - \\alpha\\beta$.\n\nIn both cases, $k(k-1)/2 = -\\alpha\\beta r s = -\\alpha\\beta n = (i-1)n$. Straightforward calculation (using $r > 1$, $\\alpha > 0$, $s \\ge 3$) shows that $3 \\le k \\le n-2$ and $2 \\le i \\le n/2$. Thus we have a fixed point of $f$ in either case.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 41044,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a regular cube with side length $2$. Let $A$ and $B$ be two vertices that are furthest apart. Construct a sequence of points on the surface of the cube $A_1, A_2, \\dots, A_k$ so that $A_1 = A$, $A_k = B$ and for any $i = 1, \\dots, k-1$, the distance from $A_i$ to $A_{i+1}$ is $3$. Find the minimum value of $k$.",
    "options": [],
    "answer": "7",
    "solution": "The sphere with centre $A$ and radius $3$ intersects the three edges at $B$ in three points $K, L, N$. By straightforward calculation using Pythagoras' Theorem, it's easy to show that they are the midpoints of the edges. Let $M$ be an interior point of any of the arcs $KL, LN, KN$ arising from the intersection with the sphere. The sphere with centre $M$ and radius $3$ contains the point $A$. All the other points of the cube are in the interior of this sphere since the distance from $M$ to all the other vertices are $< 3$. Thus $A_2$ must be one of $K, L, N$.\n\nFrom each of $K, L, N$ we can reach one of the vertices adjacent to $A$. Thus $A_3$ is a vertex connected to $A$ by a single edge. Now $A_4$ must be a midpoint of a side and $A_5$ is a vertex connected to $A$ by at most $2$ edges. Since $A$ is connected to $B$ by a sequence of $3$ edges, we see that $k \\ge 7$.\n\nIt is easy to construct a sequence of length $7$ that works.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41045,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $\\triangle ABC$, let $I$, $O$, $\\Gamma$ denote its incenter, circumcenter and circumcircle respectively. Let $AI$ intersect $\\Gamma$ at $M\\ (\\neq A)$. Circle $\\omega$ is tangent to $AB$, $AC$ and $\\Gamma$ internally at $T$ (i.e. the mixtilinear incircle opposite $A$). Let the tangents at $A$ and $T$ to $\\Gamma$ meet at $P$, and let $PI$ and $TM$ intersect at $Q$. Prove that $QA$ and $MO$ intersect at a point on $\\Gamma$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41046,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be a point inside a triangle $ABC$ such that $\\angle BOC = 90^\\circ$ and $\\angle BAO = \\angle BCO$. Prove that $\\angle OMN = 90^\\circ$, where $M$, $N$ are the midpoints of $AC$ and $BC$ respectively.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Extend $CO$ to a point $D$ such that $CO = OD$. Since $\\angle BOC = 90^\\circ$, the triangles $BOC$ and $BOD$ are congruent. Let $\\omega_1$ be the circumcircle of the triangle $BOD$. Since $\\angle BDO = \\angle BCO = \\angle BAO$, the point $A$ lies on $\\omega_1$. As $M$, $O$, $N$ are the midpoints of $CA$, $CD$, $CB$ respectively, the homothety centered at $C$ with similitude ratio $\\frac{1}{2}$ maps $\\omega_1$ to the circumcircle $\\omega_2$ of the triangle $MON$. As $DB$ is a diameter of $\\omega_1$, $ON$ is a diameter of $\\omega_2$. Thus $\\angle OMN = 90^\\circ$.\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41047,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $AB = AC = 14\\sqrt{2}$, $D$ is the midpoint of $CA$ and $E$ is the midpoint of $BD$. Suppose $\\triangle CDE$ is similar to $\\triangle ABC$. Find the length of $BD$.",
    "options": [],
    "answer": "14",
    "solution": "Let $\\ell = AB = AC = 14\\sqrt{2}$ and $BC = x$. The 4 angles marked in the figure are all equal. This implies that $\\triangle ABC$, $\\triangle BCD$, $\\triangle CDE$ are all similar. Thus $BD = BC = x$.\n\nAlso $AB/BC = BC/CD$. That is $\\ell/x = x/(\\ell/2)$. From this we get $x = \\ell/\\sqrt{2}$. Since $\\ell = 14\\sqrt{2}$, we have $x = 14$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41048,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a convex quadrilateral $ABCD$, $\\angle A < 90^\\circ$, $\\angle B < 90^\\circ$ and $AB > CD$. Points $P$ and $Q$ are on the segments $BC$ and $AD$ respectively. Suppose the triangles $APD$ and $BQC$ are similar. Prove that $AB$ is parallel to $CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet the lines $BC$ and $AD$ intersect at $M$. The conditions $\\angle A < 90^\\circ$, $\\angle B < 90^\\circ$ and $AB > CD$ imply that $C$ lies between $MB$ and $D$ lies between $MA$. Given the triangles $APD$ and $BQC$ are similar, we have $\\angle BCQ = \\angle ADP$ so that $\\angle MCQ = \\angle MDP$. From this, we see that the triangles $MCQ$ and $MDP$ are similar. Thus $MC/MD = MQ/MP$. Using the given condition that the triangles $APD$ and $BQC$ are similar, we have $\\angle PAD = \\angle QBC$ so that $\\angle PAM = \\angle QBM$. It follows that the triangles $PAM$ and $QBM$ are similar. Thus $MQ/MP = MB/MA$. Therefore, $MB/MA = MC/MD$. This implies that $AB$ is parallel to $CD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41049,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Show that there exists an integer $m$ such that\n$$\n2018m^2 + 20182017m + 2017\n$$\nis divisible by $2^n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We shall show more generally that $am^2 + bm + c \\equiv 0 \\pmod{2^n}$ has a solution for all $n$ whenever $b$ is odd and $a$ or $c$ is even. For $n = 1$, take $m = 0$ if $c$ is even and $m = 1$ if $c$ is odd. Now suppose the claim is true for $n$. If $c$ is even, then by assumption, the congruence $2at^2 + bt + c/2 \\equiv 0 \\pmod{2^n}$ has a solution in $t$. Letting $m = 2t$, we get $am^2 + bm + c = 2(2at^2 + bt + c/2) \\equiv 0 \\pmod{2^{n+1}}$. If $c$ is odd, then $a$ is even and thus $a+b+c$ is even; hence by assumption, the congruence $2at^2 + (2a+b)t + (a+b+c)/2 \\equiv 0 \\pmod{2^n}$ has a solution in $t$. Letting $m = 2t+1$, we get $am^2 + bm + c = 2(2at^2 + (2a+b)t + (a+b+c)/2) \\equiv 0 \\pmod{2^{n+1}}$. This completes the proof of the claim by induction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41050,
    "subject": "Mathematics (Multi-modal)",
    "question": "One hundred balls labelled $1$ to $100$ are to be put into two identical boxes so that each box contains at least one ball and the greatest common divisor of the product of the labels of all the balls in one box and the product of the labels of all the balls in the other box is $1$. Determine the number of ways that this can be done.",
    "options": [],
    "answer": "2047",
    "solution": "First let's assume the two boxes are different. The balls labelled $1$ and $2$ can be put into either box. All balls with even labels must then be put into the same box that the ball labelled $2$ goes. Thus any ball with label which has a common factor greater than $1$ with any one of those even labels must be put into the same box too. Therefore, we only need to consider those balls with labels greater than $50$. That leaves with $10$ balls labelled by the primes: $53$, $59$, $61$, $67$, $71$, $73$, $79$, $83$, $89$, $97$. Thus we have twelve entities, namely the ten prime numbers, the number $1$ and all the balls that must go with $2$. Each can be put into any of the two boxes. Taking away the $2$ cases where one of the boxes is empty, the number of ways is $2^{12} - 2 = 4094$. Since the two boxes are identical, we need to divide this number by $2$. Therefore the answer is $2047$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41051,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest positive integer $n$ such that there exist $n$ real polynomials where the sum of any two has no real roots but the sum of any three does.",
    "options": [],
    "answer": "3",
    "solution": "When $n = 3$, we can take the constant polynomials $f, g, h = -1, -2, 3$ which clearly satisfy the problem conditions.\n\nNow assume that $n = 4$ and let our polynomials be $f_1, f_2, f_3, f_4$.\nNote that for any $i, j$, we must have either $f_i(x) + f_j(x) > 0$ or $f_i(x) + f_j(x) < 0$ for all $x$ as otherwise it must have a real root. If there exist indices $i, j, k$ such that $f_i(0) + f_j(0), f_i(0) + f_k(0), f_j(0) + f_k(0)$ all have the same sign, say positive, then for all $x$\n$$\nf_i(x) + f_j(x) > 0, \\quad f_i(x) + f_k(x) > 0, \\quad f_j(x) + f_k(x) > 0\n$$\n$$\n\\therefore f_i(x) + f_j(x) + f_k(x) > 0.\n$$\nwhich is a contradiction as the sum $f_i(x) + f_j(x) + f_k(x)$ would have no real roots. We shall show that such a triple of indices must exist.\n\nWLOG let $|f_1(0)| \\ge |f_i(0)|$ for $i = 2, 3, 4$ and that $f_1(0) > 0$. Then $f_1(0) + f_i(0) > 0$ for all $i$. If there exist $i, j$ chosen from 2, 3, 4 such that $f_i(0) + f_j(0) > 0$, then $1, i, j$ is such a triple. If not, then 2, 3, 4 is such a triple. Thus $n$ cannot be four. Therefore the only answer is $n = 3$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41052,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $f: N \\to N$ is a function such that\n$$\nf^n(n) = 2n\n$$\nfor all $n \\in N$. Must $f(n) = n + 1$ for all $n$?",
    "options": [],
    "answer": "Yes",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41053,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{N} \\setminus \\{1\\} \\rightarrow \\mathbb{N}$ such that for all distinct $x, y \\in \\mathbb{N}$ with $y \\ge 2018$,\n$$\n\\gcd(f(x), y) \\cdot \\operatorname{lcm}(x, f(y)) = f(x)f(y)\n$$",
    "options": [],
    "answer": "f(n) = n for all n ≥ 2",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41054,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ with at least 4 factors such that $n$ is the sum of the squares of its 4 smallest factors.",
    "options": [],
    "answer": "130",
    "solution": "Let $a < b < c < d$ be the 4 smallest factors of $n$. Then $n = a^2 + b^2 + c^2 + d^2$.\n\nIf $n$ is odd, then $a, b, c, d$ are all odd, impossible. Thus $n$ is even. Hence $a = 1, b = 2$.\n\nIf $4 \\mid n$, then one of $c, d$ is $4$ and the other is odd. By taking mod $4$, we see the LHS is $0$ while the RHS is $2$, a contradiction. Thus $4$ is not a factor.\n\nNote that $c$ cannot be even otherwise $c/2$ is also a factor which means $c$ has to be $2$. Thus $c$ is odd and $d$ is even. Therefore $d = 2c$.\n\nSo we have $n = 1 + 4 + c^2 + 4c^2 = 5 + 5c^2$. Hence $5$ is a factor.\n\nIf $3$ is a factor, then $c = 3$ and $d = 5$, a contradiction. Therefore $c = 5$ and $n = 5(1 + 5^2) = 130$.\n\nIt is easily checked that the smallest 4 factors are $1, 2, 5, 10$ and $130$ is the sum of their squares.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41055,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c, d$ be positive integers such that $a + c = 20$ and $\\frac{a}{b} + \\frac{c}{d} < 1$. Find the maximum possible value of $\\frac{a}{b} + \\frac{c}{d}$.",
    "options": [],
    "answer": "1385/1386",
    "solution": "Therefore the maximum value of the sum of the fractions is attained when $y$ is minimum, which is $1$. When $y = 1$, since $ac < xy$, the optimum value for $x$ is $ac + 1$ (we want to choose $x$ so that it is as small as possible). Thus for fixed $a, c$, the maximum value is\n$$\n\\frac{a}{a+ac+1} + \\frac{c}{c+1} = 1 - \\frac{1}{(a+ac+1)(c+1)}.\n$$\nBy checking all pairs $(a, c) = (1, 19), (2, 18), \\dots, (10, 10)$, we find that the maximum value of $(a + ac + 1)(c + 1)$ is $1386$, attained when $(a, c) = (7, 13)$. Therefore the maximum value is $1 - \\frac{1}{1386} = \\frac{1385}{1386}$.\n\nWe shall use the following result. Let $p, q, k$ be positive integers. Then $\\frac{p}{p+k} < \\frac{q}{q+k}$ iff $p < q$.\n\nFirst fix $a, c$. Without loss of generality, we assume that $a \\le c$. Let $b = a + x$ and $d = c + y$ where $x, y$ are positive integers. Then it is easy to see that\n$$\n\\frac{a}{a+x} + \\frac{c}{c+y} \\ge \\frac{a}{a+y} + \\frac{c}{c+x}\n$$\nif $y \\le x$. Thus we may assume that $y \\le x$. Also\n$$\n1 > \\frac{a}{b} + \\frac{c}{d} = \\frac{a}{a+x} + \\frac{c}{c+y} = \\frac{ac + ay + ac + cx}{ac + ay + xy + cx}\n$$\nThus $ac < xy$ and $xy - ac$ is the difference of the denominator and numerator of the fraction on the RHS. Thus if $x', y'$ are positive integers such that $y' \\le x'$ and $xy = x'y'$, then $|y - y'| \\le |x - x'|$. Therefore\n$$\n\\begin{align*}\n\\frac{a}{a+x} + \\frac{c}{c+y} &\\le \\frac{a}{a+x'} + \\frac{c}{c+y'} \\\\\n\\Leftrightarrow (a+x)(c+y) &\\le (a+x')(c+y') \\\\\n\\Leftrightarrow a(y - y') &\\le c(x' - x) \\\\\n\\Leftrightarrow y &\\ge y \\quad \\text{and} \\quad x' \\ge x\n\\end{align*}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41056,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each of the squares in a $2 \\times 2018$ grid of squares is to be coloured black or white such that in any $2 \\times 2$ block, at least one of the $4$ squares is white. Let $P$ be the number of ways colouring the grid. Find the largest $k$ so that $3^k$ divides $P$.",
    "options": [],
    "answer": "1009",
    "solution": "Let $P_n$ be the number of colourings of a $2 \\times n$ grid, $A_n$ be the number of colourings in which the first two squares are coloured black and $B_n$ be the number of colourings in which at least one of the first two squares is coloured white. Then $P_1 = 4$, $P_2 = 15$ and for $n \\ge 3$,\n$$\n\\begin{align*}\nP_n &= A_n + B_n \\\\\nA_n &= B_{n-1} \\\\\nB_n &= 3(B_{n-1} + A_{n-1}) \\\\\n\\therefore A_n &= 3(B_{n-2} + A_{n-2}) \\\\\n\\therefore B_n &= 3(B_{n-1} + B_{n-2}) \\\\\n\\therefore P_n &= 3(P_{n-1} + P_{n-2})\n\\end{align*}\n$$\nLet $v(n)$ denote the highest power of $3$ that divides $P_n$. We have that $v(1) = 0$, $v(2) = 1$, $v(3) = 1$, $v(4) = 3$, $v(5) = 2$, $v(6) = 3$, etc. It is easy to see by induction that $v(2k-1) \\ge k-1$, $v(2k) \\ge k$. That is $v(n) \\ge \\lfloor \\frac{n}{2} \\rfloor$.\nLet $Q_n = \\frac{P_n}{3^{\\lfloor \\frac{n}{2} \\rfloor}}$. Then $Q_n$ satisfies the recurrence relations:\n$$\n\\begin{align*}\nQ_{2k} &= Q_{2k-1} + Q_{2k-2}, \\\\\nQ_{2k-1} &= 3Q_{2k-2} + Q_{2k-3}.\n\\end{align*}\n$$\nFrom this we have $Q_1 = 4$, $Q_2 = 5$, $Q_3 = 19$, $Q_4 = 24$, $Q_5 = 91$, etc. Taking mod $3$ of the above relations, we have\n$$\n\\begin{align*}\nQ_{2k} &\\equiv Q_{2k-1} + Q_{2k-2} \\pmod{3}, \\\\\nQ_{2k-1} &\\equiv Q_{2k-3} \\pmod{3}.\n\\end{align*}\n$$\nThe second equation gives $Q_n \\equiv Q_1 \\equiv 1 \\pmod{3}$ for all $n$ odd. Therefore $Q_{2k} \\equiv 1+Q_{2k-2} \\pmod{3}$. Inductively, $Q_{2k} \\equiv k-1+Q_2 \\pmod{3}$. Since $Q_2 = 5$, we have $Q_{2k} \\equiv k+1 \\pmod{3}$. Thus $Q_{2018} \\equiv 1010 \\equiv 2 \\pmod{3}$. Therefore $Q_{2018}/3^{1009} = Q_{2018}$ is not divisible by $3$. In other words, $v(2018) = 1009$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 41057,
    "subject": "Mathematics (Multi-modal)",
    "question": "You are given some equilateral triangles and squares, all with side length $1$, and asked to form convex $n$ sided polygons using these pieces. If both types must be used, what are the possible values of $n$, assuming that there is sufficient supply of the pieces?",
    "options": [],
    "answer": "5, 6, 7, 8, 9, 10, 11, 12",
    "solution": "Thus $n \\le 12$. It's easy to construct such polygons with $5$-$12$ sides.\n\n![](attached_image_1.png)\n\nTo complete the proof, we now show that $n$ cannot be $3$ or $4$. An *interior square* is a square with none of its vertices lie on the boundary (of the convex polygon). At a vertex of an interior square, there are either $2$ squares and $3$ triangles or $4$ squares. Thus there are at least $2$ squares at every vertex of an interior square. A *boundary square* is a square that has at least one vertex on the boundary. It is also easy to see that a boundary square must have one side on the boundary. We first show that the convex polygon must have a boundary square. Suppose on the contrary, all the squares are interior. Consider a line $L$ containing one side of the polygon. Let $v$ be a vertex belonging to a square so that $v$ is closest to $L$. At $v$ there are two squares. Since between $L$ and the line $M$ containing $v$ and parallel to $L$, there are no vertices belonging to a square, one side of each of the two squares containing $v$ must lie on $M$, i.e., the $2$ squares share a common side. Let $w$ be the last vertex on the 'left' side of $v$ on the line $M$ belonging to a square. By applying the same argument, there must be another vertex to the left of $w$ that belongs to a square, contradicting the choice of $w$. Therefore, there are boundary squares. If the boundary of the polygon formed is formed only by the edges of squares, we get a rectangle. This means the boundary is a cycle of squares. Removing these squares either leaves nothing or another rectangle. This means that no triangle is used, a contradiction. Thus on the boundary, there are edges belonging to triangles and squares. When an edge belonging to a triangle meets an edge belonging to a square, they meet at a vertex of the polygon and the angle at the vertex is $150^\\circ$. There are at least two such vertices and therefore there are two angles which are $150^\\circ$. This means $n$ cannot be $3$ or $4$.\n\n1. The possible internal angles are $60^\\circ$, $90^\\circ$, $120^\\circ$, $150^\\circ$. Let their respective numbers be $a, b, c, d$. Then we have\n$$\n\\begin{aligned}\na+b+c+d &= n \\\\\n60a+90b+120c+150d &= (n-2)180\n\\end{aligned}\n$$\nEliminating $n$, we obtain\n$$\n4a+3b+2c+d=12.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41058,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a polynomial $P(x, y, z)$ in three variables with integer coefficients such that for any real numbers $a, b, c$,\n$$\nP(a, b, c) = 0 \\Leftrightarrow a = b = c.\n$$\nFind the largest integer $r$ such that for all such polynomials $P(x, y, z)$ and integers $n, m$,\n$$\nm^r \\mid P(n, n+m, n+2m).\n$$",
    "options": [],
    "answer": "2",
    "solution": "Consider $P(x, y, z) = (y-x)^2 + (z-y)^2$. Then, $P(n, n+m, n+2m) = 2m^2$. Clearly, $r > 2$ does not work. We show that $r = 2$.\n\nIf there exist $(a_1, b_1, c_1)$, $(a_2, b_2, c_2)$ such that $P(a_1, b_1, c_1) > 0$ and $P(a_2, b_2, c_2) < 0$, draw a continuous path from $(a_1, b_1, c_1)$ to $(a_2, b_2, c_2)$, not passing through the line $x = y = z$.\nSince $P$ is continuous, there exists $(a_3, b_3, c_3)$ on the path such that $P(a_3, b_3, c_3) = 0$. This is a contradiction as the condition that $a_3 = b_3 = c_3$ is not satisfied. Hence, either all $P \\le 0$ or $P \\ge 0$.\n\nLet $u = y - x$, $v = z - y$. $P(x, y, z) = Q(x, u, v) = uvQ_1(x, u, v) + u^2Q_2(x, u, v) + v^2Q_3(x, u, v) + uR_1(x) + vR_2(x) + R_3(x)$. Substituting $u, v = 0$, $x = a$, we have $R_3(a) = 0$ for all real $a \\Rightarrow R_3(x) = 0$. Substituting $v = 0$, $x = a$, we have $Q(a, u, 0) = u^2Q_2(a, u, 0) + uR_1(a)$. However, since $P \\ge 0$ or $P \\le 0$, for any fixed real $a$, there must be a double root at $u = 0$. Hence, $R_1(a) = 0$ for all $a \\Rightarrow R_1(x) = 0$. Similarly, $R_2(x) = 0$.\n\nSince $m \\mid u, v$, we have that $m^2$ divides $P(n, n+m, n+2m) = Q(n, m, m) = m^2(Q_1(n, m, m) + Q_2(n, m, m) + Q_3(n, m, m))$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41059,
    "subject": "Mathematics (Multi-modal)",
    "question": "Starting with any $n$-tuple $R_0$, $n > 1$, of symbols from $A$, $B$, $C$, we define a sequence $R_0, R_1, R_2, \\dots$, according to the following rule: If $R_j = (x_1, x_2, \\dots, x_n)$, then $R_{j+1} = (y_1, y_2, \\dots, y_n)$, where $y_i = x_i$ if $x_i = x_{i+1}$ (taking $x_{n+1} = x_1$) and $y_i$ is the symbol other than $x_i, x_{i+1}$ if $x_i \\neq x_{i+1}$. Find all positive integers $n > 1$ for which there exists some integer $m > 0$ such that $R_m = R_0$.",
    "options": [],
    "answer": "All odd integers n greater than one",
    "solution": "Replace $A$, $B$, $C$ by $0$, $1$, $2$. Then $y_i \\equiv -x_i - x_{i+1} \\pmod{3}$. We first show that such an $m$ does not exist for even $n$ as seen from the following $n$-tuple\n$$\n(1, 0, 1, 0, \\dots, 1, 0).\n$$\nNext we show that $m$ exists when $n$ is odd. Since the total number of tuples that can be formed is $3^n$, there exist indices $s < t$ such that $R_s = R_t$. Without loss of generality, we may assume that $R_s$ is the only term in the sequence from $R_0$ to $R_t$ that is the same as $R_t$. If we can show that for any $i \\ge 0$, $R_i$ can be obtained from $R_{i+1}$, then the sequence going backwards from $R_s$ is the same as the sequence going backwards from $R_t$. Since going backwards from $R_s$, we can get $R_0$, then going backwards from $R_t$, before reaching $R_s$, we can get an $n$-tuple which is the same as $R_0$. In other words, there is an index $m$, $s < m < t$, such that $R_0 = R_m$ and we are done.\nTo show that the sequence can be reversed, all we need to show is if two $n$-tuples $(a_1, a_2, \\dots, a_n)$ and $(b_1, b_2, \\dots, b_n)$ give rise to the same $n$-tuple, then they are in fact the same. We have from the given rule:\n$$\n\\begin{aligned}\n-a_1 - a_2 &= -b_1 - b_2 \\\\\na_2 + a_3 &= b_2 + b_3 \\\\\n\\multicolumn{4}{c}{\\dots} \\\\\n-a_{n-2} - a_{n-1} &= -b_{n-2} - b_{n-1} \\\\\na_{n-1} + a_n &= b_{n-1} + b_n \\\\\n-a_n - a_1 &= -b_n - b_1\n\\end{aligned}\n$$\nSumming up, we get $2a_1 = 2b_1$ and therefore $a_1 = b_1$. Repeat this starting with the index pair $(2, 3)$, we can show that $a_2 = b_2$. Continuing this way, we have $(a_1, \\dots, a_n) = (b_1, \\dots, b_n)$:",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41060,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1$, $x_2$, $x_3$, $y_1$, $y_2$, $y_3$ be real numbers in $[-1, 1]$. Find the maximum value of\n$$\n(x_1y_2 - x_2y_1)(x_2y_3 - x_3y_2)(x_3y_1 - x_1y_3)\n$$",
    "options": [],
    "answer": "84*sqrt(3) - 144",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41061,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, let $O$, $H$, and $N$ be its circumcenter, orthocenter, and nine-point center respectively. Let $AN$ meet the circumcircle of $\\triangle ABC$ at $S$. Let the tangents to the circumcircle of $ABC$ at $B$ and $C$ meet at $D$. Show that $\\angle DSH = \\angle DOA$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41062,
    "subject": "Mathematics (Multi-modal)",
    "question": "Anana has an ordered $n$-tuple $(a_1, a_2, ..., a_n)$ of integers. Banana may make a guess on Anana's ordered integer $n$-tuple $(x_1, x_2, ..., x_n)$, upon which Anana will reveal the product of differences $(a_1 - x_1)(a_2 - x_2)...(a_n - x_n)$. How many guesses does Banana need to figure out Anana's $n$-tuple for certain?",
    "options": [],
    "answer": "n + 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41063,
    "subject": "Mathematics (Multi-modal)",
    "question": "Simon plays a game on an $n \\times n$ grid of cells. Initially, each cell is filled with an integer. Every minute, Simon picks a cell satisfying the following:\n(a) The magnitude of the integer in the chosen cell is less than $n^n$\n(b) The sum of all the integers in the neighboring cells (sharing one side with the chosen cell) is non-zero\nSimon then adds each integer in a neighboring cell to the chosen cell.\nShow that Simon will eventually not be able to make any valid moves.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 41064,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a scalene triangle $ABC$ with centroid $G$ and circumcircle $\\omega$ centred at $O$, the extension of $AG$ meets $\\omega$ at $M$; lines $AB$ and $CM$ intersect at $P$; and lines $AC$ and $BM$ intersect at $Q$. Suppose the circumcentre $S$ of the triangle $APQ$ lies on $\\omega$ and $A$, $O$, $S$ are collinear. Prove that $\\angle AGO = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AD$, $BE$ and $CF$ be the medians of the triangle $ABC$. Since $A$, $O$, $S$ are collinear, $AS$ is a diameter of $\\omega$ with $AO = OS$. The triangles $AOB$ and $ASP$ are similar isosceles triangles. Thus $OB \\parallel SP$. Since $O$ is the midpoint of $AS$, we have $B$ is the midpoint of $AP$. Similarly, $C$ is the midpoint of $AQ$. Since $F$ is the midpoint of $AB$ and $C$ is the midpoint of $AQ$, we have $FC \\parallel BQ$. Thus $GC \\parallel BM$. Since $F$ is the midpoint of $AB$, $G$ is the midpoint of $AM$. Therefore, $GO \\parallel MS$. Then $\\angle AGO = \\angle AMS = 90^\\circ$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41065,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square, $E$ be a point on the side $DC$, $F$ and $G$ be the feet of the altitudes from $B$ to $AE$ and from $A$ to $BE$, respectively. Suppose $DF$ and $CG$ intersect at $H$. Prove that $\\angle AHB = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $A$, $F$, $G$, $B$ are concyclic, since they lie on the circle $\\Gamma$ with diameter $AB$. Let $DF$ intersect $\\Gamma$ again at $H'$; it suffices to show that $H = H'$.\n\n**Solution 1:**\nLet the diagonals $AC$ and $BD$ intersect at $O$; note that $O$ also lies on $\\Gamma$. Let $H'G \\cap AO = C'$. Now Pascal's theorem on $FAOBGH'$ shows that $C'$, $D$, $E$ are collinear, so $C = C'$ and $H = H'$.\n\n**Solution 2:**\nBy power of point, $DF \\times DH' = DA^2 = DC^2$. Therefore $\\triangle DH'C \\sim \\triangle DCF$. Therefore, since $\\angle ECB = \\angle EFB = 90^\\circ$ implies that $ECBF$ is cyclic,\n$$\n\\angle DH'C = \\angle DCF = \\angle ECF = \\angle EBF = \\angle EBF\n$$\nHence if $CH'$, $EB$ intersect at $G'$ then $FG'BH'$ is cyclic, which implies $G'$ lies on $\\Gamma$, so $G' = G$ and thus $H' = H$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41066,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a convex quadrilateral $ABCD$, the diagonals intersect at $O$, $M$ and $N$ are points on the segments $OA$ and $OD$ respectively. Suppose $MN$ is parallel to $AD$ and $NC$ is parallel to $AB$. Prove that $\\angle ABM = \\angle NCD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use the notation $[PQR]$ to denote the area of $\\triangle PQR$.\n$$\nMN \\parallel AD \\Rightarrow [AMN] = [DMN] \\Rightarrow [AON] = [DOM].\n$$\n$$\nNC \\parallel AB \\Rightarrow [ACN] = [BCN] \\Rightarrow [AON] = [BOC].\n$$\n$$\n\\text{Thus } [DOM] = [BOC] \\Rightarrow [MCD] = [BCD] \\Rightarrow MB \\parallel DC.\n$$\n$$\n\\text{Finally } AB \\parallel NC, MB \\parallel DC \\Rightarrow \\angle ABM = \\angle NCD.\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41067,
    "subject": "Mathematics (Multi-modal)",
    "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": "59",
    "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 $< 32$ between a pair of such numbers. Also $1$ cannot be chosen. Also the two neighbours of $31$ are $32$ or less. So the maximum number of chosen integers is $\\le 30 \\times 2 - 1 = 59$. This bound can be achieved by the following 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": 41068,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a positive integer. For any integer $a$, let $Q_a(x)$ denote the polynomial $x^n + ax$. Let $p$ be a prime number and $S_a$ be the set\n$$\nS_a = \\{b \\mid 0 \\le b \\le p-1, \\exists c \\in \\mathbb{Z}, Q_a(c) \\equiv b \\pmod{p}\\}.\n$$\nShow that the expression $\\frac{1}{p-1} \\sum_{a=1}^{p-1} |S_a|$ is an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "All congruences are taken modulo $p$. We count pairs $(a, b)$ taken modulo $p$ where $a \\neq 0$ and the equation $x^n + ax \\equiv b$ has a solution in $x$. Then $\\sum_{a=1}^{p-1} |S_a|$ is the total number of such pairs.\n\nFirst, consider all pairs $(a, b)$ where $a \\neq 0$. Defined the relation $(a, b) \\sim (a', b')$ if there exists $k \\neq 0$ such that $a \\equiv a'k^{n-1}$ and $b \\equiv b'k^n$. It follows that if $(a, b) \\sim (a', b')$ and $(a', b') \\sim (a'', b'')$, then $(a, b) \\sim (a'', b'')$. Also if $(a, b) \\sim (a', b')$ then $(a', b') \\sim (a, b)$. Hence this relation partitions our pairs into subsets, where each subset consists of pairs that are all related to each other. Let A be such a subset and $(a, b) \\in A$. Then all other pairs may be described as $(ak^{n-1}, bk^n)$ where $x^n+ak^{n-1}x \\equiv bk^n$ has a solution. But since $x^n+ax \\equiv b$ has a solution, say $x_0$, one may check that $kx_0$ is a solution to $x^n+ak^{n-1}x \\equiv bk^n$ for each non-zero $k$. Thus the pair $(ak^{n-1}, bk^n) \\in A$. This gives us exactly $p-1$ distinct pairs in A since if $a, b \\neq 0$,\n$$\nai^{n-1} \\equiv aj^{n-1}, bi^n \\equiv bj^n \\Rightarrow i^{n-1} \\equiv j^{n-1}, in \\equiv j^n \\Rightarrow i \\equiv j.\n$$\n\nNow when $b \\equiv 0$, the equation $x^n + ax \\equiv 0$ has a solution for all $a$. This gives us an additional $p-1$ pairs $(a, 0)$ as we want $a \\neq 0$. Adding the sizes of the subsets up, we conclude that $\\sum_{a=1}^{p-1} |S_a|$ is divisible by $p-1$ as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41069,
    "subject": "Mathematics (Multi-modal)",
    "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? (Example: In the $3 \\times 3$ grid shown below, John could move 2 dominoes: D, followed by A.)\n\n| A | B | C |\n|---|---|---|\n| D |   |   |\n\n",
    "options": [],
    "answer": "1024143",
    "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 be also tiled with dominoes that do not move throughout.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41070,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ such that $f(x+y)[(f(x)-f(y))^2+f(xy)] = f(x^3)+f(y^3)$.",
    "options": [],
    "answer": "All solutions are: (1) f(x) = x for all integers x; (2) f(x) = 0 for all integers x; (3) f(x) = 2 for all integers x; (4) f(x) = 1 for odd integers x and f(x) = 2 for even integers x.",
    "solution": "There are 4 solutions, namely\n$$\nf(x) = x, \\quad f(x) = 0, \\quad f(x) = 2, \\quad f(x) = \\begin{cases} 1 & \\text{if } x \\text{ odd} \\\\ 2 & \\text{if } x \\text{ even} \\end{cases}\n$$\nLet (1) be the original equation. The first 3 solutions can be easily verified. The last solution can be verified by considering the cases when $x, y$ have the same parity, and when they don't. We proceed to prove these are the only solutions.\n\nPut $x = y = 0$: $f(0)^2 = 2f(0) \\Rightarrow f(0) = 0$ or $f(0) = 2$\n\nPut $y = 0$: $f(x)^3 - 2f(x)^2f(0) + f(x)f(0)^2 + f(x)f(0) = f(x^3) + f(0)$ (2)\n\nPut $y = -x$: $f(0)[(f(x) - f(-x))^2 + f(-x^2)] = f(x^3) + f(-x^3)$ (3)\n\nCase 1. $f(0) = 0$. From (2) and (3) we get $f(x)^3 = f(x^3)$ and $f(x^3) = -f(-x^3)$. Therefore $f(-x) = \\sqrt[3]{f(-x^3)} = \\sqrt[3]{-f(x^3)} = -f(x)$, so $f$ is an odd function. Putting $x = 1$ in (2) yields $f(1)^3 = f(1)$, so $f(1) = 1, 0$ or $-1$. Putting $y = 1$ in (1), yields\n$$\nf(x+1)((f(x)-f(1))^2 + f(x)) = f(x^3) + f(1) \\quad (4)\n$$\nCase 1.1. $f(1) = 0$. (4) becomes $f(x+1)(f(x)^2 + f(x)) = f(x)^3$. Thus $f(x+1) = 0$ implies $f(x) = 0$. Since $f(1) = 0$, $f(x) = 0$ for all $x \\le 1$. Since $f$ is odd, $f(x) = 0$ for all $x \\in \\mathbb{Z}$ which is the second solution.\n\nCase 1.2. $f(1) = 1$. (4) becomes $f(x+1)(f(x)^2 - f(x) + 1) = f(x)^3 + 1$. Since $f(x)^2 - f(x) + 1 > 0$, $f(x+1) = \\frac{f(x)^3+1}{f(x)^2-f(x)+1} = f(x)+1$. From here we can easily show $f(x) = x$ for all $x \\in \\mathbb{Z}$ which is the first solution.\n\nCase 1.3. $f(1) = -1$. (4) becomes $f(x+1)(f(x)^2 + 3f(x) + 1) = f(x)^3 - 1$. Since $f(x)^2 + 3f(x) + 1 > 0$, $f(x+1) = \\frac{f(x)^3-1}{f(x)^2+3f(x)+1}$. Putting $x = 1$ and then $x = 2$ yield $f(2) = 2$ and $f(3) = \\frac{7}{11}$ which is not an integer. Thus there is no solution.\n\nCase 2. $f(0) = 2$. Substituting $x = 1$ in (2) yields $f(1)^3 - 4f(1)^2 + 5f(1) - 2 = 0$ which implies $(f(1) - 2)(f(1) - 1)^2 = 0$. Thus $f(1) = 1$ or $f(1) = 2$. Similarly, substituting $x = -1$ in (2) yields $f(-1) = 1$ or $2$. Thus $(f(1), f(-1))$ can be either $(1, 1)$, $(1, 2)$, $(2, 1)$, or $(2, 2)$. Putting $x = 1$ in (3), we get $2[(f(1) - f(-1))^2 + f(-1)] = f(1) + f(-1)$ implying $f(1) + f(-1)$ is even which rules out $(1, 2)$ and $(2, 1)$.\n\nPutting $x = -1$ in (1), we get\n$$\nf(x-1)((f(x)-f(-1))^2 + f(-x)) = f(x^3) + f(-1) \\quad (5)\n$$\nCase 2.1. $f(1) = f(-1) = 1$. We will inductively prove this leads to the function $f(\\text{odd}) = 1$, $f(\\text{even}) = 2$. It suffices to show the function is periodic with period 2. For $x \\in [-1, 1]$ this is already true.\n\nSuppose we have proven the periodicity for all $x \\in [-n, n]$, where $n$ is a positive integer. This implies $f(n) = f(-n) > 0$. Since $f(1) = f(-1)$, combining (4) and (5) and substituting $x = n$, we get\n$$\nf(n+1) = \\frac{f(n)^3 + f(1)}{(f(n) - f(1))^2 + f(n)} = \\frac{f(n)^3 + f(1)}{(f(n) - f(-1))^2 + f(-n)} = f(n-1)\n$$\nSimilarly, substituting $x = -n$, we get\n$$\nf(-n-1) = \\frac{f(-n)^3 + f(-1)}{(f(-n) - f(-1))^2 + f(n)} = \\frac{f(-n)^3 + f(1)}{(f(-n) - f(1))^2 + f(-n)} = f(-n+1)\n$$\nThus the function is periodic with period 2 in the range $[-n - 1, n + 1]$ as well. The induction is complete and we have found our fourth solution.\n\n**Case 2.2.** $f(1) = f(-1) = 2$. Similarly to Case 2.1, we can inductively prove the function is periodic with period 2, thus yielding the function $f(x) = 2$, the third solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41071,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $m, n$ satisfying $n! + 2^{n-1} = 2^m$.",
    "options": [],
    "answer": "(m, n) = (1, 1), (2, 2), (5, 4)",
    "solution": "We first check that $n = 1, 2, 4$ are solutions among $n \\le 4$, with $m = 1, 2, 5$ respectively.\n\nWe now assume that $n \\ge 5$. By Legendre's formula, we know that $v_2(n!) = n - s_2(n)$ where $s_2(n)$ is the number of non-zero digits in the binary representation of $n$. Thus $v_2(n!) \\le n - 1$.\n\nIf $v_2(n!) = a < n - 1$, letting odd($x$) denote the odd part of $x$, we have\n$$\n\\mathrm{odd}(n!) + 2^{n-1-a} = 2^{m-a}\n$$\nwhich cannot happen as $\\mathrm{odd}(n!)$ is odd and the other two terms are even. Hence $v_2(n!) = n - 1$ and we must have $s_2(n) = 1 \\Rightarrow n$ is a power of 2.\n\nLet $n = 2^a$. Then $a \\ge 3$. We claim that $\\mathrm{odd}(n!) \\equiv 3 \\pmod 8$. To see so, we pair up $i$ with $n-i = 2^a-i$. If $v_2(i) < a-2$, then $\\mathrm{odd}(2^a-i) \\equiv -\\mathrm{odd}(i) \\pmod 8$ and so we have\n$$\n\\mathrm{odd}(i) \\cdot \\mathrm{odd}(2^a - i) \\equiv -\\mathrm{odd}(i)^2 \\equiv -1 \\pmod 8\n$$\nand we have an even number of such pairs. This gives us a product of 1. The remaining $i$'s are of simply $2^{a-2}, 2^{a-1}$ and $3 \\cdot 2^{a-2}$, whose odd part multiply to 3 mod 8. Hence the total product is 3 mod 8 as desired. Then\n$$\nn! + 2^{n-1} = 2^m \\Rightarrow \\mathrm{odd}(n!) + 1 = 2^{m-n-1} \\Rightarrow 2^{m-n-1} \\equiv 4 \\pmod 8 \\Rightarrow 2^{m-n-1} \\le 4\n$$\nThus $\\mathrm{odd}(n!) \\le 4$ which leads to $n \\le 4$. So there are no other solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41072,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two distinct 2-digit prime numbers $p$, $q$ can be written one after the other in 2 different ways to form two 4-digit numbers. For example, $11$ and $13$ yield $1113$ and $1311$. If the two 4-digit numbers formed are both divisible by the average value of $p$ and $q$, find all possible pairs \\{$p, q$\\}.\n",
    "options": [],
    "answer": "{13,53}, {19,47}, {23,43}, {29,37}",
    "solution": "The average of $p$, $q$ is $\\frac{p+q}{2}$. Therefore $2(100p + q) = (p+q)k$ for some integer $k$. Thus\n$$\n200p + 2q = (p+q)k \\Rightarrow 198p = (p+q)(k-2).\n$$\nSince $p$, $q$ are primes, $(p+q) \\mid 198$. Now $11 + 13 = 24 \\leq p+q \\leq 89 + 97 = 186$. Since $198 = 2 \\times 3^2 \\times 11$, the only even factor of $198$ that lies between $24$ and $186$ is $66$. Thus $p+q = 66$ and there are 4 pairs: $(13, 53)$, $(19, 47)$, $(23, 43)$, $(29, 37)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41073,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all real numbers $x$ between $0$ and $180$ such that it is possible to partition an equilateral triangle into finitely many triangles, each of which has an angle of $x^\\circ$.",
    "options": [],
    "answer": "0 < x ≤ 120",
    "solution": "Consider the sum of the angles of the triangles at each type of the vertices. At type 0, the sum is $180^\\circ$. At type 1, the sum is $180t_1$. At type 2, the sum is $360t_2$. Therefore\n$$\n180 + 180t_1 + 360t_2 = 180n \\Rightarrow 1 + t_1 + 2t_2 = n.\n$$\nNow each Type 1 vertex has at most 1 angle of $x^\\circ$ and each Type 2 vertex has at most 2 which each Type 0 vertex has, none. So the number of angles of $x^\\circ$ is $\\le t_1 + 2t_2 < n$, a contradiction.\n\nNow we construct for each $x \\in (0, 120]$. It is trivial for $x = 120$. Take the centre of the triangle and join it to the vertices to obtain a partition into 3 congruent triangles. Now assume $x \\in (0, 120)$.\n\nDefine an $(a, b)$-trapezium to be the trapezium $ABCD$ with $\\angle A = \\angle D = 60^\\circ$, $AB = CD = a$, $BC = b$. Then it follows that $AD = a + b$. We call a polygon $x$-good if it can be partitioned into triangles each with one $x^\\circ$ angle.\n\n**Lemma 1:** There exists $R(a)$ such that an $(a, b)$-trapezium is $x$-good when $b > R(a)$.\n**Proof:** The proof is clear from Fig. A. The second, third and fourth triangles are isosceles. The fifth and sixth triangles form a parallelogram with arbitrarily long horizontal side.\n\n![](attached_image_1.png)\n\nFig.A\n\n![](attached_image_2.png)\nFig.B\n\n**Lemma 2:** An $(a, b)$-trapezium is $x$-good.\n**Proof:** It follows from the fact that the trapezium can be sliced into arbitrarily thin $(a', b')$-trapezium so that $b' > R(a')$. (See Fig. B.)\nSince an equilateral triangle can be partitioned in $(a, b)$-trapeziums, the proof is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41074,
    "subject": "Mathematics (Multi-modal)",
    "question": "Colour a $20000 \\times 20000$ square grid using $2000$ different colours with $1$ colour in each square. Two squares are neighbours if they share a vertex. A path is a sequence of squares so that $2$ successive squares are neighbours. Mark $k$ of the squares. For each unmarked square $x$, there is exactly $1$ marked square $y$ of the same colour so that $x$ and $y$ are connected by a path of squares of the same colour. For any $2$ marked squares of the same colour, any path connecting them must pass through squares of all the colours. Find the maximum value of $k$.",
    "options": [],
    "answer": "20000",
    "solution": "The maximum value is $k = 20000$. Define a region as a maximal set of same-colour squares such that there is a path of the same colour between any $2$ squares in the region. From the first condition, there is exactly one marked square in a region and so there are $k$ regions. Two regions are neighbours if they have a vertex in common.\n\nNow suppose that $k > 20000$. Form a graph whose vertices are the regions and whose edges join neighbouring regions. Since there are only $2000$ colours, there are $2$ regions $P, Q$ with the same colour. Thus there is a path $P = A_0, A_1, \\dots, A_i = Q$, with $i \\ge 2000$ by the second condition. Consider the longest path in the graph: $B_0, B_1, \\dots, B_j$. Then $j \\ge i \\ge 2000$. Suppose there is a vertex $C$ not in the path but is adjacent to a vertex $B_k$ in the path. Then $k \\ne 0, j$, else there is a longer path. Consider the $2001$ vertices $C, B_p, B_{p+1}, \\dots, B_{p+1999}$ where $p < k < p+1999 \\le j$. Two of them have the same colour but their distance apart is $< 2000$, contradicting the second condition. So $B_0, B_1, \\dots, B_j$ are all the regions and $j = k-1$. Thus to move from a square in $B_m$ to a square in $B_{m+x}$, where the index is taken mod $k$, at least $x$ moves or $k-x$ moves (if $m+x > k$) are needed. (Here a move is moving from a square to a neighbouring square.) Also $2$ neighbouring squares belong to the same region or $2$ adjacent regions.\n\nLet $X$ be the $2 \\times 2$ square in the centre of the grid. Since the $4$ squares in $X$ are mutually adjacent, $X \\subseteq B_m \\cup B_{m+1}$ for some $m$. Take $Y \\in B_{m+\\lfloor(k+1)/2\\rfloor}$. Suppose the square in $X$ that is nearest to $Y$ can move to $Y$ in a minimum number of $s$ moves. Then, since that square can be in $B_m$ or $B_{m+1}$, $9999 \\ge m \\ge \\lfloor(k-1)/2\\rfloor$ or $20000 \\ge k$.\n\nThere is a colouring that gives $20000$ regions. Colour all the squares in row $i$ with colour $j$ if $i \\equiv j \\pmod{2000}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 41075,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $k$ such that there exist positive integers $a$, $b$ such that\n$$\na^2 + 4 = (k^2 - 4)b^2.\n$$",
    "options": [],
    "answer": "k = 3",
    "solution": "Suppose $k, a, b$ satisfy the equation. Rewrite the equation as\n$$\na = \\sqrt{(k^2 - 4)b^2 - 4} = \\sqrt{(kb)^2 - 4(b^2 + 1)} \\quad (1)\n$$\nConsider the quadratic equation\n$$\nx^2 - kbx + (b^2 + 1) = 0 \\quad (2)\n$$\nIts solutions are\n$$\nc = \\frac{kb \\pm \\sqrt{(kb)^2 - 4(b^2 + 1)}}{2} = \\frac{kb \\pm a}{2} \\quad (3)\n$$\nwhere $a > 0$, $k \\ge 3$. Since $a$ and $kb$ have the same parity, $c \\in \\mathbb{Z}^+$. If one of the roots is $b$, then (2) becomes $(2-k)b^2 + 1 = 0$ which gives $k = 3$, $b = 1$ and hence $a = 1$.\n\nLet the 2 roots of (2) be $c_1, c_2$ with $c_2 < c_1$. Now assume that $b \\neq c_1, c_2$. Since the product of the roots is $b^2 + 1$, we have $c_2 < b < c_1$. Now replace $b$ in (2) by $c_2$ to get $x^2 - kc_2x + (c_2^2 + 1) = 0$. If one of the roots of the equation is $c_2$, we have $k = 3$. If not we can repeat the argument to get another equation with $c_2$ replaced by another positive integer $c_3 < c_2$. Eventually, we must reach that case where one of the roots is equal to $c_i$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41076,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$, $AC > AB$, $D$ is the point on $BC$ such that $AD = AB$. Let $\\omega_1$ be the circle through $C$ tangent to $AD$ at $D$, and $\\omega_2$ the circle through $C$ tangent to $AB$ at $B$. Let $F (\\neq C)$ be the second intersection of $\\omega_1$ and $\\omega_2$. Prove that $F$ lies on $AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n![](attached_image_2.png)\nLet $\\omega_2$ intersect $AC$ at $F'$. We shall prove $F'$ lies on $\\omega_1$. Thus $F = F'$ lies on $AC$.\n\nReferring to the figure on the right. First $\\angle ABF' = \\angle BCA = \\alpha$ as $AB$ is tangent to $\\omega_2$ at $B$. Let $\\angle DBF' = \\beta$. Then $\\angle AF'B = \\angle ABC = \\angle ADB = \\alpha + \\beta$ so that the quadrilateral $ABDF'$ is cyclic. Thus $\\angle DF'C = \\angle ABC = \\alpha + \\beta$.\n\nNow refer to the figure on the left. Since $AD$ is tangent to $\\omega_1$, $\\angle DFC = \\angle ADB = \\alpha + \\beta = \\angle DF'C$. Thus $F'$ is on $\\omega_1$ and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41077,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, $\\angle B = 90^\\circ$, $AB > BC$, and $P$ is the point such that $BP = BC$ and $\\angle APB = 90^\\circ$, where $P$ and $C$ lie on the same side of $AB$. Let $Q$ be the point on $AB$ such that $AP = AQ$, and let $M$ be the midpoint of $QC$. Prove that the line through $M$ parallel to $AP$ passes through the midpoint of $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nLet $\\angle BPC = \\angle BCP = \\alpha$, $\\angle CBP = \\beta$. As $\\angle QAP = 90^\\circ - \\angle PBA = \\angle CBP = \\beta$. Therefore $\\angle APQ = \\angle AQP = \\alpha$, since $AP = AQ$.\n\nAlso $\\angle CPQ = \\alpha + \\angle BPQ = \\angle APB = 90^\\circ$. Therefore the points $B$, $C$, $P$, $Q$ all lie on a circle with centre $M$ which is the midpoint of $QC$. Thus $\\angle QBM = \\angle BQC = \\angle BPC = \\alpha$.\n\nSince $MN \\parallel PA$, $\\angle BNM = \\angle QAP = \\beta$. Therefore $\\angle BMN = \\alpha$ and it follows that $BN = MN$.\n\nSince $AP = AQ$ and $MP = MQ$, $AM$ bisects $\\angle QAP = \\beta$. Since $\\angle BNM = \\beta$, $\\angle AMN = \\angle MAN = \\beta/2$. Therefore $MN = AN$. Thus $N$ is the midpoint of $AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41078,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles right-angled triangle of area $1$. Find the length of the shortest segment that divides the triangle into $2$ parts of equal area.",
    "options": [],
    "answer": "sqrt(2*sqrt(2) - 2)",
    "solution": "There are $2$ cases as shown in the figures below:\n![](attached_image_1.png)\nFig.1\n![](attached_image_2.png)\nFig.2\n\nIn Fig. 1,\n$$\nXY^2 = AX^2 + AY^2 = (AX - AY)^2 + 2AX \\cdot AY = (AX - AY)^2 + 2.\n$$\nThus $XY$ is minimum when $AX = AY$ and so $XY^2 = 2$.\n\nIn Fig. 2, let $B$ be the origin and $BA$ be the $x$-axis, $X$ be $(x, 0)$ and $Y$ be $(y, y)$. Then $2[BXY] = xy = 1$ and\n$$\nXY^2 = (x - y)^2 + y^2 = x^2 + 2y^2 - 2xy = x^2 + 2y^2 - 2 = (x - \\sqrt{2}y)^2 + 2\\sqrt{2} - 2.\n$$\nSo $XY$ is minimum when $x = \\sqrt{2}y$ and the minimum value is $2\\sqrt{2} - 2$.\nSince $2\\sqrt{2} - 2 < 2$, the length of the shortest segment is $\\sqrt{2\\sqrt{2} - 2}$.\n\nAlternatively, we have $2[BXY] = BX \\cdot BY \\sin 45^\\circ = BX \\cdot BY \\cos 45^\\circ = 1$. Therefore $BX \\cdot BY = \\sqrt{2}$. Using the cosine rule,\n$$\nXY^2 = BX^2 + BY^2 - 2BX \\cdot BY \\cos 45^\\circ = BX^2 + BY^2 - 2 = (BX - BY)^2 + 2\\sqrt{2} - 2.\n$$\nTherefore $XY$ is minimum when $BX = BY$ and $XY^2 = 2\\sqrt{2} - 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41079,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. There are $n$ islands with $n - 1$ bridges connecting them such that one can travel from any island to another. One afternoon, a fire breaks out in one of the islands. Every morning, it spreads to all neighbouring islands. (Two islands are neighbours if they are connected by a bridge.) To control the spread, one bridge is destroyed every night until the fire has nowhere to spread the next day. Let $X$ be the minimum possible number of bridges one has to destroy before the fire stops spreading. Find the maximum possible value of $X$ over all possible configurations of bridges and islands where the fire starts at.",
    "options": [],
    "answer": "floor(sqrt(n - 1))",
    "solution": "Solution:\n\nSuppose that there are integer solutions. By completing squares, the equation becomes\n$$\n(y+1)^2 = (x^2 + 10x + 2)^2 + 2000.\n$$\nLet $a = y + 1$, $b = x^2 + 10x + 2$, $u = a - b$, $v = a + b$. Then $uv = 2000$ and $b = \\frac{v-u}{2}$.\nThe equation $x^2 + 10x + 2 - b = x^2 + 10x + 2 + \\frac{u-v}{2} = 0$ in the unknown $x$ has integer solutions. Therefore the discriminant\n$$\n\\Delta = 100 - 4\\left(2 + \\frac{u-v}{2}\\right) = 92 + 2(u-v)\n$$\nis a square. Since $v-u$ is even and $uv = 2000$, $u, v$ are both even and we have\n$$\n\\{u, v\\} = \\{2, 1000\\}, \\{4, 500\\}, \\{8, 250\\}, \\{10, 200\\}, \\{20, 100\\}, \\{40, 50\\}\n$$\nand the negative counterparts. $\\Delta$ is a square only when $(u, v) = (8, 250), (-250, -8)$. Thus $(x, y) = (7, 128), (7, -130), (-17, -128), (-17, -130)$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 41080,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a parallelogram and the points $E$, $F$ are in the exterior. If triangles $BCF$ and $DEC$ are similar, i.e., $\\triangle BCF \\sim \\triangle DEC$, prove that triangle $AEF$ is similar to these two triangles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AB = x$, $BF = x'$, $AF = x'$, $AD = y$, $DE = y'$, $AE = y'$, $CF = z$, $CE = z'$, $EF = z'$. Then $x : y' : z' = x : y : z'$.\nLet $\\angle CDE = \\angle FBC = a$, $\\angle DCE = \\angle BFC = b$, $\\angle DEC = \\angle BCF = c$ and $\\angle ADC = \\angle ABC = d$. Then $\\angle BCD = \\angle BAD = 180^\\circ - d$.\n$$\n\\therefore \\angle ECF = 360^\\circ - b - c - (180^\\circ - d) = a + d = \\angle ABF = \\angle EDA.\n$$\n$$\n\\text{Also } \\frac{x'}{x} = \\frac{y'}{y} = \\frac{z'}{z} \\text{ (since } \\triangle BCF \\sim \\triangle DEC.)\n$$\n$$\n\\therefore \\triangle ABF \\sim \\triangle ADE \\sim \\triangle FCE.\n$$\n$$\n\\therefore \\angle DEA = \\angle CEF.\n$$\n$$\n\\therefore \\angle AEF = \\angle DEC = c. \\text{ Similarly } \\angle AFE = b.\n$$\n$$\n\\therefore \\triangle AFE \\sim \\triangle DCE \\sim \\triangle BFC.\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41081,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the minimum value of\n$$\n\\frac{x_1^3 + \\cdots + x_n^3}{x_1 + \\cdots + x_n}\n$$\nwhere $x_1, x_2, \\dots, x_n$ are distinct positive integers.",
    "options": [],
    "answer": "n(n+1)/2",
    "solution": "The minimum value $\\frac{1}{2}n(n+1)$ is achieved by letting $x_k = k$ for $1 \\le k \\le n$. To prove the inequality, it suffices to prove that\n$$\nx_1^3 + \\cdots + x_n^3 \\ge (x_1 + \\cdots + x_n)^2,\n$$\nsince $x_1 + \\cdots + x_n \\ge 1 + 2 + \\cdots + n = n(n+1)/2$.\nWe may assume that $x_1 < x_2 < \\cdots < x_n$. Let's prove the above inequality by induction on $n$. It is clear that the inequality holds when $n=1$. Let $n>1$, and assume $x_1^3 + \\cdots + x_n^3 \\ge (x_1 + \\cdots + x_n)^2$. Now\n$$\n\\sum_{k=1}^{n} x_k^3 - \\left( \\sum_{k=1}^{n} x_k \\right)^2 = \\left[ \\sum_{k=1}^{n-1} x_k^3 - \\left( \\sum_{k=1}^{n-1} x_k \\right)^2 \\right] + x_n \\left\\{ x_n^2 - x_n - 2 \\sum_{k=1}^{n-1} x_k \\right\\}.\n$$\nThe induction hypothesis implies that the first bracket on the right hand side of this equality is nonnegative.\nNext observe that $x_n \\ge n$. As $x_k < x_{k+1}$, we have $x_k \\le x_{k+1} - 1 \\le \\cdots \\le x_n - (n-k)$, when $1 \\le k < n$. Thus the term inside the second bracket on the right hand side of this equality satisfies\n$$\nx_n^2 - x_n - 2 \\sum_{k=1}^{n-1} x_k \\ge x_n^2 - x_n - 2 \\sum_{k=1}^{n-1} (x_n - n + k) = x_n^2 - x_n - 2 \\sum_{k=1}^{n-1} x_n + n(n-1) \\\\\n\\ge x_n^2 - x_n - 2(n-1)x_n + n(n-1) = (x_n - n + 1)(x_n - n) \\ge 0\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41082,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $n$ for which there exist integers $x_1 < x_2 < \\dots < x_n$ such that every integer from $1000$ to $2000$ can be written as a sum of some of the integers from $x_1, x_2, \\dots, x_n$, without repetition.",
    "options": [],
    "answer": "11",
    "solution": "The answer is $n = 11$. The $11$ integers $2^0, 2^1, \\dots, 2^{10}$ certainly work. Now we show that $n = 10$ does not work.\n\nSuppose on the contrary that the $10$ integers $x_1 < x_2 < \\dots < x_{10}$ satisfy the requirement. They can form $2^{10} = 1024$ sums, not necessarily distinct. There are $1001$ integers from $1000$ to $2000$. So at most $23$ of these sums can be $< 1000$. Now $x_1, \\dots, x_9$ can pair up to form $\\binom{9}{2} = 36$ sums. So at least one of the sums is $\\ge 1000$. Therefore $x_9 > 500$.\n\nLet $s$ be the sum of some of the integers in $x_1, \\dots, x_8$. Since $x_9 + x_{10} \\ge 1003$, at most one of $s \\ge 1000$ and $s + x_9 + x_{10} \\le 2000$ can hold. Therefore at most $3$ of the $4$ sums\n$$\ns,\\ s+x_9,\\ s+x_{10},\\ s+x_9+x_{10}\n$$\ncan be between $1000$ and $2000$ (both inclusive). The integers $x_1, \\dots, x_8$ can form at most $2^8$ different sums. Therefore $x_1 \\dots x_{10}$ can form at most $3 \\times 2^8 < 1000$ sums between $1000$ and $2000$. So $n = 10$ is impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41083,
    "subject": "Mathematics (Multi-modal)",
    "question": "Seven triangles of area $7$ lie inside a square of area $27$. Prove that among the $7$ triangles there are $2$ that intersect in a region of area not less than $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the triangular regions be $P_i$, $i = 1, \\dots, 7$, with $A(P_i) = 7$ and suppose that $A(P_i \\cap P_j) < 1$ if $i \\neq j$. Then\n$$\n\\begin{align*}\nA(P_1 \\cup P_2) &= A(P_1) + A(P_2) - A(P_1 \\cap P_2) > 14 - 1 = 13 \\\\\nA(P_1 \\cup P_2 \\cup P_3) &= A(P_1 \\cup P_2) + A(P_3) - A((P_1 \\cup P_2) \\cap P_3) \\\\\n&> 13 + 7 - A((P_1 \\cap P_3) \\cup (P_2 \\cap P_3)) \\\\\n&\\geq 13 + 7 - A(P_1 \\cap P_3) - A(P_2 \\cap P_3) \\geq 18 \\\\\nA(P_1 \\cup P_2 \\cup P_3 \\cup P_4) &= A(P_1 \\cup P_2 \\cup P_3) + A(P_4) - A((P_1 \\cup P_2 \\cup P_3) \\cap P_4) \\\\\n&> 18 + 7 - 3 > 22 \\\\\n\\dots \\\\\nA(P_1 \\cup P_2 \\cup \\dots \\cup P_7) > 27 + 7 - 6 > 28\n\\end{align*}\n$$\n\nBut the union of all $7$ triangles must be contained in the square, whose area is $27$. This contradiction shows that our assumption is false, so there must exist $i \\neq j$ such that $A(P_i \\cap P_j) \\geq 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41084,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every positive integer $n$, there is an unique $n$-digit integer $A(n)$ which is a multiple of $5^n$ and whose digits are all odd.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We shall prove by induction.\nThe assertion is true for $n=1$ as $A(1) = 5$.\nAssume that the assertion is true when $n = m$.\nLet $A(m) = \\overline{a_1a_2\\cdots a_m}$ be the unique integer with all digits odd such that $5^m \\mid A(m)$.\nWhen $n = m+1$, let $A(1, m) = \\overline{1a_1a_2\\cdots a_m}$, $A(3, m) = \\overline{3a_1a_2\\cdots a_m}$, ..., $A(9, m) = \\overline{9a_1a_2\\cdots a_m}$.\nTheir pairwise differences are $2 \\times 10^m$, $4 \\times 10^m$, $6 \\times 10^m$, $8 \\times 10^m$ and none is divisible by $5^{m+1}$.\nHence the remainders of $A(1, m)$, $A(3, m)$, ..., $A(9, m)$ are different when divided by $5^{m+1}$.\nSince $A(m)$ is divisible by $5^m$, these 5 numbers are also divisible by $5^m$.\nHence the remainders of $\\overline{A_i(m)}$, $i=1, 3, 5, 7, 9$, when divided by $5$ are also distinct and they are $0, 1, 2, 3, 4$.\nIt follows that exactly one of the remainders is $0$.\nTherefore one of $A(1, m)$, $A(3, m)$, ..., $A(9, m)$ is a multiple of $5^{m+1}$.\n\nNow we prove uniqueness.\nLet $\\overline{b_1b_2\\cdots b_{m+1}}$ be a multiple of $5^{m+1}$ with all digits odd.\nThen $\\overline{b_1b_2\\cdots b_{m+1}} = b_1 \\times 10^m + \\overline{b_2\\cdots b_{m+1}}$.\nThe first term is a multiple of $5^m$ and so the second term is also a multiple of $5^m$.\nSince its digits are all odd, it is the unique such multiple of $5^m$, i.e., $\\overline{b_2\\cdots b_{m+1}} = \\overline{a_1a_2\\cdots a_m} = A(m)$.\nThis proves uniqueness.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41085,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose for some positive integer $n$, the numbers $2^n$ and $5^n$ have equal first digit. What are the possible values of this first digit?",
    "options": [],
    "answer": "3",
    "solution": "From $2^1 = 2$, $2^2 = 4$, $2^3 = 8$, $2^4 = 16$, $2^5 = 32$, $5^1 = 5$, $5^2 = 25$, $5^3 = 125$, $5^4 = 625$, $5^5 = 3125$, $3$ is a possible value. We shall show that there are no others.\n\nSuppose the first digits of $2^n$ and $5^n$ are both $a$ and that they have $s$ and $t$ digits, respectively.\n\nWhen $n > 3$, we have\n$$\na \\cdot 10^s < 2^n < (a+1) \\cdot 10^s, \\quad a \\cdot 10^t < 5^n < (a+1) \\cdot 10^t.\n$$\n$$\n\\therefore a^2 \\cdot 10^{s+t} < 2^n \\cdot 5^n = 10^n < (a+1)^2 \\cdot 10^{s+t}\n$$\n$$\n\\therefore a^2 < 10^{n-s-t} < (a+1)^2\n$$\nSince $a > 1$ and $a+1 < 10$, $n-s-t = 1$, $a^2 < 10$ and $(a+1)^2 > 10$ and so $a = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41086,
    "subject": "Mathematics (Multi-modal)",
    "question": "Alice and Bob play a game. Bob starts by picking a set $S$ consisting of $M$ vectors of length $n$ with entries either $0$ or $1$. Alice picks a sequence of numbers $y_1 \\le y_2 \\le \\cdots \\le y_n$ from the interval $[0, 1]$, and a choice of real numbers $x_1, \\dots, x_n \\in \\mathbb{R}$. Bob wins if he can pick a vector $(z_1, z_2, \\dots, z_n) \\in S$ such that\n$$\n\\sum_{i=1}^{n} x_i y_i \\le \\sum_{i=1}^{n} x_i z_i,\n$$\notherwise Alice wins. Determine the minimum value of $M$ so that Bob can guarantee a win?",
    "options": [],
    "answer": "n + 1",
    "solution": "The answer is $M = n + 1$.\n\n$|S| \\ge n + 1$, since if Alice selects\n$$\n(y_1, y_2, \\dots, y_n) = \\underbrace{(0, 0, \\dots, 0)}_{k\\ 0's}, 1, 1, \\dots, 1)\n$$\n$$\n(x_1, x_2, \\dots, x_n) = \\underbrace{(-1, -1, \\dots, -1)}_{k\\ (-1)'s}, 1, 1, \\dots, 1),\n$$\nwhere $0 \\le k \\le n$, then Bob can only win if $(z_1, z_2, \\dots, z_n) = (y_1, y_2, \\dots, y_n)$, so $S$ contains $(0, 0, \\dots, 0, 1, 1, \\dots, 1)$. Now, suppose $S$ only contains elements of that form. It remains to show that Bob can win. Rewrite the inequality as:\n$$\n\\sum_{i=1}^{n} (y_i - y_{i-1})(x_i + \\dots + x_n) \\le \\sum_{i=1}^{n} (z_i - z_{i-1})(x_i + \\dots + x_n),\n$$\nwhere $y_0 = z_0 = 0$ for convenience. Note that from the choice of $S$, $z_i - z_{i-1} = 1$ for exactly one index $i$, and $0$ for the others. However,\n$$\n\\sum_{i=1}^{n} (y_i - y_{i-1})(x_i + \\dots + x_n) \\le \\left( \\sum_{i=1}^{n} (y_i - y_{i-1}) \\right) \\max_{1 \\le i \\le n} (x_i + \\dots + x_n) \\le \\max_{1 \\le i \\le n} (x_i + \\dots + x_n).\n$$\nWhatever Alice's choices are, Bob can always pick precisely the $i$ that attains that maximum and wins.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 41087,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $p$ is a prime number and $x, y, z$ are integers satisfying $0 < x < y < z < p$. If $x^3, y^3, z^3$ have equal remainders when divided by $p$, prove that $x^2 + y^2 + z^2$ is divisible by $x + y + z$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $p > 3$ and $p \\mid x^3 - y^3 = (x-y)(x^2 + xy + y^2)$. Since $x, y < p$, $p \\nmid x-y$. Therefore $p \\mid x^2 + xy + y^2$. Similarly $p \\mid x^2 + xz + z^2$, $p \\mid y^2 + yz + z^2$.\n\nThus $p \\mid (x^2 + xy + y^2) - (y^2 + yz + z^2) = (x-z)(x+y+z)$ and so $p \\mid x+y+z$.\n\nSince $x, y, z < p$ and $p$ is a prime, $x+y+z = p$ or $2p$.\n\nSince the parity of $x+y+z$ and $x^2+y^2+z^2$ are the same, it suffices to prove that $p \\mid x^2+y^2+z^2$.\n\nNow $p \\mid x^2 + xy + y^2 = x(x+y+z) + y^2 - xz$ and so $p \\mid y^2 - xz$. Therefore $p \\mid (x^2 + xy + y^2) + (y^2 - xz) = x^2 + y^2 + z^2$ and we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41088,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number. Determine the largest possible $n$ such that the following holds. It is possible to fill an $n \\times n$ table with integers $a_{ik}$ in the $i$-th row and $k$-th column, for $1 \\le i, k \\le n$, such that for any quadruple $i, j, k, l$ with $1 \\le i < j \\le n$ and $1 \\le k < l \\le n$, the number $a_{ik}a_{jl} - a_{il}a_{jk}$ is not divisible by $p$.",
    "options": [],
    "answer": "n = p + 1",
    "solution": "The answer is $n = p + 1$. We first show that $n \\le p + 1$. Since we are interested only in divisibility by $p$, we assume that $0 \\le a_{ik} \\le p - 1$. Note that each row and each column can have at most one zero. Also notice that we can scale each row or column by scalar not divisible by $p$ without affecting the given property of the table. Consider the first row, which must have at least $n-1$ non-zero values, say in the first $n-1$ columns. By scaling those columns by an appropriate scalar, we may assume that the first $n-1$ values in the first row are all 1's. Now in the second row, the first $n-1$ values must be distinct, hence $n-1 \\le p$, as desired.\n\nWe now construct for $n = p + 1$. Take the following grid. For $2 \\le i, k \\le p + 1$,\n$$\na_{11} = 0,\\ a_{1k} = 1,\\ a_{i1} = 1,\\ a_{ik} = k - i.\n$$\nFor a valid quadruple $i, j, k, l$ as described in the problem, it is simple to check it works when $i = 1$ or $k = 1$. When $i \\ge 2$ and $k \\ge 2$, the number $a_{ik}a_{jl} - a_{il}a_{jk}$ turns out to be $(i-j)(k-l) \\ne 0 \\pmod{p}$.\n\n$$\n\\begin{pmatrix} 0 & 1 & 1 & 1 & 1 & 1 \\\\ 1 & 0 & 1 & 2 & 3 & 4 \\\\ 1 & 4 & 0 & 1 & 2 & 3 \\\\ 1 & 3 & 4 & 0 & 1 & 2 \\\\ 1 & 2 & 3 & 4 & 0 & 1 \\\\ 1 & 1 & 2 & 3 & 4 & 0 \\end{pmatrix}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41089,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integer solutions of the equation\n$$\ny^2 + 2y = x^4 + 20x^3 + 104x^2 + 40x + 2003.\n$$",
    "options": [],
    "answer": "(7, 128), (7, -130), (-17, -128), (-17, -130)",
    "solution": "Suppose that there are integer solutions. By completing squares, the equation becomes\n$$\n(y+1)^2 = (x^2 + 10x + 2)^2 + 2000.\n$$\nLet $a = y + 1$, $b = x^2 + 10x + 2$, $u = a - b$, $v = a + b$. Then $uv = 2000$ and $b = \\frac{v-u}{2}$.\nThe equation $x^2 + 10x + 2 - b = x^2 + 10x + 2 + \\frac{u-v}{2} = 0$ in the unknown $x$ has integer solutions. Therefore the discriminant\n$$\n\\Delta = 100 - 4\\left(2 + \\frac{u-v}{2}\\right) = 92 + 2(u-v)\n$$\nis a square. Since $v-u$ is even and $uv = 2000$, $u, v$ are both even and we have\n$$\n\\{u, v\\} = \\{2, 1000\\}, \\{4, 500\\}, \\{8, 250\\}, \\{10, 200\\}, \\{20, 100\\}, \\{40, 50\\}\n$$\nand the negative counterparts. $\\Delta$ is a square only when $(u, v) = (8, 250), (-250, -8)$. Thus $(x, y) = (7, 128), (7, -130), (-17, -128), (-17, -130)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41090,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots$ be a sequence of positive numbers satisfying, for any positive integers $k, l, m, n$ such that $k + n = m + l$,\n$$\n\\frac{a_k + a_n}{1 + a_k a_n} = \\frac{a_m + a_l}{1 + a_m a_l}.\n$$\nShow that there exist positive numbers $b, c$ so that $b \\le a_n \\le c$ for any positive integer $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_{k+n} = \\frac{a_k+a_n}{1+a_k a_n}$. So for any $n$, $A_n = \\frac{a_1+a_{n-1}}{1+a_1 a_{n-1}}$. Consider the function\n$$\nf(x) = \\frac{a_1 + x}{1 + a_1 x}\n$$\nwhere $x > 0$. Since $f(x) = \\frac{1}{a_1} \\frac{a_1^2+a_1x}{1+a_1x} = \\frac{a_1(a_1+a_1x)}{a_1+a_1^2x}$, it follows that\n$$\nf(x) \\geq \\begin{cases} 1/a_1 & \\text{if } a_1 \\geq 1 \\\\ a_1 & \\text{if } 0 < a_1 < 1. \\end{cases}\n$$\nSo for any value of $a_1$, there exists $0 < t \\le 1$ so that $f(x) \\ge t$. (In fact just take $t$ to be the smaller of $a_1$ and $1/a_1$.)\nThus for any $n$, $A(n) \\ge t$ and so\n$$\nA_{2n} = A_{n+n} = \\frac{2a_n}{1+a_n^2} \\ge t \\\\\n\\therefore ta_n^2 - 2a_n + t \\le 0\n$$\nThus $a_n$ lies between the 2 roots of the equation $tx^2 - 2x + t = 0$. The roots are $\\frac{1\\pm\\sqrt{1-t^2}}{t}$. Letting $b = \\frac{1-\\sqrt{1-t^2}}{t}$ and $c = \\frac{1+\\sqrt{1-t^2}}{t}$ and we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41091,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$, $\\angle B > 90^\\circ$, the incircle touches the sides $BC$ and $CA$ at $D$ and $E$, respectively. The lines $ED$ and $AB$ intersect at $P$. The incircle of the triangle $AEP$ touches the sides $PE$ and $AP$ at $D_1$ and $E_1$, respectively. The lines $E_1D_1$ and $AE$ intersect at $P_1$. Suppose $P, C, E, B$ are concyclic. Prove that $BE$ is parallel to $PP_1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\angle EPB = x$. Since $P, C, E, B$ are concyclic, $x = \\angle EPB = \\angle BCA$ and therefore $\\triangle AEP \\sim \\triangle ABC$. Thus the points $(A, B, C, D, E, P)$ correspond to the points $(A, E, P, D_1, E_1, P_1)$ in 2 similar configurations associated with triangles $ABC, AEP$. Therefore $\\angle EPA = \\angle E_1 P_1 A = x$ and $\\angle CBE = \\angle PEE_1$.\nThus $P, P_1, E, E_1$ are concyclic and $E_1 P_1 \\parallel BC$ and so $\\angle PP_1 E_1 = \\angle PEE_1 = \\angle CBE$, $\\angle E_1 P_1 B = \\angle CBP_1$.\n\nTherefore $\\angle PP_1B = \\angle PP_1E_1 - \\angle E_1P_1B = \\angle CBE - \\angle CBP_1 = \\angle P_1BE$. Hence $BE \\parallel PP_1$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41092,
    "subject": "Mathematics (Multi-modal)",
    "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": "Detailed solution",
    "solution": "Let the centres be $P$, $Q$ 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![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 41093,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all 6-digit integers $n$ such that $n$ is a perfect square and that 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": "183184, 328329, 528529, 715716",
    "solution": "Let $n = y^2$ and $x$ be the number formed by the first 3 digits of $n$. Note that $y$ is a 3-digit number. Then\n$$\ny^2 = 1000x + x + 1 = 1001x + 1 \\quad \\Rightarrow \\quad (y-1)(y+1) = 7 \\times 11 \\times 13x.\n$$\n\nSince $y \\le 999$, 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 $y+1 = 77\\alpha \\le 1000$ and so $\\alpha \\le 12$. Testing $\\alpha = 1, 2, \\dots, 12$, we see that only $\\alpha = 11$ works. We get $y = 846$ and $n = 715716$. The other 5 cases are $7 \\times 13 \\mid y+1, 11 \\mid y-1$; $11 \\times 13 \\mid y+1, 7 \\mid y-1$; $7 \\mid y+1, 11 \\times 13 \\mid y-1$; $11 \\mid y+1, 7 \\times 13 \\mid y-1$; $13 \\mid y+1, 7 \\times 11 \\mid y-1$. They yield the solutions $n = 528529, 183184, 328329, 75076, 24025$. The 2 5-digit numbers are discarded and we have 4 solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41094,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incenter of $\\triangle ABC$ and $\\omega$ be its incircle. Let $D$ be the intersection of $\\omega$ with $BC$. Suppose $X$, $Y$ are points on $\\omega$ such that lines $IX$, $IY$ and $AD$ are tangent to the circumcircle of $\\triangle AXY$. Prove that line $XY$ bisects segment $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet $M$ be the midpoint of $BC$. Denote $D'$ and $E$ as the reflection of $D$ across $I$ and $M$ respectively. It is well known that $E$ is the intersection of the $A$-excircle with $BC$. Hence, the homothety centered at $A$ sending the incircle to the $A$-excircle also sends $D'$ to $E$. Thus, $A$, $D'$ and $E$ are collinear.\n\nConsider $\\triangle ADE$. The $E$-midline $MI$ passes through the midpoint of $AD$, say $N$. Note that $NA^2 = ND^2$ and $IX^2 = ID^2$. Hence, both $I$ and $D$ have equal powers with respect to the circle $(AXY)$ and the point circle $D$. This means that $M$ lies on the radical axis of the two circles, and thus $\\mathrm{Pow}(M, (AXY)) = MD^2 = \\mathrm{Pow}(M, \\omega)$.\n\nThis means that $M$ also lies on the radical axis of $(AXY)$ and the incircle, and thus lies on $XY$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41095,
    "subject": "Mathematics (Multi-modal)",
    "question": "Jack and Jill play the following game: Jack throws 3 dice and Jill can select some of them, possibly none, and turn each of them to the opposite side. Jill wins if the sum of the values on the dice is a multiple of 4. Can Jill always win? (Note the game is played with standard dice where the sum of the numbers on opposite sides is 7.)",
    "options": [],
    "answer": "Yes, Jill can always win.",
    "solution": "Jill can always turn the dice so that the numbers are $2a$, $2b$, $2c$, i.e., all even. On the other hand she can also achieve $7 - 2a$, $7 - 2b$, $2c$.\nLet $S = 2a + 2b + 2c$ and $T = (7 - 2a) + (7 - 2b) + 2c$. Then both $S$, $T$ are even and $S + T = 14 + 4c$. Since $S + T$ is even and not a multiple of 4, one of $S$, $T$ is a multiple of 4. Therefore Jill can always win.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41096,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any 4-digit positive integer $n$, define $f(n) = (a+b)^2$, where $a, b$ are the numbers formed by the first two and last two digits of $n$, respectively (leading zeroes are allowed). Find all 4-digit positive integers $n$ such that $f(n) = n$.",
    "options": [],
    "answer": "2025, 3025, 9801",
    "solution": "Since $(a+b)^2 = f(n) = n = 100a + b$, we have\n$$\n(a+b)(a+b-1) = 99a.\n$$\nBy Euclid's lemma, one of the factors in the LHS is divisible by 11. Also, note that the two factors are coprime, so one of them is not divisible by 3, and thus the other must be divisible by 9.\nHence $a+b \\equiv 0 \\text{ or } 1 \\pmod{9}$ and $a+b \\equiv 0 \\text{ or } 1 \\pmod{11}$, so by the Chinese Remainder Theorem we have the following possibilities for $a+b \\pmod{99}$:\n\n$$\na+b \\equiv 0 \\pmod{9},\\ a+b \\equiv 0 \\pmod{11} \\Rightarrow a+b \\equiv 0 \\pmod{99},\n$$\n$$\na+b \\equiv 1 \\pmod{9},\\ a+b \\equiv 0 \\pmod{11} \\Rightarrow a+b \\equiv 55 \\pmod{99},\n$$\n$$\na+b \\equiv 0 \\pmod{9},\\ a+b \\equiv 1 \\pmod{11} \\Rightarrow a+b \\equiv 45 \\pmod{99}, \\text{ or}\n$$\n\n$a+b \\equiv 1 \\pmod{9}$, $a+b \\equiv 1 \\pmod{11} \\Rightarrow a+b \\equiv 1 \\pmod{99}$.\nBut $10^3 \\le n < 10^4$ and $n = f(n) = (a+b)^2$ implies $32 \\le a+b \\le 99$. Hence the only possible values of $a+b$ are 45, 55, 99, so the only possible values of $n$ are $45^2 = 2025$, $55^2 = 3025$, and $99^2 = 9801$. It is easy to check that $f(n) = n$ in each of these three cases, so the complete set of solutions is $n = 2025, 3025, 9801$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41097,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{Z}^+ \\to \\mathbb{Z}^+$, where $\\mathbb{Z}^+$ is the set of all positive integers, such that for any positive integers $m, n$, $f^{f(m)}(n) + n = f^m(n) + f(n)$.",
    "options": [],
    "answer": "The only functions are f(n) = n for all positive integers n and f(n) = n + 1 for all positive integers n.",
    "solution": "We claim that the only solutions are:\n1. $f(n) \\equiv n + 1$. This works because\n$$\nf^{f(m)}(n) + n = f^{m+1}(n) + n = m + 1 + 2n\n$$\nand\n$$\nf^m(n) + f(n) = m + n + n + 1 = m + 1 + 2n.\n$$\n2. $f(n) \\equiv n$. This works because\n$$\nf^{f(m)}(n) + n = 2n = f^m(n) + f(n).\n$$\nWe now show these are the only solutions. Note that\n$$\nf^{f^k(m)}(n) - f^{f^{k-1}(m)}(n) = f(n) - n.\n$$\nHence, by induction,\n$$\nf^{f^k(m)}(n) = k(f(n) - n) + f^m(n).\n$$\nThis means that $f(n) \\ge n$ for all positive integers $n$, else by choosing a sufficiently large $k$, $f^{f^k(m)}(n)$ will be negative. If there exists some positive integer $c$ such that $f(c) = c$, then\n$$\nf^{f(c)}(n) = f^c(n) \\Rightarrow f(n) = n \\text{ for all positive integers } n.\n$$\nElse, assume $f(n) \\ne n$, so that $f(n) > n$ for all positive integers $n$. Pick $A$ such that $f(A) - A = t$ is minimal. For any positive integer $m$, consider the numbers $f^m(A), f^{m+1}(A), \\dots, f^{f(m)}(A)$. This sequence of numbers is strictly increasing, with consecutive differences at least $t$. Hence,\n$$\nt \\le t(f(m) - m) \\le f^{f(m)}(A) - f^m(A) = f(A) - A = t.\n$$\nThus, equality must hold everywhere, and in particular $f(m) = m + 1$ for all positive integers $m$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41098,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be positive numbers and $A$ their arithmetic mean. Prove that\n$$\nA^{n-1} \\geq \\frac{1}{n} \\sum_{i=1}^{n} a_1 \\cdots \\hat{a}_i \\cdots a_n, \\quad \\text{where } \\hat{a}_i \\text{ denotes the term } a_i \\text{ is omitted.}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use induction on $n$. When $n=2$, it is an equality. Assume $a_1 \\le a_2 \\le \\dots \\le a_n$. Then $a_1 \\le A \\le a_n$. Let $A_o = (a_1 + \\dots + a_{n-1})/(n-1)$. Then $A = A_o + (a_n - A_o)/n$ and $a_1 \\le A_o \\le A_{n-1} \\le a_n$. Since $a_n - A_o \\ge 0$, we use binomial expansion:\n$$\nA^{n-1} = \\left(A_o + \\frac{a_n - A_o}{n}\\right)^{n-1} \\ge A_o^{n-1} + (n-1)A_o^{n-2} \\left(\\frac{a_n - A_o}{n}\\right) = \\frac{n-1}{n} A_o^{n-2} a_n + \\frac{1}{n} A_o^{n-1}.\n$$\nBy induction hypothesis, $A_o^{n-2} \\ge \\frac{1}{n-1} \\sum_{i=1}^{n-1} a_i \\cdots \\hat{a}_i \\cdots a_{n-1}$.\nBy $AM \\ge GM$, $A_o^{n-1} \\ge a_1 a_2 \\cdots a_{n-1}$.\nTherefore, $A^{n-1} \\ge \\frac{1}{n} \\sum_{i=1}^{n-1} a_i \\cdots \\hat{a}_i \\cdots a_{n-1} a_n + \\frac{1}{n} a_1 a_2 \\cdots a_{n-1} = \\frac{1}{n} \\sum_{i=1}^{n} a_i \\cdots \\hat{a}_i \\cdots a_n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41099,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $n$, or show no such $n$ exists, such that one can find infinitely many distinct $n$-tuplets of positive rationals $(a_1, a_2, \\dots, a_n)$ satisfying:\n$$\n\\sum_{i=1}^{n} a_i \\in \\mathbb{Z}, \\quad \\sum_{i=1}^{n} \\frac{1}{a_i} \\in \\mathbb{Z}.\n$$",
    "options": [],
    "answer": "3",
    "solution": "The smallest such $n$ is $n=3$. For $n=2$, let $a_1 = \\frac{a}{b}$, $a_2 = \\frac{c}{d}$ where $\\gcd(a,b) = \\gcd(c,d) = 1$. Then we must have $bd \\mid ad+bc$, $ac \\mid ad+bc$. Taking mod $d$ in the first condition, we get that $d \\mid bc$ but since $c,d$ are coprime we have that $d|b$. Taking mod $b$, we get $b \\mid ad$ and since $a,b$ are coprime we get $b \\mid d$ which tells us that $b=d$. Similarly we get $a=c$ and hence we have $a_1=a_2$. Clearly only $a_1=a_2=1$ or $\\frac{1}{2}$ works and we only have finitely many such duplets.\n\nNow for $n=3$, we will look for triplets of the form $(\\frac{a}{a+b+c}, \\frac{b}{a+b+c}, \\frac{c}{a+b+c})$ where $a,b,c$ are positive integers. Fixing $a=1$, it suffices to find infinitely many pairs of $(b,c)$ such that $\\frac{1}{b}+\\frac{1}{c}+\\frac{b}{c}+\\frac{c}{b}$ is an integer. We shall show that there are infinitely many solutions to $\\frac{1}{b}+\\frac{1}{c}+\\frac{b}{c}+\\frac{c}{b}=3$ by Vieta Jumping. If we let $c \\ge b$, the equation rearranges into $b^2-b(3c-1)+c^2+c=0$. Viewing it as a quadratic in $b$, we get that there exists another $b'$ satisfying $\\frac{1}{b'}+\\frac{1}{c}+\\frac{b'}{c}+\\frac{c}{b'}=3$ and that this $b'$ satisfies $b'+b' = 3c-1$, $bb' = c^2+c$. Since $b$ is a positive integer, we get that $b' = 3c-1-b$ is also a positive integer. Furthermore, $b' = \\frac{c^2+c}{b} \\ge \\frac{c^2+c}{c} > c$. Hence our triplet $(1,b,c)$ will transform into another triplet $(1,c,b')$ where $b' > c$ and we can then jump again but this time with $c$ as the subject of the quadratic. Starting with $b=3, c=6$, this algorithm will generate infinitely many distinct triplets of desired positive integers $(a,b,c)$ which will then give us infinitely many distinct triplets of rationals satisfying the problem conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41100,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let *A*, *B*, *C*, *D* be the 4 squares in the centre of a $4 \\times 4$ grid of squares. These 4 squares form the centre of the grid. A frog can jump from one of the 4 squares in the centre 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": "If n is even: 0. If n = 1: 2. If n is odd and at least 3: 4^{(n-1)/2}.",
    "solution": "Let $A, B, C, D$ be labeled in the clockwise manner. Let $a_n, b_n, c_n, d_n$ be respectively the number of ways to start $A, B, C, D$ 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 $n \\ge 3, a_n = 2b_{n-1}$.\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$$\na_n = 4a_{n-2}.\n$$\nConsequently, $a_{2m+1} = 4^{m-1}a_3 = 4^m$ for $m \\ge 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41101,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive $m$ so that it is possible to choose 8 vertices from a regular polygon with $m$ sides such that no three of them are the vertices of an isosceles triangle.",
    "options": [],
    "answer": "18",
    "solution": "Let $P_1, P_2, \\dots, P_m$ be the vertices of a regular $m$-gon. Let $P_{t_1}, P_{t_2}, \\dots, P_{t_8}$ be 8 distinct vertices so that no 3 of them are the vertices of an isosceles triangle. Without loss of generality, we may assume that $1 = t_1 < t_2 < \\dots < t_8$. We'll show that $m \\ge 18$.\n\nLet $d_1 = t_2 - t_1, d_2 = t_3 - t_2, \\dots, d_7 = t_8 - t_7, d_8 = m + 1 - t_8$. Then $d_1 \\ne d_2, d_2 \\ne d_3, \\dots, d_7 \\ne d_8$. In general, if $1 \\le i < j < k \\le 8$, then $t_j - t_i \\ne t_k - t_j$ or equivalently $d_i + d_{i+1} + \\dots + d_{j-1} \\ne d_j + d_{j+1} + \\dots + d_{k-1}$. This fact, no 2 consecutive distances are equal, is essential.\n\nFrom this we have $d_1 + d_2 \\ge 3$, $d_3 + d_4 \\ge 3$. Since $d_1 + d_2 \\ne d_3 + d_4$, $d_1 + \\dots + d_4 \\ge 7$. If $d_1 + \\dots + d_4 = 7$, then the values of $d_1, \\dots, d_4$ are 1, 1, 2, 3. In all arrangements of these 4 numbers, one of the blocks 213 and 312 is present and this contradicts the essential fact. So $d_1 + \\dots + d_4 \\ge 8$. For equality, it is not hard to check $d_1 d_2 d_3 d_4$ can only be 1421, 1241, 1412 or 2141. So we conclude that the sum of any 4 consecutive distances is at least 8.\n\n![](attached_image_1.png)\n\nSuppose $m \\le 17$. Then $d_1 + d_2 + \\dots + d_8 \\le 17$. Since the sum of any 4 consecutive distances is at least 8, the sum of any 4 consecutive distances is at most 9. Since we can't have $d_1 + \\dots + d_4 = 9$ and $d_5 + \\dots + d_8 = 9$ at the same time, one of the sums is 8. Without loss of generality, we assume that $d_1 + \\dots + d_4 = 8$. Since the sum of 4 consecutive distances is at most 9, $d_5 = 1$. Then the first 5 distances are 14121, the other 3 cases are not possible. This is not possible since $4 = 1 + 2 + 1$. Therefore $m \\ge 18$.\n\nFor $m = 18$, $(p_{t_1}, p_{t_2}, \\dots, p_{t_8}) = (1, 2, 5, 7, 10, 11, 14, 16)$ works. So the answer is 18.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41102,
    "subject": "Mathematics (Multi-modal)",
    "question": "Alice starts with some coins in a box. On the $n^{\\text{th}}$ minute (for $n = 1, 2, 3, \\dots$), she can choose to put an additional coin into the box or do nothing, and then she writes the number $2c^2 - n^2$ on the board where $c$ is the number of coins in the box. Show that for any positive integer $m$, there will eventually be two numbers on the board that sum to a multiple of $m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $c_n$ be the number of coins on the $n^{\\text{th}}$ minute. Consider the ordered pair $(x_n, y_n) = (c_n, n - c_n)$, noting that each minute either $x_n$ or $y_n$ increases by 1. The number written on the $n^{\\text{th}}$ minute is\n$$\n2c_n^2 - n^2 = 2x_n^2 - (x_n + y_n)^2 = x_n^2 - 2x_ny_n - y_n^2.\n$$\nIt thus suffices to prove the slightly more general problem:\nAlice starts at an arbitrary integer lattice point $(x,y) = (x_0, y_0)$. Every minute, she walks one unit right or up (corresponding to increasing $x$ or $y$ by 1), then writes down the number $f(x,y) = x^2 - 2xy - y^2$. Show that for any positive integer $m$, there will eventually be two numbers on the board that sum to a multiple of $m$.\nThe function $f$ has the special property of $90^\\circ$ rotational anti-symmetry, i.e., taken mod $m$:\n$$\nf(x, y) + f(y, -x) \\equiv 0\n$$\n\n**Claim:** Among the numbers written over $4m$ minutes, there is either (a) two numbers that sum to a multiple of $m$, or (b) a number which is a multiple or half-multiple of $m$.\n**Proof of Claim:** Suppose Bob starts at integer lattice point $(u,v) = (u_0, v_0) \\equiv (y_0, -x_0) \\pmod{m}$. Each minute, if Alice moves right, Bob moves down; if Alice moves up, Bob moves right.\nIn this manner, Alice's and Bob's coordinates are always related by $(u,v) \\equiv (y,-x) \\pmod{m}$. In particular, whenever Alice writes down the number $f(x,y)$, Bob writes down the number $f(u,v) \\equiv -f(x,y) \\pmod{m}$.\n\nIt suffices to show that Alice's and Bob's paths coincide on some lattice point $(x^*, y^*)$. If this occurs, then either\n1. (a) Alice and Bob reach $(x^*, y^*)$ at different times, thus Alice has written down $f(x^*, y^*)$ and $-f(x^*, y^*)$ (mod $m$) at two different times; or\n2. (b) Alice and Bob reach $(x^*, y^*)$ at the same time; thus $f(x^*, y^*) \\equiv -f(x^*, y^*)$ (mod $m$), so Alice has written down a multiple or half-multiple of $m$.\nWLOG among Alice's first $2m-1$ moves, she moves right at least $m$ times (the alternative is that she moves up at least $m$ times, in which case a similar argument holds by swapping the axes).\nThen, within Alice's first $2m-1$ moves, she passes through some lattice point $(u_0, w_0)$, where $x_0 < u_0 \\le x_0 + m$ and $u_0 \\equiv y_0 \\pmod m$. We then select Bob's starting point as $(u_0, v_0)$ where $w_0 < v_0 \\le w_0 + m$ and $v_0 \\equiv -x_0 \\pmod m$.\nConsequently, the path $\\mathcal{L}_{\\text{Bob}}$ traversed by Bob within his first $2m-1$ moves starts vertically above $(u_0, w_0)$, and at distance $\\le m$ from $(u_0, w_0)$. Since Bob moves down at least $m$ times and moves right at most $m-1$ times within his first $2m-1$ moves, his path $\\mathcal{L}_{\\text{Bob}}$ traverses within the upper-right quadrant $\\{(x, y) : x \\ge u_0, y \\ge w_0\\}$ with respect to $(u_0, w_0)$, and leaves the quadrant from the bottom edge at a distance $\\le m$ from $(u_0, w_0)$ (see Figure 1).\nThus, a further $2m$ moves by Alice starting from $(u_0, w_0)$ will take her from below $\\mathcal{L}_{\\text{Bob}}$ to above it, guaranteeing a lattice point intersection with $\\mathcal{L}_{\\text{Bob}}$.\nSince two multiples of $m$ or two half-multiples of $m$ sum to a multiple of $m$, thus by the claim, Alice is guaranteed to write two numbers that sum to a multiple of $m$ within $3 \\cdot 4m = 12m$ moves.\n\n![](attached_image_1.png)\nFigure 1: Sample game for $m=6$. Alice's path is lightly shaded and starts at $(x_0, y_0) = (2, 1)$ which is at the bottom. Bob's path is heavily shaded and starts at $(u_0, v_0) = (7, 10)$ at the top. Alice's path after $(u_0, w_0)$ which is at the bottom left corner of the box, is guaranteed to intersect Bob's path; the intersection is the square with 2. When Bob is at the intersection, Alice has also made 5 moves is at the square with 4. The numbers written by Alice in these 2 squares sum to a multiple of $m=6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41103,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral and $M$ be the intersection of its diagonals. Through $M$ draw a line meeting the side $AB$ at $P$ and the side $CD$ at $Q$. Find all the quadrilaterals so that there exists the segment $PQ$ that divides the triangles $ABM$ and $CDM$ into 4 similar triangles.",
    "options": [],
    "answer": "All trapezoids with AB ∥ CD for which either (i) the legs are equal (isosceles trapezoid), or (ii) the diagonals are perpendicular.",
    "solution": "Suppose $\\angle APM > 90^\\circ$. Then in $\\triangle BPM$, $\\angle BPM < 90^\\circ$. Since $\\angle PBM + \\angle BMP = \\angle APM$, $\\angle PBM, \\angle BMP < \\angle APM$. So none of the angles of $\\triangle BPM$ can be equal to $\\angle APM$ of $\\triangle APM$. Therefore $\\triangle APM$ cannot be similar to $\\triangle BPM$. Thus $\\angle APM = 90^\\circ$ and $PQ \\perp AB$. Similarly $PQ \\perp CD$ and it follows that $AB \\parallel DC$.\n\n![](attached_image_1.png)\n\nIt then follows that $\\triangle APM \\sim \\triangle CQM$ and $\\triangle BMP \\sim \\triangle DQM$.\nIf $\\triangle APM \\sim \\triangle BPM$, then $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$. Therefore $ABCD$ is an isosceles trapezium.\nIf $\\triangle APM \\sim \\triangle MPB$, then $\\angle MAP = \\angle BMP$ and $\\angle PMA = \\angle PBM$. It follows that $\\angle AMB = 90^\\circ$. So the diagonals are perpendicular. The quadrilaterals are either isosceles trapezia or trapezia in which the diagonals are perpendicular.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41104,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa naravna števila $a$, $b$ in $c$, ki zadoščajo enačbi\n$$\n\\frac{a+b}{a+c}=\\frac{b+c}{b+a}\n$$\nin za katere velja, da je $ab+ac+bc$ praštevilo.",
    "options": [],
    "answer": "(1, 1, 1)",
    "solution": "Solution:\n\nOznačimo praštevilo $ab+bc+ca$ s $p$. Iz zveze dobimo, da je $(a+b)^2 = c^2 + p$ oziroma $p = (a+b-c)(a+b+c)$. Ker je $p$ praštevilo in je $0 < a+b-c < a+b+c$, mora biti $a+b-c = 1$ in $a+b+c = p$ oziroma $c = \\frac{p-1}{2}$ in $a+b = \\frac{p+1}{2}$. Po drugi strani pa mora biti $ab = p - bc - ca = p - c(a+b) = p - \\frac{p-1}{2} \\cdot \\frac{p+1}{2} = -\\frac{p^2 - 4p - 1}{4}$ pozitivno število. Ničli kvadratne funkcije $f(x) = x^2 - 4x - 1$, ki ima pozitiven vodilni koeficient, sta $2 - \\sqrt{5}$ in $2 + \\sqrt{5} < 5$, torej je $p$ lahko le $2$ ali $3$. Ker $2$ ne more biti, saj $c = \\frac{p-1}{2}$ ne bi bilo celo število, je $p = 3$. Tako je $c = 1$, $a+b = 2$ in ker sta $a$ in $b$ naravni števili, je edina rešitev $a = b = c = 1$. Trojica $(1, 1, 1)$ seveda ustreza vsem pogojem naloge.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41105,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bodo $k$, $m$ in $n$ poljubna naravna števila, ki niso deljiva s $5$. Dokaži, da je vsaj eno izmed števil $k^{2}-m^{2}$, $m^{2}-n^{2}$ in $n^{2}-k^{2}$ deljivo s $5$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nKer števila $k$, $m$ in $n$ niso deljiva s $5$, so oblike $5t \\pm 1$ ali $5t \\pm 2$. Iz $(5t \\pm 1)^{2} = 5t' + 1$ in $(5t \\pm 2)^{2} = 5t' + 4$ pa vidimo, da imajo števila $k^{2}$, $m^{2}$ in $n^{2}$ le dva možna ostanka pri deljenju s $5$. Torej je vsaj ena izmed razlik $k^{2}-m^{2}$, $m^{2}-n^{2}$ in $n^{2}-k^{2}$ deljiva s $5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41106,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bodo $a, b, c, d, e$ in $f$ taka pozitivna realna števila, da je zaporedje $a, b, c, d$ aritmetično, zaporedje $a, e, f, d$ pa geometrično. Dokaži, da velja\n$$\nb c \\geq e f\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nKer je $a, b, c, d$ aritmetično zaporedje, velja $b = a + \\frac{1}{3}(d - a)$ in $c = a + \\frac{2}{3}(d - a)$. Podobno je $e = a \\left(\\frac{d}{a}\\right)^{1/3}$ in $f = a \\left(\\frac{d}{a}\\right)^{2/3}$. Sledi\n$$\nbc - ef = \\left(\\frac{2}{3}a + \\frac{1}{3}d\\right)\\left(\\frac{1}{3}a + \\frac{2}{3}d\\right) - a^2 \\frac{d}{a} = \\frac{2}{9}(a - d)^2 \\geq 0\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41107,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. Dokaži, da za vsako naravno število $n$ velja\n$$\n\\sqrt{n+1}-\\sqrt{n}<\\frac{1}{2 \\sqrt{n}}<\\sqrt{n}-\\sqrt{n-1}\n$$\n\nb. Dokaži, da je celi del izraza\n$$\n1+\\frac{1}{\\sqrt{2}}+\\frac{1}{\\sqrt{3}}+\\cdots+\\frac{1}{\\sqrt{m^{2}-1}}+\\frac{1}{\\sqrt{m^{2}}}\n$$\nkjer je $m$ naravno število, enak $2 m-2$ ali $2 m-1$.",
    "options": [],
    "answer": "[S] ∈ {2 m - 2, 2 m - 1}",
    "solution": "Solution:\n\na. Ko neenakost $\\sqrt{n+1}-\\sqrt{n}<\\frac{1}{2 \\sqrt{n}}$ pomnožimo s $\\sqrt{n+1}+\\sqrt{n}$, dobimo $1< \\frac{\\sqrt{n+1}+\\sqrt{n}}{2 \\sqrt{n}}$, kar očitno drži. Ko pa neenakost $\\frac{1}{2 \\sqrt{n}}<\\sqrt{n}-\\sqrt{n-1}$ pomnožimo s $\\sqrt{n}+\\sqrt{n-1}$, dobimo $\\frac{\\sqrt{n}+\\sqrt{n-1}}{2 \\sqrt{n}}<1$, kar spet očitno drži.\n\nb. Označimo $S=\\sum_{n=1}^{m^{2}} \\frac{1}{\\sqrt{n}}$. Ko seštejemo vse neenakosti\n$$\n\\sqrt{n+1}-\\sqrt{n}<\\frac{1}{2 \\sqrt{n}}<\\sqrt{n}-\\sqrt{n-1}\n$$\nod $n=1$ do $n=m^{2}$, dobimo\n$$\n\\sqrt{m^{2}+1}-\\sqrt{1}<\\frac{1}{2} S<\\sqrt{m^{2}}-\\sqrt{1-1}\n$$\nin od tod\n$$\n2\\left(\\sqrt{m^{2}+1}-1\\right)<S<2 m\n$$\nKer je $2 m-2\\left(\\sqrt{m^{2}+1}-1\\right)=2\\left(m-\\sqrt{m^{2}+1}+1\\right)<2$, velja ocena $2 m-2<S<2 m$. Torej za celo število $S$ velja $2 m-2 \\leq[S]<2 m$ oz. $[S] \\in\\{2 m-2,2 m-1\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41108,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTina je na pet listkov zapisala po eno naravno število, a ni hotela izdati, katera števila je zapisala. Prebrisani Žan jo je prepričal, da mu je povedala vse vsote po dveh števil. Zvedel je, da so bile vsote 17, 20, 28, $14, 42, 36, 28, 39, 25$ in 31. Katera števila je zapisala Tina?",
    "options": [],
    "answer": "3, 11, 14, 17, 25",
    "solution": "Solution:\n\nOznačimo števila z $a, b, c, d$ in $e$. Predpostavimo lahko, da velja $a \\leq b \\leq c \\leq d \\leq e$. Uredimo vsote po velikosti: $14, 17, 20, 25, 28, 28, 31, 36, 39, 42$. Gotovo je $a+b=14$, $a+c=17$, $d+e=42$ in $c+e=39$. Če seštejemo vse vsote, se vsako od števil pojavi v tej vsoti natanko štirikrat, zato je $4(a+b+c+d+e)=280$ ali $a+b+c+d+e=70$. Ker je $a+b=14$ in $d+e=42$, je $c=14$. Sedaj ni težko poiskati še drugih štirih števil, tako da je rešitev: $3, 11, 14, 17, 25$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41109,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa praštevila $p$, za katera je število $3^{p}-(p+2)^{2}$ praštevilo.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nZa $p=2$ je število $3^{2}-4^{2}$ negativno.\n\nČe je $p$ liho število, je $3^{p}-(p+2)^{2}$ sodo; torej enako 2. Enačba $3^{p}-(p+2)^{2}=2$ ima očitno rešitev $p=3$.\n\nDa ni drugih, dokažemo z indukcijo.\n\nNatančneje: Z indukcijo dokažemo, da za $n>3$ velja $3^{n}-(n+2)^{2}>2$ oziroma $3^{n}>n^{2}+4 n+6$.\n\nPri $n=4$ ocena velja, v dokazu induktivnega koraka pa induktivni hipotezi $3^{n}>n^{2}+4 n+6$ prištejemo neenakost $2 \\cdot 3^{n}>2 n+5$ (velja za $n \\geq 2$ ) in dobimo $3^{n+1}>(n+1)^{2}+4(n+1)+6$.\n\n(Dokaz neenakosti $2 \\cdot 3^{n}>2 n+5$ za $n \\geq 2$ s pomočjo indukcije prepustimo bralcu.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41110,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nPoišči vsa racionalna števila $r$, za katera so vse rešitve enačbe\n$$\nr x^{2}+(r+1) x+r=1\n$$\ncela števila.",
    "options": [],
    "answer": "r = 1 and r = -1/7",
    "solution": "Solution:\nPišimo $r=\\frac{a}{b}$, kjer sta $a$ in $b$ tuji števili, $a>0$, in predpostavimo, da ima enačba celoštevilske rešitve. Dobimo $a x^{2}+(a+b) x+(a-b)=0$, od koder sledi $a \\mid (a-b)$, zato je $a=1$.\n\nDa bi imela kvadratna enačba $x^{2}+(1+b) x+(1-b)=0$ kakšno celoštevilsko rešitev, mora biti njena diskriminanta popolni kvadrat. Torej $(b+3)^{2}-12=c^{2}$, kar lahko preoblikujemo v $(b+3-c)(b+3+c)=12$.\n\nKer sta števili $b+3-c$ in $b+3+c$ iste parnosti, sta $2 \\cdot 6=12$ in $(-2) \\cdot (-6)=12$ edina možna razcepa. Sledi $b=-3 \\pm 4 \\in\\{1,-7\\}$.\n\nZa $b=1$ imamo enačbo $x^{2}+2 x=0$, ki ima celoštevilski rešitvi $x=0$ in $x=-2$, za $b=-7$ pa dobimo enačbo $x^{2}-6 x+8=0$, ki ima celoštevilski rešitvi $x=4$ in $x=2$.\n\nTorej sta $r=1$ in $r=-\\frac{1}{7}$ edini rešitvi.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41111,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa poljubno naravno število $n$ označimo s $p(n)$ produkt števk tega naravnega števila, zapisanega desetiško. Izračunaj vsoto\n$$\np(1)+p(2)+\\cdots+p(2001)\n$$",
    "options": [],
    "answer": "184320",
    "solution": "Solution:\n\nIzračunamo $p(1)+p(2)+\\cdots+p(9)=45$.\n\nČe označimo s $k$ stotice in z $j$ desetice, potem je\n$$\n\\begin{aligned}\np(j 0)+p(j 1)+\\cdots+p(j 9) & =j \\cdot 45 \\\\\np(k j 0)+p(k j 1)+\\cdots+p(k j 9) & =k \\cdot j \\cdot 45 \\text{ in } \\\\\np(1 k j 0)+p(1 k j 1)+\\cdots+p(1 k j 9) & =k \\cdot j \\cdot 45\n\\end{aligned}\n$$\nSledi $p(1)+\\cdots+p(99)+p(100)+\\cdots+p(999)+p(1000)+\\cdots+p(1999)+p(2000)+p(2001)=45(1+$ $1+2+\\cdots+9)+45(1+2+\\cdots+9)(1+2+\\cdots+9)+45(1+2+\\cdots+9)(1+2+\\cdots+9)+0+0=184320$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41112,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $D$ nožišče višine na stranico $BC$ trikotnika $ABC$. Simetrala notranjega kota pri $C$ seka nasprotno stranico v točki $E$. Koliko meri kot $\\Varangle EDB$, če je $\\Varangle CEA = \\frac{\\pi}{4}$?",
    "options": [],
    "answer": "pi/4",
    "solution": "Solution:\n\nOznačimo kote trikotnika z $\\alpha$, $\\beta$ in $\\gamma$ na običajen način. V trikotniku $AEC$ velja $\\alpha + \\frac{\\pi}{4} + \\frac{\\gamma}{2} = \\pi$, od koder z upoštevanjem $\\gamma = \\pi - \\alpha - \\beta$ izpeljemo $\\alpha = \\beta + \\frac{\\pi}{2}$.\n\nKer je $AD \\perp DB$, je $\\frac{|AD|}{|DB|} = \\operatorname{tg} \\beta$. Ker je $CE$ simetrala kota, je $|AE| : |EB| = |AC| : |BC|$. Po sinusnem izreku je $\\frac{|AC|}{|BC|} = \\frac{\\sin \\beta}{\\sin \\alpha} = \\frac{\\sin \\beta}{\\sin \\left(\\beta + \\frac{\\pi}{2}\\right)} = \\frac{\\sin \\beta}{\\cos \\beta} = \\operatorname{tg} \\beta$. Torej je $|AE| : |EB| = |AD| : |DB|$ in je $ED$ simetrala pravega kota\n\n![](attached_image_1.png)\n\n$\\Varangle ADB$, zato je $\\Varangle EDB = \\frac{\\pi}{4}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41113,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bosta $E$ in $F$ taki notranji točki na stranici $AB$ pravokotnika $ABCD$, da je $|AE| = |EF|$. Pravokotnica na $AB$ skozi točko $E$ seka diagonalo $AC$ v točki $G$, daljici $FD$ in $BG$ pa se sekata v točki $H$. Dokaži, da imata trikotnika $FBH$ in $GHD$ enaki ploščini.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNaj $p_{XYZ}$ označuje ploščino trikotnika $XYZ$. Računajmo\n$$\n\\begin{aligned}\np_{DGH} - p_{HFB} = & \\left(p_{DAF} - p_{DAG} - p_{AEG} - p_{EFG} - p_{GFH}\\right) \\\\\n& - \\left(p_{EBG} - p_{EFG} - p_{GFH}\\right) = \\\\\n= & p_{DAF} - p_{DAG} - p_{AEG} - p_{EBG}\n\\end{aligned}\n$$\nOznačimo z $a = |AB|$, $b = |BC|$, $x = |AE| = |EF|$ in $y = |EG|$.\n\n![](attached_image_1.png)\n\nSledi\n$$\n\\begin{aligned}\np_{DAF} - p_{DAG} - p_{AEG} - p_{EBG} & = b x - \\frac{1}{2} b x - \\frac{1}{2} x y - \\frac{1}{2} (a - x) y = \\\\\n& = \\frac{1}{2}(b x - a y) = 0\n\\end{aligned}\n$$\nsaj zaradi podobnosti trikotnikov $AEG$ in $ABC$ velja $x : y = a : b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41114,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa stranici $BC$ ostrokotnega trikotnika $ABC$ leži taka točka $D$, da je $|AB| = |AD|$. Naj bo $E$ taka točka na višini iz $C$ trikotnika $ABC$, da se krožnica $\\mathcal{K}_1$ s središčem v $E$ dotika premice $AD$ v točki $D$. Označimo s $\\mathcal{K}_2$ krožnico skozi $C$, ki se dotika premice $AB$ v točki $B$. Dokaži, da leži točka $A$ na premici, ki poteka skozi presečišči krožnic $\\mathcal{K}_1$ in $\\mathcal{K}_2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNarišimo dovolj veliko skico in privzemimo običajne oznake kotov trikotnika $ABC$. Označimo presečišče krožnice $\\mathcal{K}_2$ z daljico $AC$ z $F$. Kot med tetivo $FB$ in tangento $AB$ je enak obodnemu kotu nad to tetivo. Sledi $\\angle FBA = \\gamma$, kar nam da $\\angle AFB = \\beta$. Torej je $ABDF$ tetivni štirikotnik in $\\angle DFC = \\beta$.\n\nPo definiciji krožnice $\\mathcal{K}_1$ je $ED \\perp AD$, zato $\\angle EDC = \\pi/2 - \\beta = \\angle ECD$. Krožnica $\\mathcal{K}_1$ torej poteka skozi točko $C$ in $\\angle DEC = 2\\beta$. Zaradi $\\angle DFC = \\beta$ je $F \\in \\mathcal{K}_1$.\n\nKrožnici $\\mathcal{K}_1$ in $\\mathcal{K}_2$ se tako sekata v točkah $F$ in $C$, točka $F$ pa po konstrukciji leži na $AC$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41115,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana je daljica $AB$. Na tej daljici izberimo poljubno točko $P$, različno od $A$ in $B$, ter konstruirajmo enakokraka pravokotna trikotnika $APQ$ in $PBR$, katerih vrhova $Q$ in $R$ ležita na istem bregu premice skozi $A$ in $B$. Naj bo točka $M$ razpolovišče daljice $QR$. Dokaži, da oddaljenost točke $M$ od premice skozi $A$ in $B$ ni odvisna od izbire točke $P$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNalogo bomo rešili nekoliko splošneje. Privzeli bomo, da sta $APQ$ in $PBR$ podobna enakokraka trikotnika (z vrhoma $Q$ oz. $R$).\n\nOznačimo s $C$ presečišče premic $AQ$ in $BR$. Potem je $QPRC$ paralelogram in $M$ razpolovišče njegovih diagonal.\n\nČe sedaj označimo z $A'$ in $B'$ razpolovišči daljic $AC$ in $BC$, leži točka $M$ na daljici $A'B'$. Trditev je tako dokazana, saj je $A'B' \\parallel AB$, točke $A'$, $B'$ in $C$ pa so\n\n![](attached_image_1.png)\n\nneodvisne od izbire točke $P$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41116,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $n \\geq 4$. Oštevilčimo $n$ točk na krožnici s številkami od 1 do $n$. Rečemo, da sta nesosednji točki, oštevilčeni z $a$ in $b$, pravilen par, če so vsaj na enem izmed lokov, ki ju določata $a$ in $b$, vse točke ostevilčene s števili, manjšimi od $a$ in $b$. Dokaži, da je število pravilnih parov enako $n-3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNalogo bomo dokazali z indukcijo. Dokaz začnemo pri $n=4$. Sosedi od točke 1 sta pravilen par. Edini drug kandidat vsebuje točko 1, ki pa nikoli ni v pravilnem paru. Torej imamo le en pravilen par.\n\nRecimo, da formula velja za $n$. Oglejmo si $n+1$ točk na krožnici. Označimo sosedi točke 1 z $a$ in $b$. Par $(a, b)$ je očitno pravilen par. Sedaj pa izbrišimo točko 1 in vsem ostalim številom odštejmo 1. Dobimo $n$ točk. Ker točka 1 ni bila v nobenem pravilnem paru, vsi pravilni pari razen $(a, b)$, ki sta sedaj sosedi, ostanejo. Torej imamo pri $n+1$ točkah natanko en pravilen par več kot pri $n$ točkah.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41117,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoloči najmanjše število polj na tabli razsežnosti $8 \\times 8$, ki jih moramo pobarvati, da bo na vsakem kosu tabele v obliki črke L (glej sliko) vsaj eno polje pobarvano.\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "32",
    "solution": "Solution:\n\nNajmanjše število polj, ki jih moramo pobarvati, je 32. Tablo najprej razdelimo na 16 kvadratov s po 4 polji. Če bi pobarvali samo 31 polj table, potem je vsaj v enem od teh kvadratov največ eno polje pobarvano. V tem kvadratu je potem lik oblike črke L brez pobarvanih polj. Če pobarvamo tablo kot šahovnico, pa vidimo, da 32 pobarvanih polj zadošča.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41118,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAndraž in Breda sta iz časopisa odrezala dva dolga trakova dolžin $a$ in $b$, da bi se z njima igrala. Pri tej igri odreže igralec, ki je na vrsti, od poljubnega traku kos dolžine $d$. Igro izgubi igralec, ki prvi ne more odrezati kosa dolžine $d$. Andraž kot kavalir prepusti Bredi, da začne igro. Ugotovi, kako dolžini trakov vplivata na to, kdo bo zmagal.",
    "options": [],
    "answer": "Breda wins if and only if floor(a divided by d) plus floor(b divided by d) is odd; otherwise Andraž wins.",
    "solution": "Solution:\n\nIz traku dolžine $a$ lahko zaporedoma odrežemo $\\left[\\frac{a}{d}\\right]$ kosov dolžine $d$, iz traku dolžine $b$ pa $\\left[\\frac{b}{d}\\right]$ kosov dolžine $d$. Igre bo torej konec po $n=\\left[\\frac{a}{d}\\right]+\\left[\\frac{b}{d}\\right]$ potezah. Ker začne igrati Breda, dela lihe reze, Andraž pa sode. Torej, če je število $n$ liho, bo zmagala Breda, ki bo naredila zadnji rez, če pa je število rezov sodo, bo zmagal Andraž.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41119,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $k$ tako naravno število, da ima kvadratna enačba\n$$\nk x^{2}-(1-2 k) x+k-2=0\n$$\nracionalni rešitvi. Dokaži, da je $k$ zmnožek 2 zaporednih celih števil.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDa bosta rešitvi kvadratne enačbe racionalni, mora biti diskriminanta $(1-2 k)^{2}-4 k(k-2)=1+4 k$ enaka $m^{2}$, $m \\in \\mathbb{N}$. Torej je\n$$\nk=\\frac{(m-1)(m+1)}{4}.\n$$\nKer je $k$ naravno število, mora biti $m$ liho število, večje od $1$, torej obstaja tako naravno število $n$, da je $m=2 n+1$, sledi $k=n(n+1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41120,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNajmanjše naravno število, katerega kvadrat se konča s tremi štiricami, je 38, saj je $38^{2}=1444$. Katero je naslednje najmanjše naravno število s to lastnostjo?",
    "options": [],
    "answer": "462",
    "solution": "Solution:\n\nNaj bo $38+n$ iskano število. Tedaj je $(38+n)^{2}=1444+n(76+n)$, kjer se število $n(76+n)$ konča s tremi ničlami, oziroma je večkratnik števila $1000$. Ker je $1000=5^{3} \\cdot 2^{3}$, mora biti ali $n$ ali $76+n$ deljivo s $5$. Toda $n$ in $76+n$ nista hkrati deljivi s $5$, zato mora biti eno izmed teh dveh števil deljivo s $125=5^{3}$ in torej oblike $125k$.\n\nOglejmo si še faktor $2^{3}$. Iz $76=2^{2} \\cdot 19$ sledi, da imata številki $n$ in $76+n$ enak ostanek pri deljenju s $4$. Ker mora biti zmnožek $n(76+n)$ deljiv z $8$, mora biti vsaj eno od števil $n$ ali $76+n$ deljivo s $4$. Torej sta s $4$ deljivi obe in je eno od njiju oblike $125 \\cdot 4 \\cdot m=500~m$, kjer je $m \\in \\mathbb{N}$. Ker iščemo najmanjšo možno vrednost, izberemo $m=1$ in iz $76+n=500$ izračunamo $38+n=462$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41121,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nReši enačbo:\n$$\n\\left(2\\left(2^{\\sqrt{x}+3}\\right)^{\\frac{1}{2 \\sqrt{x}}}\\right)^{\\frac{2}{\\sqrt{x}-1}}=4\n$$",
    "options": [],
    "answer": "9",
    "solution": "Solution:\n\nPotenciranje potence\n$$\n\\left(2 \\cdot 2^{\\frac{\\sqrt{x}+3}{2 \\sqrt{x}}}\\right)^{\\frac{2}{\\sqrt{x}-1}}=4\n$$\n\nZapis števila $4$ kot potenca\n$$\n4 = 2^2\n$$\n\nMnoženje potenc\n$$\n\\left(2^{\\frac{2 \\sqrt{x}+\\sqrt{x}+3}{2 \\sqrt{x}}}\\right)^{\\frac{2}{\\sqrt{x}-1}}=2^{2}\n$$\n\nPotenciranje potenc\n$$\n2^{\\frac{(3 \\sqrt{x}+3) \\cdot 2}{2 \\sqrt{x}(\\sqrt{x}-1)}}=2^{2}\n$$\n\nZapis enačbe\n$$\n\\frac{(3 \\sqrt{x}+3) \\cdot 2}{2 \\sqrt{x}(\\sqrt{x}-1)}=2\n$$\n\nOdprava ulomkov\n$$\n3 \\sqrt{x}+3=2 \\sqrt{x}(\\sqrt{x}-1)\n$$\n\nUreditev enačbe\n$$\n5 \\sqrt{x}=2 x-3\n$$\n\nKvadriranje\n$$\n25 x=4 x^{2}-12 x+9\n$$\n\nUreditev enačbe\n$$\n4 x^{2}-37 x+9=0\n$$\n\nObe rešitvi\n$$\nx_{1}=\\frac{1}{4}\\quad \\text{in}\\quad x_{2}=9\n$$\n\nPreiskus in izločitev rešitve $\\frac{1}{4}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41122,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa katere vrednosti realnega parametra $a$ ima sistem enačb\n$$\nx + y = a^{3} - a \\quad \\text{in} \\quad x y = a^{2}\n$$\nrealni rešitvi $x$ in $y$?",
    "options": [],
    "answer": "a = 0 or a^2 >= 3",
    "solution": "Solution:\n\nIzrazimo $x = a^{3} - a - y$, vstavimo $x$ v drugo enačbo in dobimo $(a^{3} - a - y) y = a^{2}$ oziroma $y^{2} + y(a - a^{3}) + a^{2} = 0$.\n\nKvadratna enačba ima realni rešitvi, če je njena diskriminanta nenegativna, torej $(a - a^{3})^{2} - 4 a^{2} \\geq 0$.\n\nNeenačbo poenostavimo in dobimo $a^{2}(a^{2} + 1)(a^{2} - 3) \\geq 0$.\n\nNeenačba velja, če je $a = 0$ ali $a^{2} \\geq 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41123,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana je funkcija $f(x) = \\frac{x-2}{x^{2} + x - 2}$. Za katere vrednosti $x$ bo graf funkcije $f(x)$ ležal nad premico z enačbo $y = 1$?",
    "options": [],
    "answer": "(-2,0) ∪ (0,1)",
    "solution": "Solution:\n\nZapis pogoja $\\frac{x-2}{x^{2} + x - 2} > 1$\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41124,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOče želi razdeliti trem sinovom 14560 SIT tako, da vsak naslednji sin dobi $20 \\%$ večji znesek kot njegov mlajši brat. Koliko dobi vsak sin? Zapiši odgovor.",
    "options": [],
    "answer": "4000 SIT, 4800 SIT, 5760 SIT",
    "solution": "Solution:\n\nUgotovitev: sinovi dobijo $x$, $1{,}2x$, $1{,}44x$.\n\n$\\begin{align*}\nx + 1{,}2x + 1{,}44x &= 14560 \\\\\n3{,}64x &= 14560 \\\\\nx &= \\frac{14560}{3{,}64} = 4000 \\text{ SIT}\n\\end{align*}$\n\nDrugi sin dobi $1{,}2 \\times 4000 = 4800$ SIT.\n\nTretji sin dobi $1{,}44 \\times 4000 = 5760$ SIT.\n\nOdgovor:\n\nPrvi sin dobi 4000 SIT, drugi sin dobi 4800 SIT, tretji sin dobi 5760 SIT.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41125,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nAli obstaja funkcija $f: \\mathbb{N} \\rightarrow \\mathbb{N}$, da bo\n$$\nf(f(2002))=17, \\quad f(m n)=f(m) f(n) \\quad \\text{ in } \\quad f(n) \\leq n\n$$\nza vsaka $m, n \\in \\mathbb{N}$ ?",
    "options": [],
    "answer": "No, such a function does not exist.",
    "solution": "Solution:\nKer je $2002=2 \\cdot 7 \\cdot 11 \\cdot 13$, z upoštevanjem zveze $f(x y)=f(x) f(y)$ izpeljemo\n$$\nf(f(2002))=f(f(2)) \\cdot f(f(7)) \\cdot f(f(11)) \\cdot f(f(13))=17\n$$\nKer je $f(x) \\leq x$, je $f(f(x)) \\leq f(x) \\leq x$, in zato $f(f(2)) \\leq 2, f(f(7)) \\leq 7, f(f(11)) \\leq 11$ in $f(f(13)) \\leq 13$. Na levi strani enačbe je zmnožek števil, ki so največ 13 , število 17 pa je praštevilo in takšna funkcija torej ne obstaja.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41126,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči najmanjše naravno število, ki ga lahko zapišemo kot vsoto 9, 10 in 11 zaporednih naravnih števil.",
    "options": [],
    "answer": "495",
    "solution": "Solution:\n\nKer lahko število zapišemo kot vsoto 9 zaporednih naravnih števil, je enako devetkratniku srednjega števila v tem zaporedju 9 števil. Podobno sklepamo, da je enako enajstkratniku srednjega števila v zaporedju 11 zaporednih števil. Ker se da število zapisati tudi kot vsoto 10 zaporednih naravnih števil, je enako petkratniku vsote srednjih dveh števil v tem zaporedju. Tako torej vemo, da je iskano število deljivo z $9$, $11$ in $5$, najmanjše tako število pa je $9 \\cdot 11 \\cdot 5 = 495$.\n\nSrednje število v vsoti 9 zaporednih naravnih števil je enako $\\frac{495}{9} = 55$, srednje število v vsoti 11 zaporednih števil je $\\frac{495}{11} = 45$, vsota srednjih dveh števil v vsoti 10 zaporednih naravnih števil pa je enaka $\\frac{495}{5} = 99 = 49 + 50$. Tako je:\n$$\n\\begin{aligned}\n495 & = 51 + 52 + 53 + 54 + 55 + 56 + 57 + 58 + 59 \\\\\n& = 45 + 46 + 47 + 48 + 49 + 50 + 51 + 52 + 53 + 54 \\\\\n& = 40 + 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 + 50\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41127,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $S=\\{a_{1}, \\ldots, a_{n}\\}$, kjer so $a_{i}$ različna naravna števila. Vsota števil iz nobene prave podmnožice množice $S$ ni deljiva z $n$. Dokaži, da je vsota vseh števil iz množice $S$ deljiva z $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo $s_{j}=a_{1}+\\ldots+a_{j}$. Če za $j<k$ velja $s_{j} \\equiv s_{k} \\pmod{n}$, je $a_{j+1}+\\ldots+a_{k} \\equiv 0$ $\\pmod{n}$, kar je v protislovju s predpostavko naloge. Torej dajo števila $s_{1}, \\ldots, s_{n}$ vse možne ostanke pri deljenju z $n$. Ker po predpostavki nobeno od števil $s_{1}, \\ldots, s_{n-1}$ ni deljivo z $n$, mora biti z $n$ deljivo število $s_{n}=a_{1}+\\ldots+a_{n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41128,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKje dosežeta funkciji $f(x)=\\sin 4 x$ in $g(x)=-\\cos 2 x$ enako vrednost?",
    "options": [],
    "answer": "x = π/4 + k·π/2, x = −π/12 + k·π, or x = 7π/12 + k·π, where k ∈ ℤ",
    "solution": "Solution:\n\nZapis $f(x)=g(x)$ ali $\\sin 4 x=-\\cos 2 x$\n\n$2 \\sin 2 x \\cos 2 x+\\cos 2 x=0$\n\nZapis v obliki produkta: $\\cos 2 x(2 \\sin 2 x+1)=0$\n\nZapis enačb:\n$\\cos 2 x=0$\n$\\sin 2 x=-\\frac{1}{2}$\n\nRešitev:\n$x_{1}=\\frac{\\pi}{4}+\\frac{k \\pi}{2}$\n$x_{2}=-\\frac{\\pi}{12}+k \\pi$\n$x_{3}=\\frac{7 \\pi}{12}+k \\pi,\\ k \\in \\mathbf{Z}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41129,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTine je zbiral znamke. Za rojstni dan je dobil nov album, v katerega bo lahko spravil veliko znamk. Iz hranilnika je vzel $2002$ tolarja in sklenil, da bo ves denar porabil za nakup znamk. Prijatelj mu je ponudil manjše znamke po $10$ tolarjev in večje po $28$ tolarjev. Tine se je odločil, da bo kupil čim večje število znamk. Koliko znamk bo lahko kupil?",
    "options": [],
    "answer": "193",
    "solution": "Solution:\n\nDenimo, da bo Tine kupil $x$ znamk po $10$ tolarjev in $y$ znamk po $28$ tolarjev. Tedaj velja $10x + 28y = 2002$ oziroma $5x + 14y = 1001$, od tod pa $5x + 5y = 1001 - 9y$ oziroma $x + y = \\frac{1001 - 9y}{5}$. Vrednost vsote bo tem večja, čim manjši bo $y$. Ker je $y$ naravno število, lahko poskušamo z vrednostmi $y = 1, y = 2, y = 3 \\ldots$, dokler ne dobimo cele vrednosti za vsoto $x + y$. Sicer pa vemo, da bo vsota $x + y$ celo število, ko bodo enice v razliki $1001 - 9y$ enake $0$ ali $5$, to je, ko bodo enice v zmnožku $9y$ enake $1$ ali $6$. Najmanjši $y$, ko to velja, je $y = 4$. Tedaj je $x + y = \\frac{1001 - 36}{5} = \\frac{965}{5} = 193$. Tine bo kupil $193$ znamk.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41130,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nImamo $n \\geq 3$ listov, ki jih oštevilčimo od 1 do $n$. Liste nato razdelimo na 2 kupa in ugotavljamo, ali sta v vsaj 1 kupu lista, označena s številoma, katerih vsota je popolni kvadrat. Dokaži, da\n\na) Če je $n \\geq 15$, taka lista obstajata ne glede na to, kako liste razdelimo.\n\nb) Če je $n \\leq 14$, taka lista ne obstajata pri vsaki porazdelitvi.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nRešimo najprej prvi del naloge. Dovolj je dokazati trditev za $n=15$. Če namreč taka lista obstajata za $n=15$, obstajata tudi za $n \\geq 16$, saj lahko liste s številkami od 16 dalje izločimo in sklepamo kot prej. Če pri tem en kup v celoti izpraznimo, ostaneta na drugem kupu denimo števili 1 in 3.\n\nPredpostavimo, da to ne bi bilo res in da se torej pri $n=15$ to ne bi dalo narediti. To pomeni, da bi morala biti 1 in 15 na različnih kupih, prav tako 1 in 3. Tako bi bila 3 in 15 na istem kupu. Na tem kupu ne bi smelo biti števila 6, saj bi sicer imeli $6+3=9=3^{2}$. Niti števila 10 ne bi smelo biti na tem kupu, saj bi imeli $10+15=25=5^{2}$. To pa pomeni, da bi bila lista s številoma 6 in 10 oba na drugem kupu in bi imeli protislovje, saj je $6+10=16=4^{2}$. Za $n=15$ torej vedno najdemo vsaj v enem kupu lista s številoma, katerih vsota je popolni kvadrat.\n\nSedaj rešimo še drugi del naloge. Vzemimo najprej $n=14$ in poiščimo porazdelitev, ko vsota nobenih dveh števil s posameznega kupa ni popolni kvadrat: na en kup postavimo liste s števili $1,2,4,6,9,11$ in 13, na drugi pa liste s števili $3,5,7,8,10,12$ in 14. Če je $3 \\leq n<14$, lahko uporabimo kar rešitev za $n=14$, le liste s števili, večjimi od $n$, odstranimo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41131,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOd 25 učencev so pri pisni nalogi štirje dobili oceno 5, pet oceno 4 in pet oceno 2. Koliko učencev je dobilo oceno 1 in koliko učencev oceno 3, če je bila povprečna ocena natanko 3? Zapiši odgovor.",
    "options": [],
    "answer": "Grade 1: 4; Grade 3: 7",
    "solution": "Solution:\n\nZapis enačbe\n$$\n\\frac{4 \\cdot 5 + 5 \\cdot 4 + 2 \\cdot 5 + x + 3y}{25} = 3\n$$\n\n$x + y = 11$\n\nReševanje sistema - postopek\n\nRešitvi: $x = 4$\n\n$y = 7$\n\nOdgovor: Oceno 3 je dobilo 7 učencev, oceno 1 pa 4 učenci.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41132,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSkupina geometrov želi določiti višino hriba. Iz kraja $A$ v ravnini vznožja vidijo vrh hriba pod kotom $\\alpha=19^\\circ 40^\\prime$, ko pa se po ravnem približajo hribu za $250 \\mathrm{~m}$, vidijo vrh pod kotom $\\beta=24^\\circ$. Kako visok je hrib? Vmesne rezultate zaokroži na štiri decimalke, rezultat pa na celo število. Zapiši odgovor.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "453 m",
    "solution": "Solution:\n\nZapisana zveza $\\operatorname{tg} \\alpha=\\frac{h}{250+x}$\n\nZapisana zveza $\\operatorname{tg} \\beta=\\frac{h}{x}$\n\nZapisana ugotovitev $x \\cdot \\operatorname{tg} \\beta=(x+250) \\cdot \\operatorname{tg} \\alpha$\n\nIzražena spremenljivka $x: x=\\frac{250 \\cdot \\operatorname{tg} \\alpha}{\\operatorname{tg} \\beta-\\operatorname{tg} \\alpha}$\n\nIzračunan $x=1017,6538$\n\nIzračunan $h=x \\cdot \\operatorname{tg} \\beta=453,0595$\n\nZapisan odgovor: $453 \\mathrm{~m}$\n\n![](attached_image_2.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41133,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $M$ razpolovišče osnovnice $AB$ trapeza $ABCD$. V notranjosti daljice $AC$ leži taka točka $E$, da se premici $BC$ in $ME$ sekata v točki $F$, premici $FD$ in $AB$ se sekata v točki $G$ ter premici $DE$ in $AB$ v točki $H$. Dokaži, da je $M$ razpolovišče daljice $GH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo z $N$ presečišče premic $CD$ in $EF$.\n\n![](attached_image_1.png)\n\nKer je $DC \\parallel GB$ in so točke $M$, $N$ in $F$ kolinearne, je $$\\frac{|GM|}{|MB|} = \\frac{|DN|}{|NC|}.$$ Ker je $AH \\parallel DC$ in so točke $M$, $E$ ter $N$ kolinearne, je $$\\frac{|AM|}{|MH|} = \\frac{|NC|}{|DN|}.$$ Sledi $$\\frac{|GM|}{|MB|} \\cdot \\frac{|AM|}{|MH|} = 1,$$ od koder zaradi $|AM| = |MB|$ sledi $|GM| = |MH|$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41134,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa enotski krožnici s središčem v koordinatnem izhodišču izberemo krožni lok s krajiščema $A$ in $B$, ki leži v prvem kvadrantu. Naj bo $p_{1}$ ploščina lika pod krožnim lokom in nad abcisno osjo, $p_{2}$ pa naj bo ploščina lika levo od krožnega loka in desno od ordinatne osi (glej sliko). Dokaži, da je vsota $p_{1}+p_{2}$ odvisna le od dolžine krožnega loka, ne pa tudi od njegove lege.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo točke, kot kaže slika. Ploščina krožnega izseka, ki pripada loku $\\widehat{A B}$, je enaka $p_{\\widehat{A B}}=\\frac{\\beta-\\alpha}{2}$, kjer smo kota $\\alpha$ in $\\beta$ merili v radianih. Sledi\n$$\n\\begin{aligned}\n& p_{1}=p_{\\widehat{A B}}+p_{O B B^{\\prime \\prime}}-p_{O A A^{\\prime \\prime}}=\\frac{\\beta-\\alpha}{2}+\\frac{1}{2} \\sin \\beta \\cos \\beta-\\frac{1}{2} \\sin \\alpha \\cos \\alpha \\\\\n& p_{2}=p_{\\widehat{A B}}+p_{O A^{\\prime} A}-p_{O B^{\\prime} B}=\\frac{\\beta-\\alpha}{2}+\\frac{1}{2} \\sin \\alpha \\cos \\alpha-\\frac{1}{2} \\sin \\beta \\cos \\beta\n\\end{aligned}\n$$\nTorej je $p_{1}+p_{2}=\\beta-\\alpha$, kar je enako dolžini loka $\\widehat{A B}$.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41135,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $\\mathcal{K}$ krožnica v ravnini, $\\mathcal{K}_1$ in $\\mathcal{K}_2$ pa disjunktni krožnici, ki se od znotraj dotikata krožnice $\\mathcal{K}$ v točkah $A$ in $B$. Naj bo $t$ skupna tangenta krožnic $\\mathcal{K}_1$ in $\\mathcal{K}_2$, ki se ju dotika v točkah $C$ in $D$ tako, da sta $\\mathcal{K}_1$ in $\\mathcal{K}_2$ obe na istem bregu premice $t$, središče krožnice $\\mathcal{K}$ pa na njenem drugem bregu. Označimo z $E$ presek premic $A C$ in $B D$. Dokaži, da leži točka $E$ na krožnici $\\mathcal{K}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo z $O$, $O_1$ in $O_2$ središča krožnic, presečišči premice $t$ z daljicama $O A$ in $O B$ pa označimo z $A'$ in $B'$. Naj bo $\\alpha = \\Varangle C A O_1$. Ker je trikotnik $A O_1 C$ enakokrak, je $\\Varangle C O_1 A' = 2 \\alpha$. Ker je $C D \\perp O_1 C$, je $\\Varangle O_1 A' C = \\pi / 2 - 2 \\alpha$. Če označimo $\\beta = \\Varangle O_2 B D$, lahko podobno izpeljemo $\\Varangle D B' O_2 = \\pi / 2 - 2 \\beta$. V trikotniku $O B' A'$ tako velja $\\Varangle B' O A' = \\pi - (\\pi / 2 - 2 \\beta) - (\\pi / 2 - 2 \\alpha) = 2 \\alpha + 2 \\beta$. V trikotniku $A E B$ velja $\\Varangle B E A = \\pi - \\Varangle E A B - \\Varangle A B E$. Ker je $\\Varangle E A B + \\Varangle A B E = \\alpha + \\beta + (\\pi - \\Varangle B O A) = \\pi - \\alpha - \\beta$, sledi $\\Varangle B E A = \\alpha + \\beta$. Torej je $\\Varangle B O A = 2 \\Varangle B E A$, zato točka $E$ res leži na krožnici $\\mathcal{K}$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41136,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKvadratna funkcija $f(x)$ doseže minimalno vrednost $-2$ za $x=1$. Določi $f(x)$ tako, da bo veljalo $f(-3)+4 f(0)=0$. Izračunaj $f(1+\\sqrt{5})$.",
    "options": [],
    "answer": "f(x) = (1/2)(x - 1)^2 - 2; f(1 + sqrt(5)) = 1/2",
    "solution": "Solution:\n\nZapisana enačba kvadratne funkcije $y = a(x-1)^2 - 2$\n\nZapisani in vstavljeni podatki: za $f(-3) = 16 \\cdot a - 2$\nza $4 \\cdot f(0) = 4(a - 2)$\n\nZapisana enačba $16 \\cdot a - 2 + 4(a - 2) = 0$\n\nIzračun $a = \\frac{1}{2}$\n\nZapisana kvadratna funkcija $f(x) = \\frac{1}{2}(x-1)^2 - 2$ ali $f(x) = \\frac{1}{2} x^2 - x - \\frac{3}{2}$\n\nIzračunano $f(1+\\sqrt{5}) = \\frac{1}{2}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41137,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPri deljenju števila $a$ s $7$ dobimo ostanek $3$, pri deljenju števila $b$ s $7$ pa ostanek $4$. Kolikšen je ostanek pri deljenju kvadrata vsote števil $a$ in $b$ s $7$? Odgovor utemelji.",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n\nZapis: $a = 7x + 3$\n\nZapis $b = 7y + 4$\n\n$$(a + b)^2 = (7x + 3 + 7y + 4)^2 = (7x + 7y + 7)^2 = (7(x + y + 1))^2 = 7 \\cdot (7(x + y + 1)^2)$$\n\nZapisana ugotovitev: Ostanek je $0$, saj je $(a + b)^2$ večkratnik števila $7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41138,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $A'$ nožišče višine na stranico $BC$ ostrokotnega trikotnika $ABC$. Krožnica s premerom $AA'$ seka stranico $AB$ v točkah $A$ in $D$, stranico $AC$ pa v točkah $A$ in $E$. Dokaži, da leži središče očrtane krožnice trikotnika $ABC$ na nosilki višine na $DE$ trikotnika $ADE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo z $A''$ nožišče višine iz $A$ trikotnika $ADE$. Središče trikotniku $ABC$ očrtane krožnice leži na premici $AA''$, če je $\\angle BAA'' = \\frac{\\pi}{2} - \\gamma$, kjer je $\\gamma = \\angle ACB$. Ker je $AA'' \\perp DE$ in $AB \\perp A'D$ ter je štirikotnik $ADA'E$ tetiven, je $\\angle BAA'' = \\angle A'DE = \\angle A'AE$. Ker pa je $AA' \\perp BC$, je tako res $\\angle A'AE = \\frac{\\pi}{2} - \\gamma = \\angle BAA''$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41139,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV živalski vrt naselijo družino risov. Število risov $N$ po $t$ letih $(t \\geq 0)$ določa funkcija $N=10 \\cdot e^{\\frac{2}{5} t}$.\n\na) Koliko risov šteje družina ob naselitvi?\n\nb) Koliko let bi potrebovali v živalskem vrtu, da bi družina risov štela 100 članov? Rezultat zaokroži na celo število.\n\nZapiši odgovora.",
    "options": [],
    "answer": "a) 10; b) 6",
    "solution": "Solution:\n\na)\nZa $t=0$:\n$$\nN = 10 \\cdot e^{\\frac{2}{5} \\cdot 0} = 10 \\cdot e^{0} = 10 \\cdot 1 = 10\n$$\nOdgovor: 10 risov\n\nb)\nPoiščemo $t$, da bo $N=100$:\n$$\n100 = 10 \\cdot e^{\\frac{2}{5} t}\n$$\n$$\n10 = e^{\\frac{2}{5} t}\n$$\n$$\n\\ln 10 = \\ln e^{\\frac{2}{5} t}\n$$\n$$\n\\ln 10 = \\frac{2}{5} t \\ln e\n$$\nKer je $\\ln e = 1$:\n$$\n\\ln 10 = \\frac{2}{5} t\n$$\n$$\nt = \\frac{5}{2} \\ln 10 \\approx 5{,}75646\n$$\nZaokrožimo na celo število:\n$$\nt \\approx 6\n$$\nOdgovor: 6 let",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41140,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDani so štirje pravokotniki z dolžino $a=18$. Njihove širine tvorijo geometrijsko zaporedje. Obseg drugega pravokotnika je 60, tretji pravokotnik je kvadrat. Določi širine pravokotnikov.",
    "options": [],
    "answer": "8, 12, 18, 27",
    "solution": "Solution:\n\nOznačimo širine geometrijskega zaporedja z $b$, $bq$, $bq^2$, $bq^3$.\n\nObseg drugega pravokotnika je $2a + 2bq = 60$.\n\n$a + bq = 30 \\Rightarrow bq = 12$\n\nTretji pravokotnik je kvadrat: $a = bq^2 \\Rightarrow bq^2 = 18$\n\nRešimo sistem:\n\n$12 \\cdot q = 18 \\Rightarrow q = \\frac{3}{2}$\n\n$b = 8$\n\nŠirine pravokotnikov: $8$, $12$, $18$, $27$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41141,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV škofjeloški grajski kleti 7 palčkov hrani svoj zaklad. Zaklad je za 12 vrati, vsaka vrata pa so zaklenjena z 12 ključavnicami. Vse ključavnice so različne. Vsak palček ima ključe za nekaj ključavnic. Katerikoli 3 palčki imajo skupaj ključe za vse ključavnice. Dokaži, da imajo palčki skupaj vsaj 333 (ne nujno različnih) ključev.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nZagotovo obstajajo 4 palčki, od katerih ima vsak vsaj 48 ključev (sicer bi lahko izbrali 3, ki bi skupaj imeli manj kot $3 \\cdot 48=144$ ključev in ne bi mogli odpreti vseh ključavnic). Preostali 3 palčki imajo skupaj vsaj 144 ključev. Torej imamo 4 palčke z vsaj 48 ključi in trojico z vsaj 144 ključi, skupno vsaj $4 \\cdot 48+144=336$ ključev.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41142,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAli lahko pokrijemo šahovnico velikosti $15 \\times 15$ s $55$ dominami velikosti $4 \\times 1$ tako, da ostanejo središčno polje in vsa polja ob ogliščih šahovnice nepokrita? (Domine morajo v celoti ležati na šahovnici in se ne smejo prekrivati.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPobarvajmo šahovnico, kot kaže slika. Vsaka domina velikosti $4 \\times 1$ pokrije po $2$ polji vsake barve. Ker število belih polj ni enako številu osenčenih polj, šahovnice ne moremo pokriti na predpisani način.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41143,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNajmanj koliko zvezdic moramo narisati v tabelo velikosti $4 \\times 4$, da bo po brisanju poljubnih 2 stolpcev in poljubnih 2 vrstic ostala v tabeli vsaj 1 zvezdica?",
    "options": [],
    "answer": "7",
    "solution": "Solution:\n\nSlika kaže, da zadošča 7 zvezdic. Dokažimo, da 6 zvezdic ne zadošča. V tem primeru je vsaj v 2 stolpcih največ 1 zvezdica. Če zbrišemo preostala 2 stolpca, ostaneta le še 2 zvezdici, ki ju lahko zbrišemo, če izpraznimo vrstici, v katerih ležita.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41144,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKateta $a$ in hipotenuza pravokotnega trikotnika sta v razmerju $12:13$. Če skrajšamo hipotenuzo za $23~\\mathrm{cm}$ in kateto $a$ za $27~\\mathrm{cm}$, dobimo nov pravokotni trikotnik, ki se mu dolžina druge katete ni spremenila. Izračunaj stranice prvotnega trikotnika.",
    "options": [],
    "answer": "a = 48 cm, b = 20 cm, c = 52 cm",
    "solution": "Solution:\n\nZapisane stranice prvega trikotnika: $a = 12x$, $b = b$, $c = 13x$\n\nZapisane stranice drugega trikotnika: $a' = 12x - 27$\n\n$$\n\\begin{aligned}\nb' &= b \\\\\nc' &= 13x - 23\n\\end{aligned}\n$$\n\nZapisan Pitagorov izrek $b^2 = c^2 - a^2 = c'^2 - a'^2$\n\nVstavljeni podatki $(13x)^2 - (12x)^2 = (13x - 23)^2 - (12x - 27)^2$\n\nPreoblikovana in urejena enačba $50x = 200$\n\nIzračunan $x = 4$\n\nZapisane stranice prvega trikotnika: $a = 48~\\mathrm{cm}$, $b = 20~\\mathrm{cm}$, $c = 52~\\mathrm{cm}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41145,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPremer prednjega kolesa je $1,1~\\mathrm{m}$, zadnjega pa $0,8~\\mathrm{m}$. Kolikšno razdaljo smo prevozili, če je prvo kolo naredilo 69 obratov manj kot zadnje? Rezultat zaokroži na centimeter natančno. Za $\\pi$ uporabi približek $\\frac{22}{7}$. Zapiši odgovor.",
    "options": [],
    "answer": "636.11 m",
    "solution": "Solution:\n\nPo enem obratu prevozimo $o = 2 \\pi r$.\n\nS prednjim kolesom prevozimo $x \\cdot 1{,}1 \\pi$.\n\nZ zadnjim kolesom prevozimo $(x + 69) \\cdot 0{,}8 \\pi$.\n\nZapisana enačba:\n$$\nx \\cdot 1{,}1 \\pi = (x + 69) \\cdot 0{,}8 \\pi\n$$\n\nIzračunan $x = 184$.\n\nRešitev in zapisan odgovor:\n$$\no = 636{,}11~\\mathrm{m}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41146,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIzdelati moramo 1320 parov smuči. Pri izdelavi s strojem $A$ bi porabili 2 uri manj kot pri uporabi stroja $B$. Stroj $B$ naredi 5 parov smuči manj na uro kot stroj $A$. Izračunaj čas izdelave smuči, če uporabimo oba stroja.",
    "options": [],
    "answer": "1320/115 hours (approximately 11.48 hours)",
    "solution": "Solution:\n\nNaj stroj $A$ na uro naredi $x$ parov smuči, za 1320 parov porabi $t$ ur.\nStroj $B$ na uro naredi $(x-5)$ parov smuči, za 1320 parov porabi $(t+2)$ ur.\n\nIzražena spremenljivka: $t = \\frac{1320}{x}$\n\nZapišemo enačbo:\n$$\nt \\cdot x = (t+2)(x-5)\n$$\n\nUrejena enačba:\n$$\nx^2 - 5x - 3300 = 0\n$$\n\nIzračunamo $x$:\n$$\nx = 60\n$$\n\nIzračunan čas:\n$$\nt_1 = \\frac{1320}{60} = 22~\\mathrm{h}\n$$\n\nČe uporabimo oba stroja skupaj, je skupna hitrost $60 + 55 = 115$ parov na uro:\n$$\nt = \\frac{1320}{115} \\approx 11,48~\\mathrm{h}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41147,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNariši graf funkcije s predpisom $f(x) = -\\sqrt{x^{2} - 6x + 9}$ in izračunaj ploščino lika, ki ga oklepa graf dane funkcije s koordinatnima osema.",
    "options": [],
    "answer": "9/2",
    "solution": "Solution:\n\nZapis $f(x) = -\\sqrt{(x-3)^{2}}$\n\nZapis $f(x) = -|x-3|$\n\nIzračunana ničla $(m)$, začetna vrednost $(n): m = 3, n = -3$\n\nNarisan graf funkcije $g(x) = -(x-3)$\n\nNarisan graf funkcije $f(x) = -|x-3|$\n\nIzračunana ploščina trikotnika $S = \\frac{9}{2}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41148,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrije Butalci, Bingo, Bunko in Balko, so oropali banko in odnesli 22 vreč z bankovci. Postavili so jih v vrsto tako, da je bilo v prvi vreči najmanj denarja, v vsaki naslednji pa en šop bankovcev več kot v tisti pred njo. Šef Bingo je ukradene vreče denarja delil po načelu: prva meni (Bingu), druga tebi (Bunku), tretja meni (Bingu), četrta tebi (Balku), peta meni (Bingu), šesta tebi (Bunku) ... Nato so denar prešteli. Bunko in Balko sta ugotovila, da imata skupaj bajno vsoto 6710000 SIT. Veselila sta se tudi dejstva, da imata skupaj 110000 SIT več kot njun šef. Tvoja naloga je, da ugotoviš, koliko denarja je bilo v prvi vreči. Zapiši odgovor.",
    "options": [],
    "answer": "500000 SIT",
    "solution": "Solution:\n\nŠopi šefa: $x + (x + 2) + \\cdots + (x + 20) = \\frac{6600000}{k}$\n\nŠopi Bunka in Balka: $(x + 1) + (x + 3) + \\cdots + (x + 21) = \\frac{6710000}{k}$\n\nUreditev obeh enačb:\n$$\n\\begin{aligned}\n& 11(x + 10) = \\frac{6600000}{k} \\quad \\text{in} \\\\\n& 11(x + 11) = \\frac{6710000}{k}\n\\end{aligned}\n$$\n\nRešitev sistema $x = 50$\n\n$k = 10000$\n\nOdgovor: V prvi vreči je $x \\cdot k = 500000$ SIT.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41149,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj za polinom $p(x)$ s celimi koeficienti velja $p(3)=2$. Ali je število $p(2003)$ lahko popolni kvadrat?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nKer ima polinom $p$ cele koeficiente, $x-y$ deli $p(x)-p(y)$. Potem je $p(2003)-p(3) = (2003-3)k$ in $p(2003) = 2000k + 2$. Torej da $p(2003)$ pri deljenju s $4$ ostanek $2$ in ne more biti popolni kvadrat.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41150,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoenostavi izraz\n$$\n\\left(x+\\sqrt[3]{3 \\cdot \\sqrt{\\frac{x^{3}-1}{9}+\\frac{x-x^{2}}{3}}}\\right) \\cdot (x-\\sqrt{x-1})\n$$",
    "options": [],
    "answer": "x^2 - x + 1",
    "solution": "Solution:\n\nIzraz poenostavimo:\n$$\n\\left(x+\\sqrt[3]{3 \\cdot \\sqrt{\\frac{x^{3}-1}{9}+\\frac{x-x^{2}}{3}}}\\right) \\cdot (x-\\sqrt{x-1}) = \\left(x+\\sqrt[3]{3 \\cdot \\sqrt{\\frac{x^{3}-1+3x-3x^{2}}{9}}}\\right) \\cdot (x-\\sqrt{x-1})\n$$\n\nOpazimo, da je:\n$$\n\\frac{x^{3}-1}{9}+\\frac{x-x^{2}}{3} = \\frac{x^{3}-1+3x-3x^{2}}{9}\n$$\n\ntorej:\n$$\n\\sqrt[3]{3 \\cdot \\sqrt{\\frac{x^{3}-1+3x-3x^{2}}{9}}}\n$$\n\nSedaj zapišimo $x-\\sqrt{x-1}$ kot $x-\\sqrt{x-1}$.\n\nOpazimo, da je:\n$$\n\\sqrt[6]{(x-1)^{3}} = \\sqrt{x-1}\n$$\n\nTorej lahko izraz zapišemo kot:\n$$\n(x+\\sqrt{x-1}) \\cdot (x-\\sqrt{x-1}) = x^{2}-x+1\n$$\n\nKončni rezultat je:\n$$\nx^{2}-x+1\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41151,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVemo, da je $A=\\frac{a^{-2}-b^{-2}}{a^{-1}-b^{-1}}$\n$$\n\\text{in}\\quad B=\\left(\\frac{a^{-1}}{a^{-1}-b^{-1}}-\\frac{b^{-1}}{a^{-1}+b^{-1}}\\right) \\cdot\\left(a^{-1}-b^{-1}\\right) \\cdot\\left(a^{-2}+b^{-2}\\right)^{-1}\n$$\nDokaži, da je $A=B^{-1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNajprej poenostavimo izraz $A$ :\n$$\n\\frac{\\frac{b^{2}-a^{2}}{a^{2} b^{2}}}{\\frac{b-a}{a b}}=\\frac{(b-a)(b+a)}{(b-a) a b}=\\frac{b+a}{a b}\n$$\n\nNato pa še izraz $B$ :\n$$\n\\left(\\frac{\\frac{1}{a}}{\\frac{b-a}{a b}}-\\frac{\\frac{1}{b}}{\\frac{b+a}{a b}}\\right) \\cdot \\frac{b-a}{a b} \\cdot \\frac{1}{\\frac{b^{2}+a^{2}}{a^{2} b^{2}}}\n$$\n$$\n=\\left(\\frac{b}{b-a}-\\frac{a}{b+a}\\right) \\cdot \\frac{a b(b-a)}{b^{2}+a^{2}}\n$$\n$$\n=\\frac{b^{2}+a b-a b+a^{2}}{(b-a)(b+a)} \\cdot \\frac{a b(b-a)}{b^{2}+a^{2}}\n$$\n$$\n=\\frac{a b}{b+a}\n$$\n\nVidimo, da res velja $A=B^{-1}$.\n\nZapis: $A=\\frac{\\frac{b^{2}-a^{2}}{a^{2} b^{2}}}{\\frac{b-a}{a b}}$ 1 točka\nZapis: $A=\\frac{b+a}{a b}$ 1 točka\nPoenostavljanje prvega oklepaja do oblike: $\\frac{b}{b-a}-\\frac{a}{a+b}$ 1 točka\nZapis zveze: $\\left(a^{-2}+b^{-2}\\right)^{-1}=\\frac{a^{2} b^{2}}{a^{2}+b^{2}}$. 1 točka\nZapis: $B=\\frac{a b}{b+a}$ 1 točka\nSklep: $A=B^{-1}$ 1 točka",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41152,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDano je aritmetično zaporedje $a_{1}, a_{2}, a_{3}, \\ldots$ Označimo z $s_{i}$ vsoto prvih $i$ členov tega zaporedja, z $s_{j}$ vsoto prvih $j$ členov in z $s_{k}$ vsoto prvih $k$ členov. Dokaži, da vrednost izraza\n$$\n\\frac{s_{i}}{i}(j-k)+\\frac{s_{j}}{j}(k-i)+\\frac{s_{k}}{k}(i-j)\n$$\nni odvisna niti od izbire števil $i, j$ in $k$, niti od zaporedja.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nVsota prvih $i$ členov zaporedja je $s_{i}=a_{1}+a_{2}+a_{3}+\\ldots+a_{i}=a_{1}+(a_{1}+d)+(a_{1}+2 d)+\\ldots+(a_{1}+(i-1) d)=i \\cdot a_{1}+\\frac{(i-1) i}{2} \\cdot d$\nin podobno $s_{j}=j \\cdot a_{1}+\\frac{(j-1) j}{2} \\cdot d$ ter $s_{k}=k \\cdot a_{1}+\\frac{(k-1) k}{2} \\cdot d$. Vstavimo v izraz in izračunamo\n$$\n\\begin{gathered}\n\\frac{i \\cdot a_{1}+\\frac{(i-1) i}{2} \\cdot d}{i}(j-k)+\\frac{j \\cdot a_{1}+\\frac{(j-1) j}{2} \\cdot d}{j}(k-i)+\\frac{k \\cdot a_{1}+\\frac{(k-1) k}{2} \\cdot d}{k}(i-j)= \\\\\n=a_{1}(j-k+k-i+i-j)+\\frac{d}{2}((i-1)(j-k)+(j-1)(k-i)+(k-1)(i-j))= \\\\\n=\\frac{d}{2}(i j-i k-j+k+j k-j i-k+i+k i-k j-i+j)=0 .\n\\end{gathered}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41153,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa ulomek $\\frac{m}{n}$, kjer sta $m$ in $n$ naravni števili, velja $\\frac{1}{3}<\\frac{m}{n}<1$. Če števcu prištejemo naravno število, imenovalec pa s tem številom pomnožimo, se vrednost ulomka ne spremeni. Poišči vse take ulomke $\\frac{m}{n}$.",
    "options": [],
    "answer": "2/3, 2/4, 2/5",
    "solution": "Solution:\n\nIz $\\frac{m}{n}=\\frac{m+k}{n \\cdot k}$ izrazimo $m=\\frac{k}{k-1}$. Ker je $m$ naravno število, mora biti $k=2$, tako da je tudi $m=2$. Zaradi $\\frac{1}{3}<\\frac{2}{n}<1$ mora biti $2<n<6$. Vse možne rešitve so $\\frac{2}{3}, \\frac{2}{4}$ in $\\frac{2}{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41154,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDano je praštevilo $p$. Poišči vsa naravna števila $x$ in $y$, ki zadoščajo enačbi $p \\cdot (x-5) = x \\cdot y$.",
    "options": [],
    "answer": "All solutions in positive integers are: (x, y) = (p, p − 5) for primes p greater than five, and (x, y) = (5p, p − 1) for every prime p.",
    "solution": "Solution:\n\nKer je $p$ praštevilo, ločimo 2 možnosti: $p$ deli $x$ ali pa $y$.\n\nOglejmo si najprej možnost, ko $p$ deli $x$. Tedaj lahko zapišemo $x = p x'$, kjer je $x'$ pozitivno število, in tako dobimo $p x' - 5 = x' y$, kar zapišemo kot $x'(p - y) = 5$.\n\nPotem je $x' = 1$ in $y = p - 5$ ali $x' = 5$ in $y = p - 1$.\n\nToda $x = p$ in $y = p - 5$ je rešitev le takrat, ko je $p > 5$, $x = 5p$ in $y = p - 1$ pa pri vsakem $p$.\n\nPoglejmo še primer, ko $p$ deli $y$. Tedaj je $y = p y'$, in tako $x - 5 = x y'$, oziroma $x(1 - y') = 5$. Ta enačba nima rešitev, saj sta $x$ in $y'$ pozitivni števili.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41155,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMetka stoji $60~\\mathrm{m}$ vzhodno in $80~\\mathrm{m}$ južno od kraja, kjer stoji Tine. Oba sta enako oddaljena od lipe v mestnem parku, ki je naravnost vzhodno od kraja, kjer je Tine. Sočasno se vsak s svojega kraja odpravita naravnost proti lipi. Koliko metrov poti bo vsak izmed njiju prehodil do njunega srečanja pod lipo?",
    "options": [],
    "answer": "83 1/3 m",
    "solution": "Solution:\n\nKer sta Metka in Tine enako oddaljena od lipe, velja $60 + x = \\sqrt{80^{2} + x^{2}}$. Če obe strani enačbe kvadriramo in nato enačbo uredimo, dobimo $120x = 2800$, od tod pa izrazimo $x = \\frac{70}{3}~\\mathrm{m}$. Vsak izmed njiju bo do srečanja prehodil $83\\,\\frac{1}{3}~\\mathrm{m}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41156,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $X$ presečišče diagonal konveksnega štirikotnika $ABCD$. Dokaži, da se trikotnikoma $ABX$ in $CDX$ očrtani krožnici dotikata natanko tedaj, ko je $AB \\parallel CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo $\\angle BAX = \\alpha$ in $\\angle DCX = \\beta$. Potem je $\\angle BO_1X = 2\\alpha$ in $\\angle DO_2X = 2\\beta$, od koder izračunamo\n$$\n\\angle O_1XB = \\frac{\\pi}{2} - \\alpha \\text{ in } \\quad \\angle O_2XD = \\frac{\\pi}{2} - \\beta,\n$$\nsaj je $O_1BX$ enakokraki trikotnik z vrhom $O_1$ in $O_2DX$ enakokraki trikotnik z vrhom $O_2$. Torej je $\\alpha = \\beta$ natanko tedaj, ko je $\\angle O_1XB = \\angle O_2XD$. Iz konstrukcije pa vidimo, da je $\\angle O_1XB = \\angle O_2XD$ natanko tedaj, ko so točke $O_1$, $X$ in $O_2$ kolinearne. Ta pogoj pa je ekvivalenten\n\n![](attached_image_1.png)\n\npogoju, da se krožnici s polmeroma $O_1X$ in $O_2X$ v točki $X$ dotikata.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41157,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVsota in zmnožek 2 ulomkov sta celi števili. Eden izmed ulomkov ima imenovalec 2003. Dokaži, da sta tudi oba ulomka celi števili.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNaj bosta ulomka $\\frac{r}{2003}$ in $\\frac{p}{q}$, pri čemer smemo predpostaviti, da je ulomek $\\frac{p}{q}$ okrajšan. Če zapišemo\n$$\n\\frac{r}{2003} + \\frac{p}{q} = a \\quad \\text{in} \\quad \\frac{r}{2003} \\cdot \\frac{p}{q} = b\n$$\nsta $a$ in $b$ celi števili. Izrazimo\n$$\nq \\cdot r + 2003 \\cdot p = 2003 \\cdot q \\cdot a, \\quad r \\cdot p = 2003 \\cdot q \\cdot b\n$$\nIz prve enačbe razberemo, da $2003$ deli $r$ ali $q$, ker je $2003$ praštevilo. Če $2003$ deli $r$, potem je $\\frac{r}{2003}$ celo število in prav tako $\\frac{p}{q}$, saj je $\\frac{p}{q} = a - \\frac{r}{2003}$. Obravnavajmo še primer, ko $2003$ ne deli $r$. Tedaj $2003$ deli $p$ zaradi druge enačbe, zaradi prve enačbe pa deli tudi $q$. To je nemogoče, ker smo predpostavili, da je $\\frac{p}{q}$ okrajšan ulomek. Torej primer, ko $2003$ ne deli $r$, sploh ni možen.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41158,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV letalu, ki ima 62 vrst s po 6 sedeži v vsaki vrsti, so se potniki posedli tako, da v nobenih 2 vrstah niso zasedeni sedeži na istih mestih. Največ koliko potnikov je lahko v letalu?",
    "options": [],
    "answer": "191",
    "solution": "Solution:\n\nKer je v vsaki vrsti 6 sedežev, imamo $2^{6} = 64$ različnih možnosti posedanja potnikov v posamezni vrsti. Ker lahko za vsakega izmed 64 načinov posedanja najdemo njemu komplementarnega (tj. takega, ki ima zasedene natanko tiste sedeže, ki so v dani vrsti nezasedeni), lahko iz 32 komplementarnih parov sestavimo 32 polno zasedenih vrst. Torej bi bilo v letalu s 64 vrstami $6 \\cdot 32 = 192$ potnikov. Ker pa ima naše letalo le 62 vrst, izpustimo najmanj zasedeni vrsti: prazno vrsto in vrsto z 1 potnikom. V letalu z 62 vrstami je lahko največ $192 - 1 = 191$ potnikov.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41159,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDani sta premici z enačbama $(1-a) x - 2 a y - 2 = 0$ in $-2 x + a y - 1 = 0$. Določi $a$ tako, da se bosta premici sekali na simetrali lihih kvadrantov.",
    "options": [],
    "answer": "a = 1",
    "solution": "Solution:\n\nČe je $a = 0$, sta premici med seboj vzporedni (njuni enačbi sta tedaj $x - 2 = 0$ in $2 x + 1 = 0$), zato privzemimo, da $a \\neq 0$.\n\nIzrazimo $y = \\frac{(1-a)x - 2}{2a}$ iz prve in $y = \\frac{2x + 1}{a}$ iz druge enačbe.\n\nIzenačimo dobljeni desni strani:\n$$\n\\frac{(1-a)x - 2}{2a} = \\frac{2x + 1}{a}\n$$\nin enačbo preuredimo v\n$$\n(-a - 3)x = 4\n$$\nod koder izrazimo\n$$\nx = -\\frac{4}{a + 3}\n$$\nče je $a \\neq -3$ (prepričamo se lahko, da predstavljata dani enačbi dve vzporedni premici, če upoštevamo $a = -3$).\n\nIzrazimo še ordinato presečišča:\n$$\ny = \\frac{a - 5}{a(a + 3)}\n$$\n\nVsaka točka na simetrali lihih kvadrantov ima absciso enako ordinati, zato mora veljati\n$$\n-\\frac{4}{a + 3} = \\frac{a - 5}{a(a + 3)}\n$$\nod tod pa končno dobimo $a = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41160,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj točka $X$ leži na simetrali daljice $AB$ in ne na premici $AB$. Naj bo $C$ poljubna točka v notranjosti daljice $AB$. Dokaži, da imata trikotnikoma $ACX$ in $CBX$ očrtani krožnici enaka polmera ne glede na izbiro točke $C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nKer leži točka $X$ na simetrali daljice $AB$, je $|AX| = |BX|$ in je zato $\\angle CAX = \\angle XBC$. Torej sta enaka tudi središčna kota $\\angle CO_1X = \\angle XO_2C$. Trikotnika $XO_1C$ in $CO_2X$ sta enakokraka, z osnovnico $XC$ in enakima kotoma pri vrhovih $O_1$ in $O_2$. Torej sta skladna in je zato $|O_1C| = |O_2C|$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41161,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa voljo imamo 6 različnih barv in veliko kock. Posamezno kocko pobarvamo z vsemi 6 barvami, in sicer vsako mejno ploskev z 1 barvo. Največ koliko kock lahko pobarvamo, če naj bo vsaka izmed njih drugače pobarvana? (Če lahko 1 izmed pobarvanih kock zasučemo tako, da so barve mejnih ploskev enako razporejene kot na drugi kocki, sta kocki enako pobarvani.)",
    "options": [],
    "answer": "30",
    "solution": "Solution:\n\nDenimo, da imamo na voljo rdečo, modro, zeleno, rumeno, oranžno in belo barvo. Kocke postavimo tako, da imajo spodnjo mejno ploskev pobarvano z belo barvo. Za zgornjo ploskev imamo na voljo katero koli izmed preostalih 5 barv.\n\nSedaj se vprašamo, koliko kock ima lahko zgornjo ploskev pobarvano z neko izbrano barvo (denimo oranžno). Te kocke postavimo tako, da imajo nepobarvano prednjo, zadnjo, levo in desno mejno ploskev. Z rumeno barvo pobarvamo zadnjo mejno ploskev posamezne kocke. Vsako izmed teh kock namreč lahko zavrtimo tako, da ima zadnjo ploskev rumeno, ne da bi pri tem spremenili barvo zgornje in spodnje ploskve. Preostale 3 mejne ploskve lahko pobarvamo na 6 načinov (npr. 3 barve imamo na voljo za desno mejno ploskev in 2 barvi za sprednjo mejno ploskev).\n\nVseh možnosti je torej $5 \\cdot 6=30$. Pobarvamo lahko največ 30 kock, če želimo, da bo vsaka izmed njih drugačne barve.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41162,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $D$ razpolovišče hipotenuze $AB$ pravokotnega trikotnika $ABC$. Označimo z $O_1$ in $O_2$ središči trikotnikoma $ADC$ in $DBC$ očrtanih krožnic. Dokaži, da je $AB$ tangenta na krožnico s premerom $O_1O_2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo s $k$ krožnico, ki se dotika stranice $AB$ v $D$ in na kateri leži točka $C$. Naj bo $O$ središče krožnice $k$. Potem so točke $O$, $O_1$ in $O_2$ kolinearne, saj ležijo na simetrali daljice $CD$.\n\nOznačimo $\\angle BAC = \\alpha$. Potem je središčni kot $\\angle DO_1C$ krožnice $k$ enak $2\\alpha$. Ker je trikotnik $ADC$ enakokrak, je $\\angle BDC = 2\\alpha$, zato po izreku o kotu med tetivo $DC$ in tangento $AB$ krožnice $k$ točka $O_1$ leži na tej krožnici. Podobno tudi točka $O_2$ leži na krožnici $k$.\n\nDokazali smo že, da so točke $O$, $O_1$ in $O_2$ kolinearne, zato je $O_1O_2$ res premer krožnice $k$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41163,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV jami pod Krimom spi grozna pošast. Ko postane lačna, se zbudi in požre toliko ovc, kolikor je vsota števk tistega leta. Potem spet zaspi za toliko let, kolikor ovc je pojedla. Vemo, da se je zbudila 12. aprila leta 354. Ali je pošast lahko pred vrati?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nKer je število $354$ deljivo s $3$, je tudi vsota njegovih števk in zato število požrtih ovc deljivo s $3$. Tudi vsako naslednje leto, ko se pošast zbudi, je torej deljivo s $3$. Pošast se leta $2003$ ne more zbuditi, saj $2003$ ni deljivo s $3$; torej tudi danes, $12.$ aprila $2003$, ne more biti pred vrati.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41164,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV jami pod Krimom spi grozna pošast. Ko postane lačna, se zbudi in požre toliko ovc, kolikor je vsota števk tistega leta. Potem spet zaspi za toliko let, kolikor ovc je pojedla. Vemo, da se je zbudila 12. aprila leta 666. Ali je pošast lahko pred vrati? Ali se bo lahko zbudila leta 3003?",
    "options": [],
    "answer": "No; No",
    "solution": "Solution:\n\nKer je število $666$ deljivo z $9$, je tudi vsota njegovih števk in zato število požrtih ovc deljivo z $9$. Tudi vsako naslednje leto, ko se pošast zbudi, je torej deljivo z $9$. Pošast se ne more zbuditi ne leta $2003$ ne leta $3003$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41165,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMaček iz sosednje vasi hodi v Butale dražit vaške pse. Vsak večer, ko že vsi spijo, se prikrade v Butale, na ves glas zamijavka, nato pa jo ucvre nazaj domov. Ko maček zamijavka, zalajajo vsi psi, ki so od njega oddaljeni do $90~\\mathrm{m}$. Ker so Butale majhna vas, sta vsaka 2 psa v vasi med seboj oddaljena do $100~\\mathrm{m}$. Ali se lahko maček postavi tako, da nanj zalajajo vsi vaški psi hkrati?",
    "options": [],
    "answer": "Yes",
    "solution": "Solution:\n\nMaček se lahko postavi tako, da nanj zalajajo vsi psi hkrati.\n\nNaj bosta $A$ in $B$ psa, ki sta med seboj najbolj oddaljena. Področje na sliki je presek krogov s središčema v $A$ in $B$ in polmerom $|AB|$. Izven tega območja ni nobenega psa. Če se maček postavi v točko $M$, ki je razpolovišče daljice $AB$, so vsi psi od njega oddaljeni za največ $\\frac{\\sqrt{3}}{2} \\cdot |AB| \\leq 50 \\cdot \\sqrt{3} < 90~\\mathrm{m}$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41166,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZapiši enačbo polinoma 3. stopnje (lahko tudi v razstavljeni obliki), katerega graf poteka skozi točke $A(4,-5)$, $B(-1,0)$, $C(0,5)$ in $D(5,0)$. Skiciraj graf polinoma.",
    "options": [],
    "answer": "y = 1/2 (x+1)(x-5)(x-2)",
    "solution": "Solution:\n\nTočki $B$ in $D$ sta ničli polinoma, zato lahko zapišemo $y = a(x+1)(x-5)(x-x_3)$. Polinom seka ordinatno os v točki $C(0,5)$, torej je $5 = a \\cdot 1 \\cdot (-5) \\cdot (-x_3)$ oziroma $5 a x_3 = 5$, od koder dobimo $x_3 = \\frac{1}{a}$, saj je $a \\neq 0$.\n\nKončno upoštevamo, da gre polinom skozi točko $A$, pa imamo $-5 = a \\cdot 5 \\cdot (-1) \\cdot \\left(4 - \\frac{1}{a}\\right)$ oziroma $a \\cdot \\left(4 - \\frac{1}{a}\\right) = 1$, od koder izrazimo $a = \\frac{1}{2}$.\n\nPolinom ima enačbo\n$$\ny = \\frac{1}{2}(x+1)(x-5)(x-2).\n$$\n\nZapis: $y = a(x-x_1)(x-x_2)(x-x_3)$ ali $y = a(x+1)(x-5)(x-x_3)$\n\nTočka $C$ - zapisana enačba $a \\cdot x_3 = 1$\n\nTočka $A$ - zapisana enačba $1 = 4a - a x_3$\n\nRešitvi $a = \\frac{1}{2}$, $x = 2$\n\nZapis polinoma: $y = \\frac{1}{2}(x+1)(x-5)(x-2)$\n\nSkiciran graf\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41167,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKrt je izkopal podzemne sobe in jih povezal z rovi tako, da iz vsake sobe vodijo natanko 3 rovi v 3 različne sobe. Rovi se med seboj ne sekajo. Med poljubnimi 3 sobami vedno obstajata 2, ki nista povezani z rovom. Dokaži, da je krt izkopal vsaj 6 sob. Ali je možno, da je krt izkopal natanko 6 sob?",
    "options": [],
    "answer": "At least six rooms; yes, it is possible to have exactly six rooms.",
    "solution": "Solution:\n\nIzberimo 1 sobo. Iz nje vodijo 3 rovi v 3 nove sobe, iz vsake od njih pa vodita še 2 rova, kot kaže prva slika na desni. Če bi povezali 2 nezaključena rova na tej sliki, bi dobili 3 sobe, ki so paroma povezane. Torej obstajata vsaj\n\n![](attached_image_1.png)\n\nše 2 sobi. Dokazali smo, da je sob vsaj 6. Druga slika na\n\n![](attached_image_2.png)\n\ndesni pa potrjuje, da jih je lahko natanko 6 .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41168,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse celoštevilske rešitve enačbe $x^{2}+73=y^{2}$.",
    "options": [],
    "answer": "[(36, 37), (-36, -37), (-36, 37), (36, -37)]",
    "solution": "Solution:\n\nEnačbo preoblikujemo v $y^{2}-x^{2}=73$ oziroma v $(y+x)(y-x)=73$. Število $73$ je praštevilo, zato ga lahko razcepimo le na štiri načine: $73 \\cdot 1$, $1 \\cdot 73$, $-73 \\cdot (-1)$ in $-1 \\cdot (-73)$. Tako pridemo do štirih sistemov enačb, ki jih rešimo:\n$$\n\\begin{array}{rrrr}\ny+x=73 & y+x=-73 & y+x=1 & y+x=-1 \\\\\ny-x=1 & y-x=-1 & y-x=73 & y-x=-73 \\\\\n\\end{array}\n$$\n\nRešimo posamezne sisteme:\n\n- Če $y+x=73$ in $y-x=1$, potem $y=37$, $x=36$.\n- Če $y+x=-73$ in $y-x=-1$, potem $y=-37$, $x=-36$.\n- Če $y+x=1$ in $y-x=73$, potem $y=37$, $x=-36$.\n- Če $y+x=-1$ in $y-x=-73$, potem $y=-37$, $x=36$.\n\nTorej so vse celoštevilske rešitve pari $(x, y)$: $(36, 37)$, $(-36, -37)$, $(-36, 37)$ in $(36, -37)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41169,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa petmestna števila $\\overline{abcde}$, ki so deljiva z $9$ in za katera velja $\\overline{ace} - \\overline{bda} = 760$.",
    "options": [],
    "answer": "81828, 91269",
    "solution": "Solution:\n\nEnačbo $\\overline{ace} - \\overline{bda} = 760$ preoblikujemo v $100a + 10c + e - 100b - 10d - a = 760$, od koder sledi, da je $e = a$. Zdaj lahko enačbo delimo z $10$ in dobimo $10(a - b) + (c - d) = 76$. Ločimo 2 možnosti, in sicer $c - d = 6$ ali $c - d = -4$.\n\nV prvem primeru je $c = d + 6$ in $a = b + 7$. Upoštevamo pogoj, da je petmestno število deljivo z $9$. To pomeni, da je $a + b + c + d + e = b + 7 + b + d + 6 + d + b + 7 = 3b + 2d + 20$ deljivo z $9$. Zato je $d + 1$ deljivo s $3$ in zaradi $c - d = 6$ sledi $d = 2$. Potem je $c = 8$ in velja, da $9$ deli $3(b + 2)$, od koder zaradi $a = b + 7$ sledi $b = 1$. V tem primeru je petmestno število enako $81828$.\n\nV drugem primeru je $d = c + 4$ in $a = b + 8$. Zato je $a = 8$ in $b = 0$ ali $a = 9$ in $b = 1$. Če je $a = 8$, iz pogoja $9 \\mid (a + b + c + d + e) = 8 + 2c + 4 + 8$ sledi, da $9$ deli $2c + 2$. Torej je $c = 8$, vendar potem $d = c + 4 = 12$ ni števka. Če pa je $a = 9$, iz pogoja $9 \\mid (a + b + c + d + e) = 10 + 2c + 4 + 9$ sledi, da $9$ deli $2c + 5$. Torej je $c = 2$ in je petmestno število enako $91269$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41170,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAli obstaja naravno število $n$, za katerega velja: če število $n$ pomnožimo z vsoto njegovih števk, je vsota števk dobljenega zmnožka enaka 3?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo s $s(m)$ vsoto števk naravnega števila $m$. Naloga sprašuje, ali obstaja naravno število $n$, da je $s(n \\cdot s(n))=3$. Ker je naravno število deljivo s 3 natanko tedaj, ko je s 3 deljiva vsota njegovih števk, mora biti s 3 deljiv zmnožek $n \\cdot s(n)$. Torej mora biti s 3 deljivo vsaj eno izmed števil $n$ in $s(n)$. Če pa je eno deljivo s 3, je tudi drugo deljivo s 3 in je zmnožek $n \\cdot s(n)$ deljiv z 9. Tedaj bi moralo biti število $s(n \\cdot s(n))$ deljivo z 9, zato enačba $s(n \\cdot s(n))=3$ ni rešljiva. Iskano naravno število ne obstaja.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41171,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoenostavi:\n$$\n\\frac{a^{3}-1}{1+\\frac{1}{a-\\frac{a}{a+1}}}\n$$",
    "options": [],
    "answer": "a^2(a-1)",
    "solution": "Solution:\n\nIzraz poenostavimo:\n$$\n\\frac{a^{3}-1}{1+\\frac{1}{a-\\frac{a}{a+1}}} = \\frac{a^{3}-1}{1+\\frac{1}{\\frac{a^{2}}{a+1}}} = \\frac{a^{3}-1}{\\frac{a^{2}+a+1}{a^{2}}} = \\frac{(a-1)\\left(a^{2}+a+1\\right) a^{2}}{a^{2}+a+1} = a^{2}(a-1)\n$$\n\nPoenostavljeno do oblike: $\\frac{a^{3}-1}{1+\\frac{1}{\\frac{a^{2}}{a+1}}}$\n\nPoenostavljeno do oblike: $\\frac{a^{3}-1}{\\frac{a^{2}+a+1}{a^{2}}}$\n\nOdprava dvojnih ulomkov: $\\frac{\\left(a^{3}-1\\right) a^{2}}{a^{2}+a+1}$\n\nRazstavljanje števca: $\\frac{(a-1)\\left(a^{2}+a+1\\right) a^{2}}{\\left(a^{2}+a+1\\right)}$\n\nKrajšanje\n\nRezultat: $a^{2}(a-1)$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41172,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nČleni neskončnega geometrijskega zaporedja so naravna števila, od katerih vsaj dve nista deljivi s $4$. Zapiši splošni člen tega zaporedja, če veš, da je eden izmed členov enak $2004$.",
    "options": [],
    "answer": "a_n = 501 * 2^n",
    "solution": "Solution:\n\nZaporedje je oblike $a_{n} = a \\cdot r^{n}$, $n \\geq 0$. Najprej dokažemo, da so vsi členi zaporedja cela števila. Ker je $r = \\frac{a_{n}}{a_{n-1}}$, je $r$ racionalno število. Zapišimo $r = \\frac{\\alpha}{\\beta}$, $\\beta > 0$, kjer sta si števili $\\alpha$ in $\\beta$ tuji. Dokazati je potrebno, da je $\\beta = 1$. Recimo, da obstaja praštevilo $p$, ki deli $\\beta$. Tedaj število $a_{n} = a \\frac{\\alpha^{n}}{\\beta^{n}}$ za dovolj velike $n$ ne bo celo.\n\nČe je člen zaporedja deljiv s $4$, so deljivi s $4$ tudi vsi členi, ki mu sledijo, zato $a_{0}$ in $a_{1}$ nista deljiva s $4$. Prav tako je $r \\neq 1$, saj bi bili sicer vsi členi enaki $2004$. Denimo, da je $k$-ti člen enak $2004$. Tedaj je $a_{k} = a \\cdot r^{k} = 2004 = 2 \\cdot 2 \\cdot 3 \\cdot 167$. Ker je $2004$ sodo število, je $k \\geq 2$. Če bi bil $k > 2$, bi od tod sledilo $r = 1$, to pa ni mogoče. Zato je $k = 2$ in $r > 1$. Edina možnost je $r = 2$, $a = 3 \\cdot 167 = 501$. Iskano zaporedje je $a_{n} = 501 \\cdot 2^{n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41173,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV razredu je 25 dijakov. Rok je računal, koliko točk je v povprečju dosegel posamezen dijak pri šolski nalogi. Najprej je izračunal povprečje 74,5 točk, a se je spomnil, da je pozabil upoštevati svoj dosežek. Ko ga je upošteval, je izračunal povprečje 75 točk. Koliko točk je dosegel Rok pri šolski nalogi? Zapiši odgovor.",
    "options": [],
    "answer": "87",
    "solution": "Solution:\n\nKer je vsak izmed 24 dijakov dosegel v povprečju $74,5$ točke, so vsi skupaj dosegli $24 \\cdot 74,5 = 1788$ točk. Ko je Rok upošteval tudi svoje točke, je izračunal povprečje $75$ točk, zato je $25$ dijakov doseglo skupaj $25 \\cdot 75 = 1875$ točk. Razlika $1875 - 1788$ predstavlja število točk, ki jih je dosegel Rok. Rok je dosegel $87$ točk.\n\nPri upoštevanju $24$ nalog je vsota vseh točk $24 \\cdot 74,5 = 1788$ točk.\n\nPri upoštevanju $25$ nalog je vsota vseh točk $25 \\cdot 75 = 1875$ točk.\n\nRazlika je $87$ točk.\n\nOdgovor: $87$ točk.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41174,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse celoštevilske rešitve enačbe $\\sqrt{x} + \\sqrt{y} = \\sqrt{2004}$.",
    "options": [],
    "answer": "[(2004, 0), (501, 501), (0, 2004)]",
    "solution": "Solution:\n\nKvadrirajmo enačbo $\\sqrt{x} = \\sqrt{2004} - \\sqrt{y}$ in izrazimo $2 \\sqrt{y \\cdot 2004} = 2004 + y - x$. Od tod sledi, da mora biti $2 \\sqrt{y \\cdot 2004} = 4 \\sqrt{y \\cdot 501}$ celo število, zato mora biti $y = 501 \\cdot k^{2}$, kjer je $k$ nenegativno celo število. Če sedaj $y$ vstavimo v prvotno enačbo, dobimo $\\sqrt{x} = (2 - k) \\sqrt{501}$, od koder sledi, da je $k$ lahko $0$, $1$ ali $2$. Vsi možni pari $(x, y)$ so tako $(2004, 0)$, $(501, 501)$ in $(0, 2004)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41175,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV trgovini Moda je stal moški suknjič po $30\\%$ pocenitvi $24500$ SIT. Pred koncem razprodaje so ga pocenili še za $20\\%$. Koliko tolarjev znaša razlika med začetno ceno in ceno po drugi pocenitvi?\n\nV trgovini Obleka je imel tak suknjič enako začetno ceno kot v trgovini Moda. Pocenili so ga le enkrat in takoj prodajali po ceni, ki je veljala v trgovini Moda šele po drugi pocenitvi. Za koliko odstotkov so suknjič pocenili v trgovini Obleka?\n\nZapiši odgovora.",
    "options": [],
    "answer": "15400 SIT, 44%",
    "solution": "Solution:\n\nČe označimo začetno ceno suknjiča z $x$, velja $0,7x = 24500$, od koder izračunamo $x = 35000$.\n\nPo drugi pocenitvi je suknjič stal $0,8 \\cdot 24500 = 19600$ SIT.\n\nRazlika med začetno ceno in ceno po drugi pocenitvi je $35000 - 19600 = 15400$ SIT.\n\nV trgovini Obleka so pocenili suknjič, ki je stal $35000$ SIT, za $15400$ SIT, to je za $\\frac{15400}{35000} = 44\\%$.\n\nOdgovora: $15400$ SIT, $44\\%$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41176,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDano je pozitivno realno število $p$. Med vsemi pari pozitivnih realnih števil $(x, y)$, ki ustrezajo enačbi $x y(x+y)=p$, poišči tistega, za katerega je vrednost izraza $x^{3}+y^{3}$ najmanjša.",
    "options": [],
    "answer": "x = y = (p/2)^{1/3}",
    "solution": "Solution:\n\nKer je $x^{3}+y^{3}-p=x^{3}+y^{3}-x^{2} y-x y^{2}=x^{2}(x-y)-y^{2}(x-y)=(x-y)^{2}(x+y) \\geq 0$, je vrednost izraza zagotovo večja ali enaka $p$. Enakost je dosežena le za $x=y=\\left(\\frac{p}{2}\\right)^{\\frac{1}{3}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41177,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse celoštevilske rešitve enačbe $a^{b}=a b+2$.",
    "options": [],
    "answer": "(-1, 3), (1, -1), (2, 3)",
    "solution": "Solution:\n\nČe je $b<0$, število $a^{b}$ ni celo, če je $|a|>1$. Pri $a=1$ dobimo enačbo $1=b+2$ in od tod $b=-1$. Pri $a=-1$ dobimo enačbo $(-1)^{b}=-b+2$, ki pa nima rešitve: ker je $b<0$, je $-b+2 \\geq 3$, medtem ko je $(-1)^{b}= \\pm 1$.\n\nČe je $b=0$, dana enačba ni rešljiva.\n\nČe pa je $b>0$, število $a$ deli $a^{b}$. Zato $a$ deli tudi število 2, saj je $a^{b}-a b=2$. Torej je $a$ lahko enak $-2,-1,1$ ali $2$. Oglejmo si vsakega izmed teh primerov posebej.\n\nČe je $a=-2$, je $(-2)^{b}=-2 b+2$ oziroma $(-2)^{b-1}=(b-1)$. Tej enačbi ne ustreza nobeno naravno število $b$. Če število $b$ ustreza pogoju, je liho; torej $b=2 s+1$. Enačbo tedaj preoblikujemo v $4^{s}=s$, kar pa ni rešljivo, saj za vsako realno število $s$ velja $4^{s}>s$.\n\nČe je $a=-1$, je $(-1)^{b}=-b+2$. To pomeni, da je $|-b+2|=1$, kar nam da $b=1$ ali $b=3$, vendar le $b=3$ ustreza enačbi $(-1)^{b}=-b+2$.\n\nČe je $a=1$, dobimo enačbo $1=b+2$, ki pri pogoju $b>0$ nima rešitev.\n\nČe je $a=2$, je $2^{b}=2(b+1)$ oziroma $2^{b-1}=b+1$. Tej enačbi zadošča le $b=3$. Za $b>3$ pa velja $2^{b-1}>b+1$, kar preverimo z indukcijo. (Pri $b=4$ imamo $2^{4-1}=8>4+1=5$. V dokazu induktivnega koraka pa $2^{(b+1)-1}=2 \\cdot 2^{b-1}>2(b+1)>(b+1)+2$.)\n\nPari $(a, b)$, ki rešijo enačbo, so: $(2,3)$, $(1,-1)$ in $(-1,3)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41178,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $n$ naravno število, ki je enako vsoti svojih od $n$ manjših deliteljev. Tako število je denimo število $28$. Kolikšna je vsota recipročnih vrednosti vseh njegovih deliteljev?",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nNaj bodo $d_{1}, d_{2}, \\ldots, d_{k}$ delitelji števila $n$ in naj velja $1 = d_{1} < d_{2} < \\ldots < d_{k} = n$. Potem je $d_{i} \\cdot d_{k+1-i} = n$, zato je\n$$\n\\frac{1}{d_{1}} + \\frac{1}{d_{2}} + \\ldots + \\frac{1}{d_{k}} = \\frac{1}{d_{1}} + \\frac{1}{d_{2}} + \\ldots + \\frac{1}{d_{k}} = \\frac{d_{k}}{n} + \\frac{d_{k-1}}{n} + \\ldots + \\frac{d_{1}}{n} = \\frac{d_{k} + d_{k-1} + \\ldots + d_{1}}{n} = \\frac{n + n}{n} = 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41179,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa vsaki ploskvi kocke je napisano naravno število, v vsakem oglišču pa je napisan zmnožek števil na 3 ploskvah, ki se stikajo v tem oglišču. Vsota števil v ogliščih kocke je 70. Kolikšna je vsota števil na ploskvah kocke?",
    "options": [],
    "answer": "14",
    "solution": "Solution:\n\nOznačimo števila na ploskvah kocke z $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$, $a_{5}$ in $a_{6}$. V ogliščih kocke so zapisana števila $a_{1} a_{2} a_{5}$, $a_{2} a_{3} a_{5}$, $a_{3} a_{4} a_{5}$, $a_{4} a_{1} a_{5}$, $a_{1} a_{2} a_{6}$, $a_{2} a_{3} a_{6}$, $a_{3} a_{4} a_{6}$ in $a_{4} a_{1} a_{6}$, zato je\n\n$$\n\\begin{aligned}\n70 & = a_{1} a_{2} a_{5} + a_{2} a_{3} a_{5} + a_{3} a_{4} a_{5} + a_{4} a_{1} a_{5} + a_{1} a_{2} a_{6} + a_{2} a_{3} a_{6} + a_{3} a_{4} a_{6} + a_{4} a_{1} a_{6} = \\\\\n& = (a_{5} + a_{6})(a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + a_{4} a_{1}) = \\\\\n& = (a_{1} + a_{3})(a_{2} + a_{4})(a_{5} + a_{6})\n\\end{aligned}\n$$\n\nKer je $70 = 2 \\cdot 5 \\cdot 7$ in so vsi 3 faktorji večji od 1, je eden izmed faktorjev enak 2, eden je enak 5 in eden je enak 7, njihova vsota pa je enaka $a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} = 2 + 5 + 7 = 14$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41180,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je ostrokotni trikotnik $ABC$. Naj bo $C'$ nožišče višine na $AB$, $D$ in $E$ pa različni točki na daljici $CC'$. Naj bosta $F$ in $G$ pravokotni projekciji točke $D$ na stranici $AC$ oziroma $BC$. Dokaži, da je trikotnik $ABC$ enakokrak, če je štirikotnik $DGEF$ paralelogram.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n1. način\nOznačimo s $F'$ presečišče premice $FE$ s stranico $BC$ ter z $G'$ presečišče premice $GE$ s stranico $AC$. Naj bo $\\angle BAC = \\alpha$. Zaradi tetivnosti štirikotnika $FDGC$ je $\\angle CGF = \\angle CDF = \\alpha$. Predpostavimo, da je $DGEF$ paralelogram. Ker je $DF \\parallel GG'$, je tudi $\\angle CEG' = \\alpha$, zaradi tetivnosti štirikotnika $G'EF'C$ pa je $\\angle CEG' = \\angle CF'G' = \\alpha$. Nazadnje upoštevamo še, da je zaradi pravih kotov pri $F'$ in $G'$ tudi štirikotnik $FGF'G'$ tetiven in izpeljemo\n![](attached_image_1.png)\n$\\angle GFG' = \\angle CF'G' = \\alpha$. Torej smo dokazali, da je $\\angle CGF = \\angle GFC = \\alpha$. Povsem analogno lahko sklepamo, da je $\\angle CGF = \\angle GFC = \\beta$, kjer smo označili $\\beta = \\angle CBA$. Torej res velja $\\alpha = \\beta$ in je trikotnik $ABC$ enakokrak.\n\n\n2. način\nOznačimo $F'$ in $G'$ kot v 1. načinu in naj bo $DGEF$ paralelogram. Ker je $GG' \\perp FC$ in $FF' \\perp BC$, je $E$ višinska točka trikotnika $FGC$ in zato je $CC' \\perp FG$. Ker se diagonali v paralelogramu razpolavljata, je trikotnik $FGC$ enakokrak. Ker pa velja $FG \\parallel AB$, je tudi trikotnik $ABC$ enakokrak.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41181,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $ABCD$ tetivni štirikotnik, pri katerem si nobeni 2 nasprotni stranici nista vzporedni. Presečišče premic $AB$ in $CD$ označimo z $E$, presečišče premic $AD$ in $BC$ pa s $F$. Simetrala kota $\\angle AFB$ seka daljico $AB$ v točki $P$, daljico $CD$ pa v točki $R$. Simetrala kota $\\angle BEC$ seka daljico $BC$ v točki $Q$, daljico $AD$ pa v točki $S$. Dokaži, da je štirikotnik $PQRS$ romb.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nZaradi simetrije lahko predpostavimo, da leži točka $A$ med $E$ in $B$, točka $C$ pa med $F$ in $B$. Presečišče obeh simetral označimo s $T$, notranje kote štirikotnika $ABCD$ pa z $\\alpha, \\beta, \\gamma$ in $\\delta$. Potem je\n$$\n\\begin{aligned}\n\\angle FTE &= \\angle TFQ + \\angle FQT = \\\\\n&= \\frac{1}{2} \\angle AFB + \\left(\\frac{1}{2} \\angle BEC + \\angle FBE\\right) = \\\\\n&= \\frac{1}{2}(\\pi - \\alpha - \\beta) + \\frac{1}{2}(\\pi - \\beta - \\gamma) + \\beta = \\\\\n&= \\pi - \\frac{1}{2}(\\alpha + \\gamma) = \\frac{1}{2} \\pi\n\\end{aligned}\n$$\n\n![](attached_image_1.png)\n\nsaj je $\\alpha + \\gamma = \\pi$. Torej je $ET \\perp FT$. Ker je simetrala kota $\\angle PER$ hkrati tudi višina na stranico $PR$, je $PER$ enakokrak trikotnik z osnovnico $PR$, točka $T$ pa njeno razpolovišče. Podobno je točka $T$ razpolovišče osnovnice $SQ$ enakokrakega trikotnika $SQF$. Torej je štirikotnik $PQRS$ romb, saj ima diagonali, ki se pravokotno razpolavljata.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41182,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bosta $E$ in $F$ razpolovišči stranic $AD$ in $DC$ pravokotnika $ABCD$. Označimo z $G$ presečišče daljic $AF$ in $EC$. Dokaži, da je $\\Varangle CGF = \\Varangle FBE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nKer je $\\Varangle FBA = \\Varangle BAF = \\Varangle DFA$ in $\\Varangle EBA = \\Varangle DCE$, je\n$$\n\\begin{aligned}\n\\Varangle CGF & = \\pi - \\Varangle FCG - \\Varangle GFC = \\\\\n& = \\pi - \\Varangle DCE - (\\pi - \\Varangle DFG) = \\\\\n& = -\\Varangle DCE + \\Varangle DFA = \\\\\n& = \\Varangle FBA - \\Varangle EBA \\\\\n& = \\Varangle FBE.\n\\end{aligned}\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41183,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa krožnici $k$ s središčem v točki $O$ izberemo točki $A$ in $B$ tako, da izbrani točki nista krajišči premera. Na trikotniku $OAB$ očrtani krožnici izberemo točko $C$, ki ne sovpada niti z $A$ niti z $B$. Premica $AC$ seka krožnico $k$ v točkah $A$ in $D$. Dokaži, da je trikotnik $DCB$ enakokrak.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nČe točka $C$ leži na istem loku nad $AB$ kot točka $O$, je $\\angle AOB = \\angle ACB = \\angle CDB + \\angle DBC$. Ker je $\\angle CDB = \\frac{1}{2} \\angle AOB$, je tudi $\\angle DBC = \\frac{1}{2} \\angle AOB$. Torej je $DCB$ enakokrak trikotnik z vrhom $C$.\n\n![](attached_image_1.png)\n\nČe pa točki $C$ in $O$ ne ležita na istem loku nad $AB$, je $\\angle BDA = \\pi - \\frac{1}{2} \\angle AOB = \\angle DBC + \\angle BCD$. Zaradi $\\angle BCD = \\pi - \\angle AOB$ je $\\angle DBC = \\frac{1}{2} \\angle AOB$. Ker je $\\angle CDB = \\pi - \\angle BDA = \\frac{1}{2} \\angle AOB$, je tudi v tem primeru $DCB$ enakokrak trikotnik z vrhom $C$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41184,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV trikotniku je $\\beta=74^{\\circ} 18^{\\prime}$ in $\\gamma=38^{\\circ} 46^{\\prime}$ ter $|AC|-|AB|=2,5~\\mathrm{cm}$. Izračunaj dolžini stranic $|AB|$ in $|AC|$ ter rezultat zaokroži na dve mesti natančno. Nariši skico.",
    "options": [],
    "answer": "AB = 4.6 cm, AC = 7.1 cm",
    "solution": "Solution:\n\nZvezo $c = b - 2,5$ vstavimo v obrazec za sinusni izrek.\n\nIzrazimo $b = -\\frac{2,5 \\sin \\beta}{\\sin \\gamma - \\sin \\beta}$ in izračunamo $b = 7,1~\\mathrm{cm}$ ter $c = 4,6~\\mathrm{cm}$.\n\nSkica.\n\nZapis ali uporaba sinusnega izreka.\n\nZveza $c = b - 2,5$ vstavljena v sinusni izrek.\n\nIzražen $b = -\\frac{2,5 \\sin \\beta}{\\sin \\gamma - \\sin \\beta}$.\n\nPravilen izračun $b = 7,1~\\mathrm{cm}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41185,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nČe zmnožek treh zaporednih naravnih števil $n-1$, $n$ in $n+1$ povečamo za srednje število, dobimo število med 3000 in 4000. Določi ta števila.",
    "options": [],
    "answer": "14, 15, 16",
    "solution": "Solution:\n\nNajprej ugotovimo, da je $(n-1) n (n+1) + n = n^{3}$.\n\nZapišemo neenačbo $3000 < n^{3} < 4000$.\n\nSklepamo, da je $\\sqrt[3]{3000} < n$, oziroma $14,42 < n$, in da je $n^{3} < 4000$, oziroma $n < 15,87$.\n\nTako je $n = 15$.\n\nIskana zaporedna naravna števila so $14$, $15$ in $16$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41186,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa racionalno funkcijo $f(x)=\\frac{a x+b}{c x+1}$ velja: $f(1)=\\frac{3}{4}$, $f(2)=1$ in $f(-1)=-\\frac{1}{2}$. Določi realne parametre $a, b$ in $c$ ter zapiši funkcijo $f(x)$. Zapis funkcije poenostavi.",
    "options": [],
    "answer": "a=2/3, b=1/3, c=1/3; f(x) = (2x+1)/(x+3)",
    "solution": "Solution:\n\nUpoštevamo zapisane pogoje in zapišemo enačbe $\\frac{a+b}{c+1}=\\frac{3}{4}$, $\\frac{2 a+b}{2 c+1}=1$ in $\\frac{-a+b}{-c+1}=-\\frac{1}{2}$. Odpravimo ulomke in rešimo sistem treh enačb s tremi neznankami. Dobimo rešitev $a=\\frac{2}{3}$, $b=c=\\frac{1}{3}$. Zapišemo funkcijo $f(x)=\\frac{\\frac{2}{3} x+\\frac{1}{3}}{\\frac{1}{3} x+1}$ in zapis poenostavimo $f(x)=\\frac{2 x+1}{x+3}$.\n\nNastavljene enačbe: $\\frac{a+b}{c+1}=\\frac{3}{4}$, $\\frac{2 a+b}{2 c+1}=1$, $\\frac{-a+b}{-c+1}=-\\frac{1}{2}$\n\nRešitev: $a=\\frac{2}{3}$, $b=c=\\frac{1}{3}$\n\nZapisana funkcija: $f(x)=\\frac{2 x+1}{x+3}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41187,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV telenoveli o dogodkih v zarotniškem mestecu nastopa $n$ meščanov, $n \\geq 4$. Vsaka skupina 3 meščanov kuje zaroto proti enemu izmed ostalih meščanov. Dokaži, da obstaja tak meščan, da je vsaj $\\sqrt[3]{(n-1)(n-2)}$ meščanov vpletenih v zaroto proti njemu.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nVseh (neurejenih) trojic meščanov je $\\frac{n(n-1)(n-2)}{6}$. Vsako trojico označimo s tistim izmed $n-3$ meščanov, proti kateremu trojica kuje zaroto. Potem obstaja $m$ trojic, ki so vse označene z istim meščanom, in je $m \\geq \\frac{(n-1)(n-2)}{6}$. Denimo, da v teh $m$ trojicah sodeluje $k$ meščanov. Tedaj velja $\\frac{k(k-1)(k-2)}{6} \\geq m$, torej je $k^{3}>k(k-1)(k-2) \\geq 6 m \\geq(n-1)(n-2)$ in zato $k^{3} \\geq(n-1)(n-2)$. Res je, da obstaja tak meščan, proti kateremu se je zarotilo $k \\geq \\sqrt[3]{(n-1)(n-2)}$ meščanov.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41188,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAna je izbrala števke $1,2,3,4,5,6,7$ in $9$. Odločila se je, da bo oblikovala skupine s po 4 dvomestnimi praštevili in da bo za vsako skupino praštevil uporabila vse izbrane števke. Kolikšna je vsota praštevil posamezne skupine?",
    "options": [],
    "answer": "190",
    "solution": "Solution:\n\nAna lahko z izbranimi števkami oblikuje skupino praštevil $\\{23,41,59,67\\}$. Da ne bi iskali vseh možnih skupin praštevil, razmišljajmo drugače. Dvomestno praštevilo se ne more končati z nobeno izmed števk $2,4,5$ oziroma $6$, zato te števke nastopajo na mestu desetic. Kakor koli Ana oblikuje skupino 4 praštevil, je njihova vsota enaka $10 \\cdot (2+4+5+6) + (1+3+7+9) = 170 + 20 = 190$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41189,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIgralca imata vsak po $2004$ žetonov. Izmenoma mečeta neobičajno igralno kocko, na kateri je napisanih prvih $6$ praštevil. Igralec, ki je na potezi, vrže kocko, drugi pa mu da toliko žetonov, kolikor je ostanek pri deljenju števila $2004$ s številom, ki je padlo pri metu kocke. Ali je mogoče, da bi imel eden izmed igralcev v nekem trenutku igre $7$-krat toliko žetonov kot drugi?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nČe bi imel eden izmed igralcev $7$-krat toliko žetonov kot drugi, bi jih imel eden $3507$, drugi pa $501$, kajti žetonov je ves čas igre $4008$. Prvih $6$ praštevil je $2$, $3$, $5$, $7$, $11$ in $13$, število $2004$ pa da pri deljenju s temi praštevili zaporedoma ostanke $0, 0, 4, 2, 2, 2$. Pri vsakem metu kocke se torej število žetonov vsakega igralca spremeni za sodo število. Ker ima na začetku vsak igralec sodo število žetonov, ni mogoče, da bi jih imel eden izmed njiju v nekem trenutku igre $7$-krat toliko kot drugi, saj bi to pomenilo, da bi tedaj vsak igralec imel liho število žetonov.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41190,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOsnovna ploskev pokončne prizme je deltoid, ki ima krajšo diagonalo dolgo $e$. Notranja kota deltoida z vrhoma v krajiščih daljše diagonale merita $90^{\\circ}$ in $60^{\\circ}$. Višina prizme je enaka daljši diagonali deltoida. Izrazi prost notino prizme z $e$. Rezultat naj bo točen.",
    "options": [],
    "answer": "V = \\frac{e^3(2+\\sqrt{3})}{4}",
    "solution": "Solution:\n\nProstornina prizme je $V = \\frac{e \\cdot f}{2} \\cdot f$. Daljša diagonala je razdeljena na dela $f_1$ in $f_2$. Oba dela izrazimo z dolžino $e$ krajše diagonale. Pravokotni trikotnik $DBC$ je enakokrak, zato je $f_2 = \\frac{e}{2}$. Trikotnik $BDA$ je enakostraničen, zato je $f_1 = \\frac{e \\sqrt{3}}{2}$, saj je to višina enakostraničnega trikotnika. Torej je dolžina diagonale $f$ enaka $\\frac{e \\sqrt{3}}{2} + \\frac{e}{2} = \\frac{e}{2}(\\sqrt{3} + 1)$. Prostornina prizme je\n\n![](attached_image_1.png)\n\n$\\frac{e \\cdot \\frac{e}{2}(\\sqrt{3} + 1)}{2} \\cdot \\frac{e}{2}(\\sqrt{3} + 1) = \\frac{e^3(2 + \\sqrt{3})}{4}$.\n\nIzražena prostornina $V = \\frac{e \\cdot f}{2} \\cdot f$.\n\nIzražena $f_1 = \\frac{e \\sqrt{3}}{2}$.\n\n$\\displaystyle f_2 = \\frac{e}{2}$\n\nIzražen $f = \\frac{e \\sqrt{3}}{2} + \\frac{e}{2} = \\frac{e}{2}(\\sqrt{3} + 1)$\n\nIzračun $V = \\frac{e \\cdot \\frac{e}{2}(\\sqrt{3} + 1)}{2} \\cdot \\frac{e}{2}(\\sqrt{3} + 1) = \\frac{e^3(3 + 2\\sqrt{3} + 1)}{8}$.\n\nRezultat $V = \\frac{e^3(2 + \\sqrt{3})}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41191,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV telenoveli o dogodkih $v$ zarotniškem mestecu nastopa $n$ meščanov, $n \\geq 3$. Vsaka 2 meščana skupaj kujeta zaroto proti enemu izmed ostalih meščanov. Dokaži, da obstaja tak meščan, da je vsaj $\\sqrt{n}$ meščanov vpletenih v zaroto proti njemu.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nVseh (neurejenih) parov meščanov je $\\frac{n(n-1)}{2}$. Vsak par označimo s tistim izmed $n-2$ meščanov, proti kateremu par kuje zaroto. Potem obstaja $m$ parov, ki so vsi označeni z istim meščanom, in je $m \\geq \\frac{n-1}{2}$. Denimo, da v teh $m$ parih sodeluje $k$ meščanov. Tedaj velja $\\frac{k(k-1)}{2} \\geq m \\geq \\frac{n-1}{2}$, torej je $k^{2}>k(k-1) \\geq n-1$. Res je, da obstaja tak meščan, proti kateremu se je zarotilo $k \\geq \\sqrt{n}$ meščanov.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41192,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je pravokotnik $ABCD$ z oglišči $A(-2,-1)$, $B(1,-1)$, $C(1,3)$, $D(-2,3)$. Izračunaj koordinati središča $S$ in polmer $R$ pravokotniku očrtane krožnice. Nariši sliko.",
    "options": [],
    "answer": "S = (-1/2, 1), R = 5/2",
    "solution": "Solution:\n\nSredišče $S$ pravokotniku očrtane krožnice je hkrati razpolovišče diagonale $AC$, zato je $S\\left(\\frac{x_{A}+x_{C}}{2}, \\frac{y_{A}+y_{C}}{2}\\right)$ oziroma $S\\left(-\\frac{1}{2}, 1\\right)$. S skice je razvidno, da je polmer $R$ enak polovici dolžine diagonale $AC$. Ker je $|AC|=\\sqrt{(x_{C}-x_{A})^{2}+(y_{C}-y_{A})^{2}}=\\sqrt{9+16}=5$, je $R=\\frac{5}{2}$.\n\n![](attached_image_1.png)\n\nSkica\n\nIzračunani koordinati središča: $S\\left(-\\frac{1}{2}, 1\\right)$\n\nSklep $R=\\frac{1}{2} \\cdot d(A, C)$ oziroma $R=\\frac{1}{2} \\cdot|AC|$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41193,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nReši enačbo:\n$$\n\\log \\left(\\frac{1}{x}\\right) \\cdot \\log \\left(\\frac{4}{x}\\right) = \\frac{3}{4} \\cdot (\\log 4)^2\n$$",
    "options": [],
    "answer": "8, 1/2",
    "solution": "Solution:\n\nNajprej uporabimo pravilo za logaritmiranje količnika:\n$$(\\log 1 - \\log x) \\cdot (\\log 4 - \\log x) = \\frac{3}{4}(\\log 4)^2.$$\nKer je $\\log 1 = 0$, dobimo:\n$$(-\\log x) \\cdot (\\log 4 - \\log x) = \\frac{3}{4}(\\log 4)^2.$$\nRazpremo oklepaj:\n$$-\\log x \\cdot \\log 4 + (\\log x)^2 = \\frac{3}{4}(\\log 4)^2.$$\nPreuredimo enačbo:\n$$(\\log x)^2 - \\log x \\cdot \\log 4 - \\frac{3}{4}(\\log 4)^2 = 0.$$\nTo je kvadratna enačba oblike $a^2 - ab - c = 0$ za $a = \\log x$.\n\nIzračunamo diskriminanto:\n$$D = (\\log 4)^2 - 4 \\cdot 1 \\cdot \\left(-\\frac{3}{4}(\\log 4)^2\\right) = (\\log 4)^2 + 3(\\log 4)^2 = 4(\\log 4)^2.$$\n\nRešitvi kvadratne enačbe sta:\n$$\\log x_{1,2} = \\frac{\\log 4 \\pm 2\\log 4}{2} = \\frac{1}{2}\\log 4 \\pm \\log 4.$$\nTorej:\n$$\\log x_1 = \\frac{1}{2}\\log 4 + \\log 4 = \\frac{3}{2}\\log 4 = \\log 8,$$\n$$\\log x_2 = \\frac{1}{2}\\log 4 - \\log 4 = -\\frac{1}{2}\\log 4 = \\log \\frac{1}{2}.$$\n\nIz teh rešitev dobimo:\n$$x_1 = 8,$$\n$$x_2 = \\frac{1}{2}.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41194,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nZa realna števila $x$, $y$ in $z$ velja $x y z = 1$. Izračunaj vrednost izraza\n$$\n\\frac{x+1}{x y + x + 1} + \\frac{y+1}{y z + y + 1} + \\frac{z+1}{z x + z + 1}\n$$",
    "options": [],
    "answer": "2",
    "solution": "Solution:\nRačunajmo:\n$$\n\\frac{x+1}{x y + x + 1} + \\frac{y+1}{y z + y + 1} + \\frac{z+1}{z x + z + 1} = \\frac{x+1}{x y + x + x y z} + \\frac{y+1}{y z + y + 1} + \\frac{z+1}{z x + z + 1}\n$$\n$$\n= \\frac{x+1 + x(y+1)}{x(y z + y + 1)} + \\frac{z+1}{z x + z + 1}\n$$\n$$\n= \\frac{x y + 2x + 1}{x y (z+1 + x z)} + \\frac{z+1}{z+1 + z x}\n$$\n$$\n= \\frac{x y + 2x + 1 + z x y + x y}{x y (z+1 + x z)}\n$$\n$$\n= \\frac{2(x y + x^2 y z + x y z)}{x y (z+1 + x z)}\n$$\n$$\n= \\frac{2 x y (1 + x z + z)}{x y (z+1 + x z)} = 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41195,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIzračunaj vsoto $[\\log_{2} 1]+[\\log_{2} 2]+[\\log_{2} 3]+\\cdots+[\\log_{2} 256]$. (Izraz $[x]$ pomeni največje celo število, ki ni večje od $x$.)",
    "options": [],
    "answer": "1546",
    "solution": "Solution:\nVemo, da je\n$$\n[\\log_{2} n]= \\begin{cases}0, & n=1 \\\\ 1, & 2 \\leq n<2^{2} \\\\ 2, & 2^{2} \\leq n<2^{3} \\\\ \\vdots \\\\ 7, & 2^{7} \\leq n<2^{8} \\\\ 8, & n=2^{8}=256\\end{cases}\n$$\nZato je iskana vsota enaka\n$$\n\\begin{aligned}\n& 1 \\cdot (2^{2}-2)+2 \\cdot (2^{3}-2^{2})+\\cdots+7 \\cdot (2^{8}-2^{7})+8= \\\\\n= & 8+7 \\cdot 2^{8}-(2+2^{2}+\\cdots+2^{6}+2^{7})= \\\\\n= & 8+7 \\cdot 256-(2+4+8+16+32+64+128)= \\\\\n= & 8+1792-254= \\\\\n= & 1546\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41196,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $[x]$ največje celo število, ki ni večje od $x$, in naj bo $\\{x\\}=x-[x]$. Poišči vsa pozitivna realna števila $x$, za katera velja\n$$\n20\\{x\\}+0.5[x]=2005\n$$",
    "options": [],
    "answer": "All solutions are x = 4010 − a + a/40 for integers a = 0, 1, …, 39.",
    "solution": "Solution:\n\nPomnožimo enačbo z $2$. Tedaj jo lahko zapišemo kot $40\\{x\\}=4010-[x]=a$, kjer je $a$ neko celo število. Zato je $\\{x\\}=\\frac{a}{40}$ za $0 \\leq a \\leq 39$. Velja $[x]=4010-a$, zato imamo $40$ rešitev oblike $x=4010-a+\\frac{a}{40}$, kjer je $a=0,1, \\ldots, 39$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41197,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa realna števila $x$ in $y$, za katera je\n$$\nx^{3}-y^{3}=7(x-y) \\quad \\text{ in } \\quad x^{3}+y^{3}=5(x+y)\n$$",
    "options": [],
    "answer": "(0,0), (sqrt(5), sqrt(5)), (-sqrt(5), -sqrt(5)), (sqrt(7), -sqrt(7)), (-sqrt(7), sqrt(7)), (sqrt(2)+1, sqrt(2)-1), (sqrt(2)-1, sqrt(2)+1), (-sqrt(2)-1, -sqrt(2)+1), (-sqrt(2)+1, -sqrt(2)-1)",
    "solution": "Solution:\n\nEnačbi preoblikujemo v $(x-y)\\left(x^{2}+x y+y^{2}\\right)=7(x-y)$ in $(x+y)\\left(x^{2}-x y+y^{2}\\right)=5(x+y)$. Če je $x=y$, iz druge enačbe dobimo $2 x^{3}=10 x$, torej $x=0$ ali $x= \\pm \\sqrt{5}$, v nasprotnem primeru pa lahko prvo enačbo delimo z $x-y$ in dobimo\n$$\nx^{2}+x y+y^{2}=7\n$$\nČe je $x=-y$, iz prve enačbe dobimo $2 x^{3}=14 x$, torej je $x=0$ ali $x= \\pm \\sqrt{7}$, v nasprotnem primeru pa lahko drugo enačbo delimo z $x+y$ in dobimo\n$$\nx^{2}-x y+y^{2}=5\n$$\nVsota tako dobljenih enačb nam da $x^{2}+y^{2}=6$, razlika pa $x y=1$, torej je $(x+y)^{2}=$ $x^{2}+2 x y+y^{2}=8$ in zato $x+y= \\pm \\sqrt{8}$. Upoštevamo, da je $x=\\frac{1}{y}$ in dobimo kvadratno enačbo za $y$, se pravi $y^{2} \\mp \\sqrt{8} y+1=0$, od koder sledi $y= \\pm(\\sqrt{2}+1)$ ali $y= \\pm(\\sqrt{2}-1)$. V prvem primeru je potem $x= \\pm(\\sqrt{2}-1)$, v drugem pa $x= \\pm(\\sqrt{2}+1)$. Vse rešitve $(x, y)$ so tako pari $(0,0),(\\sqrt{5}, \\sqrt{5}),(-\\sqrt{5},-\\sqrt{5}),(\\sqrt{7},-\\sqrt{7}),(-\\sqrt{7}, \\sqrt{7}),(\\sqrt{2}+1, \\sqrt{2}-1),(-\\sqrt{2}-1,-\\sqrt{2}+1)$, $(\\sqrt{2}-1, \\sqrt{2}+1)$ in $(-\\sqrt{2}+1,-\\sqrt{2}-1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41198,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa praštevila $p$, za katera ima število $p^{2}+11$ manj kot 11 pozitivnih deliteljev.",
    "options": [],
    "answer": "2, 3, 5",
    "solution": "Solution:\n\nČe je $p=2$, je $p^{2}+11=15=3 \\cdot 5$ in to število ima 4 delitelje (1, 3, 5 in 15).\n\nČe je $p=3$, je $p^{2}+11=17$ in ima 2 delitelja.\n\nZa $p>3$ je praštevilo $p$ liho, zato da $p^{2}$ pri deljenju s 4 ostanek 1 in je število $p^{2}+11$ deljivo s 4.\n\nPraštevilo $p>3$ ni deljivo s 3, torej da število $p^{2}$ pri deljenju s 3 ostanek 1, zato je število $p^{2}+11$ deljivo tudi s 3.\n\nTako je $p^{2}+11=2^{2} \\cdot 3 \\cdot a$.\n\nČe je $p \\geq 11$, je $p^{2}+11 \\geq 132$ in je $a>11$, zato so delitelji števila $p^{2}+11$ vsaj števila $1,2,3,4,6,12, a, 2a, 3a, 4a, 6a$ in $12a$, torej jih je vsaj 11 različnih (ker je $a>11$, je namreč lahko $a=12$).\n\nPosebej moramo izračunati še za $p=5$, ko dobimo $p^{2}+11=36=2^{2} \\cdot 3^{2}$, ki ima 6 deliteljev, in za $p=7$, ko dobimo $p^{2}+11=60=2^{2} \\cdot 3 \\cdot 5$, ki ima 12 deliteljev.\n\nLe za praštevila 2, 3 in 5 ima število $p^{2}+11$ manj kot 11 pozitivnih deliteljev.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41199,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa katera praštevila $p$ in $q$ je število $(p+1)^{q}$ popolni kvadrat?",
    "options": [],
    "answer": "All prime pairs (p, q) with q = 2 (any prime p), or p = 3 (any prime q).",
    "solution": "Solution:\n\nČe je $q=2$, je $(p+1)^{2}$ popolni kvadrat za vsako praštevilo $p$.\n\nSicer pa je $q$ liho, torej je $q=2k+1$ za neko naravno število $k$. Tedaj je $(p+1)^{q} = (p+1)^{2k+1} = (p+1)(p+1)^{2k}$, zato mora biti število $(p+1)$ popolni kvadrat, torej $p+1 = n^{2}$, od koder sledi $p = (n-1)(n+1)$. Ker je $p$ praštevilo, je to možno le, če je $n=2$, tedaj je $p=3$.\n\nTorej je $4^{q}$ popolni kvadrat za vsako praštevilo $q$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41200,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDano je geometrijsko zaporedje $\\left(a_{n}\\right)$ s pozitivnimi členi. Označimo\n$$\nS_{n}=\\log a_{1}+\\log a_{2}+\\ldots+\\log a_{n-1}+\\log a_{n}\n$$\nDokaži: če za nek $m \\neq n$ velja $S_{n}=S_{m}$, je $S_{n+m}=0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nČlene geometrijskega zaporedja lahko zapišemo v obliki $a_{n}=a_{1} q^{n-1}$. Ker so členi pozitivni, je $a_{1}>0$ in $q>0$. Potem je\n$$\nS_{n}=\\log a_{1}+\\log a_{1} q+\\ldots+\\log a_{1} q^{n-1}=\\log a_{1}^{n} q^{\\frac{n(n-1)}{2}}\n$$\nČe je $S_{n}=S_{m}$, sledi\n$$\na_{1}^{n} q^{\\frac{n(n-1)}{2}}=a_{1}^{m} q^{\\frac{m(m-1)}{2}}\n$$\ntorej dobimo\n$$\na_{1}^{n-m}=q^{\\frac{(m-n)(m+n-1)}{2}}\n$$\nin ker je $m \\neq n$ dobimo $a_{1}=q^{\\frac{1-m-n}{2}}$. Sledi\n$$\nS_{n+m}=\\log a_{1}^{m+n} q^{\\frac{(m+n)(m+n-1)}{2}}=\\log q^{\\frac{(m+n)(1-m-n)}{2}} q^{\\frac{(m+n)(m+n-1)}{2}}=\\log q^{0}=0",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41201,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči najmanjše praštevilo $p$, za katerega ima število $p^{3}+2 p^{2}+p$ natanko 42 pozitivnih deliteljev.",
    "options": [],
    "answer": "23",
    "solution": "Solution:\n\nNajprej zapišemo $p^{3}+2 p^{2}+p = p(p+1)^{2}$. Ker $p$ in $p+1$ nimata skupnih deliteljev (razen 1), je vsak delitelj števila $p(p+1)^{2}$ enak bodisi 1-krat neki delitelj števila $(p+1)^{2}$ bodisi $p$-krat ta delitelj. Ker ima število $p(p+1)^{2}$ natanko 42 deliteljev, ima $(p+1)^{2}$ natanko 21 deliteljev. Iz praštevilskega razcepa\n$$\n(p+1)^{2} = \\left(p_{1}^{\\alpha_{1}} \\cdot p_{2}^{\\alpha_{2}} \\cdots p_{k}^{\\alpha_{k}}\\right)^{2} = p_{1}^{2 \\alpha_{1}} \\cdot p_{2}^{2 \\alpha_{2}} \\cdots p_{k}^{2 \\alpha_{k}}\n$$\nugotovimo, da ima $(p+1)^{2}$ ravno $\\left(2 \\alpha_{1}+1\\right)\\left(2 \\alpha_{2}+1\\right) \\cdots\\left(2 \\alpha_{k}+1\\right)$ deliteljev. To pomeni, da je $2 \\alpha_{1}+1=3$ in $2 \\alpha_{2}+1=7$ ter $\\alpha_{i}=0$ za $i>2$. Velja torej $(p+1)^{2}=p_{1}^{2} \\cdot p_{2}^{6}$ oziroma $p+1=p_{1} \\cdot p_{2}^{3}$. Odtod je $p=p_{1} \\cdot p_{2}^{3}-1$. Najmanj dobimo, če izberemo $p_{1}=3$ in $p_{2}=2$. Tedaj je namreč $p=3 \\cdot 8-1=23$, ki je res praštevilo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41202,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je enakokrak trikotnik $A B C$ z vrhom $A$. Naj bo $D$ razpolovišče stranice $A C$ in $E$ pravokotna projekcija točke $D$ na stranico $B C$. Označimo z $F$ razpolovišče daljice $D E$. Dokaži, da se daljici $B F$ in $A E$ sekata pravokotno natanko tedaj, ko je trikotnik $A B C$ enakostraničen.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPredpostavimo najprej, da se daljici $A E$ in $B F$ sekata pravokotno. Naj bo $A'$ nožišče višine na stranico $B C$ ter $G$ presečišče daljic $A E$ in $B F$. Potem je $A G A' B$ tetiven štirikotnik in zato je $\\angle A' A E = \\angle A' A G = \\angle A' B G = \\angle E B F$. Tako sta trikotnika $B E F$ in $A A' E$ podobna. Ker je $F$ razpolovišče daljice $D E$ in $E$ razpolovišče daljice $A' C$, sta tudi trikotnika $B E D$ in $A A' C$ podobna. Torej imamo $\\angle G B D = \\angle F B D = \\angle E A C = \\angle G A D$. Štirikotnik $A D G B$ je tetiven in je $\\angle A D B = \\angle A G B = \\frac{\\pi}{2}$. Ker je $D$ razpolovišče stranice $A C$ in $B D \\perp A C$, je $|A B| = |C B|$ in trikotnik $A B C$ je enakostraničen.\n\n![](attached_image_1.png)\n\nObratno, če je trikotnik $A B C$ enakostraničen, sta trikotnika $A A' C$ in $B E D$ podobna. Daljica $A E$ je težiščnica prvega, daljica $B F$ pa drugega. Torej sta tudi trikotnika $B F E$ in $A E A'$ podobna. Tako je $\\angle A' B G = \\angle E B F = \\angle A' A E = \\angle A' A G$, zato je $B A' G A$ tetiven štirikotnik in je $\\angle B G A = \\frac{\\pi}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41203,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTangenti iz točke $P$ na krožnico $k$ se krožnice dotikata v točkah $A$ in $B$. Naj bo $X$ poljubna točka na krajšem loku $\\widehat{A B}$. Označimo s $C$ pravokotno projekcijo točke $P$ na premico $A X$ in z $D$ pravokotno projekcijo točke $P$ na premico $B X$. Dokaži, da premica $C D$ poteka skozi neko točko $Y$, ki je neodvisna od izbire točke $X$ na loku $\\widehat{A B}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNaj bo $O$ središče krožnice $k$, $A'$ razpolovišče daljice $P A$, $B'$ razpolovišče daljice $P B$ in $Y$ razpolovišče $A' B'$. Označimo še $\\angle A P B = \\varphi$ in $\\angle X A P = \\alpha$. Potem je $\\angle O A X = \\frac{\\pi}{2} - \\alpha$, $\\angle B O A = \\pi - \\varphi$, $\\angle A X B = \\frac{\\pi \\mp \\varphi}{2}$, $\\angle X B O = \\alpha + \\frac{\\varphi}{2}$, $\\angle P B X = \\frac{\\pi}{2} - \\frac{\\varphi}{2} - \\alpha$, $\\angle A A' C = \\pi - 2\\alpha$ in $\\angle D B' B = \\varphi + 2\\alpha$.\n\nTorej premici $C A'$ in $D B'$ oklepata s premico $A P$ kot $2\\alpha$ in ker leži točka $C$ na istem bregu premice $A P$ kot točka $B$, točka $D$ pa na istem bregu premice $B P$ kot točka $A$, sta premici $C A'$ in $D B'$ vzporedni, točki $C$ in $D$ pa ležita na različnih bregovih premice $A' B'$ (razen v primeru $C D \\parallel A' B'$, kjer pa $Y$ očitno leži na $C D$).\n\n![](attached_image_1.png)\n\nKer pa po Talesovem izreku točka $C$ leži na krožnici s središčem $A'$ in polmerom $|A' A|$, točka $D$ pa na krožnici s središčem $B'$ in polmerom $|B' B|$, je $|B' D| = |B' B| = |A' A| = |A' C|$, zato je $C B' D A'$ paralelogram. Ker se diagonali paralelograma razpolavljata, premica $C D$ poteka skozi razpolovišče diagonale $A' B'$, torej skozi točko $Y$, ki pa je neodvisna od izbire točke $X$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41204,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $T$ poljubna točka znotraj kvadrata $A B C D$, točke $A^{\\prime}, B^{\\prime}, C^{\\prime}$ in $D^{\\prime}$ pa druga presečišča premic $T A, T B, T C$ in $T D$ z očrtano krožnico kvadrata $A B C D$. Dokaži, da je $|A^{\\prime} B^{\\prime}| \\cdot |C^{\\prime} D^{\\prime}| = |A^{\\prime} D^{\\prime}| \\cdot |B^{\\prime} C^{\\prime}|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nZaradi tetivnosti imamo mnogo enakih obodnih kotov in zato tudi mnogo parov podobnih trikotnikov: $\\triangle A B T \\sim \\triangle B^{\\prime} A^{\\prime} T$, $\\triangle B C T \\sim \\triangle C^{\\prime} B^{\\prime} T$, $\\triangle C D T \\sim \\triangle D^{\\prime} C^{\\prime} T$ in $\\triangle D A T \\sim \\triangle A^{\\prime} D^{\\prime} T$. Iz podobnosti sledi\n\n$\\frac{|A B|}{|A^{\\prime} B^{\\prime}|} = \\frac{|B T|}{|A^{\\prime} T|}$,\n\n$\\frac{|B C|}{|B^{\\prime} C^{\\prime}|} = \\frac{|B T|}{|C^{\\prime} T|}$,\n\n$\\frac{|C D|}{|C^{\\prime} D^{\\prime}|} = \\frac{|D T|}{|C^{\\prime} T|}$,\n\n$\\frac{|D A|}{|D^{\\prime} A^{\\prime}|} = \\frac{|D T|}{|A^{\\prime} T|}$,\n\nod koder izpeljemo\n\n$$\n\\frac{|A B|}{|A^{\\prime} B^{\\prime}|} \\cdot \\frac{|C D|}{|C^{\\prime} D^{\\prime}|} = \\frac{|B T|}{|A^{\\prime} T|} \\cdot \\frac{|D T|}{|C^{\\prime} T|} = \\frac{|B C|}{|B^{\\prime} C^{\\prime}|} \\cdot \\frac{|D A|}{|D^{\\prime} A^{\\prime}|}\n$$\n\n![](attached_image_1.png)\n\nKer je $|A B| = |C D| = |B C| = |D A|$, od tod sledi želena enakost.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41205,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV trikotniku $ABC$, katerega središče včrtane krožnice je $I$, velja\n$$\n|CA| + |AI| = |BC|\n$$\nPoišči razmerje med velikostima kotov $\\Varangle BAC$ in $\\Varangle CBA$.",
    "options": [],
    "answer": "1:2",
    "solution": "Solution:\n\nIzberimo tako točko $D$, $D \\notin AC$, na poltraku $CA$, da je $|AD| = |AI|$. Tedaj je $DBC$ enakokrak trikotnik z vrhom $C$ in ker $I$ leži na simetrali kota $\\Varangle DCB$, je $\\Varangle IDC = \\Varangle CBI$. Ker je $ADI$ enakokrak trikotnik z vrhom $A$, je $\\Varangle IAC = 2 \\Varangle IDC$. Velja $\\Varangle IAC = \\frac{1}{2} \\Varangle BAC$ in $\\Varangle CBI = \\frac{1}{2} \\Varangle CBA$, zato je $\\Varangle BAC = 2 \\Varangle CBA$ oz. $\\Varangle BAC : \\Varangle CBA = 1 : 2$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41206,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLukcu je bilo med uro matematike dolgčas, zato je najprej narisal krog in nato naokrog po obodu še $n$ praznih polj, kjer je $n \\geq 3$, ter vanje zapisal po 1 pozitivno število. Kasneje je ta števila pobrisal, v vsako polje pa zapisal kvadratni koren zmnožka dveh števil, ki sta prej ležali na temu polju sosednjih poljih. Pokaži, da obstaja polje, v katerem je zapisano število manjše ali enako tistemu, ki je bilo zapisano prej.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n1. način\nOznačimo z $a_{1}, a_{2}, \\ldots, a_{n}$ števila, ki so v poljih ležala na začetku, z $b_{1}, b_{2}, \\ldots, b_{n}$ pa tista, ki ležijo na koncu. Po neenakosti med aritmetično in geometrijsko sredino velja\n$$\na_{1}+a_{2}+\\cdots+a_{n}=\\frac{a_{1}+a_{3}}{2}+\\frac{a_{2}+a_{4}}{2}+\\cdots+\\frac{a_{n}+a_{2}}{2} \\geq \\sqrt{a_{1} a_{3}}+\\sqrt{a_{2} a_{4}}+\\cdots+\\sqrt{a_{n} a_{2}} = b_{2}+b_{3}+\\cdots+b_{n}+b_{1}.\n$$\nKer se je vsota vseh števil v poljih zmanjšala, mora obstajati število $k$, za katerega velja $a_{k} \\geq b_{k}$.\n\n\n2. način\nTrditev bomo dokazali s protislovjem. Števila na začetku naj bodo $a_{1}, a_{2}, \\ldots, a_{n}$, na koncu pa $\\sqrt{a_{1} a_{3}}, \\sqrt{a_{2} a_{4}}, \\ldots, \\sqrt{a_{n} a_{2}}$. Privzemimo torej, da je $a_{1}<\\sqrt{a_{1} a_{3}}, a_{2}<\\sqrt{a_{2} a_{4}}, \\ldots, a_{n}<\\sqrt{a_{n} a_{2}}$. Torej je tudi\n$$\na_{1} a_{2} \\cdots a_{n}<\\sqrt{a_{1} a_{3}} \\cdot \\sqrt{a_{2} a_{4}} \\cdots \\sqrt{a_{n} a_{2}}=\\sqrt{a_{1}^{2} a_{2}^{2} \\cdots a_{n}^{2}},\n$$\nkar pa seveda ne drži.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41207,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVaške obrekljivke si svoje čenče vsak dan izmenjujejo po telefonu, in sicer tako, da vsaki dve med seboj govorita natanko enkrat. Nekega dne je vsaka obrekljivka poklicala vsaj eno izmed ostalih. Dokaži, da obstajajo tri, za katere velja, da je prva poklicala drugo, druga tretjo in tretja prvo.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNaj bo $n$ obrekljivk. Potem je vsaka poklicala med $1$ in $n-1$ ostalih, zato po Dirichletovem načelu obstajata dve, ki sta poklicali enako število obrekljivk, recimo $k$. Brez izgube splošnosti lahko predpostavimo, da je prva poklicala drugo. Prva je poklicala še $k-1$ ostalih, druga pa $k$. Med preostalimi obrekljivkami obstaja tretja obrekljivka, ki jo je poklicala druga, prva pa ne. Tretja obrekljivka je zato poklicala prvo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41208,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAleš, Brane in Cvetka so pripravili veliko kartončkov, na katere so zapisali po eno izmed števil $2,3,4,5,6,7$ in $8$. Maja, ki se jim je kasneje pridružila, je vzela tri kartončke in vsakemu prilepila po enega na čelo. Aleš, Brane in Cvetka niso videli, katero število imajo na čelu, vsak izmed njih je videl le kartončka drugih dveh. Maja jim je povedala: \"Na čelih nimate različnih števil, zmnožek vseh treh števil je popolni kvadrat.\" Aleš, Brane in Cvetka so nato vsak zase ugotavljali, katero število imajo na čelu. Ali so to lahko vsi trije ugotovili?",
    "options": [],
    "answer": "Yes",
    "solution": "Solution:\n\nKer Aleš, Brane in Cvetka niso imeli različnih števil, sta bili vsaj dve številki enaki. Denimo, da so imeli števila $x, x$ in $y$. Zmnožek $x \\cdot x \\cdot y = x^{2} y$ je popolni kvadrat, zato je tudi $y$ popolni kvadrat. Med števili, ki so jih zapisali na kartončke, je bilo le število $4$ popolni kvadrat, zato je $y = 4$. Imamo dve možnosti: vsi trije so imeli na kartončku zapisano število $4$ ali pa je eden imel število $4$, druga dva pa neko drugo (a seveda isto) število.\n\nV prvem primeru je vsak izmed njih videl, da imata druga dva zapisano število $4$. Ker pa je vsak izmed njih sklepal, da lahko število $4$ nastopi enkrat ali trikrat, je lahko ugotovil, da ima tudi sam zapisano število $4$.\n\nV drugem primeru je eden izmed njih videl, da imata druga dva zapisano isto, od $4$ različno število, in je sklepal, da ima sam zapisano število $4$. Druga dva sta videla na kartončkih zapisani različni števili, in sicer $4$ in $x \\neq 4$, zato sta sklepala, da imata na svojem čelu zapisano število $x$, pri čemer je $x$ lahko $2,3,5,6,7$ ali $8$.\n\nVsi trije so lahko ugotovili, katero število imajo na čelu.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41209,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa nogometnem turnirju so sodelovale le ekipe iz Malega mesta in Velikega mesta. Iz Velikega mesta je bilo 9 ekip več kot iz Malega mesta. Vsaki ekipi sta se srečali natanko enkrat, pri čemer je zmagovalna ekipa dobila 1 točko, poražena 0 točk, neodločen izid pa ni bil možen. Ekipe iz Velikega mesta so osvojile 9 krat toliko točk kot ekipe iz Malega mesta. Določi največje možno število zmag najboljše ekipe iz Malega mesta.",
    "options": [],
    "answer": "11",
    "solution": "Solution:\n\nOznačimo število ekip iz Malega mesta z $x$. Potem je število ekip iz Velikega mesta enako $x+9$. Ekipe iz Malega mesta so med seboj igrale $\\frac{x(x-1)}{2}$ tekem in so zato osvojile $\\frac{x(x-1)}{2}+k$ točk, kjer je $k$ število zmag, osvojenih nad ekipami iz Velikega mesta. Ekipe iz Velikega mesta so tako osvojile\n$$\n\\frac{(x+8)(x+9)}{2}+x(x+9)-k\n$$\ntočk. Veljati mora\n$$\n9\\left(\\frac{(x-1) x}{2}+k\\right)=\\frac{(x+8)(x+9)}{2}+x(x+9)-k\n$$\ntorej $3 x^{2}-22 x+10 k-36=0$. Ker je $x$ naravno število, mora biti diskriminanta $4(229-30 k)$ popolni kvadrat. Potem je $k \\leq 7$ in zato je $k=2$ ali $k=6$. V prvem primeru je $x=8$ in najboljša ekipa iz Malega mesta je lahko osvojila $7+2=9$ točk. V drugem primeru pa je $x=6$ in najboljša ekipa iz Malega mesta lahko osvojila $5+6=11$ točk.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41210,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana je funkcija $f(x)=3^{x}$. Določi funkciji $g(x)=f(x+1)$ in $h(x)=4 f(x)-3$ ter izračunaj presečišče njunih grafov.",
    "options": [],
    "answer": "g(x) = 3^{x+1}, h(x) = 4·3^{x} − 3; intersection point: (1, 9)",
    "solution": "Solution:\n\nKer je $g(x)=f(x+1)$, dobimo $g(x)$ tako, da v $f(x)$ vstavimo $x+1$ namesto $x$. Tako dobimo $g(x)=3^{x+1}$.\n\nPodobno dobimo $h(x)=4 \\cdot 3^{x}-3$.\n\nPresečišča grafov funkcij $g(x)$ in $h(x)$ dobimo tako, da enačimo enačbi obeh funkcij. Dobimo enačbo\n$$\n3^{x+1}=4 \\cdot 3^{x}-3.\n$$\nEnačbo uredimo:\n$$\n3^{x+1}=3 \\cdot 3^{x}=4 \\cdot 3^{x}-3\n$$\n$$\n3 \\cdot 3^{x}=4 \\cdot 3^{x}-3\n$$\n$$\n3=3^{x}\n$$\n$$\nx=1\n$$\nNato izračunamo še $y$:\n$$\ng(1)=3^{1+1}=3^{2}=9\n$$\nTorej je presečišče $P(1,9)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41211,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n1. Dani sta enačbi $2 \\log_{3}(x+2) - \\log_{3} x = 2$ in $25^{x} = 0{,}008$.\n\na) Reši enačbi.\n\nb) Rešitve enačb so ničle polinoma $p$ tretje stopnje. Graf tega polinoma poteka tudi skozi točko $(2,7)$. Zapiši predpis polinoma $p$ v ničelni obliki.",
    "options": [],
    "answer": "Equation 1 solutions: x = 1 and x = 4. Equation 2 solution: x = -3/2. The polynomial is p(x) = -(x - 4)(x - 1)(x + 3/2).",
    "solution": "Solution:\n\na)\nLogaritemsko enačbo preoblikujemo:\n\n$$\n2 \\log_{3}(x+2) - \\log_{3} x = 2\n$$\n\nUporabimo lastnosti logaritmov:\n$$\n\\log_{3}(x+2)^2 - \\log_{3} x = 2\n$$\n$$\n\\log_{3} \\frac{(x+2)^2}{x} = 2\n$$\n$$\n\\frac{(x+2)^2}{x} = 3^2\n$$\n$$\n\\frac{x^2 + 4x + 4}{x} = 9\n$$\n$$\nx^2 + 4x + 4 = 9x\n$$\n$$\nx^2 - 5x + 4 = 0\n$$\nRazstavimo:\n$$\n(x-4)(x-1) = 0\n$$\nDobimo rešitvi:\n$$\nx_1 = 4, \\quad x_2 = 1\n$$\n\nEksponentno enačbo preuredimo:\n$$\n25^x = 0{,}008\n$$\n$$\n(5^2)^x = 8 \\cdot 10^{-3}\n$$\n$$\n5^{2x} = 8 \\cdot 10^{-3}\n$$\nKer $0{,}008 = 8 \\cdot 10^{-3} = 2^3 \\cdot 10^{-3}$, vendar je bolj smiselno zapisati $0{,}008 = 8/1000 = 1/125 = 5^{-3}$.\nTorej:\n$$\n5^{2x} = 5^{-3}\n$$\n$$\n2x = -3\n$$\n$$\nx = -\\frac{3}{2}\n$$\n\nb)\nIz vseh rešitev danih enačb zapišemo polinom:\n$$\np(x) = a(x-4)(x-1)\\left(x + \\frac{3}{2}\\right)\n$$\nUporabimo koordinate točke $(2,7)$ in izračunamo vodilni koeficient $a$:\n\n$$\np(2) = a(2-4)(2-1)\\left(2 + \\frac{3}{2}\\right) = 7\n$$\n$$\na(-2)(1)\\left(\\frac{7}{2}\\right) = 7\n$$\n$$\na \\cdot (-2) \\cdot 1 \\cdot \\frac{7}{2} = 7\n$$\n$$\na \\cdot (-7) = 7\n$$\n$$\na = -1\n$$\nTorej je polinom:\n$$\np(x) = -(x-4)(x-1)\\left(x + \\frac{3}{2}\\right)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41212,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana je funkcija $f(x)=m x-(3 m-20)$, kjer je $m$ rešitev enačbe\n$$\n\\frac{(\\sqrt[3]{2})^{6} \\cdot 2^{-6}}{0,5^{2} \\cdot 8^{-\\frac{2}{3}}} = m \\cdot \\frac{1}{(\\sqrt[3]{2})^{9}}\n$$\nNariši graf funkcije $f$.",
    "options": [],
    "answer": "y = 8x - 4",
    "solution": "Solution:\n\nNajprej rešimo enačbo\n$$\n\\frac{(\\sqrt[3]{2})^{6} \\cdot 2^{-6}}{0,5^{2} \\cdot 8^{-\\frac{2}{3}}} = m \\cdot \\frac{1}{(\\sqrt[3]{2})^{9}}\n$$\nLeva stran enačbe je enaka\n$$\n\\frac{2^{2} \\cdot 2^{-6}}{2^{-2} \\cdot 2^{-2}} = 1\n$$\ndesna stran pa $m \\cdot 2^{-3}$. Iz enakosti $1 = m \\cdot 2^{-3}$ izračunamo $m = 8$.\n\nVstavimo $m$ v predpis za funkcijo in dobimo enačbo premice\n$$\ny = 8x - 4\n$$\nNarišemo njen graf.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41213,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je trikotnik $ABC$ z oglišči $A(-1,2)$, $B(-2,-3)$ in $C(2,-1)$. Iz točke $A$ konstruiramo pravokotnico na stranico $BC$. Pravokotnica seka $BC$ v točki $E$. Natančno izračunaj dolžino daljice $AE$.",
    "options": [],
    "answer": "9√5/5",
    "solution": "Solution:\n\nUgotovimo, da je v danem trikotniku $v_{a} = d(A, E)$. Izračunamo ploščino trikotnika $ABC$: $S = 9$.\n\nDolžina stranice $a$ je enaka $d(B, C) = 2 \\sqrt{5}$.\n\nNato izrazimo $v_{a}$ iz ploščine:\n$$\nv_{a} = \\frac{2 \\cdot S}{a} = \\frac{9 \\sqrt{5}}{5}\n$$\n\nDolžina $v_{a} = \\frac{9 \\sqrt{5}}{5}$.\n\nUgotovitev, da je $d(A, E) = v_{a}$.\n\nUporaba ustreznega obrazca za ploščino:\n\n![](attached_image_1.png)\n\nIzračunana $a = d(B, C) = 2 \\sqrt{5}$.\n\nPravilno vstavljeni podatki v obrazec $S = \\frac{a v_{a}}{2}$.\n\nRezultat $v_{a} = \\frac{9 \\sqrt{5}}{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41214,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa krožnico s premerom $10~\\mathrm{cm}$ sta narisani obe tangenti iz točke $A$ izven krožnice. Točka $A$ je od središča krožnice oddaljena $1{,}3~\\mathrm{dm}$. Izračunaj razdaljo med dotikališčema tangent na krožnico in jo izrazi v centimetrih. Rezultat zaokroži na eno decimalno mesto natančno.",
    "options": [],
    "answer": "9.2 cm",
    "solution": "Solution:\n\nNarišemo skico: središče krožnice $S$, točka $A$ izven krožnice, dotikališči $B$ in $C$. Označimo kot $\\angle B S A$ z $\\alpha$, razdaljo od $S$ do $A$ z $x$ in razdaljo od $B$ do $C$ z $2y$. Ker je trikotnik $A S B$ pravokoten, uporabimo kotno funkcijo $\\cos \\alpha = \\frac{r}{x}$, iz česar izračunamo $\\alpha = 67,38^{\\circ}$. Tudi trikotnik $D S B$ je pravokoten ($D$ je presečišče daljic $S A$ in $B C$). Z uporabo kotne funkcije $\\sin \\alpha = \\frac{y}{r}$, izračunamo $y = r \\cdot \\sin \\alpha = 4,6$.\n\n![](attached_image_1.png)\n\nNato izračunamo razdaljo $2y = 9,2$.\n\nNalogo lahko rešimo drugače: ploščina pravokotnega trikotnika $A S C$ je enaka $\\frac{1}{2}|C S| \\cdot |C A| = \\frac{1}{2}|A S| \\cdot |C D|$, od koder izrazimo $|B C| = 2|C D| = \\frac{2 \\cdot |S C| \\cdot \\sqrt{|S A|^{2} - |S C|^{2}}}{|S A|} = \\frac{2 r \\sqrt{x^{2} - r^{2}}}{x} = \\frac{2 \\cdot 5 \\cdot 12}{13} \\approx 9,2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41215,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoenostavi izraz $\\left((1-x) \\sqrt{\\frac{1+x}{1-x}}+1-x\\right) \\cdot\\left(\\sqrt{\\frac{1+x}{1-x}}-1\\right)$, če je $-1 \\leq x<1$.",
    "options": [],
    "answer": "2x",
    "solution": "Solution:\n\nČe uporabimo v prvem oklepaju pravilo $a \\cdot \\sqrt{b}=\\sqrt{a^{2} b}$, ki velja za $a \\geq 0$, dobimo\n$$\n((1-x) \\sqrt{\\frac{1+x}{1-x}}+1-x) = \\left(\\sqrt{\\frac{(1-x)^{2}(1+x)}{1-x}}+1-x\\right)\n$$\nkrajšamo in dobimo\n$$\n(\\sqrt{(1-x)(1+x)}+1-x) \\left(\\sqrt{\\frac{1+x}{1-x}}-1\\right)\n$$\nNato odpravimo oklepaja:\n$$\n\\sqrt{\\frac{(1-x)(1+x)(1+x)}{1-x}} + \\sqrt{\\frac{(1-x)(1+x)}{1-x}} - \\sqrt{(1-x)(1+x)} - (1-x)\n$$\nzopet krajšamo ter dobimo\n$$\n\\sqrt{(1+x)^{2}} - 1 + x\n$$\nUpoštevamo pogoj $\\sqrt{a^{2}}=|a|, a \\in \\mathbb{R}$. Ker pa je $1+x \\geq 0$, imamo rešitev:\n$$\n1+x-1+x=2x\n$$\n\n1. način:\nUreditev izraza v prvem oklepaju\n$$\n\\left((1-x) \\sqrt{\\frac{1+x}{1-x}}+1-x\\right)=\\left(\\sqrt{\\frac{(1-x)^{2}(1+x)}{1-x}}+1-x\\right)\n$$\nin krajšanje.\n\nPravilna odprava oklepaja:\n$$\n\\sqrt{\\frac{(1-x)(1+x)(1+x)}{1-x}}+\\sqrt{\\frac{(1-x)(1+x)}{1-x}}-\\sqrt{(1-x)(1+x)}-(1-x)\n$$\nUreditev izraza do $\\sqrt{(1+x)^{2}}-1+x$.\nUporaba pogoja $|1+x|$.\n\nUtemeljitev, da druge rešitve ni, saj $1+x$ ne more biti negativen.\n\n2. način:\n\nNalogo lahko rešimo drugače: ker je\n$$\n\\left((1-x) \\sqrt{\\frac{1+x}{1-x}}+1-x\\right)=(1-x)\\left(\\sqrt{\\frac{1+x}{1-x}}+1\\right)\n$$\nlahko zapišemo\n$$\n\\left((1-x) \\sqrt{\\frac{1+x}{1-x}}+1-x\\right) \\cdot\\left(\\sqrt{\\frac{1+x}{1-x}}-1\\right)=(1-x)\\left(\\sqrt{\\frac{1+x}{1-x}}-1\\right)\n$$\n$=(1-x)\\left(\\left|\\frac{1+x}{1-x}\\right|-1\\right)=(1-x)\\left(\\frac{1+x}{1-x}-1\\right)=2x$\nkjer smo lahko opustili absolutno vrednost zaradi predpostavke $-1 \\leq x<1$.\n\nIzpostavljen skupni faktor $(1-x)\\left(\\sqrt{\\frac{1+x}{1-x}}+1\\right)\\left(\\sqrt{\\frac{1+x}{1-x}}-1\\right)$.\n\nPoenostavitev do $(1-x)\\left(\\sqrt{\\left(\\frac{1+x}{1-x}\\right)^{2}}-1\\right)$.\n\nUpoštevanje $\\sqrt{a^{2}}=|a|, a \\in \\mathbb{R}$.\n\nZapis pogoja in izračun:\n$$\n(1-x)\\left(\\frac{1+x}{1-x}-1\\right)=1+x-1+x=2x,\n$$\nker je $1+x \\geq 0$.\n\nUtemeljitev, da druge rešitve ni, saj $1+x$ ne more biti negativen.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41216,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nPoišči vsa realna števila $x$, ki zadoščajo enačbi $\\sqrt{x^{2}-2 x+17}=3 \\sqrt{3}-1$, in jih zapiši v obliki $x=m+n \\sqrt{3}$, kjer sta $m$ in $n$ celi števili.",
    "options": [],
    "answer": "x = 4 - sqrt(3) or x = -2 + sqrt(3)",
    "solution": "Solution:\n1. Ker sta obe strani enačbe pozitivni, lahko enačbo kvadriramo in dobimo\n$$\nx^{2}-2 x+17=27-6 \\sqrt{3}+1=28-6 \\sqrt{3}\n$$\nNato levo stran dopolnimo do popolnega kvadrata\n$$\n(x-1)^{2}+16=28-6 \\sqrt{3}\n$$\nin enačbo preuredimo do\n$$\n(x-1)^{2}=12-6 \\sqrt{3}\n$$\nOd tod po korenjenju dobimo $x-1= \\pm \\sqrt{12-6 \\sqrt{3}}$ oziroma $x=1 \\pm \\sqrt{12-6 \\sqrt{3}}$. Da bomo rešitvi lahko zapisali v predpisani obliki, preoblikujmo izraz pod korenom v popoln kvadrat\n$$\nx=1 \\pm \\sqrt{9-6 \\sqrt{3}+3}=1 \\pm \\sqrt{(3-\\sqrt{3})^{2}}=1 \\pm(3-\\sqrt{3}) \\text{.}\n$$\nRešitvi sta torej $x=1+(3-\\sqrt{3})=4-\\sqrt{3}$ in $x=1-(3-\\sqrt{3})=-2+\\sqrt{3}$. To sta tudi res rešitvi, ker je izraz pod korenom dane enačbe avtomatično pozitiven, saj je enak $28-6 \\sqrt{3}$.\n\n\n2. način. Ker sta obe strani enačbe pozitivni, lahko enačbo kvadriramo in dobimo\n$$\nx^{2}-2 x+17=27-6 \\sqrt{3}+1=28-6 \\sqrt{3}\n$$\nPišimo $x=m+n \\sqrt{3}$, kjer sta $m$ in $n$ celi števili. Ko slednje vstavimo v enačbo, dobimo\n$$\nm^{2}+2 m n \\sqrt{3}+3 n^{2}-2 m-2 n \\sqrt{3}+17=28-6 \\sqrt{3}\n$$\nEnačbo preoblikujemo do\n$$\n\\left(m^{2}-2 m+3 n^{2}+17\\right)-(2 n-2 m n) \\sqrt{3}=28-6 \\sqrt{3}\n$$\nKer sta $m$ in $n$ celi števili, od tod sledi\n$$\n\\begin{aligned}\nm^{2}-2 m+3 n^{2}+17 & =28 \\\\\n2 n-2 m n & =6\n\\end{aligned}\n$$\nDrugo enačbo preoblikujemo v $n(1-m)=3$. Ker sta $m$ in $n$ celi števili, so edine možne rešitve $n=1$ in $m=-2$, $n=-1$ in $m=4$, $n=3$ in $m=0$ ter $n=-3$ in $m=2$. Zlahka preverimo, da je le pri prvih dveh možnostih izpolnjena tudi prva enačba sistema. Torej dobimo rešitvi $x=-2+\\sqrt{3}$ in $x=4-\\sqrt{3}$. To sta tudi res rešitvi, ker je izraz pod korenom dane enačbe avtomatično pozitiven, saj je enak $28-6 \\sqrt{3}$. Ker ima vsaka kvadratna enačba največ dve realni rešitvi, sta dobljeni dve rešitvi hkrati edini rešitvi.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41217,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa katere kote $\\alpha$ z lastnostjo $0<\\alpha<2 \\pi$ velja neenakost\n$$\n\\frac{5 \\sin \\alpha-2}{\\sin \\alpha} \\geq 2 \\sin \\alpha ?\n$$",
    "options": [],
    "answer": "[π/6, 5π/6] ∪ (π, 2π)",
    "solution": "Solution:\n\n1. Očitno mora biti $\\alpha \\neq \\pi$, saj sicer ulomek na levi strani neenakosti ne bi bil dobro definiran. Obravnavajmo dva primera.\n\nČe je $0<\\alpha<\\pi$, tedaj je $\\sin \\alpha>0$. Torej če neenakost pomožimo s $\\sin \\alpha$, se neenačaj ohrani in dobimo $5 \\sin \\alpha-2 \\geq 2 \\sin ^{2} \\alpha$. Neenakost preuredimo do $2 \\sin ^{2} \\alpha-5 \\sin \\alpha+2 \\leq 0$ in levo stran razstavimo $(2 \\sin \\alpha-1)(\\sin \\alpha-2) \\leq 0$. Ker je $\\sin \\alpha-2<0$, mora biti $2 \\sin \\alpha-1 \\geq 0$ oziroma $\\sin \\alpha \\geq \\frac{1}{2}$. Od tod sledi $\\frac{\\pi}{6} \\leq \\alpha \\leq \\frac{5 \\pi}{6}$ in vsak tak $\\alpha$ ustrezajo tudi pogoju $0<\\alpha<\\pi$.\n\nČe pa je $\\pi<\\alpha<2 \\pi$, tedaj je $\\sin \\alpha<0$. Pri množenju neenakosti s $\\sin \\alpha$ se neenačaj obrne, zato po preureditvi dobimo $(2 \\sin \\alpha-1)(\\sin \\alpha-2) \\geq 0$. Ker je $\\sin \\alpha<0$, je $2 \\sin \\alpha-1<0$ in $\\sin \\alpha-2<0$, torej je neenakost v tem primeru avtomatično izpolnjena.\n\nRešitev naloge je torej $\\alpha \\in\\left[\\frac{\\pi}{6}, \\frac{5 \\pi}{6}\\right] \\cup(\\pi, 2 \\pi)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41218,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse pare naravnih števil $a$ in $b$, ki zadoščajo enačbi $a^{2}-5 a b+24=0$.",
    "options": [],
    "answer": "Pairs: (1, 5), (4, 2), (6, 2), (24, 5)",
    "solution": "Solution:\n\n1. Enačbo preoblikujemo v $a(a-5 b)=-24$ in jo pomnožimo z $-1$, da dobimo $a(5 b-a)=24$. Ker sta $a$ in $5 b-a$ celi števili, je $a$ pozitiven delitelj števila $24$. Torej imamo naslednje možnosti: $a=1$ in $5 b-a=24$, $a=2$ in $5 b-a=12$, $a=3$ in $5 b-a=8$, $a=4$ in $5 b-a=6$, $a=6$ in $5 b-a=4$, $a=8$ in $5 b-a=3$, $a=12$ in $5 b-a=2$ ter $a=24$ in $5 b-a=1$. Pri vsaki možnosti izračunamo še $b$ in po vrsti dobimo $b=5$, $b=\\frac{14}{5}$, $b=\\frac{11}{5}$, $b=2$, $b=2$, $b=\\frac{11}{5}$, $b=\\frac{14}{5}$ ter $b=5$. Le v štirih primerih je $b$ naravno število. Pari naravnih števil, ki zadoščajo dani enačbi, so torej $a=1$ in $b=5$, $a=4$ in $b=2$, $a=6$ in $b=2$ ter $a=24$ in $b=5$.\n\n\n2. način. Iz enačbe izrazimo $b=\\frac{a^{2}+24}{5 a}$. Ker sta $a$ in $b$ naravni števili, mora $a$ deliti $24$ in $5$ deliti $a^{2}+24$. Pozitivni delitelji števila $24$ so $1,2,3,4,6,8,12$ in $24$, toda število $a^{2}+24$ je deljivo s $5$ le, ko je $a$ enak $1$, $4$, $6$ ali $24$. V teh primerih je $b$ po vrsti enak $5$, $2$, $2$ in $5$. Pari naravnih števil, ki zadoščajo dani enačbi, so torej $a=1$ in $b=5$, $a=4$ in $b=2$, $a=6$ in $b=2$ ter $a=24$ in $b=5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41219,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bosta $F_{1}$ in $F_{2}$ gorišči elipse $4 x^{2}+9 y^{2}=36$ in $K$ taka točka na tej elipsi, da je $\\left|K F_{1}\\right| : \\left|K F_{2}\\right|=2: 1$. Izračunaj ploščino trikotnika $K F_{1} F_{2}$.",
    "options": [],
    "answer": "4",
    "solution": "![](attached_image_1.png)\n\nEnačbo elipse delimo s $36$, da dobimo $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$. Dana elipsa je v središčni legi, njena velika polos je enaka $a=3$, njena mala polos pa $b=2$. Ker točka $K$ leži na elipsi, velja $\\left|K F_{1}\\right|+\\left|K F_{2}\\right|=2 a=6$. Ker pa je $\\left|K F_{1}\\right|:\\left|K F_{2}\\right|=2$, sledi $\\left|K F_{1}\\right|=4$ in $\\left|K F_{2}\\right|=2$. Linearna ekscentričnost elipse je enaka $e=\\sqrt{a^{2}-b^{2}}=\\sqrt{5}$, zato je $\\left|F_{1} F_{2}\\right|=2 e=2 \\sqrt{5}$. Opazimo, da velja $\\left|K F_{1}\\right|^{2}+\\left|K F_{2}\\right|^{2}=20=\\left|F_{1} F_{2}\\right|^{2}$, torej je trikotnik $K F_{1} F_{2}$ pravokoten s pravim kotom pri $K$. Njegova ploščina je zato enaka $p=\\frac{1}{2} \\cdot\\left|K F_{1}\\right| \\cdot\\left|K F_{2}\\right|=\\frac{1}{2} \\cdot 4 \\cdot 2=4$.\n\n\n2. način. Postopamo enako kot v prvi rešitvi, ploščino trikotnika $K F_{1} F_{2}$ pa izračunamo po Heronovem obrazcu $p=\\sqrt{s\\left(s-a_{1}\\right)\\left(s-b_{1}\\right)\\left(s-c_{1}\\right)}$, kjer je $s=\\frac{a_{1}+b_{1}+c_{1}}{2}$ in $a_{1}=\\left|F_{1} F_{2}\\right|=2 \\sqrt{5}$, $b_{1}=\\left|K F_{2}\\right|=2$, $c_{1}=\\left|K F_{1}\\right|=4$. Torej je $s=3+\\sqrt{5}$ in zato je ploščina trikotnika enaka\n$$\np=\\sqrt{(3+\\sqrt{5})(3-\\sqrt{5})(\\sqrt{5}+1)(\\sqrt{5}-1)}=\\sqrt{(9-5)(5-1)}=4\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41220,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $x$ tako realno število, da je $x+\\frac{1}{x}$ celo število. Dokaži, da je tedaj za vsako naravno število $n$ tudi $x^{n}+\\frac{1}{x^{n}}$ celo število.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTrditev bomo dokazali z matematično indukcijo. V bazi indukcije preverimo primera $n=1$ in $n=2$. Števili\n$$\nx^{1}+\\frac{1}{x^{1}}=x+\\frac{1}{x}\n$$\nin\n$$\nx^{2}+\\frac{1}{x^{2}}=\\left(x+\\frac{1}{x}\\right)^{2}-2\n$$\nsta celi števili, saj je $x+\\frac{1}{x}$ po predpostavki naloge celo število. V indukcijski predpostavki predpostavimo, da za neko naravno število $n \\geq 2$ velja, da je $x^{k}+\\frac{1}{x^{k}}$ celo število za vsa naravna števila $k \\leq n$. V indukcijskem koraku moramo pokazati, da je tedaj tudi $x^{n+1}+\\frac{1}{x^{n+1}}$ celo število. Opazimo, da velja\n$$\nx^{n+1}+\\frac{1}{x^{n+1}}=\\left(x^{n}+\\frac{1}{x^{n}}\\right) \\cdot\\left(x+\\frac{1}{x}\\right)-\\left(x^{n-1}+\\frac{1}{x^{n-1}}\\right)\n$$\nKer je $n \\geq 2$, sta $n$ in $n-1$ naravni števili, zato so po indkukcijski predpostavki števila $x^{n}+\\frac{1}{x^{n}}$, $x^{n-1}+\\frac{1}{x^{n-1}}$ in $x+\\frac{1}{x}$ cela števila. Od tod sledi, da je tudi $x^{n+1}+\\frac{1}{x^{n+1}}$ celo število, saj so cela števila zaprta za množenje in odštevanje. S tem je indukcija zaključena in trditev dokazana.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41221,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse pare kompleksnih števil $z$ in $w$, ki rešijo sistem enačb\n$$\n\\begin{aligned}\nw^{2} & =63+16 i \\\\\n2 z-i|z| & =w\n\\end{aligned}\n$$",
    "options": [],
    "answer": "Two solutions: w = 8 + i with z = 4 + 3i; and w = −8 − i with z = −4 + (5/3)i.",
    "solution": "Solution:\n\nNaj bo $z=a+b i$ in $w=c+d i$. Iz prve enačbe sledi\n$$\nc^{2}+2 i c d-d^{2}=63+16 i\n$$\nS primerjavo realnih in imaginarnih delov dobimo sistem enačb\n$$\n\\begin{aligned}\nc^{2}-d^{2} & =63 \\\\\n2 c d & =16\n\\end{aligned}\n$$\nIz druge enačbe izrazimo $d=\\frac{8}{c}$ in vstavimo v prvo enačbo, da dobimo $c^{2}-\\frac{64}{c^{2}}=63$. Enačbo preuredimo do $c^{4}-63 c^{2}-64=0$ in levo stran razstavimo $(c^{2}+1)(c^{2}-64)=0$. Od tod sledi $c^{2}=64$ oziroma $c= \\pm 8$ in posledično $d= \\pm 1$. Torej je $w=8+i$ ali $w=-8-i$. Obravnavajmo oba primera.\n\nČe je $w=8+i$, iz druge enačbe v nalogi sledi\n$$\n2(a+b i)-i \\sqrt{a^{2}+b^{2}}=8+i\n$$\nZopet primerjamo realna in imaginarna dela in dobimo\n$$\n\\begin{aligned}\n2 a & =8 \\\\\n2 b-\\sqrt{a^{2}+b^{2}} & =1\n\\end{aligned}\n$$\nOd tod najprej izračunamo $a=4$, nato pa slednje vstavimo v drugo enačbo, da dobimo $2 b-\\sqrt{16+b^{2}}=1$. Enačbo preoblikujemo v $\\sqrt{16+b^{2}}=2 b-1$ in jo kvadriramo, da dobimo $16+b^{2}=4 b^{2}-4 b+1$. Enačbo sedaj uredimo $3 b^{2}-4 b-15=0$ in levo stran razstavimo $(3 b+5)(b-3)=0$. Rešitvi enačbe sta $b=-\\frac{5}{3}$ in $b=3$, od katerih pa le druga ustreza enačbi $\\sqrt{16+b^{2}}=2 b-1$. Torej je $b=3$ in $z=4+3 i$.\n\nČe je $w=-8-i$, na enak način kot v prvem primeru pridemo do sistema enačb\n$$\n\\begin{aligned}\n2 a & =-8 \\\\\n2 b-\\sqrt{a^{2}+b^{2}} & =-1\n\\end{aligned}\n$$\nIz prve enačbe izračunamo $a=-4$, zato iz druge enačbe dobimo $2 b-\\sqrt{16+b^{2}}=-1$. Po preoblikovanju tokrat dobimo $\\sqrt{16+b^{2}}=2 b+1$, po kvadriranju pa $16+b^{2}=4 b^{2}+4 b+1$. Enačbo uredimo $3 b^{2}+4 b-15=0$ in levo stran spet razstavimo $(3 b-5)(b+3)=0$. Rešitvi enačbe sta $b=\\frac{5}{3}$ in $b=-3$, od katerih le prva ustreza enačbi $\\sqrt{16+b^{2}}=2 b+1$. Sledi $b=\\frac{5}{3}$ in $z=-4+\\frac{5}{3} i$.\n\nSistem enačb ima torej dve rešitvi, to sta $w=8+i, z=4+3 i$ ter $w=-8-i, z=-4+\\frac{5}{3} i$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41222,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana je kocka $A B C D E F G H$. Na stranici $G H$ leži točka $K$, tako da velja $|G K|:|K H|=1:3$, na stranici $H E$ leži točka $L$, tako da velja $|H L|:|L E|=1:1$, na stranici $C G$ pa leži točka $M$. Kakšno mora biti razmerje $|C M|:|M G|$, da se bosta premici $A K$ in $L M$ sekali?",
    "options": [],
    "answer": "2:1",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nNaj bo $P$ presečišče premic $A K$ in $L M$. Označimo vektorje $\\vec{a}=\\overrightarrow{A B}$, $\\vec{b}=\\overrightarrow{A D}$ in $\\vec{c}=\\overrightarrow{A E}$. Naj bo $\\overrightarrow{A P}=\\alpha \\overrightarrow{A K}$, $\\overrightarrow{L P}=\\beta \\overrightarrow{L M}$ in $\\overrightarrow{G M}=\\gamma \\overrightarrow{G C}=-\\gamma \\vec{c}$, kjer so $\\alpha, \\beta, \\gamma \\in \\mathbb{R}$. Izrazimo vektor $\\overrightarrow{A P}$ z vektorji $\\vec{a}, \\vec{b}$ in $\\vec{c}$ na dva različna načina. Po eni strani je\n\n$$\n\\overrightarrow{A P}=\\alpha \\overrightarrow{A K}=\\alpha(\\overrightarrow{A E}+\\overrightarrow{E H}+\\overrightarrow{H K})=\\alpha\\left(\\vec{c}+\\vec{b}+\\frac{3}{4} \\vec{a}\\right)=\\frac{3}{4} \\alpha \\vec{a}+\\alpha \\vec{b}+\\alpha \\vec{c}\n$$\n\npo drugi strani pa\n\n$$\n\\begin{aligned}\n\\overrightarrow{A P} & =\\overrightarrow{A E}+\\overrightarrow{E L}+\\overrightarrow{L P}=\\vec{c}+\\frac{1}{2} \\vec{b}+\\beta \\overrightarrow{L M}=\\vec{c}+\\frac{1}{2} \\vec{b}+\\beta(\\overrightarrow{L H}+\\overrightarrow{H G}+\\overrightarrow{G M}) \\\\\n& =\\vec{c}+\\frac{1}{2} \\vec{b}+\\beta\\left(\\frac{1}{2} \\vec{b}+\\vec{a}-\\gamma \\vec{c}\\right)=\\beta \\vec{a}+\\left(\\frac{1}{2}+\\frac{\\beta}{2}\\right) \\vec{b}+(1-\\beta \\gamma) \\vec{c}\n\\end{aligned}\n$$\n\nKer so vektorji $\\vec{a}, \\vec{b}$ in $\\vec{c}$ linearno neodvisni, sledi $\\frac{3}{4} \\alpha=\\beta$, $\\alpha=\\frac{1}{2}+\\frac{\\beta}{2}$ in $\\alpha=1-\\beta \\gamma$. Iz prvih dveh enačb sledi $\\alpha=\\frac{1}{2}+\\frac{3}{8} \\alpha$ oziroma $\\frac{5}{8} \\alpha=\\frac{1}{2}$. Torej je $\\alpha=\\frac{4}{5}$ in zato $\\beta=\\frac{3}{5}$. Iz zadnje enačbe izrazimo še $\\gamma=\\frac{1-\\alpha}{\\beta}=\\frac{1}{3}$. Torej je $\\overrightarrow{G M}=\\frac{1}{3} \\overrightarrow{G C}$ in zato je $|C M|:|M G|=2:1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41223,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKmet Martin je pri baliranju sena izdelal 3 enake bale sena. Prečni prerez vsake bale sena je krog s polmerom $r$. Bale sena je postavil v piramido, tako da so se dotikale druga druge, in čez njih napel vrv, ki jo je na obeh koncih pritrdil v tla. Zgornji del vrvi je napet po obodu zgornje bale sena, stranska dela vrvi pa sta ravna in se dotikata spodnjih dveh bal sena (glej sliko). Izrazi dolžino vrvi s polmerom $r$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "(2*sqrt(3)+4+2*pi/3)*r",
    "solution": "Solution:\n\n![](attached_image_2.png)\n\nOznačimo nekaj točk kot je prikazano na sliki. Točke $A, B$ in $C$ so središča bal sena, točka $D$ je eden od koncev vrvi, točki $E$ in $F$ sta dotikališči leve bale sena s tlemi in z vrvjo, točki $G$ in $H$ pa sta skrajni točki, kjer se vrv še dotika zgornje bale sena. Trikotnik $A B C$ je enakostraničen z dolžino stranice $2 r$, zato je $\\angle B A C=60^\\circ$. Štirikotnik $F A C G$ pa je pravokotnik, zato je $\\angle C A F=90^\\circ$ in podobno je tudi $\\angle E A B=90^\\circ$. Od tod sledi\n$$\n\\angle F A E=360^\\circ-\\angle E A B-\\angle B A C-\\angle C A F=360^\\circ-90^\\circ-60^\\circ-90^\\circ=120^\\circ\n$$\nin posledično je $\\angle F A D=\\angle D A E=60^\\circ$. Torej sta trikotnika $A F D$ in $D E A$ polovici enakostraničnega trikotnika, zato je $|D F|=\\sqrt{3}|F A|=\\sqrt{3} r$. Ker je $F A C G$ pravokotnik, je $|F G|=|A C|=2 r$. Hkrati je tudi $\\angle H C G=120^\\circ$, zato je dolžina krožnega loka $\\widehat{G H}$ vrvi enaka $|\\widehat{G H}|=\\frac{2 \\pi r}{3}$. Dolžina vrvi je torej enaka\n$$\nd=2(|D F|+|F G|)+|\\widehat{G H}|=2 \\sqrt{3} r+4 r+\\frac{2 \\pi}{3} r.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41224,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDokaži, da za vsa realna števila $x$ in $y$ velja neenakost\n$$\n\\cos \\left(x^{2}\\right)+\\cos \\left(y^{2}\\right)-\\cos (x y)<3\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nKer za vsako realno število $\\alpha$ velja $-1 \\leq \\cos \\alpha \\leq 1$, je $\\cos x^{2} \\leq 1$, $\\cos y^{2} \\leq 1$ in $-\\cos x y \\leq 1$. Sledi\n$$\n\\cos x^{2}+\\cos y^{2}-\\cos x y \\leq 3\n$$\n\nPokazati moramo, da ne more veljati enačaj. Denimo, da velja. Potem je $\\cos x^{2}=1$, $\\cos y^{2}=1$ in $\\cos x y=-1$, zato je $x^{2}=2 k \\pi$, $y^{2}=2 l \\pi$ in $x y=\\pi+2 m \\pi$ za neka cela števila $k, l$ in $m$. Potem pa velja\n$$\n(\\pi+2 m \\pi)^{2}=(x y)^{2}=x^{2} y^{2}=2 k \\pi \\cdot 2 l \\pi\n$$\ntorej je $(1+2 m)^{2}=4 k l$. V tej enačbi je leva stran liho število, desna pa sodo. Dobili smo protislovno enačbo, kar pomeni, da enačaj ne more veljati.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41225,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse pare naravnih števil $m$ in $n$, za katere ima kvadratna enačba\n$$\n2007 x^{2} + m n x + n = 0\n$$\nsamo eno rešitev. Za vsak tak par rešitev tudi zapiši.",
    "options": [],
    "answer": "(m,n) = (1, 8028) with root -2; (2, 2007) with root -1; (3, 892) with root -2/3; (6, 223) with root -1/3.",
    "solution": "Solution:\n\nEnačba ima dvojno ničlo natanko tedaj, ko je diskriminanta enaka $0$, torej\n$$\n0 = (m n)^{2} - 4 \\cdot n \\cdot 2007 = n (m^{2} n - 4 \\cdot 2007)\n$$\nOd tod sledi $m^{2} n = 4 \\cdot 2007 = 2^{2} \\cdot 3^{2} \\cdot 223$. Ker je $223$ praštevilo, mora biti delitelj števila $n$. Pišimo $n = 223 n'$. Tedaj velja\n$$\nm^{2} n' = (2 \\cdot 3)^{2}\n$$\nLočimo štiri možnosti. Število $m$ je namreč lahko enako $1, 2, 3$ ali $6$. \n\nV prvem primeru je $n = 4 \\cdot 2007$, enačba pa $2007(x^{2} + 4x + 4) = 0$ z dvojno ničlo $x = -2$.\n\nPri $m = 2$ sledi $n = 2007$ ter $2007(x^{2} + 2x + 1) = 0$, tu dobimo $x = -1$.\n\nPri $m = 3$ je $n = 4 \\cdot 223$ ter enačba $223(9x^{2} + 12x + 4) = 0$, torej $x = -\\frac{2}{3}$.\n\nNazadnje imamo še $m = 6$, $n = 223$ ter $223(9x^{2} + 6x + 1) = 0$ in torej $x = -\\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41226,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa praštevila $p$, za katera je število $7^{p-2}+9 p^{4}$ popoln kvadrat.",
    "options": [],
    "answer": "p = 7",
    "solution": "Solution:\n\nNaj bo $7^{p-2}+9 p^{4}=m^{2}$. Potem je $7^{p-2}=m^{2}-9 p^{4}=(m-3 p^{2})(m+3 p^{2})$. Torej je $m-3 p^{2}=7^{r}$ in $m+3 p^{2}=7^{p-r-2}$ za nek $r \\geq 0$ in velja $p-r-2>r$. Enakosti odštejemo in dobimo $6 p^{2}=7^{r}(7^{p-2 r-2}-1)$. Torej je $p=7$ ali $r=0$. Prvi primer je res rešitev, saj dobimo $m=196$.\n\nV drugem primeru pa je $6 p^{2}+1=7^{p-2}$. Očitno $p=2, p=3$ in $p=5$ niso rešitve, zato naj bo $p>7$. Ker je $6 p^{2}=7^{p-2}-1=(7-1)(7^{p-3}+7^{p-4}+\\ldots+7^{1}+1)$, mora veljati $p^{2}=7^{p-3}+7^{p-4}+\\ldots+7^{1}+1$. Vemo, da se premica z enačbo $y=x$ in eksponentna funkcija $7^{x-1}$ sekata pri $x=1$. Za $x>1$ pa eksponentna funkcija narašča hitreje. Torej je $7^{x-1} \\geq x$ oziroma $7^{x} \\geq 7 x$. Zato lahko za $p>7$ ocenimo\n\n$$\n\\begin{aligned}\np^{2} & =7^{p-3}+7^{p-4}+\\ldots+7^{1}+1 \\geq \\\\\n& \\geq 7(p-3)+7(p-4)+\\ldots+7 \\cdot 1+1= \\\\\n& =7((p-3)+(p-4)+\\ldots+1)+1= \\\\\n& =7 \\frac{(p-3)(p-2)}{2}+1>p^{2}+\\frac{5\\left(p^{2}-7 p\\right)}{2}= \\\\\n& =p^{2}+\\frac{5 p(p-7)}{2}>p^{2}\n\\end{aligned}\n$$\n\nV tem primeru ni rešitev, torej je $p=7$ edino praštevilo, pri katerem je število $7^{p-2}+9 p^{4}$ popoln kvadrat.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41227,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoenostavi izraz $ (\\sin 2x - 2 \\cos x) \\frac{\\tan x}{1 - \\sin^2 x} \\cdot (1 + \\sin x) $.",
    "options": [],
    "answer": "-2 sin x",
    "solution": "Solution:\n\nUporabimo zvezo za $\\sin 2x = 2 \\sin x \\cdot \\cos x$ in nato iz prvega faktorja izpostavimo skupni faktor $2 \\sin x (\\sin x + 1)$. Uredimo drugi faktor $\\frac{\\tan x}{1 - \\sin^2 x} = \\frac{\\frac{\\sin x}{\\cos x}}{(1 - \\sin x)(1 + \\sin x)}$. Izraz krajšamo in dobimo $-2 \\sin x$.\n\nIzraz za sinus dvojnega kota $\\sin 2x = 2 \\sin x \\cdot \\cos x$\n\nIzpostavljanje skupnega faktorja $2 \\sin x (\\sin x + 1)$\n\nPoenostavitev drugega faktorja $\\frac{\\tan x}{1 - \\sin^2 x} = \\frac{\\frac{\\sin x}{\\cos x}}{(1 - \\sin x)(1 + \\sin x)}$\n\nKrajšanje\n\nRezultat $-2 \\sin x$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41228,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nZa celo število $x$ velja\n$$\n|\\ldots||| x-1|-10|-10^{2}\\left|\\ldots-10^{2006}\\right|=10^{2007}\n$$\nPoišči stoto števko števila $|x|$.",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n$$\nx_{n}=|\\ldots||| x-1|-10|-10^{2}\\left|\\ldots-10^{n-1}\\right|-10^{n}\n$$\n\nEnačba iz naloge nam torej pove, da velja $\\left|x_{2006}\\right|=10^{2007}$, torej je $x_{2006}= \\pm 10^{2007}$. Ker pa lahko ocenimo\n$$\nx_{2006}=\\left|x_{2005}\\right|-10^{2006} \\geq-10^{2006}\n$$\nod tod sledi, da je $x_{2006}=10^{2007}$, torej je $\\left|x_{2005}\\right|=10^{2006}+10^{2007}$. Nadalje iz $x_{2006}=$ $\\left|x_{2004}\\right|-10^{2005} \\geq 10^{2005}$ sledi $x_{2006}=10^{2006}+10^{2007}$ in je zato\n\n$$\n\\left|x_{2004}\\right|=10^{2005}+10^{2006}+10^{2007}\n$$\n\nPodobno sklepamo naprej. Če že vemo, da je\n$$\nx_{n}=10^{n+1}+10^{n+2}+\\ldots+10^{2007}\n$$\nsledi $x_{n-1}= \\pm\\left(10^{n+1}+10^{2007-n+1}+\\ldots+10^{2007}\\right)$. Zaradi $x_{n-1}=\\left|x_{n}\\right|-10^{n} \\geq-10^{n}$ dobimo, da je\n$$\nx_{n-1}=10^{n}+10^{n+1}+\\ldots+10^{2007}\n$$\n\nTorej je $|x-1|=10+10^{2}+\\ldots+10^{2007}$, od koder sledi\n$$\nx=1+10+10^{2}+\\ldots+10^{2007}=\\underbrace{11 \\ldots 1}_{2008}\n$$\n\nali\n$$\nx=1-10-10^{2}-\\ldots-10^{2007}=-(\\underbrace{11 \\ldots 10}_{2007}-1)=-\\underbrace{11 \\ldots 1}_{2006} 09 \\text{.}\n$$\n\nStota števka števila $|x|$ je $\\mathrm{v}$ obeh primerih enaka 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41229,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDokaži, da za nobeno realno število $x$ ne velja\n$$\n\\frac{1}{9}<\\frac{\\tan 3 x}{\\tan 2 x} \\leq \\frac{3}{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n1. način\n\nUvedimo novo spremenljivko $a=\\tan^{2} x$. Ker je $\\tan 2 x=\\frac{2 \\tan x}{1-\\tan^{2} x}$ in $\\tan 3 x=\\frac{3 \\tan x-\\tan^{3} x}{1-3 \\tan^{2} x}$, je\n$$\n\\frac{\\tan 3 x}{\\tan 2 x}=\\frac{(3-a)(1-a)}{2(1-3 a)}\n$$\nzato lahko neenakost prepišemo v\n$$\n\\frac{1}{9}<\\frac{(3-a)(1-a)}{2(1-3 a)} \\leq \\frac{3}{2}\n$$\n\nDenimo najprej, da je $1-3 a>0$. Tedaj iz neenakosti (1) sledi\n$$\n(3-a)(1-a) \\leq 3(1-3 a)\n$$\noziroma $a(a+5) \\leq 0$. Ker je $a=\\tan^{2} x$, velja $a \\geq 0$ ter $a+5>0$. Zato bi moralo veljati $a=0$, kar pa ni možno, saj je v tem primeru $\\tan 2 x=0$ in izraz $\\frac{\\tan 3 x}{\\tan 2 x}$ sploh ni definiran. Torej neenakost (1) ne more biti izpolnjena, ko je $1-3 a>0$, zato naj bo $1-3 a<0$. Tedaj dobimo\n$$\n\\frac{2(1-3 a)}{9}>(3-a)(1-a)\n$$\nkar je enakovredno $0>(5-3 a)^{2}$. Tudi ta neenakost ni izpolnjena za nobeno število $a$, torej ne obstaja tak $x$, da bi veljalo\n$$\n\\frac{1}{9}<\\frac{\\tan 3 x}{\\tan 2 x} \\leq \\frac{3}{2}\n$$\n\n\n2. način\n\nS pomočjo adicijskih izrekov lahko zapišemo $\\tan 3 x=\\frac{\\sin x\\left(3 \\cos^{2} x-\\sin^{2} x\\right)}{\\cos x\\left(\\cos^{2} x-3 \\sin^{2} x\\right)}$ in $\\tan 2 x=\\frac{2 \\sin x \\cos x}{\\cos^{2} x-\\sin^{2} x}$. Torej je\n$$\n\\frac{\\tan 3 x}{\\tan 2 x}=\\frac{\\left(2 \\cos^{2} x-1\\right)\\left(4 \\cos^{2} x-1\\right) \\sin x}{2 \\cos^{2} x\\left(4 \\cos^{2} x-3\\right) \\sin x}\n$$\nČe je $\\sin x=0$, izraz ni definiran, sicer pa lahko $\\sin x$ okrajšamo. Uvedimo novo spremenljivko $a=\\cos^{2} x$. Zaradi $\\sin x \\neq 0$ sledi $a<1$. Neenakost lahko prepišemo v\n$$\n\\frac{1}{9}<\\frac{(2 a-1)(4 a-1)}{2 a(4 a-3)} \\leq \\frac{3}{2}\n$$\nČe je $4 a-3>0$, iz neenakosti (2) sledi\n$$\n(2 a-1)(4 a-1) \\leq 3 a(4 a-3)\n$$\noziroma $0 \\leq (4 a+1)(a-1)$. Očitno je $a$ pozitivno število, ki je manjše od 1, zato je $4 a+1$ pozitivno, $a-1$ pa negativno število, torej dobljena neenakost ne drži.\n\nOstane le še primer, ko je $4 a-3<0$. Tedaj iz (2) dobimo neenakost\n$$\n\\frac{1}{9} 2 a(4 a-3)>(2 a-1)(4 a-1)\n$$\nki jo lahko preoblikujemo v $0>(8 a-3)^{2}$, kar pa seveda ni možno. Torej neenakost (2) ni izpolnjena za nobeno število $a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41230,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDokaži, da je polinom $p(x) = (x-2)^{100} + (x-1)^{50} - 1$ deljiv s polinomom $g(x) = x^{2} - 3x + 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNajprej funkcijski predpis $g(x)$ razstavimo: $g(x) = (x-2)(x-1)$. Ničli polinoma $g(x)$ sta $x_{1} = 2$ in $x_{2} = 1$. Izračunamo vrednost $p(2)$ in $p(1)$. Ugotovimo, da sta to ničli polinoma $p(x)$. Ker sta $2$ in $1$ ničli obeh danih polinomov, velja trditev, da $g(x)$ deli $p(x)$.\n\nZapis $g(x) = (x-2)(x-1)$\n\nDoločitev ničel polinoma $g(x): x_{1} = 2$ in $x_{2} = 1$\n\nIzračun $p(1)$\n\nIzračun $p(2)$\n\nSklep",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41231,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nReši enačbo $\\log_{3}(\\log_{2} x + 12) + 2 = 4$ in rešitev zapiši v obliki ulomka.",
    "options": [],
    "answer": "1/8",
    "solution": "Solution:\n\nEnačbo najprej uredimo $\\log_{3}(\\log_{2} x + 12) = 2$. Upoštevamo definicijo logaritma in zapišemo $3^{2} = \\log_{2} x + 12$. Enačbo ponovno uredimo in dobimo $\\log_{2} x = -3$. Rešimo $2^{-3} = x$. Rešitev je $x = \\frac{1}{8}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41232,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV trimestnem številu so stotice večje od desetic in desetice večje od enic. Če števke tega trimestnega števila zapišemo v obratnem vrstem redu in dobljeno število prištejemo prvotnemu, dobimo število, ki vsebuje samo lihe števke. Določi vsa trimestna števila, za katera to velja.",
    "options": [],
    "answer": "843, 932, 942",
    "solution": "Solution:\n\nOznačimo trimestno število z $\\overline{a b c}$. Veljati mora $a > b > c$, poleg tega pa je število $\\overline{a b c} + \\overline{c b a} = 10^{2}(a + c) + 10(2b) + (a + c)$ sestavljeno iz samih lihih števk. Če je $a + c < 10$, je števka na mestu desetic soda, kar ni možno. Zato mora biti $a + c \\geq 10$. Ker je še število $a + c$ liho, je torej $a + c \\geq 11$. Pišimo $a + c = 10 + l$, kjer je $l$ liho število. Tedaj je\n\n$$\n\\overline{a b c} + \\overline{c b a} = 10^{3} + 10^{2} l + 10(2b + 1) + l\n$$\n\nOd tod sledi, da je $2b + 1 < 10$, saj bo nasprotnem primeru števka na mestu stotic soda. Torej je $2b < 9$ in zato $b \\leq 4$.\n\nZaradi $c < b \\leq 4$ sledi $c \\leq 3$. Torej je $a + c \\leq 9 + 3 = 12$. Pokazali pa smo že, da je $a + c$ liho in vsaj 11, torej je $a + c = 11$. Če je $c = 2$ in $a = 9$, je $b$ lahko 3 ali 4, pri $c = 3$ in $a = 8$ pa je možno le $b = 4$.\n\nVsa takšna števila so 843, 932 in 942.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41233,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSkrči izraz $\\frac{1+9 a^{-1}+20 a^{-2}}{1+8 a^{-1}+16 a^{-2}} \\cdot\\left(a^{2}+4 a\\right) \\cdot\\left(1-25 a^{-2}\\right)^{-1}$.",
    "options": [],
    "answer": "a^3/(a-5)",
    "solution": "Solution:\n\nPotence z negativnim celim eksponentom v prvem faktorju zapišemo v obliki ulomka. V števcu in imenovalcu dvojnega ulomka poiščemo skupni imenovalec. V števcu dobimo $\\frac{a^{2}+9 a+20}{a^{2}}$, v imenovalcu pa $\\frac{a^{2}+8 a+16}{a^{2}}$. Odpravimo dvojni ulomek, števec razstavimo na $(a+4)(a+5)$ in imenovalec na $(a+4)^{2}$. Drugi faktor razstavimo na $a(a+4)$. Zadnji (tretji) faktor poenostavimo $\\left(1-\\frac{25}{a^{2}}\\right)^{-1}=\\frac{a^{2}}{a^{2}-25}$ in imenovalec razstavimo $\\frac{a^{2}}{(a+5)(a-5)}$. Ulomke okrajšamo in dobimo $\\frac{a^{3}}{a-5}$.\n\nZapis potenc z ulomki $\\frac{1+\\frac{9}{a}+\\frac{20}{a^{2}}}{1+\\frac{8}{a}+\\frac{16}{a^{2}}}$.\n\nUreditev prvega faktorja $\\frac{\\frac{a^{2}+9 a+20}{a^{2}}}{\\frac{a^{2}+8 a+16}{a^{2}}}$.\n\nRazstavitev števca in imenovalca $\\frac{(a+4)(a+5)}{(a+4)^{2}}$.\n\nRazstavitev drugega faktorja $a(a+4)$\n\nUreditev tretjega faktorja $\\left(1-\\frac{25}{a^{2}}\\right)^{-1}=\\frac{a^{2}}{a^{2}-25}=\\frac{a^{2}}{(a+5)(a-5)}$\n\nRezultat $\\frac{a^{3}}{a-5}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41234,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOznačimo z $D$ razpolovišče stranice $AB$ ostrokotnega trikotnika $ABC$. Naj bosta $A'$ in $B'$ taki točki na daljicah $AC$ in $BC$, da sta trikotnika $ADA'$ in $DBB'$ enakokraka s skupnim vrhom v $D$. Pokaži: če je premica $CD$ pravokotna na premico $A'B'$, je trikotnik $ABC$ enakokrak.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nKer je $D$ razpolovišče stranice $AB$, velja $|AD| = |DB|$. Trikotnika $ADA'$ in $DBB'$ sta enakokraka z vrhom $D$, zato je $|A'D| = |AD| = |BD| = |B'D|$ in je trikotnik $A'DB'$ enakokrak z vrhom $D$.\n\nOznačimo z $E$ presečišče $A'B'$ in $CD$. Če je $DE$ pravokotna na $A'B'$, velja $|A'E| = |EB'|$, saj je trikotnik $A'DB'$ enakokrak. V trikotniku $A'B'C$ je $CE$ višina, hkrati pa točka $E$ razpolavlja $A'B'$, zato je ta trikotnik enakokrak. Velja\n$$\n\\begin{aligned}\n\\angle BAC &= \\angle DA'A = \\pi - \\angle DB'A' - \\angle B'A'C = \\\\\n&= \\pi - \\angle DBA' - \\angle A'B'C = \\\\\n&= \\angle DB'B = \\angle ABC,\n\\end{aligned}\n$$\n\n![](attached_image_1.png)\n\nzato je trikotnik $ABC$ res enakokrak.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41235,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKrožnici $\\mathcal{K}_1$ in $\\mathcal{K}_2$ s središčema $O_1$ in $O_2$ se sekata v točkah $A$ in $B$. Razdalja med središčema je večja od polmerov obeh krožnic. Naj bosta $C_1$ in $C_2$ tisti presečišči premice $O_1 O_2$ s krožnicama $\\mathcal{K}_1$ in $\\mathcal{K}_2$, ki ne ležita na daljici $O_1 O_2$. Označimo drugo presečišče premice $C_2 A$ in krožnice $\\mathcal{K}_1$ z $D_1$, drugo presečišče premice $C_1 A$ in krožnice $\\mathcal{K}_2$ pa z $D_2$.\n\nPremici $D_1 B$ in $D_2 A$ se sekata v $E$, premici $D_1 A$ in $D_2 B$ pa v $F$. Pokaži: če je štirikotnik $A O_1 B O_2$ tetiven, je tetiven tudi štirikotnik $A E B F$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nKer je štirikotnik $A O_1 B O_2$ tetiven in leži središče njemu očrtane krožnice na simetrali tetive $A B$, torej na daljici $O_1 O_2$, po Talesovem izreku sledi $\\angle O_1 A O_2 = \\angle O_2 B O_1 = \\frac{\\pi}{2}$.\n\n![](attached_image_1.png)\n\nNaj bo $\\angle A O_2 O_1 = \\alpha$. Potem je $\\angle A O_2 B = 2\\alpha$. To pa je središčni kot nad tetivo $A B$ v krožnici $\\mathcal{K}_2$, zato je enak dvakratniku obodnega kota $\\angle A D_2 B$ oziroma $\\angle A C_2 B$ nad to tetivo. Torej je $\\angle A D_2 B = \\angle A C_2 B = \\alpha$.\n\nVelja še $\\angle A O_1 O_2 = \\frac{\\pi}{2} - \\angle O_1 O_2 A = \\frac{\\pi}{2} - \\alpha$, zato je $\\angle A O_1 B = 2 \\angle A O_1 O_2 = \\pi - 2\\alpha$ in $\\angle B C_1 A = \\angle B D_1 A = \\frac{1}{2} \\angle A O_1 B = \\frac{\\pi}{2} - \\alpha$.\n\nV deltoidu $A C_1 B C_2$ poznamo $\\angle A C_1 B = \\frac{\\pi}{2} - \\alpha$ in $\\angle A C_2 B = \\alpha$, zato lahko izračunamo\n$$\n\\angle C_2 A C_1 = \\angle C_2 B C_1 = \\frac{2\\pi - \\angle A C_1 B - \\angle A C_2 B}{2} = \\frac{3\\pi}{4}.\n$$\n\nZato je $\\angle D_1 A C_1 = \\pi - \\angle C_1 A C_2 = \\frac{\\pi}{4}$ in prav tako $\\angle C_2 A D_2 = \\angle D_1 A C_1 = \\frac{\\pi}{4}$. Z upoštevanjem enakosti obodnih kotov dobimo še $\\angle C_2 B D_2 = \\angle C_2 A D_2 = \\frac{\\pi}{4}$ in $\\angle D_1 B C_1 = \\angle D_1 A C_1 = \\frac{\\pi}{4}$, zato je\n$$\n\\angle F B E = \\angle C_2 B C_1 - \\angle C_2 B D_2 - \\angle D_1 B C_1 = \\frac{3\\pi}{4} - \\frac{\\pi}{4} - \\frac{\\pi}{4} = \\frac{\\pi}{4}\n$$\n\ntorej je $\\angle E A F + \\angle F B E = \\frac{3\\pi}{4} + \\frac{\\pi}{4} = \\pi$, zato točke $A, E, B$ in $F$ res ležijo na skupni krožnici.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41236,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je paralelogram $ABCD$. Naj bo $E$ razpolovišče daljice $CD$, $F$ razpolovišče $DA$ ter $G$ razpolovišče $AB$. Trikotniku $DFE$ očrtana krožnica se dotika daljice $AB$ v točki $G$. Dokaži, da je $|AB| = \\sqrt{2} |AD|$.",
    "options": [],
    "answer": "|AB| = sqrt(2) * |AD|",
    "solution": "Solution:\n\n1. način\nOznačimo $\\angle BAD = \\alpha$. Potem je $\\angle EGB = \\alpha$ ter $\\angle ADE = \\pi - \\alpha$. Ker je štirikotnik $DEFG$ tetiven, je $\\angle FGE = \\alpha$. Zato je $\\angle FGA = \\pi - \\angle FGE - \\angle EGB = \\pi - 2\\alpha$ in $\\angle GFA = \\pi - \\angle GAF - \\angle FGA = \\pi - \\alpha - (\\pi - 2\\alpha) = \\alpha$. Torej je trikotnik $AGF$ enakokrak z vrhom pri $G$.\n\nVemo, da je kot med tetivo $EG$ in tangento $AB$ enak\n\n![](attached_image_1.png)\n\nobodnemu kotu nad to tetivo. To pomeni, da je $\\angle GFE = \\angle BGE = \\alpha$. Zato je tudi trikotnik $EFG$ enakokrak z vrhom pri $E$. Trikotnika $EFG$ in $GAF$ sta si podobna, zato je $\\frac{|EG|}{|FG|} = \\frac{|AG|}{|AF|}$. Če označimo $|AB| = a$ in $|AD| = b$, dobimo torej $\\frac{b}{a/2} = \\frac{a/2}{b/2}$, od koder sledi $2b^2 = a^2$ oziroma $a = \\sqrt{2} b$.\n\n2. način\nNalogo lahko na kratko rešimo s pomočjo potence točke na krožnico (s katero so zagotovo seznanjeni udeleženci priprav na MMO in bralci elektronske revije Brihtnež). Če jo zapišemo za točko $A$ in trikotniku $DEF$ očrtano krožnico, dobimo\n$$\n\\overrightarrow{AG} \\cdot \\overrightarrow{AG} = \\overrightarrow{AF} \\cdot \\overrightarrow{AD}\n$$\nTorej velja $|AG|^2 = |AF| \\cdot |AD|$, od koder po definiciji točk $G$ in $F$ sledi, da je\n$$\n\\left(\\frac{|AB|}{2}\\right)^2 = \\frac{|AD|}{2} \\cdot |AD|\n$$\noindent oziroma $|AB| = \\sqrt{2} |AD|$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41237,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Pokaži, da je vrednost izraza $\\sqrt{9} \\cdot \\sqrt[3]{64} - \\sqrt[3]{27} \\cdot \\sqrt{4}$ naravno število.\n\nb) Pokaži, da je za vsako naravno število $x$ število $64^{-1} \\cdot 8^{2x+4} - 24 \\cdot 64^{x} + 4 \\cdot 32^{x} \\cdot 2^{x+3}$ deljivo s številom $\\sqrt{9} \\cdot \\sqrt[3]{64} - \\sqrt[3]{27} \\cdot \\sqrt{4}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nIzračunajmo vrednost izraza $B = \\sqrt{9} \\cdot \\sqrt[3]{64} - \\sqrt[3]{27} \\cdot \\sqrt{4}$:\n\n$\\sqrt{9} = 3$\n\n$\\sqrt[3]{64} = 4$\n\n$\\sqrt[3]{27} = 3$\n\n$\\sqrt{4} = 2$\n\nTorej:\n\n$$\nB = 3 \\cdot 4 - 3 \\cdot 2 = 12 - 6 = 6\n$$\n\nKer je $6$ naravno število, je vrednost izraza naravno število.\n\nb.\nIzraz $A = 64^{-1} \\cdot 8^{2x+4} - 24 \\cdot 64^{x} + 4 \\cdot 32^{x} \\cdot 2^{x+3}$ poenostavimo.\n\nNajprej zapišimo vse osnove kot potence števila $2$:\n\n$64 = 2^6$\n\n$8 = 2^3$\n\n$32 = 2^5$\n\nTorej:\n\n$64^{-1} = 2^{-6}$\n\n$8^{2x+4} = (2^3)^{2x+4} = 2^{6x+12}$\n\n$64^{x} = (2^6)^x = 2^{6x}$\n\n$32^{x} = (2^5)^x = 2^{5x}$\n\n$2^{x+3} = 2^x \\cdot 2^3 = 2^{x+3}$\n\nSedaj vstavimo v izraz $A$:\n\n$$\nA = 2^{-6} \\cdot 2^{6x+12} - 24 \\cdot 2^{6x} + 4 \\cdot 2^{5x} \\cdot 2^{x+3}\n$$\n\n$2^{-6} \\cdot 2^{6x+12} = 2^{6x+12-6} = 2^{6x+6}$\n\n$4 \\cdot 2^{5x} \\cdot 2^{x+3} = 4 \\cdot 2^{5x+x+3} = 4 \\cdot 2^{6x+3} = 2^2 \\cdot 2^{6x+3} = 2^{6x+5}$\n\nTorej:\n\n$$\nA = 2^{6x+6} - 24 \\cdot 2^{6x} + 2^{6x+5}\n$$\n\n$2^{6x+6} = 2^{6x} \\cdot 2^6 = 64 \\cdot 2^{6x}$\n\n$2^{6x+5} = 2^{6x} \\cdot 2^5 = 32 \\cdot 2^{6x}$\n\nTorej:\n\n$$\nA = 64 \\cdot 2^{6x} - 24 \\cdot 2^{6x} + 32 \\cdot 2^{6x} = (64 - 24 + 32) \\cdot 2^{6x} = 72 \\cdot 2^{6x}\n$$\n\nKer je $B = 6$, preverimo, ali $6$ deli $72 \\cdot 2^{6x}$:\n\n$72$ je deljivo s $6$, saj $72 = 12 \\cdot 6$.\n\nTorej je $A$ za vsako naravno število $x$ deljivo z $B$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41238,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je ostrokotni trikotnik $ABC$ in središče njemu očrtane krožnice $O$. Naj bo $O_1$ točka na simetrali daljice $AB$, ki leži na nasprotnem bregu premice $AB$ kot točka $O$. Označimo krožnico s središčem $O_1$ in polmerom $AO_1$ s $\\mathcal{K}$. Naj premici $CA$ in $CB$ sekata krožnico $\\mathcal{K}$ še v točkah $A_1$ in $B_1$. Pokaži: če se daljici $A_1B$ in $AB_1$ sekata na trikotniku $ABC$ očrtani krožnici, je štirikotnik $AO_1BO$ tetiven.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNaj bo $D$ presečišče premic $A_1B$ in $AB_1$ ter označimo $\\angle A_1AB_1 = \\alpha$. Tedaj je $\\angle CAD = \\pi - \\angle DAA_1 = \\pi - \\alpha$. Ker so točke $A, C, B$ in $D$ konciklične, je $\\angle CBD = \\pi - \\angle CAD = \\pi - (\\pi - \\alpha) = \\alpha$. Zato je $\\angle A_1BB_1 = \\pi - \\angle DBC = \\pi - \\alpha$.\n\nObodna kota $\\angle A_1AB_1$ in $\\angle A_1BB_1$ nad tetivo $A_1B_1$ v krožnici $\\mathcal{K}$ sta enaka, zato je $\\pi - \\alpha = \\angle A_1BB_1 = \\angle A_1AB_1 = \\alpha$, od koder sledi $\\alpha = \\frac{\\pi}{2}$.\n\nOznačimo še $\\angle ACB = \\gamma$. Tedaj je $\\angle AOB = 2\\angle ACB = 2\\gamma$.\n\nPo drugi strani pa je $\\angle AB_1C = \\pi - \\angle B_1AC - \\angle ACB_1 = \\pi - \\frac{\\pi}{2} - \\gamma = \\frac{\\pi}{2} - \\gamma$, zato je $\\angle AO_1B = 2\\angle AB_1B = 2\\angle AB_1C = 2\\left(\\frac{\\pi}{2} - \\gamma\\right) = \\pi - 2\\gamma$.\n\nOd tod sledi\n\n$$\n\\angle AOB + \\angle BO_1A = 2\\gamma + (\\pi - 2\\gamma) = \\pi\n$$\n\ntorej je štirikotnik $AO_1BO$ res tetiven.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41239,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV ravnini leži 16 črnih točk, kot prikazuje slika. Najmanj koliko izmed teh točk moramo pobarvati rdeče, da ne bo obstajal kvadrat z oglišči v preostalih črnih točkah in s stranicami, katerih dolžine so naravna števila ali $\\sqrt{2}$? Odgovor utemelji.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "5",
    "solution": "Solution:\n\nDokazali bomo, da moramo pobarvati pet točk.\n\n![](attached_image_2.png)\n\nKer mora biti v vsakem izmed 4 kvadratkov $A, B, C, D$, označenih na prvi sliki, vsaj eno oglišče rdeče, potrebujemo vsaj 4 rdeče točke. Denimo, da je to že dovolj. Kvadrat velikosti $3 \\times 3$ ima vsaj eno rdeče oglišče. Predpostavimo lahko, da je to oglišče $(1,1)$. Torej so preostala oglišča kvadrata $A$ črna. Oglejmo si kvadratek $E$. Ker sta oglišči, ki sta skupni $A$ in $E$, črni, mora biti rdeče eno od oglišč, skupno $B$ in $E$. Preostali oglišči v $B$ sta tako črni. Podobno sklepamo še za kvadratka $F$ in $D$.\nKvadrat $G$, ki ga prikazuje tretja skica, ima že 3 črna oglišča, zato mora biti oglišče $(4,4)$ rdeče.\n\n![](attached_image_3.png)\n\nKvadrata $H$ in $K$ že imata tri črna oglišča, tako morata biti preostali rdeči. To pa je v nasprotju s predpostavko, da je v $D$ le eno rdeče oglišče.\n\nDokazali smo torej, da mora biti vsaj 5 točk pobarvanih rdeče. Kot kaže spodnja slika, pa 5 rdečih točk tudi zadošča.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41240,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV ravnini leži 16 črnih točk, kot prikazuje slika. Najmanj koliko izmed teh točk moramo pobarvati rdeče, da ne bo obstajal kvadrat z oglišči v preostalih črnih točkah in s stranicami, vzporednimi koordinatnima osema? Odgovor utemelji.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\n1. način\n\nPobarvati moramo najmanj štiri točke. V mrežo namreč lahko vrišemo štiri disjunktne kvadrate kot na prvi sliki. Ker mora biti v vsakem vsaj eno oglišče pobarvano z rdečo, potrebujemo vsaj 4 rdeče točke.\n\nPokažimo, da je to tudi dovolj. Pobarvajmo točke kot prikazuje druga slika. V mreži imamo en sam kvadrat velikosti $3 \\times 3$, ki ima očitno rdeči dve oglišči. Poleg tega imamo štiri kvadrate velikosti $2 \\times 2$ in vsak ima natanko eno oglišče rdeče.\n\nOstane še 9 kvadratov velikosti $1 \\times 1$. Štiri smo narisali že na prvi sliki in izmed teh ima jasno vsak vsaj eno oglišče rdeče. Ostalih 5 je narisanih na tretji sliki in vsi imajo vsaj eno rdeče oglišče.\n\n![](attached_image_2.png)\n\n\n2. način\n\nNajprej vsaki točki priredimo potencial, ki naj bo število kvadratov, ki imajo eno od oglišč v tej točki. Zunanjih dvanajst točk ima potencial 3, notranje štiri pa 5:\n\n| 3 | 3 | 3 | 3 |\n| :--- | :--- | :--- | :--- |\n| 3 | 5 | 5 | 3 |\n| 3 | 5 | 5 | 3 |\n| 3 | 3 | 3 | 3 |\n\nDa je potrebno pobarvati vsaj 4 točke, sedaj s pomočjo potencialov vidimo takole. Z barvanjem treh točk (ali manj) bi izločili kvečjemu\n(a) $3+3+3=9$\n(b) $3+3+5=11$\n(c) $3+5+5=13$\n(č) $5+5+5=15$\nkvadratov. Vseh kvadratov je $1+4+9=14$, zato lahko možnosti (a), (b) in (c) takoj izločimo, možnost (č) pa je tudi neustrezna, saj nam pri barvanju treh notranjih točk ostane največji kvadrat in še en najmanjši kvadrat:\n\n![](attached_image_3.png)\n\nTorej moramo pobarvati vsaj 4 točke. Dokaz sklenemo kot v prvem načinu.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41241,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nStranica $AD$ trapeza $ABCD$ je pravokotna na osnovnici $AB$ in $CD$. Dolžine stranic tega trapeza so: $|AB|=2~\\mathrm{cm}$, $|CD|=8~\\mathrm{cm}$ in $|AD|=10~\\mathrm{cm}$. Izračunaj oddaljenost presečišča diagonal $S$ od osnovnice $AB$.",
    "options": [],
    "answer": "2 cm",
    "solution": "Solution:\n\nNarišemo ustrezno skico trapeza, kjer je $a=2$, $c=8$ in $d=10$. Na skici označimo presečišče diagonal z $S$, razdaljo presečišča od stranice $a$ z $x$ in razdaljo od stranice $d$ z $y$.\n\nIz podobnih trikotnikov $\\triangle AES \\approx \\triangle ADC$ in $\\triangle FBS \\approx \\triangle ABD$ izrazimo razmerje, npr. $y : c = x : d$ in $(a - y) : a = x : d$. Torej je $y : 8 = x : 10$ in $(2 - y) : 2 = x : 10$.\n\nIz tega izračunamo oddaljenost od presečišča diagonal $x = 2~\\mathrm{cm}$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41242,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be real numbers such that $\\frac{4a}{a+2b} - \\frac{5b}{2a+b} = 1$. Find all possible values of the expression $\\frac{a-2b}{4a+5b}$.",
    "options": [],
    "answer": "0 and -3",
    "solution": "First note that $a + 2b$ and $2a + b$ cannot both be zero, so $a$ and $b$ cannot both be zero. Eliminating the fractions in $\\frac{4a}{a+2b} - \\frac{5b}{2a+b} = 1$ we get $8a^2 + 4ab - 5ab - 10b^2 = 2a^2 + 5ab + 2b^2$, which we then rewrite as $6(a + b)(a - 2b) = 0$.\n\nIf $a = -b$, then $b$ must be non-zero and the value of $\\frac{a-2b}{4a+5b}$ is equal to $\\frac{-3b}{b} = -3$.\n\nIf, on the other hand, we have $a = 2b$, then $\\frac{a-2b}{4a+5b} = 0$ (the denominator is non-zero).\n\nThe only possible values of the expression are $0$ and $-3$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41243,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$, $C$, $D$ and $E$ be consecutive points on a circle with the centre at $O$, such that $|AC| = |BD| = |CE| = |DO|$. Let $H_1$, $H_2$ and $H_3$ be the orthocentres of the triangles $ACD$, $BCD$ and $BCE$. Prove that $H_1H_2H_3$ is a right triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "This problem has several interesting solutions. A nicely drawn figure suggests that quadrilaterals $ABH_1H_2$ and $H_2H_3DE$ are parallelograms. We will prove this fact.\n\n![](attached_image_1.png)\n\nTriangles $ACD$, $BCD$ and $BDE$ are obtuse, so let us first make a short detour examining some properties of obtuse triangles. Let $KLM$ be a triangle with the obtuse angle $\\angle KLM$ and let $H$ be the orthocentre. The point $H$ lies outside the triangle $KLM$ and on a different side of the line $KL$ than $M$. Let $L'$ and $K'$ be the feet of the altitudes from $L$ and $K$. Since $\\angle ML'H = \\angle MK'H$, points $M$, $L'$, $K'$, $H$ are concyclic and $\\angle KML = \\angle LHK$.\nThe following fact is also worth mentioning: since $K'KM$ is a right triangle the angle $K'KM$ is acute. Points $K'$, $K$, $L'$, $L$ are concyclic, so $\\angle HLM = \\pi - \\angle K'KL$, and the angle $\\angle HLM$ is obtuse.\n\n![](attached_image_2.png)\n\nLet us now return to the original problem. Here to help us are the facts stated above and a new and less cluttered figure.\nSince $|AC| = |BD|$, the inscribed angles over these two chords are equal, so $\\angle ADC = \\angle BAD$, which implies $\\angle CBA = \\pi - \\angle ADC = \\pi - \\angle BAD = \\angle DCB$, so $AD$ and $BC$ are parallel and $ABCD$ is an isosceles trapezoid.\n![](attached_image_3.png)\nPoints $A$, $B$, $C$ and $D$ are concyclic, so $\\angle CAD = \\angle CBD$. From the above consideration we see that $\\angle DH_1C = \\angle CAD$ and $\\angle DH_2C = \\angle CBD$,\nso $\\angle DH_1C = \\angle DH_2C$ and points $C$, $H_1$, $H_2$, $D$ are concyclic. But $H_1C$ is perpendicular to $AD$, $H_2D$ is perpendicular to $BC$ and lines $BC$ and $AD$ are parallel, so $H_1C$ and $H_2D$ are parallel as well. The cyclic quadrilateral $CH_1H_2D$ has a pair of parallel sides $H_1C$ and $H_2D$ and is therefore an isosceles trapezoid.\nWe have seen that $CH_1H_2D$ and $ABCD$ are isosceles trapezoids, so we have $|H_1H_2| = |CD| = |AB|$. The altitude $AH_1$ is perpendicular to the side $CD$ and so is the altitude $BH_2$, hence $AH_1$ and $BH_2$ are parallel. The quadrilateral $ABH_2H_1$ has a pair of parallel sides ($BH_2$ and $AH_1$) and a pair of sides of equal length ($AB$ and $H_1H_2$). Thus, it is either an isosceles trapezoid or a parallelogram. Let us demonstrate that it is, in fact, the latter.\nLines $AH_1$ and $BH_2$ are parallel, so the point $H_1$ lies on the side of the line $BH_2$ not containing $C$. As we have shown above the angle $\\angle DCH_1$ is obtuse, so the angle $\\angle CH_1H_2$, which is equal to $\\angle DCH_1$, is also obtuse. But $\\angle AH_1H_2 = \\angle AH_1C + \\angle CH_1H_2 > \\frac{\\pi}{2}$, so the angle $\\angle AH_1H_2$ is obtuse. On the other hand, the angle $\\angle CAH_1$ is acute, so $\\angle BAH_1 = \\angle CAH_1 - \\angle CAB$ is acute as well (points $B$ and $H_1$ lie on the same side of the line $AC$, because neither of them lies on the same side as $D$). We have shown that $\\angle BAH_1$ is acute and $\\angle AH_1H_2$ is obtuse, so the quadrilateral $ABH_2H_1$ is a parallelogram. Hence, $H_1H_2$ is parallel to $AB$.\nSimilarly, we show that $H_2H_3$ is parallel to $ED$. If we wish to prove that $\\angle H_1H_2H_3 = \\frac{\\pi}{2}$, it suffices to see that lines $AB$ and $ED$ intersect at a right angle.\nDenote the intersection of $AB$ and $ED$ by $T$. Since $\\angle BET = \\angle BED = \\frac{1}{2}\\angle BOD = 30^\\circ$ and $\\angle TBE = \\pi - \\angle EBA = \\pi - \\angle ECA = \\pi - 120^\\circ = 60^\\circ$, we have $\\angle ETB = \\pi - \\angle TBE - \\angle BET = 90^\\circ$. So $\\angle ATE = 90^\\circ$ and this concludes our proof.\n\n![](attached_image_3.png)\nSince $\\angle CH_3B = \\angle BEC = \\angle BDC = \\angle CH_2B$, points $C$, $H_3$, $H_2$, $B$ are concyclic. Similarly, we show that the quadrilateral $CH_1H_2D$ is cyclic.\n![](attached_image_4.png)\n\nWe wish to see that $\\angle H_3H_2H_1 = \\frac{\\pi}{2}$. But $\\angle H_3H_2H_1 = \\angle H_3H_2C + \\angle CH_2H_1$ so $\\angle H_3H_2C = \\angle H_3BC$ and $\\angle CH_2H_1 = CDH_1$. Let $F$ be the intersection of lines $BH_3$ and $EC$ and let $G$ be the intersection of $AC$ and $DH_2$. We wish to show that $\\angle FBC + \\angle CDG = \\frac{\\pi}{2}$.\nWe will not need the orthocentres this time, so let us draw another figure. Denote $\\alpha = \\angle FBC$ and $\\beta = \\angle CDG$. We have to show that $\\alpha + \\beta = \\frac{\\pi}{2}$. We have $\\angle ECB = \\frac{\\pi}{2} + \\alpha$ and $\\angle DCA = \\frac{\\pi}{2} + \\beta$. So, $\\alpha + \\beta = \\angle ECB + \\angle DCA - \\pi = 2\\angle ECA + \\angle DCE + \\angle ACB - \\pi = \\frac{\\pi}{3} + \\angle DCE + \\angle ACB$. In particular, $\\angle ACB = \\angle CAD$, because the quadrilateral $ABCD$ is an isosceles trapezoid, which implies $\\angle DCE + \\angle ACB = \\angle DAE + \\angle CAD = \\angle CAE = \\frac{1}{2}\\angle COE = 30^\\circ$, so $\\alpha + \\beta = 60^\\circ + 30^\\circ = 90^\\circ$ and $H_1H_2H_3$ is a right triangle.\n\n![](attached_image_5.png)\nLet the system of coordinates be centred at the circumcentre $O$. Since $H_1$ is the orthocentre of the triangle $ACD$ we have $\\overrightarrow{OH_1} = \\overrightarrow{OA} + \\overrightarrow{OC} + \\overrightarrow{OD}$. Similarly,\n$$\n\\overrightarrow{OH_2} = \\overrightarrow{OB} + \\overrightarrow{OC} + \\overrightarrow{OD} \\quad \\text{and} \\quad \\overrightarrow{OH_3} = \\overrightarrow{OB} + \\overrightarrow{OC} + \\overrightarrow{OE}.\n$$\nSo, $\\overrightarrow{H_1H_2} = \\overrightarrow{OH_2} - \\overrightarrow{OH_1} = \\overrightarrow{OB} - \\overrightarrow{OA}$ and $\\overrightarrow{H_2H_3} = \\overrightarrow{OH_3} - \\overrightarrow{OH_2} = \\overrightarrow{OE} - \\overrightarrow{OD}$.\nThe inner product of these two vectors is\n$$\n\\begin{aligned}\n\\overrightarrow{H_1H_2} \\cdot \\overrightarrow{H_2H_3} &= (\\overrightarrow{OB} - \\overrightarrow{OA})(\\overrightarrow{OE} - \\overrightarrow{OD}) \\\\\n&= \\overrightarrow{OB} \\cdot \\overrightarrow{OE} - \\overrightarrow{OB} \\cdot \\overrightarrow{OD} - \\overrightarrow{OA} \\cdot \\overrightarrow{OE} + \\overrightarrow{OA} \\cdot \\overrightarrow{OD} \\\\\n&= |\\overrightarrow{OB}| \\cdot |\\overrightarrow{OE}| \\cos(\\angle BOE) - |\\overrightarrow{OB}| \\cdot |\\overrightarrow{OD}| \\cos(\\angle BOD)\n\\end{aligned}\n$$\n$$\n\\begin{aligned}\n& - |OA| \\cdot |OE| \\cos(\\angle AOE) + |OA| \\cdot |OD| \\cos(\\angle AOD) \\\\\n&= |OA|^2(\\cos(\\angle BOE) - \\cos(\\angle BOD) - \\cos(\\angle AOE) + \\cos(\\angle AOD)).\n\\end{aligned}\n$$\nHere, we used the fact that $O$ is the centre of the circle containing points $A$, $B$, $D$ and $E$. Triangles $AOC$ and $COE$ are equilateral, so $\\angle AOE = 120^\\circ$. The triangle $BOD$ is also equilateral, so $\\angle BOD = 60^\\circ$. Thus,\n$$\n\\begin{aligned}\n\\overrightarrow{H_1H_2} \\cdot \\overrightarrow{H_2H_3} &= |OA|^2(\\cos(\\angle BOE) - \\cos(60^\\circ) - \\cos(120^\\circ) + \\cos(\\angle AOD)) \\\\\n&= |OA|^2(\\cos(\\angle BOE) + \\cos(\\angle AOD)) \\\\\n&= 2|OA|^2 \\cos \\left( \\frac{\\angle BOE + \\angle AOD}{2} \\right) \\cos \\left( \\frac{\\angle BOE - \\angle AOD}{2} \\right).\n\\end{aligned}\n$$\nVerify that $\\angle BOE + \\angle AOD = 2\\angle BOD + \\angle DOE + \\angle AOB$, which implies\n$$\n\\begin{aligned}\n\\frac{\\angle BOE + \\angle AOD}{2} &= \\angle BOD + \\frac{\\angle DOE + \\angle AOB}{2} = 60^\\circ + \\angle DCE + \\angle ACB \\\\\n&= 60^\\circ + \\angle DCB - \\angle ECA = 60^\\circ + (180^\\circ - 30^\\circ) - 120^\\circ = 90^\\circ = \\frac{\\pi}{2},\n\\end{aligned}\n$$\nso $\\cos(\\frac{\\angle BOE+\\angle AOD}{2}) = 0$ and $\\overrightarrow{H_1H_2} \\cdot \\overrightarrow{H_2H_3} = 0$. Hence $H_1H_2H_3$ is a right triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41244,
    "subject": "Mathematics (Multi-modal)",
    "question": "The terms $a_1, a_2, a_3, a_4, a_5$ of a geometric sequence are positive integers. All are less than $2008$, $a_2$ is divisible by $5$, $a_3$ is divisible by $4$, $a_4$ is divisible by $3$, $a_1$ is not divisible by $6$ and no prime number divides all five of them. Determine the terms of this sequence.",
    "options": [],
    "answer": "625, 750, 900, 1080, 1296",
    "solution": "Since $a_1, a_2, a_3, a_4, a_5$ are terms of a geometric sequence, we can write $a_i = a_1 \\cdot q^{i-1}$ for $i = 2, 3, 4, 5$ and for some real number $q$. But $q = \\frac{a_2}{a_1}$ is a quotient of two positive integers, so $q$ is rational. Write $q = \\frac{m}{n}$ as a reduced fraction. The terms of the sequence are\n$$\na_1, \\frac{a_1 \\cdot m}{n}, \\frac{a_1 \\cdot m^2}{n^2}, \\frac{a_1 \\cdot m^3}{n^3}, \\frac{a_1 \\cdot m^4}{n^4}.\n$$\nSince all terms are positive integers and $m$ and $n$ are relatively prime, we see that $n^4$ is a divisor of $a_1$. So, we can write $a_1 = dn^4$, where $d$ is some positive integer. Therefore, the terms of the sequence are\n$$\ndn^4, dmn^3, dm^2n^2, dm^3n, dm^4.\n$$\nBut since no prime number divides all five of these terms, we have $d = 1$, and the sequence becomes $n^4, mn^3, m^2n^2, m^3n, m^4$. We also know that the terms are all less than $2008$, so $m^4 < 2008$ and $n^4 < 2008$. This implies $m \\le 6$ and $n \\le 6$. But $a_1 = n^4$ is not divisible by $6$, so $n \\le 5$. We also know that $a_2 = n^3m$ is divisible by $5$, $a_3 = n^2m^2$ is divisible by $4$ and $a_4 = m^3n$ is divisible by $3$, so the product $mn$ is divisible by $2$, $3$ and $5$, i.e. by $30$. At the same time $mn \\le 6 \\cdot 5 = 30$, so this product is equal to $30$. This implies $m = 6$ and $n = 5$. The terms of the sequence are $5^4 = 625$, $5^3 \\cdot 6 = 750$, $5^2 \\cdot 6^2 = 900$, $5 \\cdot 6^3 = 1080$ and $6^4 = 1296$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41245,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest possible value of the expression $|5^{4m+3} - n^2|$, given that $m$ and $n$ are non-negative integers.",
    "options": [],
    "answer": "4",
    "solution": "Assume that $n^2 - 5^{4m+3} > 0$. If $n^2 - 5^{4m+3} = 1$, then $5^{4m+3} = n^2 - 1 = (n-1)(n+1)$. Since $n$ is obviously even the numbers $n-1$ and $n+1$ must be coprime, which implies $n-1 = 1$ and $n+1 = 5^{4m+3}$. This is not possible. If $n$ is divisible by $5$, then $n^2 - 5^{4m+3} \\ge 5$, if $n$ is of the form $n = 5k \\pm 1$, then\n\n$n^2 - 5^{4m+3} \\equiv 1 \\pmod{5}$, and if $n = 5k \\pm 2$, then $n^2 - 5^{4m+3} \\equiv 4 \\pmod{5}$. Since none of the remainders are equal to $2$ or $3$ and $n^2 - 5^{4m+3} > 1$, we conclude that $n^2 - 5^{4m+3} \\ge 4$.\n\nNow, assume $5^{4m+3} - n^2 > 0$. This number cannot be equal to either $2$ or $3$. We can see this by considering the remainders modulo $5$, because $5^{4m+3} - n^2$ is either divisible by $5$ (and hence greater than or equal to $5$), or $n$ has the form $n = 5k \\pm 1$ or $n = 5k \\pm 2$. So, either $5^{4m+3} - n^2 \\equiv -1 \\equiv 4 \\pmod{5}$ or $5^{4m+3} - n^2 \\equiv -4 \\equiv 1 \\pmod{5}$.\n\nAssume there exist $m$ and $n$ such that $5^{4m+3} - n^2 = 1$. We can rewrite the equation as\n$$\n\\begin{aligned}\nn^2 &= 5^{4m+3} - 1 = (5-1)(5^{4m+2} + 5^{4m+1} + \\dots + 5 + 1) = \\\\\n&= 4(5^{4m+2} + 5^{4m+1} + \\dots + 5 + 1).\n\\end{aligned}\n$$\nThis implies that $5^{4m+2} + 5^{4m+1} + \\dots + 5 + 1$ is a perfect square. If it is equal to $l^2$, then $(l-1)(l+1) = 5(5^{4m+1} + 5^4 + \\dots + 5 + 1)$. Let us consider the expression $5^{4m+1} + 5^4 + \\dots + 5 + 1$ modulo $4$. We have\n$$\n5^{4m+1} + 5^4 + \\dots + 5 + 1 \\equiv 1 + 1 + \\dots + 1 + 1 \\equiv (4m + 2) \\equiv 2 \\pmod{4},\n$$\nsince there are $4m + 2$ summands in the sum. This sum is therefore divisible by $2$ but not by $4$. So, $(l-1)(l+1)$ is even. On the other hand, $l-1$ and $l+1$ have the same parity, so $(l-1)(l+1)$ is divisible by $4$, a contradiction. The numbers $m$ and $n$ such that $5^{4m+3} - n^2 = 1$ do not exist.\n\nWe have shown that $n^2 - 5^{4m+3}$ and $5^{4m+3} - n^2$ cannot be equal to $0$, $1$, $2$ or $3$, so the smallest possible value of $|5^{4m+3} - n^2|$ is $4$ and this value is attained when $n=11$ and $m=0$, for example.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41246,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$ and $q$ such that $2p^2q + 45pq^2$ is a perfect square.",
    "options": [],
    "answer": "(p, q) = (47, 47) and (3, 2)",
    "solution": "First assume $p = q$. Then $47p^3$ must be a perfect square. Since $47p^3$ is divisible by $47$ which is prime, it must also be divisible by $47^2$. This implies that $47$ divides either $p^3$ or $p$. But $p$ is a prime, so this is only possible when $p$ is equal to $47$. Indeed, when $p = q = 47$ we have $2p^2q + 45pq^2 = 47^4$, which is a perfect square.\n\nNow, let $p \\neq q$. Since $2p^2q + 45pq^2 = pq(2p + 45q)$ is a perfect square divisible by $p$, it must also be divisible by $p^2$. So, $p$ divides $q(2p + 45q)$ or, since $p$ and $q$ are coprime, $p$ divides $2p + 45q$. We conclude that $p$ divides $45q$, so $p$ divides $45$. There are two possible cases: $p = 3$ or $p = 5$. A similar argument shows that if $q$ divides $2p + 45q$, then $q$ divides $2p$, which implies $q = 2$. So, $pq(2p + 45q) = 4p(p + 45)$. If $p = 3$, then this expression is equal to $4 \\cdot 3 \\cdot 48 = 24^2$. If $p = 5$, then it is equal to $4000$ and this is not a perfect square. The only two pairs of primes with the required property are $p = q = 47$ and $p = 3, q = 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41247,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle. Denote its orthocentre by $H$ and let $A'$, $B'$ and $C'$ be the feet of the altitudes from $A$, $B$ and $C$. Let $P$ be the midpoint of $AH$, let $Q$ be the intersection of lines $B'P$ and $AB$ and denote the intersection of segments $A'C'$ and $BB'$ by $R$. Prove that the line $QR$ is perpendicular to the side $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The midpoint of the hypotenuse is also the circumcentre, so $P$ is the circumcentre of the triangle $AHB'$ and $|AP| = |PH| = |PB'|$. The triangle $HPB'$ is isosceles with the apex at $P$. Let $QB'B = \\alpha$. Then $\\alpha = \\angle PB'H = \\angle B'HP = \\angle BHA'$. Since $\\angle HA'B + \\angle BC'H = \\frac{\\pi}{2} + \\frac{\\pi}{2} = \\pi$, points $A'$, $B$, $C'$ and $H$ are concyclic and $\\angle BC'A' = \\angle BHA' = \\alpha$. In the quadrilateral $B'QC'R$ we have\n\n![](attached_image_1.png)\n\n$$\n\\angle RC'Q + \\angle QB'R = \\pi - \\angle BC'A + \\angle QB'B = \\pi - \\alpha + \\alpha = \\pi,\n$$\nso $B'$, $Q$, $C'$ and $R$ are concyclic.\n\nSince $AC'B'H$ is a cyclic quadrilateral we have $\\alpha = \\angle B'HA = \\angle B'C'A$. The connection between the inscribed angles of a cyclic quadrilateral $B'QC'R$ gives us the equalities $\\angle B'RQ = \\angle B'C'Q = \\angle B'C'A = \\alpha$. So, $\\angle B'RQ = \\alpha = \\angle B'HA$ and $AH$ and $QR$ are parallel. Since $AH$ is perpendicular to $BC$ it is also perpendicular to $QR$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41248,
    "subject": "Mathematics (Multi-modal)",
    "question": "Jaka chooses a three-digit number $x$, composed of three different non-zero digits. He then takes a piece of paper and writes down all other three-digit numbers he can form from those three digits. The sum of the numbers on the paper is $3434$. Find all possible $x$.",
    "options": [],
    "answer": "784",
    "solution": "Denote the digits of $x$ by $a$, $b$ and $c$, so that $x = \\overline{abc}$. Three-digit numbers we can form from $a$, $b$ and $c$ are $\\overline{abc}$, $\\overline{acb}$, $\\overline{bac}$, $\\overline{bca}$, $\\overline{cab}$, $\\overline{cba}$ and their sum is $100(2a + 2b + 2c) + 10(2a + 2b + 2c) + (2a + 2b + 2c) = 222(a + b + c)$. Hence, the sum of the numbers written on the paper is\n$$\n3434 = 222(a + b + c) - \\overline{abc} = 122a + 212b + 221c.\n$$\nConsider this equation modulo $9$. When $3434$ is divided by $9$ the remainder is $5$ and the remainder of $122a + 212b + 221c$ is the same as that of $5a + 5b + 5c = 5(a + b + c)$. We conclude that $a + b + c$ should be congruent to $1$. But $6 = 1 + 2 + 3 \\le a + b + c \\le 7 + 8 + 9 = 24$, so $a + b + c$ can only be $10$ or $19$.\nIf $a + b + c = 10$, we have $3434 = 122a + 212b + 221c < 221(a + b + c) = 221 \\cdot 10 < 3434$, and this is not possible. So, $a + b + c = 19$ and $\\overline{abc} = 222(a + b + c) - 3434 = 222 \\cdot 19 - 3434 = 784$. The only possible solution is $x = 784$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41249,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $x$ for which the inequality\n$$\n|||2 - x| - x| - 8| \\le 2008\n$$\nholds.",
    "options": [],
    "answer": "x ≥ -1007",
    "solution": "We consider two cases depending on the sign of $2-x$.\n\nIf $x \\ge 2$, then we have $|2-x| = -(2-x)$ and the inequality can be rewritten as $||x-2-x|-8| \\le 2008$ or, equivalently, $6 \\le 2008$. So for all real numbers $x \\ge 2$ the inequality holds.\n\nNow, let $x < 2$. In this case we have\n$$\n||2 - 2x| - 8| \\le 2008.\n$$\nLet us consider two further cases depending on the sign of $2-2x$.\n\nIf $x \\ge 1$, we have $|2-2x| = -(2-2x)$ and so $|2x-10| \\le 2008$. Obviously, this inequality holds for $1 \\le x < 2$.\n\nFinally, only the case $x < 1$ remains. The condition implies $|-2x+2-8| \\le 2008$ or $|2x+6| \\le 2008$. This is equivalent to\n$$\n-2008 \\le 2x + 6 \\le 2008.\n$$\nSince $x < 1$ we have $2x + 6 < 2 + 6 = 8$, so the right inequality holds. The left inequality implies $-2014 \\le 2x$ or $-1007 \\le x$. From this we derive the condition $-1007 \\le x < 1$.\n\nHence, the inequality holds for all $x \\ge -1007$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41250,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $x$ and $y$ that satisfy the equations\n$$\n\\begin{aligned}\nx^3 + 8y^3 &= x + 2y, \\\\\n2x^2y + 4xy^2 &= x + 2y.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "All real pairs (x, y) with x + 2y = 0, together with (1, 1/2) and (-1, -1/2).",
    "solution": "We can rewrite the equations as\n$$\n\\begin{aligned}\n(x + 2y)(x^2 - 2xy + 4y^2) &= x + 2y, \\\\\n2xy(x + 2y) &= x + 2y.\n\\end{aligned}\n$$\nObviously, every pair of numbers $x$ and $y$ that satisfies $x + 2y = 0$, solves the equations. Now, assume $x + 2y \\neq 0$. We can then divide by $x + 2y$ and get $x^2 - 2xy + 4y^2 = 1$ and $2xy = 1$, so\n$$\n(x + 2y)^2 = x^2 + 4xy + 4y^2 = (x^2 - 2xy + 4y^2) + 6xy = 4.\n$$\nWe consider two cases: $x + 2y = 2$ or $x + 2y = -2$. In the first case we have $1 = 2xy = 2(2 - 2y)y = 4y - 4y^2$, so $0 = 4y^2 - 4y + 1 = (2y - 1)^2$. This implies $y = \\frac{1}{2}$ and $x = 2 - 2y = 1$. In the second case we have $x = -2 - 2y$, so $1 = 2xy = 2(-2 - 2y)y = -4y - 4y^2$ and $0 = (2y + 1)^2$. This implies $y = -\\frac{1}{2}$ and $x = -1$.\n\nSolutions are all pairs of real numbers $x$ and $y$ such that $x + 2y = 0$, as well as the pairs $x = -1$, $y = -\\frac{1}{2}$ and $x = 1$, $y = \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41251,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for all real numbers $x \\in \\left(-\\frac{3\\pi}{2}, \\frac{\\pi}{2}\\right)$ the equality\n$$\n\\frac{2}{1 - \\sin x} = \\tan^2 \\left( \\frac{x}{2} + \\frac{\\pi}{4} \\right) + 1\n$$\nholds.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the addition formula for tangents we get\n$$\n\\tan^2 \\left( \\frac{x}{2} + \\frac{\\pi}{4} \\right) = \\left( \\frac{\\tan \\frac{x}{2} + \\tan \\frac{\\pi}{4}}{1 - \\tan \\frac{x}{2} \\tan \\frac{\\pi}{4}} \\right)^2 .\n$$\nUsing the fact that $\\tan \\frac{\\pi}{4} = 1$ and expressing $\\tan \\frac{x}{2}$ in terms of sines and cosines, we see that\n$$\n\\tan^2 \\left( \\frac{x}{2} + \\frac{\\pi}{4} \\right) = \\left( \\frac{\\frac{\\sin \\frac{x}{2}}{\\cos \\frac{x}{2}} + 1}{1 - \\frac{\\sin \\frac{x}{2}}{\\cos \\frac{x}{2}}} \\right)^2 = \\left( \\frac{\\cos \\frac{x}{2} + \\sin \\frac{x}{2}}{\\cos \\frac{x}{2} - \\sin \\frac{x}{2}} \\right)^2 .\n$$\nWe now square both sides of the equation to get\n$$\n\\tan^2 \\left( \\frac{x}{2} + \\frac{\\pi}{4} \\right) = \\frac{\\cos^2 \\frac{x}{2} + 2 \\cos \\frac{x}{2} \\sin \\frac{x}{2} + \\sin^2 \\frac{x}{2}}{\\cos^2 \\frac{x}{2} - 2 \\cos \\frac{x}{2} \\sin \\frac{x}{2} + \\sin^2 \\frac{x}{2}}\n$$\nand use the relations $\\cos^2 \\frac{x}{2} + \\sin^2 \\frac{x}{2} = 1$ and $2 \\cos \\frac{x}{2} \\sin \\frac{x}{2} = \\sin x$ to show that\n$$\n\\tan^2 \\left( \\frac{x}{2} + \\frac{\\pi}{4} \\right) = \\frac{1 + \\sin x}{1 - \\sin x} = \\frac{2}{1 - \\sin x} - 1.\n$$\nThus, the equality holds.\n\n\nSolution 2:\nExpress the tangent in terms of sine and cosine\n$$\n\\tan^2 \\left( \\frac{x}{2} + \\frac{\\pi}{4} \\right) = \\left( \\frac{\\sin \\left( \\frac{x}{2} + \\frac{\\pi}{4} \\right)}{\\cos \\left( \\frac{x}{2} + \\frac{\\pi}{4} \\right)} \\right)^2 .\n$$\nNow, use the addition formulas for sine and cosine to show that\n$$\n\\tan^2 \\left( \\frac{x}{2} + \\frac{\\pi}{4} \\right) = \\left( \\frac{\\sin \\frac{x}{2} \\cos \\frac{\\pi}{4} + \\cos \\frac{x}{2} \\sin \\frac{\\pi}{4}}{\\cos \\frac{x}{2} \\cos \\frac{\\pi}{4} - \\sin \\frac{x}{2} \\sin \\frac{\\pi}{4}} \\right)^2 .\n$$\nWe know that $\\sin \\frac{\\pi}{4} = \\frac{\\sqrt{2}}{2}$ and $\\cos \\frac{\\pi}{4} = \\frac{\\sqrt{2}}{2}$, so the expression is equal to\n$$\n\\tan^2 \\left( \\frac{x}{2} + \\frac{\\pi}{4} \\right) = \\left( \\frac{\\frac{\\sqrt{2}}{2} \\sin \\frac{x}{2} + \\frac{\\sqrt{2}}{2} \\cos \\frac{x}{2}}{\\frac{\\sqrt{2}}{2} \\cos \\frac{x}{2} - \\frac{\\sqrt{2}}{2} \\sin \\frac{x}{2}} \\right)^2 .\n$$\nSquaring both sides we get\n$$\n\\tan^2 \\left( \\frac{x}{2} + \\frac{\\pi}{4} \\right) = \\frac{\\cos^2 \\frac{x}{2} + 2 \\cos \\frac{x}{2} \\sin \\frac{x}{2} + \\sin^2 \\frac{x}{2}}{\\cos^2 \\frac{x}{2} - 2 \\cos \\frac{x}{2} \\sin \\frac{x}{2} + \\sin^2 \\frac{x}{2}}\n$$\nand the equalities $\\cos^2 \\frac{x}{2} + \\sin^2 \\frac{x}{2} = 1$ and $2 \\cos \\frac{x}{2} \\sin \\frac{x}{2} = \\sin x$ imply\n$$\n\\tan^2 \\left( \\frac{x}{2} + \\frac{\\pi}{4} \\right) = \\frac{1 + \\sin x}{1 - \\sin x} = \\frac{2}{1 - \\sin x} - 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41252,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integral solutions of the equation\n$$\n\\frac{x^2}{2} + \\frac{5}{y} = 7.\n$$",
    "options": [],
    "answer": "(2, 1), (3, 2), (4, -5), (-2, 1), (-3, 2), (-4, -5)",
    "solution": "If $(x, y)$ satisfies the equation then so does $(-x, y)$. We can thus limit ourselves to only non-negative values of $x$.\n\nMultiply the equation by $2y$ to get $x^2y + 10 = 14y$. This implies $y = \\frac{10}{14 - x^2}$.\n\nWe see that $14 - x^2$ has to be a divisor of $10$, so $-10 \\leq 14 - x^2 \\leq 10$, $-24 \\leq -x^2 \\leq -4$ and $24 \\geq x^2 \\geq 4$. Since we have assumed that $x \\geq 0$, we get $5 > x \\geq 2$.\n\nWhen $x = 2$ we have $y = 1$, when $x = 3$ we have $y = 2$ and when $x = 4$ we have $y = -5$.\n\nHence, the only integral solutions of the equation are $(2, 1)$, $(3, 2)$, $(4, -5)$, $(-2, 1)$, $(-3, 2)$ and $(-4, -5)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41253,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana je družina linearnih funkcij: $f(x) = m x + r x - 2 m + r + 1$. Poišči vrednosti parametrov $m$ in $r$ tako, da bo graf izbrane linearne funkcije vzporeden premici $y = -x$ in bo potekal skozi točko $T(0, 3)$.",
    "options": [],
    "answer": "m = -1, r = 0",
    "solution": "Solution:\n\nOdčitamo smerni koeficient družine premic $f(x)$. Ker sta premici vzporedni, enačimo smerna koeficienta družine premic in simetrale sodih kvadrantov $m + r = -1$. Točka $T$ je presečišče premice z ordinatno osjo, zato iz družine funkcij $f(x)$ odčitamo začetno vrednost. Zapišemo enačbo $-2 m + r + 1 = 3$. Rešimo dobljeni sistem. Rešitvi sta $m = -1$ in $r = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41254,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJanez je porabil $\\frac{3}{5}$ denarja, ki ga je imel, nato še $\\frac{5}{9}$ ostanka ter še $\\frac{3}{8}$ novega ostanka. Ostalo mu je 80 evrov. Koliko denarja je imel na začetku?",
    "options": [],
    "answer": "720",
    "solution": "Solution:\n\nNaj bo $x$ prvotni znesek. Izračunamo delež denarja, ki ga ima Janez po prvem izdatku, to je $\\frac{2}{5} x$. Nato izračunamo nov delež denarja, ki ga ima Janez po drugem izdatku, to je $\\frac{8}{45} x$. Po tretjem izdatku ima Janez še $\\frac{1}{9} x$ denarja. Nastavimo enačbo $\\frac{1}{9} x=80$. Izračunamo iskani znasek 720 evrov.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41255,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa realna števila $x$ in $y$, ki zadoščajo enačbama\n$$\n\\begin{aligned}\nx^{3}+8 y^{3} & =x+2 y \\\\\n2 x^{2} y+4 x y^{2} & =x+2 y\n\\end{aligned}\n$$",
    "options": [],
    "answer": "{(x,y) : x + 2y = 0} ∪ {(1, 1/2), (-1, -1/2)}",
    "solution": "Solution:\n\nEnačbi lahko prepišemo v obliko\n$$\n\\begin{aligned}\n(x+2 y)\\left(x^{2}-2 x y+4 y^{2}\\right) & =x+2 y \\\\\n2 x y(x+2 y) & =x+2 y\n\\end{aligned}\n$$\nOčitno vsak par števil $x$ in $y$, ki zadošča zvezi $x+2 y=0$, reši enačbi. Naj bo sedaj $x+2 y \\neq 0$. Tedaj lahko delimo z $x+2 y$ in dobimo $x^{2}-2 x y+4 y^{2}=1$ in $2 x y=1$, torej je\n$$\n(x+2 y)^{2}=x^{2}+4 x y+4 y^{2}=\\left(x^{2}-2 x y+4 y^{2}\\right)+6 x y=4\n$$\nLočimo dve možnosti in sicer je $x+2 y=2$ ali pa $x+2 y=-2$. V prvem primeru dobimo $1=2 x y=2(2-2 y) y=4 y-4 y^{2}$, torej $0=4 y^{2}-4 y+1=(2 y-1)^{2}$. Od tod sledi $y=\\frac{1}{2}$ in $x=2-2 y=1$. V drugem primeru pa je $x=-2-2 y$ in zato velja $1=2 x y=2(-2-2 y) y=$ $-4 y-4 y^{2}$, torej je $0=(2 y+1)^{2}$. Sledi $y=-\\frac{1}{2}$ in $x=-1$.\n\nEnačbi tako zadoščajo vsa realna števila $x$ in $y$, za katera velja $x+2 y=0$, poleg teh pa še $x=-1, y=-\\frac{1}{2}$ in $x=1, y=\\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41256,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJaka si je zamislil trimestno število $x$, ki ima v zapisu različne neničelne števke. Nato je na list napisal vsa druga trimestna števila, ki jih je lahko zapisal s števkami števila $x$. Določi vsa možna števila $x$, če je vsota števil na listu enaka $3434$.",
    "options": [],
    "answer": "784",
    "solution": "Solution:\n\nOznačimo števke števila $x$ z $a, b$ in $c$, torej $x=\\overline{a b c}$. Vsa trimestna števila, sestavljena iz števk $a, b$ in $c$ so $\\overline{a b c}$, $\\overline{a c b}$, $\\overline{b a c}$, $\\overline{b c a}$, $\\overline{c a b}$, $\\overline{c b a}$, njihova vsota pa je $100(2a+2b+2c)+10(2a+2b+2c)+(2a+2b+2c)=222(a+b+c)$. Vsota števil na listu je tako enaka\n$$\n3434=222(a+b+c)-\\overline{a b c}=122a+212b+221c\n$$\nOglejmo si zgornjo enačbo glede na deljivost z $9$. Ostanek števila $3434$ pri deljenju z $9$ je enak $5$, ostanek števila $122a+212b+221c$ pa je enak ostanku števila $5a+5b+5c=5(a+b+c)$. Od tod sledi, da mora biti ostanek števila $a+b+c$ pri deljenju z $9$ enak $1$. Ker pa je $6=1+2+3 \\leq a+b+c \\leq 7+8+9=24$, je torej $a+b+c$ enako bodisi $10$ bodisi $19$.\nČe je $a+b+c=10$, lahko ocenimo $3434=122a+212b+221c<221(a+b+c)=221\\cdot 10<3434$, kar ni možno. Torej je $a+b+c=19$. Sedaj lahko izračunamo $\\overline{a b c}=222(a+b+c)-3434=222\\cdot 19-3434=784$. Edina možnost je $x=784$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41257,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa praštevila $p$ in $q$, za katera je število $2 p^{2} q + 45 p q^{2}$ popoln kvadrat.",
    "options": [],
    "answer": "p = q = 47 and p = 3, q = 2",
    "solution": "Solution:\n\nNajprej denimo, da je $p = q$. Potem mora biti število $47 p^{3}$ popoln kvadrat. Ker je deljivo s $47$ in je $47$ praštevilo, mora biti deljivo tudi s $47^{2}$, od koder sledi, da $47$ deli $p^{3}$ oziroma $47$ deli $p$. Toda $p$ je praštevilo, torej mora biti enako $47$. Res, pri $p = q = 47$ je število $2 p^{2} q + 45 p q^{2}$ enako $47^{4}$ in je torej popoln kvadrat.\n\nNaj bo sedaj $p \\neq q$. Ker je število $2 p^{2} q + 45 p q^{2} = p q (2 p + 45 q)$ popoln kvadrat deljiv s $p$, mora biti deljiv tudi s $p^{2}$. Torej $p$ deli $q (2 p + 45 q)$ oziroma, ker sta $p$ in $q$ tuji, $p$ deli $2 p + 45 q$. Od tod sledi, da $p$ deli $45 q$ oziroma, da $p$ deli $45$. Torej je $p = 3$ ali pa $p = 5$. Podobno sklepamo, da $q$ deli $2 p + 45 q$, od koder sledi, da $q$ deli $2 p$ oziroma $q = 2$. Torej je $p q (2 p + 45 q) = 4 p (p + 45)$. Če je $p = 3$, je to število enako $4 \\cdot 3 \\cdot 48 = 24^{2}$, pri $p = 5$ pa je enako $4000$ in ni popolni kvadrat. Edini rešitvi sta torej $p = q = 47$ in $p = 3, q = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41258,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV nekem podjetju je zaposlenih 150 ljudi. Direktor prejema mesečno plačo 12000 evrov, trije ožji sodelavci 5000 evrov, 12 najslabše plačanih delavcev dobi 500 evrov, preostali delavci zaslužijo bodisi 1500 bodisi polovico tega zneska. Koliko zaposlenih zasluži mesečno 1500 evrov in koliko polovico manj, če je povprečna mesečna plača 1010 evrov?",
    "options": [],
    "answer": "24 employees earn 1500 euros and 110 employees earn 750 euros.",
    "solution": "Solution:\n\nNaj bo $x$ število zaposlenih, ki zaslužijo $1500$ evrov, in $y$ število zaposlenih, ki zaslužijo $750$ evrov.\n\nZapišemo enačbi:\n\n$1 + 3 + 12 + x + y = 150$\n\n$\\frac{12000 + 3 \\cdot 5000 + 12 \\cdot 500 + x \\cdot 1500 + y \\cdot 750}{150} = 1010$\n\nEnačbi uredimo in dobimo sistem:\n\n$x + y = 134$\n\n$2x + y = 158$\n\nSistem rešimo in dobimo rešitvi $x = 24$, $y = 110$.\n\nZapišemo odgovor: $24$ delavcev zasluži $1500$ evrov, polovico manj pa $110$ delavcev.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41259,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nČleni $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ geometrijskega zaporedja so naravna števila, manjša od 2008. Število $a_{2}$ je deljivo s $5$, $a_{3}$ je deljivo s $4$, $a_{4}$ je deljivo s $3$, število $a_{1}$ pa ni deljivo s $6$. Nobeno praštevilo ne deli vseh $5$ členov zaporedja. Izračunaj člene tega zaporedja.",
    "options": [],
    "answer": "625, 750, 900, 1080, 1296",
    "solution": "Solution:\n\nKer so $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ členi geometrijskega zaporedja, jih lahko zapišemo v obliki $a_{i}=a_{1} \\cdot q^{i-1}$ za $i=2,3,4,5$, kjer je $q$ neko realno število. Toda $q=\\frac{a_{2}}{a_{1}}$ je kvocient dveh naravnih števil, torej je racionalno število. Zapišimo $q=\\frac{m}{n}$ kot okrajšani ulomek. Tedaj so členi zaporedja enaki\n$$\na_{1}, \\frac{a_{1} \\cdot m}{n}, \\frac{a_{1} \\cdot m^{2}}{n^{2}}, \\frac{a_{1} \\cdot m^{3}}{n^{3}}, \\frac{a_{1} \\cdot m^{4}}{n^{4}}\n$$\nKer so vsi členi naravna števila, $m$ in $n$ pa sta si tuji, je $n^{4}$ delitelj števila $a_{1}$. Zato lahko zapišemo $a_{1}=d n^{4}$, kjer je $d$ neko naravno število. Torej so členi tega zaporedja števila\n$$\nd n^{4}, d m n^{3}, d m^{2} n^{2}, d m^{3} n, d m^{4}\n$$\nKer pa ne obstaja praštevilo, ki bi delilo vse člene zaporedja, sledi $d=1$, zaporedje pa je oblike $n^{4}, m n^{3}, m^{2} n^{2}, m^{3} n, m^{4}$. Vemo še, da so členi zaporedja manjši od $2008$, zato je $m^{4}<2008$ in $n^{4}<2008$. Od tod sledi, da je $m \\leq 6$ in $n \\leq 6$. Toda člen $a_{1}=n^{4}$ ni deljiv s $6$, zato je $n \\leq 5$. Vemo še, da je $a_{2}=n^{3} m$ deljiv s $5$, $a_{3}=n^{2} m^{2}$ deljiv s $4$ in $a_{4}=m^{3} n$ deljiv s $3$, zato je produkt $m n$ deljiv z $2$, $3$ in $5$, torej s $30$. Hkrati pa je $m n \\leq 6 \\cdot 5=30$, torej je produkt kar enak $30$, od koder sledi $m=6$ in $n=5$. Členi zaporedja so tako $5^{4}=625$, $5^{3} \\cdot 6=750$, $5^{2} \\cdot 6^{2}=900$, $5 \\cdot 6^{3}=1080$ in $6^{4}=1296$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41260,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $x$ such that $\\sqrt{1-x^2} + \\sqrt{5x-x^2}$ is an integer.",
    "options": [],
    "answer": "0, 9/41, 1",
    "solution": "Let us find an estimate for the value of this expression. Evidently, $1 - x^2 \\le 1$ and $5x - x^2$ is bounded by $5x - x^2 = \\frac{25}{4} - (x - \\frac{5}{2})^2 \\le \\frac{25}{4}$, so\n$$\n\\sqrt{1-x^2} + \\sqrt{5x-x^2} \\le \\sqrt{1} + \\sqrt{\\frac{25}{4}} = 1 + \\frac{5}{2} = 3 + \\frac{1}{2}.\n$$\nOn the other hand, the expression is non-negative and cannot be equal to 0 because that would imply $1 - x^2 = 0$ and $5x - x^2 = 0$, which is impossible. So, the only integer values the expression can take are 1, 2 or 3. Let $\\sqrt{1-x^2} + \\sqrt{5x-x^2} = a$ and eliminate the square roots. By squaring both sides of $\\sqrt{1-x^2} = a - \\sqrt{5x-x^2}$, we get $1 - x^2 = a^2 - 2a\\sqrt{5x-x^2} + 5x - x^2$ or $2a\\sqrt{5x-x^2} = a^2 - 1 + 5x$. After squaring both sides once again, we get $4a^2(5x - x^2) = (a^2 - 1)^2 - 10x + 10a^2x + 25x^2$ or\n$$\nx^2(25 + 4a^2) + x(-10a^2 - 10) + (a^2 - 1)^2 = 0.\n$$\nIf $a = 1$ we have $x(29x - 20) = 0$, so $x = 0$ or $x = \\frac{20}{29}$. Inserting these values of $x$ into the initial expression we see that only $x = 0$ works.\n\nWhen $a = 2$ the equation becomes $(x - 1)(41x - 9) = 0$, so $x = 1$ or $x = \\frac{9}{41}$. In both cases a short calculation confirms that the value of the expression is indeed 2.\n\nFinally, let us consider the case $a = 3$. The quadratic equation $61x^2 - 100x + 64 = 0$ has a negative discriminant $D = 100^2 - 4 \\cdot 61 \\cdot 64 = 100^2 - (2 \\cdot 64)(2 \\cdot 61) = 100^2 - 128 \\cdot 122$, so there are no real solutions.\n\nWe conclude that the value of the expression is an integer only when $x = 0$, $x = \\frac{9}{41}$ or $x = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41261,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles triangle with the apex at $C$ and let $A'$ be the foot of the altitude from $A$. Assuming that $|CA'| = \\frac{1}{2}|AB|$, prove that the triangle $ABC$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $C'$ be the foot of the altitude from $C$. Since the triangle $ABC$ is isosceles with the apex at $C$, we have $|AC'| = |C'B| = \\frac{1}{2}|AB| = |CA'|$. Since $ABA'$ and $CBC'$ are right triangles and $\\angle ABA' = \\angle CBC'$, they are similar. This implies\n$$\n\\frac{|AB|}{|BA'|} = \\frac{|CB|}{|BC'|}.\n$$\nLet $c$ be the length of the side $AB$ and $x = |BA'|$. Then\n$$\n\\frac{c}{x} = \\frac{\\frac{c}{2} + x}{\\frac{c}{2}},\n$$\n![](attached_image_1.png)\nand we get the quadratic equation $2x^2 + cx - x^2 = 0$. The left-hand side can be factored as $(2x - c)(x + c) = 0$. Since $x$ and $c$ are positive, we have $x = \\frac{c}{2}$. The length of the segment $BC$ is therefore equal to $c$, $|AC| = |BC| = |AB| = c$ and the triangle $ABC$ is equilateral.\n\n![](attached_image_1.png)\n\nSecond solution\n\nDenote the length of the side $AC$ by $a$ and the length of the side $AB$ by $c$. Then $|CA'| = \\frac{c}{2}$ and $|A'B| = a - \\frac{c}{2}$. We can now calculate the length of the altitude $AA'$ in two different ways, namely as the cathetus in triangles $ACA'$ and $ABA'$, respectively. Let $|AA'| = v$. Pythagoras's theorem for the first of the triangles implies $v^2 = a^2 - (c/2)^2$ and for the second $v^2 = c^2 - (a - (c/2))^2$, so\n$$\na^2 - \\frac{c^2}{4} = c^2 - a^2 + ac - \\frac{c^2}{4}\n$$\nor $2a^2 - ac - c^2 = 0$. This quadratic equation can be factored as $(2a + c)(a - c) = 0$, which implies $a = c$ (since $a$ and $c$ are positive the condition $2a + c = 0$ is never satisfied). From $|BC| = |AC| = a = c$ we conclude that the triangle $ABC$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41262,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a(b + c) + b(c + a) + c(a + b) = ab + bc + ca$, then\n$$\n\\frac{a^2(b+c) + b^2(a+c) + c^2(a+b)}{abc}\n$$\nis an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The given equation implies $ab + bc + ca = 0$, so we can write\n$$\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$$\n\nSince $ab + ac = -bc$, $ba + bc = -ca$ and $ca + cb = -ab$ we have\n$$\n\\frac{a(ab + ac) + b(ba + bc) + c(ca + cb)}{abc} = \\frac{a(-bc) + b(-ca) + c(-ab)}{abc} = \\frac{-3abc}{abc} = -3.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41263,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for all positive integers $n$ and all real numbers $x_1, x_2, \\dots, x_n$ the inequality\n$$\n\\sqrt{x_1^2 + 1} + 2\\sqrt{x_2^2 + 1} + \\dots + n\\sqrt{x_n^2 + 1} \\ge \\sqrt{(x_1 + 2x_2 + \\dots + nx_n)^2 + \\frac{n^2(n+1)^2}{4}}\n$$\nholds.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove the inequality by induction. Obviously, for $n=1$ the equality holds and for $n=2$ we have\n$$\n\\sqrt{x_1^2 + 1} + 2\\sqrt{x_2^2 + 1} \\ge \\sqrt{(x_1 + 2x_2)^2 + 9}.\n$$\nAfter squaring both sides we rewrite it as\n$$\n\\sqrt{(x_1^2 + 1)(x_2^2 + 1)} \\ge x_1 x_2 + 1.\n$$\nIf the right-hand side is negative, then the inequality holds, if not, we can square both sides and transform it into $(x_1 - x_2)^2 \\ge 0$, which is true. So, for $n=2$ the inequality holds.\n\nLet us assume the inequality holds for $n$ and prove it for $n+1$. By induction hypothesis we have\n$$\n\\sqrt{x_1^2+1} + 2\\sqrt{x_2^2+1} + \\dots + n\\sqrt{x_n^2+1} \\ge \\sqrt{(x_1+2x_2+\\dots+nx_n)^2 + \\frac{n^2(n+1)^2}{4}},\n$$\nso it suffices to show that\n$$\n\\begin{aligned}\n& \\sqrt{(x_1 + 2x_2 + \\dots + nx_n)^2 + \\frac{n^2(n+1)^2}{4}} + (n+1)\\sqrt{x_{n+1}^2 + 1} \\ge \\\\\n& \\ge \\sqrt{(x_1 + 2x_2 + \\dots + nx_n + (n+1)x_{n+1})^2 + \\frac{(n+1)^2(n+2)^2}{4}}.\n\\end{aligned}\n$$\nThis is equivalent to\n$$\n\\begin{aligned}\n& (x_1 + 2x_2 + \\dots + nx_n)^2 + \\frac{n^2(n+1)^2}{4} + (n+1)^2 x_{n+1}^2 + (n+1)^2 + \\\\\n& + 2(n+1)\\sqrt{(x_1 + 2x_2 + \\dots + nx_n)^2 + \\frac{n^2(n+1)^2}{4}} \\sqrt{x_{n+1}^2 + 1} \\ge \\\\\n& \\ge (x_1 + 2x_2 + \\dots + nx_n)^2 + (n+1)^2 x_{n+1}^2 + 2(n+1)x_{n+1}(x_1 + 2x_2 + \\dots + nx_n) + \\\\\n& \\qquad + \\frac{(n+1)^2(n+2)^2}{4},\n\\end{aligned}\n$$\nand further to\n$$\n2\\sqrt{(x_1 + 2x_2 + \\dots + nx_n)^2 + \\frac{n^2(n+1)^2}{4}} \\sqrt{x_{n+1}^2 + 1} \\ge 2x_{n+1}(x_1 + 2x_2 + \\dots + nx_n) + n(n+1).\n$$\nIf the right-hand side is negative, then the inequality holds. If not, we can square both sides and we get $4(x_1 + 2x_2 + \\dots + nx_n)^2 x_{n+1}^2 + 4(x_1 + 2x_2 + \\dots + nx_n)^2 + n^2(n+1)^2 x_{n+1}^2 + n^2(n+1)^2 \\ge 4x_{n+1}^2(x_1 + 2x_2 + \\dots + nx_n)^2 + n^2(n+1)^2 + 4n(n+1)x_{n+1}(x_1 + 2x_2 + \\dots + nx_n)$ and finally\n$$\n(2(x_1 + 2x_2 + \\dots + nx_n) - n(n+1)x_{n+1})^2 \\ge 0.\n$$\nThis inequality obviously holds, and the equality holds when\n$$\nx_{n+1} = \\frac{2}{n(n+1)}(x_1 + 2x_2 + \\dots + nx_n).\n$$\nBy induction, the equality holds when $x_1 = x_2 = \\dots = x_n = x_{n+1}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41264,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$ such that the polynomial\n$$\nq(x) = 2x^3 - 2p x^2 + (1-p)x + p\n$$\nhas at least one rational root.",
    "options": [],
    "answer": "2",
    "solution": "If $p = 2$, we have $q(x) = 2x^3 - 4x^2 - x + 2 = (x-2)(2x^2 - 1)$ and $x = 2$ is a rational root. Now, let $p$ be an odd prime. The only possible candidates for rational roots are $\\pm 1, \\pm p, \\pm \\frac{1}{2}$ and $\\pm \\frac{p}{2}$. Let us consider all possible cases.\nSince $q(1) = 3 - 2p$ and $3 - 2p$ is odd, we have $q(1) \\neq 0$. Evidently, $q(-1) = -3 \\neq 0$. The expression $q(-p) = -p^2 + 2p = p(2-p)$ is non-zero because $p \\neq 2$. Similarly, $q(p) = -4p^3 + p^2 = p^2(1-4p) \\neq 0$ and $q(\\frac{1}{2}) = \\frac{3}{4} \\neq 0$. It is also easy to check that $q(-\\frac{1}{2}) = p - \\frac{3}{4} \\neq 0$ and $q(\\frac{p}{2}) = \\frac{-p^3-2p^2+6p}{4}$. Since $p$ is odd, the denominator of this last expression is also odd, so this expression is non-zero. The same argument shows that $q(-\\frac{p}{2}) = \\frac{-3p^3+2p^2+2p}{4} \\neq 0$.\nThus, if $p$ is an odd prime, the polynomial $q$ has no rational roots. The only solution is $p = 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41265,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a right triangle with the right angle at $C$ and let $D$ be a point on the segment $BC$. Denote the circumcircle of the triangle $ABD$ by $\\mathcal{K}$. Let $E$ be a point on $\\mathcal{K}$, such that the chord $DE$ is perpendicular to $AB$. Prove that the triangle $AEB$ is isosceles with the apex at $B$ if and only if $CA$ is tangent to $\\mathcal{K}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $T$ be the intersection of the chords $DE$ and $AB$. We know that $DTB$ is a right triangle. First, assume that the triangle $ABE$ is isosceles and write $\\angle AEB = \\angle BAE = \\alpha$. Inscribed angles $\\angle EAB$ and $\\angle EDB$ over $BE$ are equal, so $\\angle EDB = \\alpha$. Since $DE$ is perpendicular to $AB$, we have $\\angle ABD = \\frac{\\pi}{2} - \\alpha$. In the right triangle $ABC$ we have $\\angle ABC = \\frac{\\pi}{2} - \\alpha$, so $\\angle CAB = \\alpha$ and the angle $\\angle CAB$ between the line $AC$ and the segment $AB$ is equal to the angle $\\angle AEB$ over the chord $AB$. Thus, $CA$ is tangent to the circumcircle $K$.\n\nConversely, assume that $AC$ is tangent to $K$. Let $\\angle CAB = \\alpha$. The angle $\\angle BAC$ is equal to the inscribed angle $\\angle AEB$ over the chord $AB$, so $\\angle AEB = \\alpha$. Also, $\\angle ABC = \\frac{\\pi}{2} - \\alpha$, so $\\angle TDB = \\alpha$. We see that $\\angle EDB = \\alpha$ and this angle is in turn equal to $\\angle BAE$, because they are both inscribed angles over the same chord $BE$. We have $\\angle BAE = \\alpha = \\angle BEA$ and the triangle $ABE$ is isosceles with the apex at $B$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41266,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be the midpoint of the segment $AB$ and denote the centre of gravity of the triangle $ABC$ by $T$. Find the lengths of the sides given that $|AD| = 3$, $|DT| = 5$ and $|TA| = 4$.",
    "options": [],
    "answer": "AB = 6, BC = 12√2, AC = 6√5",
    "solution": "Since $D$ is the midpoint of $AB$ and $|AD| = 3$, we have $|AB| = 6$. Let $E$ be the midpoint of $BC$ and let $F$ be the midpoint of $AC$. The sides of the triangle $ADT$ satisfy Pythagoras's theorem, so $ADT$ is a right triangle. The ratio in which the centre of gravity divides the median is $2:1$ and since $|AT| = 4$ we have $|AE| = 6$. Using Pythagoras's theorem for the triangle $ABE$ we can find $|BE| = \\sqrt{|AB|^2 + |AE|^2} = \\sqrt{36+36} = 6\\sqrt{2}$. This implies $|BC| = 2|BE| = 12\\sqrt{2}$. Since $E$ and $F$ are the midpoints of $BC$ and $AC$, the segment $EF$ is parallel to $AB$.\n\n![](attached_image_1.png)\n\nSince $EA$ is perpendicular to $AB$, $EA$ is also perpendicular to $EF$. Thus, $AEF$ is a right triangle. We have already shown that $|AE| = 6$, and we have $|EF| = \\frac{|AB|}{2} = 3$. We use Pythagoras's theorem once more to find\n$$\n|FA| = \\sqrt{|EF|^2 + |AE|^2} = \\sqrt{9+36} = 3\\sqrt{5}.\n$$\nThe length of the side $AC$ is $|AC| = 2|AF| = 6\\sqrt{5}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41267,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Show that there is no positive integer $n$ such that the sum of the digits of $10^n + 9n$ is divisible by $2007$.\n\nb. Find at least one positive integer $n$ such that the sum of the digits of $10^n + 9n$ equals $2008$.",
    "options": [],
    "answer": "Part (a): No such positive integer exists. Part (b): n = 111...111 (223 ones).",
    "solution": "a. An integer is divisible by $9$ if and only if the sum of its digits is divisible by $9$. Assume that the sum of the digits of $10^n + 9n$ is divisible by $2007$. Since $2007$ is a multiple of $9$, the sum of the digits of $10^n + 9n$ must be divisible by $9$. This implies that $10^n + 9n$ is divisible by $9$, which is clearly not the case.\n\nb. Let $n = 1\\ldots1$ (223 ones). Then $9n = 9\\ldots9$ (223 nines) and $10^n + 9n$ is equal to $10\\ldots09\\ldots9$ (223 nines and $n - 223 = 1\\ldots1 - 223$ zeroes). The sum of the digits of this number is $1 + 9 \\cdot 223 = 2008$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41268,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a quadrilateral $ABCD$ let $K$ be a point inside the triangle $ABD$ such that triangles $ABD$ and $KCD$ are similar. Prove that triangles $BCD$ and $AKD$ are similar as well.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since triangles $ABD$ and $KCD$ are similar, we have $\\angle ADB = \\angle KDC$ and $\\frac{|DA|}{|DB|} = \\frac{|DK|}{|DC|}$. We see that\n$$\n\\angle ADK = \\angle ADB - \\angle BDK = \\\\\n= \\angle KDC - \\angle BDK = \\angle BDC\n$$\nand since $\\frac{|DA|}{|DK|} = \\frac{|DB|}{|DC|}$, we conclude that triangles $ADK$ and $DBC$ are also similar (matching in one angle and the ratio of the two adjacent sides).\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41269,
    "subject": "Mathematics (Multi-modal)",
    "question": "On every square of an $n \\times n$ chessboard we write one of the numbers $1$ or $-1$. Let $a_k$ be the product of all numbers in the $k^{\\text{th}}$ row and $b_l$ be the product of all numbers in the $l^{\\text{th}}$ column. Assuming $n = 2007$, can we choose the numbers in such a way that the sum\n$$\na_1 + a_2 + \\dots + a_n + b_1 + b_2 + \\dots + b_n\n$$\nwill be zero? What about $n = 2008$?",
    "options": [],
    "answer": "For n = 2008: Yes. For n = 2007: No.",
    "solution": "If $n = 2008$ fill the first row of the board with $-1$ and all other squares with $1$. Then $a_k = 1$ for all $k$ and $b_l = -1$ for all $l$, so\n$$\na_1 + a_2 + \\dots + a_{2008} + b_1 + b_2 + \\dots + b_{2008} = 1 + \\dots + 1 + (-1) + \\dots + (-1) = 0.\n$$\n\nNow, let $n = 2007$. We will show that we cannot choose the numbers so that\n$$\na_1 + a_2 + \\dots + a_{2007} + b_1 + b_2 + \\dots + b_{2007} = 0.\n$$\nAssume, on the contrary, that this can be done. Obviously, each of $a_k$ and $b_l$ is either $1$ or $-1$. Let $t$ denote the number of negative elements of the set $\\{a_1, \\dots, a_{2007}\\}$ and let $s$ be the number of negative elements in $\\{b_1, \\dots, b_{2007}\\}$. Then\n$$\na_1 + \\dots + a_{2007} = -t + (2007 - t) = 2007 - 2t\n$$\nand\n$$\nb_1 + \\dots + b_{2007} = -s + (2007 - s) = 2007 - 2s.\n$$\nThis implies\n$$\n0 = a_1 + a_2 + \\dots + a_{2007} + b_1 + b_2 + \\dots + b_{2007} = 2007 - 2t + 2007 - 2s,\n$$\n$$\n\\text{so } s + t = 2007.\n$$\nHow many $-1$s are there on the board? If $a_i = -1$, then the $i^{\\text{th}}$ row contains an odd number of $-1$s. If $a_i = 1$, then this number is even. So, the parity of the number of $-1$s is the same as the parity of $t$. A similar argument for the columns shows that the parity of the total number of $-1$s is the same as the parity of $s$. So, $s$ and $t$ have the same parity, which contradicts the equation $s + t = 2007$ from above.\n\nWhen $n = 2007$ we cannot fill the board with $1$s and $-1$s so that\n$$\na_1 + a_2 + \\dots + a_{2007} + b_1 + b_2 + \\dots + b_{2007} = 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41270,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a sequence of integers $a_1, a_2, a_3, \\dots$ such that\n$$\n0 \\le a_k \\le k-1 \\quad \\text{and} \\quad a_1 + \\dots + a_k \\equiv 0 \\pmod{k}\n$$\nfor all $k > 1$. Prove that the sequence is constant from some point on. For example, when $a_1 = 9$ the sequence is $9, 1, 2, 0, 3, 3, 3, \\dots$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Taking a look at the sequences we obtain for different values of $a_1$, we notice the following: Assume there is an index $k$ such that $a_1 + a_2 + \\dots + a_k = d \\cdot k$ and $0 \\le d < k$. Then $a_1 + a_2 + \\dots + a_k + d = d \\cdot (k+1)$ and since $a_{k+1}$ is a uniquely determined number between $0$ and $k$ such that $a_1 + a_2 + \\dots + a_{k+1}$ is divisible by $k+1$, we have $a_{k+1} = d$. We come to the same conclusion about all subsequent terms of the sequence, so this sequence is constant and equal to $d$ from $a_{k+1}$ onward.\n\nLet us show that for all choices of $a_1$ there exists an index $k$ such that $a_1 + a_2 + \\dots + a_k = d \\cdot k$ and $0 \\le d < k$. Assume, to the contrary, that this is not the case. If $a_1 < 0$ and the sequence is not constantly $0$ from some point onward, there are infinitely many positive terms, so there exists an index $k$ such that $a_1 + a_2 + \\dots + a_k \\ge 0$. If by chance we have $a_1 > 0$, then this is true for all $k$. So, for $k$ sufficiently large we have $a_1 + a_2 + \\dots + a_k = d_k \\cdot k$, $d_k \\ge 0$, and by hypothesis $a_1 + a_2 + \\dots + a_k \\ge k^2$. We can bound the terms by $a_2 \\le 1$, $a_3 \\le 2$ and $a_i \\le i-1$ for all $i > 1$. So (for $k$ sufficiently large) we have\n$$\nk^2 \\le a_1 + a_2 + \\dots + a_k \\le a_1 + 1 + 2 + \\dots + (k-1) = a_1 + \\frac{k(k-1)}{2},\n$$\nwhich means that for all $k$ from some point onward we have\n$$\na_1 \\ge \\frac{k(k+1)}{2}.\n$$\nEvidently, this last inequality is not always satisfied, for example when $k \\ge 2|a_1|$. We have arrived at a contradiction which implies that the initial assumption was incorrect.\n\nHence, we have shown there exists an index $k + 1$ such that from $k + 1$ onward all terms of the sequence are equal.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41271,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDrog se je prelomil na višini $4.2~\\mathrm{m}$ od tal pod kotom $79^\\circ$ (glej sliko).\n![](attached_image_1.png)\nKolikšna je bila višina droga preden se je prelomil? Koliko $\\mathrm{dm}^3$ lesa vsebuje spodnji del droga do preloma, če je premer droga enak $60~\\mathrm{cm}$ ?",
    "options": [],
    "answer": "26.2 m; 1187.5 dm^3",
    "solution": "Solution:\n\nUporabimo kotno funkcijo $\\cos 79^\\circ = \\frac{4,2}{y}$ in izračunamo dolžino odlomljenega dela droga $y = 22~\\mathrm{m}$. Izračunamo višino droga pred prelomom, ki je $v = 4,2 + 22 = 26,2~\\mathrm{m}$. Ugotovimo, da je polmer droga $r = 3~\\mathrm{dm}$. Izračunamo prostornino dela droga, ki je $V = \\pi \\cdot 9 \\cdot 42 = 1187,5~\\mathrm{dm}^3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41272,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDokaži: če za neničelna realna števila $a, b$ in $c$ velja\n$$\na(b+c)+b(c+a)+c(a+b)=a b+b c+c a\n$$\nje vrednost izraza $\\frac{a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b)}{a b c}$ celo število.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIz dane enačbe sledi $a b+b c+c a=0$. Zato lahko zapišemo\n$$\n\\frac{a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b)}{a b c}=\\frac{a(a b+a c)+b(b a+b c)+c(c a+c b)}{a b c}\n$$\nUpoštevamo, da je $a b+a c=-b c$, $b a+b c=-c a$ in $c a+c b=-a b$ in dobimo\n$$\n\\frac{a(a b+a c)+b(b a+b c)+c(c a+c b)}{a b c}=\\frac{a(-b c)+b(-c a)+c(-a b)}{a b c}=\\frac{-3 a b c}{a b c}=-3\n$$\n\nTorej je vrednost izraza $-3$, kar je celo število.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41273,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV produktu potenc $5^{-2} \\cdot 5^{-4} \\cdot 5^{-8} \\cdot \\ldots \\cdot 5^{-x}$, kjer eksponenti tvorijo geometrijsko zaporedje, določi $x$ tako, da bo $5^{-2} \\cdot 5^{-4} \\cdot 5^{-8} \\cdot \\ldots \\cdot 5^{-x}=5^{-16382}$.",
    "options": [],
    "answer": "8192",
    "solution": "Solution:\n\nZapišemo vsoto členov geometrijskega zaporedja $-2+(-4)+(-8)+\\ldots+(-x)=-16382$, od koder odčitamo prvi člen, količnik in $x=a_{n}$. Uporabimo obrazec za vsoto prvih členov geometrijskega zaporedja $s_{n}=a_{1} \\cdot \\frac{k^{n}-1}{k-1}$. Vstavimo podatke in izračunamo, da je $n=13$. Uporabimo zvezo, da je $-x=a_{13}=a_{1} \\cdot k^{12}=-8192$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41274,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoenostavi izraz $x^{(x+1)^2} \\cdot \\left(x^{x-1}\\right)^{x+1} : \\frac{x^{x^2}}{x^{-2x}}$ in nato izračunaj vrednost izraza za $x=\\sqrt{2}$.",
    "options": [],
    "answer": "Simplified form: x^{x^2}; at x = sqrt(2): 2",
    "solution": "Solution:\n\nV izrazu $x^{(x+1)^2} = x^{x^2 + 2x + 1}$ uporabimo kvadriranje dvočlenika v eksponentu.\n\nV potenci $\\left(x^{x-1}\\right)^{x+1}$ uporabimo pravilo za potenciranje potenc, tako dobimo $\\left(x^{x-1}\\right)^{x+1} = x^{x^2 - 1}$.\n\nUlomek odpravimo z upoštevanjem pravila za deljenje potenc.\n\nUpoštevamo pravila za računanje s potencami in izraz skrčimo do $x^{x^2}$.\n\nVstavimo $\\sqrt{2}$ in izračunamo vrednost izraza.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41275,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $D$ notranja točka stranice $BC$ pravokotnega trikotnika $ABC$ s pravim kotom pri $C$. Trikotniku $ABD$ očrtano krožnico označimo s $\\mathcal{K}$. Naj bo $E$ taka točka na $\\mathcal{K}$, da je tetiva $DE$ pravokotna na $AB$. Dokaži, da je trikotnik $AEB$ enakokrak z vrhom $B$ natanko tedaj, ko je $CA$ tangenta na krožnico $\\mathcal{K}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo s $T$ presečišče tetiv $DE$ in $AB$. Vemo, da je trikotnik $DTB$ pravokotni.\n\nDenimo najprej, da je trikotnik $ABE$ enakokrak. Označimo $\\angle AEB = \\angle BAE = \\alpha$. Obodna kota $\\angle EAB$ in $\\angle EDB$ nad tetivo $BE$ sta enaka, zato je tudi $\\angle EDB = \\alpha$. Ker pa je $DE$ pravokotna na $AB$, je zato $\\angle ABD = \\frac{\\pi}{2} - \\alpha$. V pravokotnem trikotniku $ABC$ torej velja $\\angle ABC = \\frac{\\pi}{2} - \\alpha$, zato je $\\angle CAB = \\alpha$. Torej je kot $\\angle CAB$ med premico $AC$ in tetivo $AB$ enak kotu $\\angle AEB$ nad tetivo $AB$, kar ravno pomeni, da je $CA$ tangenta na krožnico $\\mathcal{K}$.\n\nObratno, denimo, da je $AC$ tangenta na krožnico $\\mathcal{K}$. Označimo $\\angle CAB = \\alpha$. Ker je $AC$ tangenta, je torej kot $\\angle BAC$ enak kotu $\\angle AEB$ nad tetivo $AB$. Zato je $\\angle AEB = \\alpha$.\n\n![](attached_image_1.png)\n\nVelja tudi, da je $\\angle ABC = \\frac{\\pi}{2} - \\alpha$, zato je $\\angle TDB = \\alpha$. Tako je $\\angle EDB = \\alpha$ in ta je enak $\\angle BAE$, saj sta obodna kota nad tetivo $BE$. Torej je $\\angle BAE = \\alpha = \\angle BEA$, zato je trikotnik $ABE$ enakokrak z vrhom $B$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41276,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPloščina trikotnika $ABC$ z oglišči $A(-1,-6)$, $B(1,0)$, $C(3,-3)$ je $9$. Izračunaj dolžino težiščnice na stranico $b$ in razdaljo med nožiščem višine na stranico $b$ in razpoloviščem stranice $b$.",
    "options": [],
    "answer": "9/2, 27/10",
    "solution": "Solution:\n\nIzračunamo koordinati razpolovišča stranice $AC$, ki je $R\\left(\\frac{-1+3}{2}, \\frac{-6+(-3)}{2}\\right)$. Izračunamo $t_{b} = d(B, R) = \\frac{9}{2}$. Izračunamo dolžino stranice $b = d(A, C) = 5$. Uporabimo obrazec $S = \\frac{b \\cdot v_{b}}{2}$ in dobimo $v_{b} = \\frac{18}{5}$. Upoštevamo zvezo $t_{b}^{2} = v_{b}^{2} + x^{2}$, pri čemer je $x$ iskana razdalja. Ta meri $x = \\frac{27}{10}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41277,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nPoišči vsa realna števila $x$, za katera je vrednost izraza\n$$\n\\sqrt{1-x^{2}}+\\sqrt{5x-x^{2}}\n$$\ncelo število.",
    "options": [],
    "answer": "x = 0, x = 9/41, x = 1",
    "solution": "Solution:\nOcenimo vrednost izraza. Očitno je $1-x^{2} \\leq 1$, vrednost $5x-x^{2}$ pa je omejena z $5x-x^{2}=\\frac{25}{4}-\\left(x-\\frac{5}{2}\\right)^{2} \\leq \\frac{25}{4}$, zato je\n$$\n\\sqrt{1-x^{2}}+\\sqrt{5x-x^{2}} \\leq \\sqrt{1}+\\sqrt{\\frac{25}{4}}=1+\\frac{5}{2}=3+\\frac{1}{2}\n$$\nPo drugi strani pa je vrednost izraza nenegativna in ne more biti enaka 0, saj bi to pomenilo, da je $1-x^{2}=0$ in $5x-x^{2}=0$, kar pa ni možno. Tako so edine celoštevilske vrednosti, ki jih izraz lahko zavzame, enake 1, 2 ali 3.\n\nPišimo $\\sqrt{1-x^{2}}+\\sqrt{5x-x^{2}}=a$ in odpravimo korene. Raje kot izraz v taki obliki kvadrirajmo $\\sqrt{1-x^{2}}=a-\\sqrt{5x-x^{2}}$, saj potem dobimo $1-x^{2}=a^{2}-2a\\sqrt{5x-x^{2}}+5x-x^{2}$ oziroma $2a\\sqrt{5x-x^{2}}=a^{2}-1+5x$. Po ponovnem kvadriranju sledi $4a^{2}\\left(5x-x^{2}\\right)=\\left(a^{2}-1\\right)^{2}-10x+10a^{2}x+25x^{2}$ oziroma\n$$\nx^{2}\\left(25+4a^{2}\\right)+x\\left(-10a^{2}-10\\right)+\\left(a^{2}-1\\right)^{2}=0\n$$\nČe je $a=1$ dobimo $x(29x-20)=0$, od koder sledi $x=0$ ali $x=\\frac{20}{29}$. Če rešitvi vstavimo v prvotni izraz, ugotovimo, da ustreza le $x=0$.\n\nPri $a=2$ lahko kvadratno enačbo za $x$ razcepimo kot $(x-1)(41x-9)=0$, od koder dobimo, da je $x=1$ ali $x=\\frac{9}{41}$. V obeh primerih izračun pokaže, da je vrednost izraza res enaka 2.\n\nOstane še primer $a=3$, ko dobimo kvadratno enačbo $61x^{2}-100x+64=0$. Ker pa je njena diskriminanta $D=100^{2}-4 \\cdot 61 \\cdot 64=100^{2}-(2 \\cdot 64)(2 \\cdot 61)=100^{2}-128 \\cdot 122$ negativna, nima realnih rešitev.\n\nVrednost izraza je torej celo število, ko je $x=0$, $x=\\frac{9}{41}$ ali pa $x=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41278,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. Pokaži, da vsota števk števila $10^{n} + 9 n$ ni deljiva z 2007 za nobeno naravno število $n$.\n\nb. Poišči vsaj eno naravno število $n$, za katero je vsota števk števila $10^{n} + 9 n$ enaka 2008.",
    "options": [],
    "answer": "There is no natural number n for which the digit sum of 10^n + 9n is divisible by 2007. An example with digit sum 2008 is n equal to the integer consisting of 223 ones.",
    "solution": "Solution:\n\na. Število je deljivo z 9 natanko tedaj, ko je vsota njegovih števk deljiva z 9. Recimo, da je vsota števk števila $10^{n} + 9 n$ deljiva z 2007. Ker je 2007 večkratnik števila 9, je potem vsota števk števila $10^{n} + 9 n$ deljiva z 9. To pa pomeni, da je $10^{n} + 9 n$ deljivo z 9, kar pa ne velja.\n\nb. Naj bo $n = 1 \\ldots 1$, kjer v zapisu nastopa 223 enic. Tedaj je $9 n = 9 \\ldots 9$, kjer v zapisu nastopa 223 devetic. Število $10^{n} + 9 n$ je potem enako $10 \\ldots 09 \\ldots 9$, pri čemer v zapisu nastopa 223 devetic in $n - 223 = 1 \\ldots 1 - 223$ ničel. Vsota števk tega števila je enaka $1 + 9 \\cdot 223 = 2008$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41279,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$ we have $|AC| > |AB|$. Let $D$ and $E$ be two points on the sides $AC$ and $AB$ such that $|CD| = |BE|$. Denote the intersection of the segments $BD$ and $CE$ by $F$ and let $G$ be a point on the segment $AC$ such that the line $GF$ is parallel to the bisector of the angle $\\angle BAC$. Prove that $|CG| = |AB|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the intersection of lines $AB$ and $FG$ by $H$ and let $J$ be the point where the bisector of the angle $\\angle BAC$ meets the segment $BC$. Since the line $FG$ is parallel to $AJ$, we have $\\angle HGA = \\angle FGD = \\angle BAJ = \\angle AHG$, so the triangle $GAH$ is isosceles and $|AH| = |AG|$. We know that $|BE| = |CD|$ and from what we just proved $|AH| = |AG|$. It remains to be seen that $|CG| = |AB|$. The figure reveals no similar triangles, so we have to find a different way to use the equality of lengths. Lengths play an important role in Ceva's and Menelaus's theorem. The former does not apply here, but we can try to use the latter.\n\nMenelaus's theorem for the triangle $AEC$ and the line $GF$ states that\n$$\n\\frac{|AH|}{|HE|} \\cdot \\frac{|EF|}{|FC|} \\cdot \\frac{|CG|}{|GA|} = 1,\n$$\nand since $|AH| = |AG|$ we get $\\frac{|CG|}{|HE|} = \\frac{|FC|}{|EF|}$. Now, let us try to express the ratio $\\frac{|FC|}{|EF|}$ in another way. Menelaus's theorem for the triangle $AEC$ and the line $BF$ gives us\n$$\n\\frac{|AB|}{|BE|} \\cdot \\frac{|EF|}{|FC|} \\cdot \\frac{|CD|}{|DA|} = 1,\n$$\nand since $|BE| = |CD|$ we have $\\frac{|AB|}{|DA|} = \\frac{|FC|}{|EF|}$. Hence, $\\frac{|AB|}{|DA|} = \\frac{|CG|}{|HE|}$. This is equivalent to\n$$\n\\frac{|AE| + |EB|}{|AG| + |GD|} = \\frac{|CD| + |DG|}{|AH| + |AE|},\n$$\nwhich implies $|AE|^2 + |AE|(|EB| + |AH|) = |GD|^2 + |GD|(|AG| + |CD|)$. Since $|AG| + |CD| = |EB| + |AH|$, we have\n$$\n\\begin{aligned}\n0 &= |AE|^2 - |GD|^2 + |AE|(|EB| + |AH|) - |GD|(|EB| + |AH|) \\\\\n  &= (|AE| - |GD|)(|AE| + |GD| + |EB| + |AH|),\n\\end{aligned}\n$$\nso $|AE| = |GD|$, and, finally, $|AB| = |AE| + |EB| = |GD| + |DC| = |GC|$, which was to be shown.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41280,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an isosceles triangle with the apex at $B$ and choose a point $D$ on the altitude from $B$ such that $AC$ is tangent to the circumcircle $\\mathcal{K}$ of the triangle $ABD$. Let $E$ be a point on $\\mathcal{K}$ such that the chord $DE$ is perpendicular to the chord $AB$. Prove that triangles $ABE$ and $ABC$ are congruent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $T$ be the intersection of the chords $DE$ and $AB$. We know that $DTB$ is a right triangle. Let $\\angle CAB = \\angle ACB = \\alpha$. The line $AC$ is tangent to the circumcircle, so the angle $\\angle CAB$ is equal to the corresponding angle $\\angle AEB$ over the chord $AB$. Thus, $\\angle AEB = \\alpha$.\n\nWe have $\\angle ABD = \\frac{\\pi}{2} - \\alpha$, so $\\angle TDB = \\alpha$ and $\\angle EDB = \\alpha$. This angle is equal to $\\angle EAB$ because they are both the inscribed angles over $BE$. So, $\\angle BAE = \\alpha = \\angle BEA$ and the triangle $ABE$ is isosceles with the apex at $B$. Triangles $ABE$ and $ABC$ have congruent angles and a common side $AB$ next to the corresponding two angles, so they are congruent.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41281,
    "subject": "Mathematics (Multi-modal)",
    "question": "For what positive integers $n > 1$ does the expression\n$$\n\\frac{\\log_{10} 2 \\cdot \\log_{10} 3 \\cdots \\log_{10} n}{10^{n-1}}\n$$\nattain the smallest possible value? Find this value.",
    "options": [],
    "answer": "n = 10^{10} - 1 and n = 10^{10}; minimal value = (\\prod_{k=2}^{10^{10}} \\log_{10} k) / 10^{10^{10}-1}",
    "solution": "Let us compare the expressions $\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} (n-1)}{10^{n-2}}$ and $\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} n}{10^{n-1}}$. The inequality\n$$\n\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} (n-1)}{10^{n-2}} \\ge \\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} n}{10^{n-1}}\n$$\nholds if and only if $1 \\ge \\frac{1}{10} \\log_{10} n = \\log_{10} \\sqrt[10]{n}$, which is equivalent to $10 \\ge \\sqrt[10]{n}$ and $10^{10} \\ge n$. This implies\n$$\n\\begin{aligned}\n\\frac{\\log_{10} 2}{10} &> \\frac{\\log_{10} 2 \\cdot \\log_{10} 3}{10^2} > \\dots > \\frac{\\log_{10} 2 \\cdot \\log_{10} 3 \\cdots \\log_{10} (10^{10} - 1)}{10^{10^{10}-2}} \\\\\n&= \\frac{\\log_{10} 2 \\cdot \\log_{10} 3 \\cdots \\log_{10} (10^{10})}{10^{10^{10}-1}}\n\\end{aligned}\n$$\nand\n$$\n\\begin{aligned}\n\\frac{\\log_{10} 2 \\cdot \\log_{10} 3 \\cdots \\log_{10} (10^{10})}{10^{10^{10}-1}} < & \\frac{\\log_{10} 2 \\cdot \\log_{10} 3 \\cdots \\log_{10} (10^{10} + 1)}{10^{10^{10}}} \\\\\n< & \\frac{\\log_{10} 2 \\cdot \\log_{10} 3 \\cdots \\log_{10} (10^{10} + 2)}{10^{10^{10}+1}} < \\dots\n\\end{aligned}\n$$\nSo, the expression $\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} n}{10^{n-1}}$ has the smallest possible value when $n = 10^{10} - 1$ or $n = 10^{10}$. For these two values of $n$ the value of the expression is\n$$\n\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} 10^{10}}{10^{10^{10}-1}}\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41282,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be positive integers of different parity. Show that\n$$\n(m + 3n)(5m + 7n)(7m + 5n)(3m + n)\n$$\ncan never be a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us assume there exist numbers $m$ and $n$, one even and one odd, such that $A = (m + 3n)(5m + 7n)(7m + 5n)(3m + n)$ is a perfect square. Let $d$ be the greatest common divisor of $m$ and $n$, and write $m = d m_1$, $n = d n_1$. Then $m_1$ and $n_1$ are coprime and $A$ can be written as\n$$\nA = d^4 (m_1 + 3n_1)(5m_1 + 7n_1)(7m_1 + 5n_1)(3m_1 + n_1).\n$$\nLet $B = (m_1 + 3n_1)(5m_1 + 7n_1)(7m_1 + 5n_1)(3m_1 + n_1)$. Since $A$ is a perfect square, so is $B$.\n\nLet $p$ be a positive integer that divides both $m_1 + 3n_1$ and $5m_1 + 7n_1$. Since $m_1$ and $n_1$ have different parity, the numbers $m_1 + 3n_1$ and $5m_1 + 7n_1$ are odd and $p$ must be odd, too. On the other hand we have\n$$\np \\mid 5 \\cdot (m_1 + 3n_1) - (5m_1 + 7n_1) = 8n_1,\n$$\n$$\np \\mid 3 \\cdot (5m_1 + 7n_1) - 7(m_1 + 3n_1) = 8m_1,\n$$\nso $p \\mid m_1$ and $p \\mid n_1$, which implies $p = 1$. We have shown that $m_1 + 3n_1$ and $5m_1 + 7n_1$ are coprime.\n\nLet $q$ be a positive integer such that $q \\mid m_1 + 3n_1$ and $q \\mid 7m_1 + 5n_1$. Again, $q$ must be odd and\n$$\nq \\mid 7(m_1 + 3n_1) - (7m_1 + 5n_1) = 16n_1,\n$$\n$$\nq \\mid 3(7m_1 + 5n_1) - 5(m_1 + 3n_1) = 16m_1.\n$$\nSo, $q \\mid n_1$ and $q \\mid m_1$, which implies $q = 1$. We conclude that $m_1 + 3n_1$ and $(5m_1 + 7n_1)(7m_1 + 5n_1)$ are coprime.\n\nSimilarly, one can show that $3m_1 + n_1$ and $(5m_1 + 7n_1)(7m_1 + 5n_1)$ are coprime. Now, assume $r \\mid m_1 + 3n_1$ and $r \\mid 3m_1 + n_1$. Then $r \\mid 3(m_1 + 3n_1) - (3m_1 + n_1) = 8n_1$ and $r \\mid 3(3m_1 + n_1) - (m_1 + 3n_1) = 8m_1$, so $r \\mid n_1$, $r \\mid m_1$ and $r = 1$.\n\nWe have shown that the two factors $3m_1 + n_1$ and $m_1 + 3n_1$ are coprime to any other factor of $B = (m_1 + 3n_1)(3m_1 + n_1)(5m_1 + 7n_1)(7m_1 + 5n_1)$, and since $B$ is a perfect square, we see that $m_1 + 3n_1$ and $3m_1 + n_1$ must be perfect squares as well. Denote them by $a^2$ and $b^2$, respectively. Then\n$$\n(a - b)(a + b) = a^2 - b^2 = 2(n_1 - m_1).\n$$\nThe right-hand side is divisible by $2$ but not by $4$ because $n_1$ and $m_1$ have different parity. On the other hand, $a - b$ and $a + b$ have the same parity and since their product is divisible by $2$, it must also be divisible by $4$. This contradiction shows that\n$$\n(m + 3n)(5m + 7n)(7m + 5n)(3m + n)\n$$\ncannot be a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41283,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the numbers $a$ and $b$ be such that $a^3 + b^3 = 13$ and $a^9 + b^9 = -299$. Find the value of $ab$, given that $ab$ is real.",
    "options": [],
    "answer": "4",
    "solution": "**First solution**\nSince $a^9 + b^9 = (a^3 + b^3)(a^6 - a^3b^3 + b^6)$, we have $a^6 - a^3b^3 + b^6 = -\\frac{299}{13} = -23$.\n\nSubtracting this equality from $a^6 + 2a^3b^3 + b^6 = (a^3 + b^3)^2 = 13^2 = 169$ we get $3a^3b^3 = 192$, which implies $ab = 4$ since $ab$ is a real number.\n\n\n**Second solution**\nWe have $13^3 = (a^3 + b^3)^3 = a^9 + 3a^6b^3 + 3a^3b^6 + b^9 = -299 + 3a^6b^3 + 3a^3b^6$, which implies $832 = a^6b^3 + a^3b^6$.\n\nFactoring the expression on the right-hand side we find that $832 = a^3b^3(a^3 + b^3) = a^3b^3 \\cdot 13$, or $a^3b^3 = \\frac{832}{13} = 64$, so $ab = 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41284,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a village with $12k$ inhabitants each person knows $3k + 6$ other people and the acquaintances are mutual. There exists a positive integer $n$, such that for any two villagers the number of villagers who know both is $n$. How many villagers are there in the village?",
    "options": [],
    "answer": "36",
    "solution": "Consider an arbitrary villager $a$. Let $A$ denote the set of all villagers who know $a$ and let $B$ contain all the rest. There are $3k + 6$ villagers in $A$ and $9k - 7$ villagers in $B$. Let $x$ be a villager from $a$. The villagers who know $a$ as well as $x$ form a subset of $A$. Let $n$ be the number of villagers in $A$ who know $x$. Then $3k + 5 - n$ villagers from $B$ know $x$. Let $y$ be any villager from $B$. Those villagers who know both $a$ and $y$ also form a subset of $A$ and there are $n$ of them.\n\nNow, let us count the number of acquaintances amongst the villagers in $A$ and villagers in $B$. There are $3k + 6$ villagers in $A$ and each of them knows $3k + 5 - n$ villagers from $B$. On the other hand there are $9k - 7$ villagers in $B$ and each knows $n$ villagers from $A$. So,\n$$\n(3k + 6)(3k + 5 - n) = (9k - 7)n.\n$$\nWe can see right away that $n$ is divisible by 3, so we can write $n = 3m$. Thus,\n$$\nm = \\frac{3k^2 + 11k + 10}{12k - 1}.\n$$\nFor $m$ to be a positive integer we must have\n$$\n4m = k + 3 + \\frac{9k + 43}{12k - 1}.\n$$\nHence, $\\frac{9k+43}{12k-1}$ is an integer, which implies that $4 \\cdot \\frac{9k+43}{12k-1} = 3 + \\frac{175}{12k-1}$ is an integer as well. So, $12k - 1$ is a divisor of 175 and since $12k - 1 \\ge 1$, it can only be equal to 25, 35 or 175. We find only one integer solution, $k = 3$. It is easy to see that in this case we have $n = 6$, which is an integer and we conclude that there are 36 people in the village.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41285,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive real numbers $x$ and $y$ such that\n$$\nx^{x+y} = y^{x-y} \\quad \\text{and} \\quad x^2y = 1\n$$",
    "options": [],
    "answer": "Two solutions: (x, y) = (1, 1) and (x, y) = (3^{-1/3}, 3^{2/3}).",
    "solution": "The second equation implies $y = x^{-2}$, so\n$$\nx^{x + x^{-2}} = x^{-2(x - x^{-2})}.\n$$\nTaking the logarithm on both sides we get\n$$\n(x + x^{-2}) \\log x = -2(x - x^{-2}) \\log x.\n$$\nIf $\\log x = 0$, then $x = 1$ and $y = 1$. Otherwise,\n$$\nx + x^{-2} = -2x + 2x^{-2},\n$$\nso $3x^3 = 1$. This implies $x = \\frac{1}{\\sqrt{3}}$ and $y = \\sqrt{3}$. There are two solutions: $x = 1$, $y = 1$ and $x = \\frac{1}{\\sqrt{3}}$, $y = \\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41286,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n = (p^2 - 1)(p^2 - 4) + 9$. Assuming $p$ is a prime number, what is the least possible sum of the digits of $n$? Find all prime numbers $p$ for which this value is attained.",
    "options": [],
    "answer": "Least sum of digits is 9; attained for p = 2 and p = 5.",
    "solution": "Let us find the first few $n$. When $p=2$ we have $n=9$, when $p=3$ we have $n=49$ and when $p=5$ we have $n=513$. Now, let $p > 5$. Rewrite $n$ as $n = (p-2)(p-1)(p+1)(p+2)+9$. Since $(p-2), (p-1), p, (p+1), (p+2)$ are five consecutive positive integers, at least one of them is divisible by $5$. Since $p > 5$, $p$ is not divisible by $5$. Hence, $5$ divides one of the numbers $(p-2), (p-1), (p+1), (p+2)$ and must also divide their product $(p-2)(p-1)(p+1)(p+2)$. At least one of the numbers $p+1$ and $p+2$ is even, so this product is divisible by $2$. Thus, for $p > 5$ the number $(p-2)(p-1)(p+1)(p+2)$ is divisible by $10$. For all $p > 5$ the number $n$ has at least two digits and the final digit is $9$, so the sum of the digits is greater than $9$. We conclude that the least possible sum of the digits is $9$ and this value is attained only when $p = 2$ or $p = 5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41287,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest three-digit integer with the property that its triple has only even digits.",
    "options": [],
    "answer": "134",
    "solution": "Denote the three-digit number by $\\overline{abc}$. Its triple is equal to\n$$\n3 \\cdot \\overline{abc} = (3a) \\cdot 100 + (3b) \\cdot 10 + 3c.\n$$\nObviously, $a$ has to be at least 1. If $a = 1$ and we want the digit at the hundreds in $3 \\cdot \\overline{abc}$ to be even, we require $3b \\cdot 10 + 3c \\ge 100$, which implies $10b + c \\ge \\frac{100}{3} = 33 + \\frac{1}{3}$. The least possible number that satisfies this inequality is 34. The solution is 134 and its triple is 402.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41288,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$ the lengths of the sides are given: $|AB| = 15$ cm, $|BC| = 14$ cm and $|CA| = 13$ cm. Let $D$ be the foot of the altitude from $A$ and let $E$ be a point on this altitude such that $\\angle BAD = \\angle DEC$. Denote the intersection of lines $AB$ and $CE$ by $F$. Find $|EF|$.",
    "options": [],
    "answer": "10/3",
    "solution": "First, let us find the lengths of the segments $AD$ and $CD$. Write $|AD| = v$ and $|CD| = x$. By Pythagoras's theorem $v^2 = |AC|^2 - x^2 = |AB|^2 - (|BC| - x)^2$, so $13^2 - x^2 = 15^2 - (14 - x)^2$ or, equivalently, $13^2 = 15^2 - 14^2 + 2 \\cdot 14 \\cdot x$. We see that $x = 5$ and $v = 12$.\n\nTriangles $EDC$ and $ADB$ are similar because $\\angle BAD = \\angle DEC$. So, $\\frac{|EC|}{|CD|} = \\frac{|AB|}{|AD|}$ and $\\angle DBA = \\angle ECD$. This implies $|EC| = \\frac{15}{9} \\cdot 5 = \\frac{25}{3}$ and $\\angle FCB = \\angle CBF$. The triangle $BFC$ is isosceles with the apex at $B$, so $|CF| = |FB|$.\n\n![](attached_image_1.png)\n\nLet $G$ be the midpoint of $BC$. Then $FG$ is perpendicular to $BC$. The triangle $FBG$ is similar to the triangle $ABD$, so $\\frac{|FB|}{|BG|} = \\frac{|AB|}{|BD|}$ and $|FB| = \\frac{|AB| \\cdot |BG|}{|BD|} = \\frac{15 \\cdot 7}{9} = \\frac{35}{3}$.\n\nThe length of the segment $EF$ is $|EF| = |CF| - |CE| = |FB| - |CE| = \\frac{35}{3} - \\frac{25}{3} = \\frac{10}{3}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41289,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right triangle with the right angle at $C$. On the segment $BC$ choose a point $D$ different from $B$ and $C$. Denote the circumcircle of the triangle $ABD$ by $\\mathcal{K}$. Let $T$ be a point on the side $AB$ such that $DT$ is perpendicular to $AB$. Let $E$ be the second intersection of the line $DT$ with $\\mathcal{K}$, denote the intersection of lines $CT$ and $EB$ by $F$ and let the line $DF$ meet $\\mathcal{K}$ again at $G$. Prove that triangles $CEF$ and $BEG$ are similar.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The sum of the two opposite angles $\\angle ATD$ and $\\angle ACD$ in the quadrilateral $ATDC$ is $\\pi$, so this quadrilateral is cyclic. Set $\\angle TCD = \\alpha$. Then\n\n$\\angle TAD = \\angle TCD = \\alpha$. Points $A$, $B$, $E$, $D$ are concyclic, so $\\angle BED = \\angle BAD = \\angle TAD = \\alpha$. We have $\\angle FED = \\angle BED = \\alpha = \\angle TCD = \\angle FCD$, so angles $FED$ and $FCD$ in the quadrilateral $FECD$ are equal and this quadrilateral is also cyclic.\n\nLet $\\angle ECF = \\beta$. Since $FECD$ is cyclic, we have $\\angle EDF = \\angle ECF = \\beta$. Points $E$, $D$, $B$ and $G$ all lie on $\\mathcal{K}$, so $\\angle EBG = \\angle EDG = \\angle EDF = \\beta$.\n\nDenote $\\angle EFC = \\gamma$. Since the quadrilateral $CDFE$ is cyclic, we have $\\angle EDC = \\angle EFC = \\gamma$, so $\\angle EDB = \\pi - \\angle EDC = \\pi - \\gamma$. Points $D$, $B$, $G$ and $E$ are concyclic, so $\\angle EGB = \\pi - \\angle EDB = \\pi - (\\pi - \\gamma) = \\gamma$, and $\\angle EFC = \\angle EGB$. But we have already shown that $\\angle ECF = \\beta = \\angle EBG$. Triangles $ECF$ and $EBG$ have two congruent angles, so they are similar.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41290,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nZa katera naravna števila $n > 1$ doseže izraz\n$$\n\\frac{\\log_{10} 2 \\cdot \\log_{10} 3 \\cdots \\log_{10} n}{10^{n-1}}\n$$\nnajmanjšo vrednost? Kolikšna je ta vrednost?",
    "options": [],
    "answer": "The minimum occurs at n = 10^10 − 1 and n = 10^10. The minimal value is (log_{10} 2 · log_{10} 3 · … · log_{10}(10^{10})) / 10^{10^{10} − 1}.",
    "solution": "Solution:\nPrimerjajmo izraza $\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10}(n-1)}{10^{n-2}}$ in $\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} n}{10^{n-1}}$. Neenakost\n$$\n\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10}(n-1)}{10^{n-2}} \\geq \\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} n}{10^{n-1}}\n$$\nvelja natanko tedaj, ko je $1 \\geq \\frac{1}{10} \\log_{10} n = \\log_{10} \\sqrt[10]{n}$, kar je enakovredno $10 \\geq \\sqrt[10]{n}$ oziroma $10^{10} \\geq n$. Od tod sledi\n$$\n\\begin{gathered}\n\\frac{\\log_{10} 2}{10} > \\frac{\\log_{10} 2 \\cdot \\log_{10} 3}{10^{2}} > \\ldots > \\frac{\\log_{10} 2 \\cdot \\log_{10} 3 \\cdots \\log_{10}\\left(10^{10}-1\\right)}{10^{10^{10}-2}} \\\\\n= \\frac{\\log_{10} 2 \\cdot \\log_{10} 3 \\cdots \\log \\left(10^{10}\\right)}{10^{10^{10}-1}}\n\\end{gathered}\n$$\nin\n$$\n\\begin{gathered}\n\\frac{\\log_{10} 2 \\cdot \\log_{10} 3 \\cdots \\log \\left(10^{10}\\right)}{10^{10^{10}-1}} < \\frac{\\log_{10} 2 \\cdot \\log_{10} 3 \\cdots \\log \\left(10^{10}+1\\right)}{10^{10^{10}}} \\\\\n\\quad < \\frac{\\log_{10} 2 \\cdot \\log_{10} 3 \\cdots \\log \\left(10^{10}+2\\right)}{10^{10^{10}+1}} < \\ldots\n\\end{gathered}\n$$\nTorej je vrednost izraza $\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} n}{10^{n-1}}$ najmanjša pri $n = 10^{10} - 1$ in $n = 10^{10}$. Enaka je\n$$\n\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} 10^{10}}{10^{10^{10}-1}}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41291,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa višini na stranico $AC$ enakokrakega trikotnika $ABC$ z vrhom $B$ izberemo točko $D$ tako, da je premica $AC$ tangenta na očrtano krožnico $\\mathcal{K}$ trikotnika $ABD$. Naj bo $E$ taka točka na krožnici $\\mathcal{K}$, da je tetiva $DE$ pravokotna na tetivo $AB$. Dokaži, da sta trikotnika $ABE$ in $ABC$ skladna.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo s $T$ presečišče tetiv $DE$ in $AB$. Vemo, da je trikotnik $DTB$ pravokotni. Označimo $\\angle CAB = \\angle ACB = \\alpha$. Ker je $AC$ tangenta, je torej kot $\\angle CAB$ enak nepriležnemu kotu $\\angle AEB$ nad tetivo $AB$. Zato je $\\angle AEB = \\alpha$. Velja tudi, da je $\\angle ABD = \\frac{\\pi}{2} - \\alpha$, zato je $\\angle TDB = \\alpha$. Tako je $\\angle EDB = \\alpha$ in ta kot je enak $\\angle EAB$, saj sta obodna kota nad tetivo $BE$. Torej je $\\angle BAE = \\alpha = \\angle BEA$, zato je trikotnik $ABE$ enakokrak z vrhom $B$. Trikotnika $ABE$ in $ABC$ se ujemata v vseh kotih in dolžini skupne in istoležne stranice $AB$, torej sta skladna.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41292,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa stranici $BC$ pravokotnega trikotnika $ABC$ s pravim kotom pri $C$ izberemo točko $D$, različno od $B$ in $C$. Trikotniku $ABD$ očrtano krožnico označimo s $\\mathcal{K}$. Naj bo $T$ taka točka na stranici $AB$, da je $DT$ pravokotna na $AB$. Premica $DT$ seka krožnico $\\mathcal{K}$ še v točki $E$. Presečišče premic $CT$ in $EB$ označimo s $F$. Premica $DF$ seka krožnico $\\mathcal{K}$ še v točki $G$. Dokaži, da sta trikotnika $CEF$ in $BEG$ podobna.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nKer je vsota nasprotnih kotov $\\angle ATD$ in $\\angle ACD$ v štirikotniku $ATDC$ enaka $\\pi$, je ta štirikotnik tetiven. Označimo $\\angle TCD=\\alpha$. Zaradi tetivnosti sledi $\\angle TAD=\\angle TCD=\\alpha$. Ker pa so točke $A, B, E, D$ konciklične, velja tudi $\\angle BED=\\angle BAD=\\angle TAD=\\alpha$. Dobili smo torej $\\angle FED=\\angle BED=\\alpha=\\angle TCD=\\angle FCD$, torej sta v štirikotniku $FECD$ kota $FED$ in $FCD$ enaka, zato je tudi ta štirikotnik tetiven.\n\nNaj bo sedaj $\\angle ECF=\\beta$. Zaradi tetivnosti štirikotnika $FECD$ sledi $\\angle EDF=\\angle ECF=\\beta$. Ker pa točke $E, D, B$ in $G$ ležijo na krožnici $\\mathcal{K}$, sledi še $\\angle EBG=\\angle EDG=\\angle EDF=\\beta$.\n\nOznačimo še $\\angle EFC=\\gamma$. Zaradi tetivnosti štirikotnika $CDFE$ sledi $\\angle EDC=\\angle EFC=\\gamma$, torej\n\n![](attached_image_1.png)\nje $\\angle EDB=\\pi-\\angle EDC=\\pi-\\gamma$. Ker pa so točke $D, B, G$ in $E$ konciklične sledi, da je $\\angle EGB=\\pi-\\angle EDB=\\pi-(\\pi-\\gamma)=\\gamma$, torej je $\\angle EFC=\\angle EGB$. Ker pa smo že pokazali, da velja $\\angle ECF=\\beta=\\angle EBG$, se trikotnika $ECF$ in $EBG$ ujemata v dveh kotih, zato sta si podobna.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41293,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMed funkcijami $f(x)=\\frac{a x^{2}+b x+8}{(x+c)^{2}}$ izberi tisto, ki ima definicijsko območje $\\mathrm{D}_{f}=(-\\infty, 1) \\cup(1, \\infty)$, ničlo 2 in asimptoto $y=0$.",
    "options": [],
    "answer": "f(x) = (-4x + 8)/(x + 1)^2",
    "solution": "Solution:\n\nIz definicijskega območja razberemo, da je pol $x=1$, zato zapišemo enačbo $(1+c)^{2}=0$.\nIzračunamo $c=-1$. Iz enačbe asimptote razberemo, da je stopnja števca manjša od stopnje\nimenovalca, zato je $a=0$. Upoštevamo, da je število 2 ničla funkcije: $a \\cdot 0+2 b+8=0$, od\nkoder izračunamo $b=-4$. Funkcija je $f(x)=\\frac{-4 x+8}{(x+1)^{2}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41294,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRešitvi enačbe $\\frac{\\log \\left(35-x^{3}\\right)}{\\log (5-x)}=3$ sta dolžini katet pravokotnega trikotnika. Izračunaj polmer kroga, ki je temu trikotniku očrtan.",
    "options": [],
    "answer": "sqrt(13)/2",
    "solution": "Solution:\n\nEnačbo množimo z $\\log (5-x)$, antilogaritmiramo in uredimo v kvadratno $15 x^{2}-75 x+90=0$. Rešitvi kvadratne enačbe sta $x_{1}=2$, $x_{2}=3$. Rešitvi enačbe sta dolžini katet pravokotnega trikotnika. Izračunamo dolžino hipotenuze. Polovična vrednost dolžine hipotenuze je enaka polmeru $R$ temu trikotniku očrtanega kroga.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41295,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDane so dolžine stranic trikotnika $ABC$: $|AB| = 15~\\mathrm{cm}$, $|BC| = 14~\\mathrm{cm}$ in $|CA| = 13~\\mathrm{cm}$. Naj bo $D$ nožišče višine na stranico $BC$, $E$ pa taka točka na tej višini, da je $\\angle BAD = \\angle DEC$. Presečišče premic $AB$ in $CE$ označimo s $F$. Izračunaj $|EF|$.",
    "options": [],
    "answer": "10/3",
    "solution": "Solution:\n\nNajprej izračunajmo dolžini $|AD|$ in $|CD|$. Označimo $|AD| = v$ in $|CD| = x$. Po Pitagorovem izreku je $v^{2} = |AC|^{2} - x^{2} = |AB|^{2} - (|BC| - x)^{2}$, od koder dobimo $13^{2} - x^{2} = 15^{2} - (14 - x)^{2}$ oziroma $13^{2} = 15^{2} - 14^{2} + 2 \\cdot 14 \\cdot x$. Torej je $x = 5$ in potem $v = 12$.\n\nKer je $\\angle BAD = \\angle DEC$, je trikotnik $EDC$ podoben trikotniku $ADB$. Zato je $\\frac{|EC|}{|CD|} = \\frac{|AB|}{|AD|}$ in $\\angle DBA = \\angle ECD$. Sledi $|EC| = \\frac{15}{9} \\cdot 5 = \\frac{25}{3}$ in $\\angle FCB = \\angle CBF$, torej je trikotnik $BFC$ enakokrak z vrhom $B$. Zato je $|CF| = |FB|$.\n\n![](attached_image_1.png)\n\nNaj bo $G$ razpolovišče $BC$. Potem je $FG$ pravokotna na $BC$. Trikotnik $FBG$ je podoben trikotniku $ABD$, zato velja $\\frac{|FB|}{|BG|} = \\frac{|AB|}{|BD|}$, torej je $|FB| = \\frac{|AB| \\cdot |BG|}{|BD|} = \\frac{15 \\cdot 7}{9} = \\frac{35}{3}$.\n\nDolžina $|EF|$ je enaka $|EF| = |CF| - |CE| = |FB| - |CE| = \\frac{35}{3} - \\frac{25}{3} = \\frac{10}{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41296,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral such that the triangle $BCD$ is acute and $|AB| = |AD|$. Denote the intersection of the bisector of the angle $\\angle CAD$ with the side $CD$ by $K$ and the intersection of the bisector of the angle $\\angle BAC$ with the side $BC$ by $L$. Let $K'$ and $L'$ be the orthogonal projections of $K$ and $L$ onto the sides $BC$ and $CD$, respectively. Prove that $B$, $D$, $L'$ and $K'$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use the Law of sines for the triangle $BAL$,\n$$\n\\frac{\\sin(\\angle BAL)}{|BL|} = \\frac{\\sin(\\angle ALB)}{|AB|},\n$$\nand for the triangle $CAL$,\n$$\n\\frac{\\sin(\\angle LAC)}{|CL|} = \\frac{\\sin(\\angle CLA)}{|AC|}.\n$$\nSince $\\angle ALB = \\pi - \\angle CLA$, we have $\\sin \\angle ALB = \\sin \\angle CLA$. From the two equations above and $\\angle BAL = \\angle LAC$ we get\n$$\n\\frac{|AB|}{|AC|} = \\frac{|BL|}{|CL|}.\n$$\nSimilarly, $\\frac{|AD|}{|AC|} = \\frac{|DK|}{|CK|}$.\n(In any triangle the bisector of an angle divides the opposite side in the ratio equal to the ratio of the lengths of the other two sides. This is a well-known fact and the proof was not required. It was sufficient to note that since $AL$ is the bisector of $\\angle BAC$, we have $\\frac{|AB|}{|AC|} = \\frac{|BL|}{|CL|}$.)\nSince $|AB| = |AD|$, these two equalities imply $\\frac{|BL|}{|CL|} = \\frac{|DK|}{|CK|}$. So,\n$$\n\\frac{|CB|}{|CL|} = \\frac{|BL| + |CL|}{|CL|} = \\frac{|BL|}{|CK|} + 1 = \\frac{|DK|}{|CK|} + 1 = \\frac{|DK| + |CK|}{|CK|} = \\frac{|CD|}{|CK|}.\n$$\nTriangles $DBC$ and $KLC$ are similar (they share the angle at $C$ and $\\frac{|CB|}{|CL|} = \\frac{|CD|}{|CK|}$). Thus, the line $KL$ is parallel to the diagonal $BD$. Because of the right angles at $K'$ and $L'$ points $K$, $L$, $K'$ and $L'$ are concyclic. Since the triangle $BCD$ is acute, $K'$ and $L'$ lie on the same side of $KL$ and on the same side of $BD$. So,\n$$\n\\angle DL'K' = \\angle KL'K' = \\pi - \\angle K'LK = \\pi - \\angle K'BD\n$$\nand $B$, $D$, $L'$, $K'$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41297,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players share a pile of coins, alternating in taking one coin off the pile and placing it on any empty square of a $2008 \\times 2008$ chessboard. The player who puts the coin onto the board in such a way that together with three other coins already on the board it forms a non-rectangular isosceles trapezoid with bases parallel to two of the edges of the board, wins. Which of the players has the winning strategy – the first or the second?",
    "options": [],
    "answer": "the second player",
    "solution": "Let us call a non-rectangular isosceles trapezoid with the bases parallel to two of the edges of the board **regular**. We show that the second player can always win. After the first player's every move the second player should check to see if they can form a regular trapezoid by the placing of a single coin. If not, the second player should put the coin onto the board in such a way, that after their move the arrangement of the coins will have the vertical line across the middle of the board as its axis of symmetry (i.e. the second player's move mirrors the first player's move with regard to this line).\n\nAssume that the first player wins by placing a coin onto the square which we denote by $a_0$. Let $a_1, a_2$ and $a_3$ denote the other three squares that together with $a_0$ form a regular trapezoid (i.e. all four are different and each contains a coin). Let $b_0, b_1, b_2$ and $b_3$ be the corresponding mirror images. Squares $b_1, b_2$ and $b_3$ contain coins and $b_0$ is empty. First, we show that none of the squares $a_0, a_1, a_2, a_3, b_0, b_1, b_2$ and $b_3$ coincide.\n\nSince the length of the board is even, we have $a_i \\neq b_i$ for all $i = 0, 1, 2, 3$. If $a_i = b_j$ for some $i \\neq j$, then $b_i = a_j$, so the side $a_i a_j$ of the trapezoid $a_0 a_1 a_2 a_3$ is horizontal. If this is the case, the side $a_k a_l$ for $\\{k,l\\} = \\{0,1,2,3\\} \\setminus \\{i,j\\}$ is also horizontal, because a regular trapezoid has no right angles. Since this trapezoid is isosceles, we get $b_k = a_l$ and $a_l = b_k$ for $l \\neq k$. This implies $b_0 = a_m$ for some $m \\neq 0$. A contradiction, the square $b_0$ is empty and squares $a_1, a_2, a_3$ contain coins.\n\nHence, all those squares are necessarily different and before the first player made his or her last move the triples $a_1, a_2, a_3$ and $b_1, b_2, b_3$ were covered with coins. This implies that at least one of the triples was covered with coins before the second player made his or her last move, so following the strategy the second player should already have won.\n\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 41298,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$, such that\n$$\nx + f(xf(y)) = f(y) + yf(x)\n$$\nfor all real $x$ and $y$.",
    "options": [],
    "answer": "f(x) = x - 1",
    "solution": "If $x = 0$ we get $f(0) = f(y) + y f(0)$. So, $f$ is a linear function of the form $f(x) = a - a x$ for some real $a$. Inserting this into the functional equation we see that for all $x, y \\in \\mathbb{R}$ we have $x + a - a x(a - a y) = a - a y + y(a - a x)$, so\n$$\nx - a^2 x + a^2 x y = -a x y.\n$$\nNow, let $y = 0$ to see that $x = a^2 x$ for all $x$, so $a^2 = 1$. Under this condition the above equality becomes $x y = -a x y$ or, equivalently, $(1 + a) x y = 0$, which then implies $a = -1$, since this last equality has to hold for all $x$ and $y$.\nWe have shown that $f(x) = x - 1$. Using this expression for $f$ in the initial functional equation, we see that the left-hand side is equal to\n$$\nx + f(x f(y)) = x + f(x y - x) = x + x y - x - 1 = x y - 1\n$$\nand the right-hand side becomes\n$$\nf(y) + y f(x) = y - 1 + y(x - 1) = x y - 1.\n$$\nThe two sides are equal for all $x$ and $y$, so $f(x) = x - 1$ is the (only) solution to our equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41299,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players share a pile of coins, alternating in taking one coin off the pile and placing it on any empty square of a $2008 \\times 2008$ chessboard. The player who puts the coin onto the board in such a way that together with three other coins already on the board it forms a rectangle with sides parallel to the edges of the board, wins. Which of the players has the winning strategy – the first or the second?",
    "options": [],
    "answer": "the second player",
    "solution": "The second player has the winning strategy. He or she can arrange the columns into 1004 pairs with columns 1 and 2 forming the first pair, columns 3 and 4 forming the second pair, and so on, with columns 2007 and 2008 forming the final pair. Whenever the first player puts a coin into one of the columns, the second player should put a coin into the second column of that pair and on the same line as the first. After at most 1004 moves the first player will be forced to put a coin into a column that already contains one other coin. The second player will then be able to form the rectangle and win.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 41300,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $K$ be a subset of the set $\\mathbb{N}$ of all natural numbers with the property that for any two numbers $a$ and $b$ in $K$ either $a$ divides $b$ or $b$ divides $a$. Prove that every $c \\in K$ is greater than the sum of all those numbers in $K$ that are smaller than $c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the elements of the set $K$ by $a_1, a_2, a_3, \\dots$, so that $a_1 < a_2 < a_3 < \\dots$. We will prove the claim by induction. Obviously, the sum of all numbers less than $a_2$ is equal to $a_1$ and is therefore less than $a_2$.\n\nNow, assume that for a positive integer $n$ we have $a_n > a_{n-1} + a_{n-2} + \\dots + a_2 + a_1$. We show that a similar estimate holds for $n+1$.\n\nWe have $a_n < a_{n+1}$, so $a_{n+1}$ cannot be a divisor of $a_n$ and $a_n$ is a divisor of $a_{n+1}$. So, there exists a positive integer $k$ such that $a_{n+1} = k a_n$. Since $a_n$ and $a_{n+1}$ are not equal, we have $k \\ge 2$ and\n$$\na_{n+1} = k a_n \\ge 2 a_n = a_n + a_n > a_n + (a_{n-1} + a_{n-2} + \\dots + a_2 + a_1)\n$$\nby induction hypothesis. This inequality shows that the claim is true for $n+1$ and, by induction, for all positive integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41301,
    "subject": "Mathematics (Multi-modal)",
    "question": "Anja has a number of $1 \\times 1$ square tiles $\\Box$, while Bojan has the L-shaped tiles $\\Box\\Box$. They take turns putting their tiles onto a rectangular board, one tile at a time. Anja wins if Bojan cannot fit another one of his tiles onto the board when his turn comes despite there being squares left uncovered. Otherwise, Bojan wins. Prove:\n\n![](attached_image_1.png)\n\na) given a $6 \\times 9$ board, Bojan cannot win regardless of who has the first move;\n\nb) given a $8 \\times 8$ board, can Bojan put the tiles onto the board in a way that insures his winning, regardless of how Anja plays and who had the first move.\n\n(Note: All tiles have to lie entirely on the board and they must not cover any of the other tiles).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(a) A $6 \\times 9$ board has $54$ squares. If Bojan wants to win, they have to cover the entire board, because Anja can always fit in another one of her tiles as long as there are empty squares left. After Bojan and Anja each put $13$ tiles onto the board there will be two squares left uncovered. Regardless of who made the first move, the board will not be covered entirely and Anja will win.\n\n(b) An $8 \\times 8$ board can be divided into $16$ $2 \\times 2$ squares as shown in the figure.\n\nIf Anja is the one to start, she puts one of her tiles into one of these $2 \\times 2$ squares. Bojan can now use one of his tiles to fill this square. He can do this every time his turn comes, so eventually they cover the entire board and Bojan wins.\n\n![](attached_image_2.png)\n\nIf Bojan is the one to start, he puts one of his tiles into one of the $2 \\times 2$ squares. If Anja should happen to cover the remaining small square, he can choose another one of the $2 \\times 2$ squares for his next move. But if, on the other hand, Anja does not cover the remaining small square, Bojan should proceed by covering whatever other $2 \\times 2$ square she chose. After $15$ moves there is either one $2 \\times 2$ square left uncovered or of the four remaining squares three lie inside the same $2 \\times 2$ square. Either way Bojan can put one of his tiles onto the board and Anja is forced to fill the last remaining square, ensuring Bojan's victory.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41302,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sides of an equilateral triangle $ABC$ measure $4$ cm. Denote the orthogonal projections of the midpoint $D$ of the side $AB$ onto the sides $BC$ and $AC$ by $E$ and $F$. Find the area of the triangle $DEF$.",
    "options": [],
    "answer": "3*sqrt(3)/4",
    "solution": "We have $|DA| = |DB| = \\frac{|AB|}{2} = 2$. The angles of the triangle $DBE$ measure $60^\\circ$, $90^\\circ$ and $30^\\circ$, so $DBE$ is one half of an equilateral triangle. This implies $|BE| = \\frac{|BD|}{2} = 1$. A similar argument for the triangle $ADF$ shows that this triangle is congruent to the triangle $BDE$. The lengths of the sides $DE$ and $DF$ are $\\frac{\\sqrt{3}}{2} \\cdot |BD| = \\sqrt{3}$, so the area of each of the two triangles is $\\frac{\\sqrt{3}}{2}$.\n\nIn the triangle $ECF$ the lengths of the sides are $|CF| = |CA| - |AF| = 3$ and $|CE| = |CB| - |EB| = 3$, so this is an isosceles triangle with the angle of $60^\\circ$ at the apex. The triangle $ECF$ is therefore equilateral. Its area is $\\frac{3^2 \\cdot \\sqrt{3}}{4}$.\n\n![](attached_image_1.png)\n\nBy subtracting the areas of triangles $DBE$, $ADF$ and $ECF$ from the area of the triangle $ABC$ we can obtain the area of the triangle $DEF$:\n$$\n\\frac{4^2 \\cdot \\sqrt{3}}{4} - \\frac{\\sqrt{3}}{2} - \\frac{\\sqrt{3}}{2} - \\frac{3^2 \\cdot \\sqrt{3}}{4} = \\frac{3\\sqrt{3}}{4}\n$$\nWe see that $|DA| = |DB| = \\frac{|AB|}{2} = 2$. The angles of triangles $AFD$ and $BED$ are equal to $60^\\circ$, $90^\\circ$ and $30^\\circ$ and these two triangles agree in the lengths of the hypotenuses, so they are congruent. This implies $|DE| = |DF|$, so $DEF$ is an isosceles triangle. Let $G$ be the intersection of segments $CD$ and $EF$. Since $\\angle FDG = \\angle GDE = 60^\\circ$, $DG$ is the altitude in the triangle $DEF$. Triangles $DGE$ and $DGF$ are congruent and their angles are $60^\\circ$, $90^\\circ$ and $30^\\circ$.\n\nTriangles $ADC$, $AFD$ and $DGF$ are similar. From the first two we get $\\frac{|AD|}{|AC|} = \\frac{|AF|}{|AD|}$, so $|AF| = 1$. Similarly, $\\frac{|AD|}{|DC|} = \\frac{|AF|}{|FD|}$, implies $|FD| = \\sqrt{3}$.\n\nSince $ADC$ and $DGF$ are similar, we have $\\frac{|AD|}{|AC|} = \\frac{|DG|}{|DF|}$, so $|DG| = \\frac{\\sqrt{3}}{2}$. Likewise, $\\frac{|AD|}{|DC|} = \\frac{|DG|}{|GF|}$ implies $|GF| = \\frac{3}{2}$.\n\nThe area of the triangle $DEF$ is equal to the sum of the areas of triangles $FGD$ and $DGE$, so\n$$\n|DG| \\cdot |FG| = \\frac{3\\sqrt{3}}{4}.\n$$\nObviously, $|DA| = |DB| = \\frac{|AB|}{2} = 2$. Triangles $AFD$ and $BED$ are congruent since they both have angles equal to $60^\\circ$, $90^\\circ$ and $30^\\circ$ and the lengths of the two hypotenuses are also equal. So, $|DE| = |DF| = \\sqrt{3}$. Let $G$ be the midpoint of the segment $EF$. Since $EF$ is parallel to $AB$, right triangles $FGD$ and $EGD$ are congruent and $\\angle DFG = \\angle DEG = 30^\\circ$. Thus,\n$$\np_{DEF} = 2p_{DFG} = 2 \\cdot \\frac{1\\sqrt{3}^2 \\cdot \\sqrt{3}}{4} = \\frac{3\\sqrt{3}}{4}.\n$$\n\n![](attached_image_2.png)\n\n![](attached_image_3.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41303,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be a point on the side $AB$ of an acute triangle $ABC$ such that the triangle $BCD$ is also acute. Denote the orthocentre of the triangle $BCD$ by $H$. Prove: if points $A$, $D$, $H$ and $C$ are concyclic, then the triangle $ABC$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\angle BAC = \\alpha$. Since $A$, $D$, $H$ and $C$ are concyclic, we have $\\angle DHC = \\pi - \\angle BAC = \\pi - \\alpha$.\n\nLet $E$ be the foot of the altitude from $C$ to the side $BD$. Since $\\angle DHE = \\pi - \\angle DHC = \\alpha$, we have $\\angle HDB = \\frac{\\pi}{2} - \\alpha$.\n\nSince the line $DH$ is perpendicular to the side $BC$, we have $\\angle CBA = \\frac{\\pi}{2} - \\angle BDH = \\alpha$, so the triangle $ABC$ is isosceles.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41304,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIgralca imata kup enakih žetonov, s katerega izmenično jemljeta po enega in ga postavljata na poljubno prazno polje kvadratne tabele velikosti $2008 \\times 2008$. Zmaga tisti, ki prvi postavi žeton tako, da skupaj s tremi drugimi tvori oglišča pravokotnika, ki ima stranice vzporedne stranicam tabele. Kateri igralec ima zmagovalno strategijo - tisti, ki je igro začel, ali njegov soigralec?",
    "options": [],
    "answer": "second player",
    "solution": "Solution:\n\nZmagovalno strategijo ima drugi igralec. Stolpce razdeli v $1004$ parov in sicer sta v prvem paru stolpca $1$ in $2$, v drugem stolpca $3$ in $4$, in tako naprej, do zadnjega para, v katerem sta stolpca $2007$ in $2008$. Kadarkoli da prvi igralec žeton v enega izmed stolpcev, da drugi igralec žeton v isto vrstico drugega stolpca iz para. Najkasneje po $1004$ potezah bo moral prvi igralec postaviti žeton v stolpec, kjer se žeton že nahaja. Tedaj bo drugi igralec s postavitvijo žetona tvoril pravokotnik in zmagal.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41305,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIgralca imata kup enakih žetonov, s katerega izmenično jemljeta po enega in ga postavljata na poljubno prazno polje kvadratne tabele velikosti $2008 \\times 2008$. Zmaga tisti, ki prvi postavi žeton tako, da skupaj s tremi drugimi tvori oglišča enakokrakega trapeza, ki ni pravokotnik, in katerega osnovnici sta vzporedni enemu izmed robov tabele. Kateri igralec ima zmagovalno strategijo - tisti, ki je igro začel, ali njegov soigralec?",
    "options": [],
    "answer": "second player",
    "solution": "Solution:\n\nPoimenujmo enakokrak trapez, ki ni pravokotnik, in ima osnovnici vzporedni enemu izmed robov tabele, pravilen trapez. Zmaga drugi igralec in to ne glede na to kako igra prvi. Drugi igralec po vsaki potezi prvega preveri, ali lahko s postavitvijo svojega žetona tvori pravilen trapez. Če tega ne more narediti, postavi svoj žeton na tabelo tako, da bo po njegovi potezi simetrična glede na navpično simetralo (t.j. njegova poteza je simetrična potezi prvega igralca glede na navpično simetralo tabele).\n\nDenimo, da je zmagal prvi igralec, ki je v zadnji potezi postavil žeton na polje $a_{0}$. Naj bodo $a_{1}, a_{2}$ in $a_{3}$ preostala tri polja, ki s poljem $a_{0}$ tvorijo pravilen trapez (torej so pokrita z žetoni in so med seboj različna). Označimo z $b_{0}, b_{1}, b_{2}$ in $b_{3}$ njim simetrična polja glede na navpično simetralo tabele. Polja $b_{1}, b_{2}$ in $b_{3}$ so pokrita z žetoni, polje $b_{0}$ pa je prazno. Pokažimo, da so polja $a_{0}, a_{1}, a_{2}, a_{3}, b_{0}, b_{1}, b_{2}$ in $b_{3}$ vsa med seboj različna.\n\nKer je dolžina tabele soda, je $a_{i} \\neq b_{i}$ za vsak $i=0,1,2,3$. Če bi veljalo $a_{i}=b_{j}$ za $i \\neq j$, potem bi bilo tudi $b_{i}=a_{j}$, torej bi bila stranica $a_{i} a_{j}$ trapeza $a_{0} a_{1} a_{2} a_{3}$ vodoravna. Vendar potem bi morala biti tudi stranica $a_{k} a_{l}$ za $\\{k, l\\}=\\{0,1,2,3\\} \\backslash\\{i, j\\}$, vodoravna, ker pravilen trapez nima pravih kotov. Od tod pa bi zaradi enakokrakosti sledilo, da je tudi $b_{k}=a_{l}$ in $a_{l}=b_{k}$ za $l \\neq k$. Zato bi bilo $b_{0}=a_{m}$ za nek $m \\neq 0$, kar pa ni mogoče, saj je polje $b_{0}$\n\n![](attached_image_1.png)\n\nsimetrala prazno, na poljih $a_{1}, a_{2}$ in $a_{3}$ pa so žetoni.\n\nVsa polja so med seboj različna, zato sta bili pred zadnjo potezo prvega igralca trojici $a_{1}$, $a_{2}, a_{3}$ in $b_{1}, b_{2}, b_{3}$ pokriti z žetoni, kar pomeni, da je bila pred zadnjo potezo drugega igralca vsaj ena izmed njiju pokrita z žetoni. Drugi igralec je imel priložnost zmagati, a ni zmagal, torej ni upošteval strategije.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41306,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMarko je postavil kola navpično na vodoravna tla tako, da je eden segal $1~\\mathrm{m}$ v višino, drugi pa $2~\\mathrm{m}$ v višino. Od vrha vsakega kola do točke, kjer je bil drugi kol zabit v vodoravna tla, je napel vrv. Na kateri višini od tal sta se vrvi križali?",
    "options": [],
    "answer": "2/3 m",
    "solution": "Solution:\n\nOb narisani skici ugotovimo podobnost trikotnikov. Zapišemo razmerji, npr: $h : 2 = y : x$ in $h : 1 = (x - y) : x$. Izračunamo $h = \\frac{2}{3}~\\mathrm{m}$, kar je višina, na kateri se sekata vrvici.\n\n$h =$ iskana višina,\n$y =$ vodoravna razdalja med nižjim kolom in nožiščem $h$-ja,\n$x =$ vodoravna razdalja med koloma,\n$(x - y) =$ vodoravna razdalja med višjim kolom in nožiščem $h$-ja",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41307,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $K$ podmnožica naravnih števil. Za vsaki dve števili $a$ in $b$ iz množice $K$ velja, da $a$ deli $b$ ali $b$ deli $a$. Dokaži, da je tedaj vsako število $c$ iz množice $K$ večje od vsote vseh tistih števil iz množice $K$, ki so manjša od $c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nElemente množice $K$ označimo z $a_{1}, a_{2}, a_{3}, \\ldots$ tako, da velja $a_{1}<a_{2}<a_{3}<\\ldots$. Dokažimo trditev z indukcijo. Očitno je vsota vseh števil, ki so manjša od $a_{2}$, enaka $a_{1}$ in torej manjša od $a_{2}$. Denimo sedaj, da za naravno število $n$ velja\n$$\na_{n}>a_{n-1}+a_{n-2}+\\ldots+a_{2}+a_{1}\n$$\nDokažimo, da taka ocena velja tudi za $n+1$.\nKer je $a_{n}<a_{n+1}$, $a_{n+1}$ ne more biti delitelj števila $a_{n}$, zato je $a_{n}$ delitelj $a_{n+1}$. Torej obstaja naravno število $k$, da velja $a_{n+1}=k a_{n}$. Ker sta $a_{n}$ in $a_{n+1}$ različni, je $k \\geq 2$. Zato velja\n$$\na_{n+1}=k a_{n} \\geq 2 a_{n}=a_{n}+a_{n}>a_{n}+\\left(a_{n-1}+a_{n-2}+\\ldots+a_{2}+a_{1}\\right) \\text{,}\n$$\nkjer smo upoštevali indukcijsko predpostavko. To pa ravno pomeni, da trditev velja tudi za $n+1$ in zato tudi za vsa ostala števila.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41308,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana je funkcija $f(x) = a + b c^{x}$. Določi realna števila $a$, $b$, $c$, če je $f(0) = 5$, $f(1) = 14$ in $f(2) = 50$.",
    "options": [],
    "answer": "a = 2, b = 3, c = 4",
    "solution": "Solution:\n\nVstavimo dane podatke $f(0) = 5$, $f(1) = 14$ in $f(2) = 50$ v predpis funkcije $f(x)$. Dobimo sistem enačb:\n$$\n5 = a + b,\n$$\n$$\n14 = a + b c,\n$$\n$$\n50 = a + b c^{2}.\n$$\nUporabimo zamenjalni način reševanja in dobimo rešitvi za $c$: $c_{1} = 4$ in $c_{2} = 1$. Rešitev $c_{2}$ odpade zaradi nedefiniranosti ulomka $\\frac{9}{c-1}$. Izračunamo preostale vrednosti $a = 2$, $b = 3$ ter zapišemo predpis funkcije:\n$$\nf(x) = 2 + 3 \\cdot 4^{x}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41309,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIzračunaj $\\sin 2x$, če je $\\sin x - \\cos x = \\frac{1}{2}$.",
    "options": [],
    "answer": "3/4",
    "solution": "Solution:\n\nZapišemo zvezo za $\\sin 2x = 2 \\sin x \\cos x$ in $(\\sin x - \\cos x)^2 = \\sin^2 x - 2 \\sin x \\cos x + \\cos^2 x$,\nkamor vstavimo $\\sin x - \\cos x = \\frac{1}{2}$. Dobimo enačbo $\\left(\\frac{1}{2}\\right)^2 = 1 - 2 \\sin x \\cos x$. Izračunamo\n$2 \\sin x \\cos x = \\frac{3}{4} = \\sin 2x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41310,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOd poljubno izbranega trimestnega števila odštejemo število, ki ima števke v obratnem vrstnem redu. Pokaži, da je razlika vedno deljiva z $11$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nObe števili zapišemo z desetiškim mestnim zapisom in ju odštejemo. Izpostavimo skupni faktor in iz razstavljene oblike ugotovimo, da je izraz deljiv z $11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41311,
    "subject": "Mathematics (Multi-modal)",
    "question": "On every face of a cube we draw one of the two diagonals. Some of these diagonals share a vertex. Let $N$ denote the number of pairs of diagonals that have a vertex in common (each diagonal can occur in several different pairs). Find the greatest and the smallest possible values of $N$.",
    "options": [],
    "answer": "Minimum N = 4, Maximum N = 12",
    "solution": "The first figure is for $N = 12$ and the second is for $N = 4$. Let us prove that these are the greatest and the smallest possible values of $N$.\n\n![](attached_image_1.png)\n\nA cube has $8$ vertices. We draw $6$ diagonals, so there are $12$ endpoints altogether. There are $0$, $1$, $2$ or $3$ of the chosen diagonals coming from every vertex of the cube. So, for every vertex we get $0$, $0$, $1$ or $3$ different pairs of diagonals.\n\nAssume that it is possible to get $N \\le 3$. If there exists a vertex with $3$ diagonals, then the remaining vertices can have at most one diagonal. This would imply that there are at most $1 \\cdot 3 + 7 \\cdot 1 = 10$ endpoints, a contradiction.\n\n![](attached_image_1.png)\n\nThus, no vertex has $3$ chosen diagonals meeting at it and there can be at most $3$ vertices where two of the diagonals meet. Again, every other vertex can be the endpoint of at most one diagonal and there are at most $3 \\cdot 2 + 5 \\cdot 1 = 11$ endpoints in this case. Once more, a contradiction. We have shown that $N \\ge 4$.\n\nThere are $\\frac{6 \\cdot 5}{2} = 15$ different pairs of diagonals. No two diagonals belonging to the opposite faces of the cube intersect, so there are at most $15 - 3 = 12$ different pairs of intersecting diagonals, $N \\le 12$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 41312,
    "subject": "Mathematics (Multi-modal)",
    "question": "At the National Mathematical Olympiad the students were given 4 problems. Each solution was awarded with an integral number of points between 0 and 7. There were 42 participants and exactly half of them achieved at least 50% of the points. To win the award one had to get at least 22 points and one sixth of the contestants achieved this. Together, the students who did not get the award had three times as many points as those who did. Prove that there were at least 6 contestants such that each of them got at least 25 points but none of them got more than 50%.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First note that 7 contestants won the award (one sixth of 42). The upper half consisted of 21 contestants, so $21 - 7 = 14$ got between 14 and 21 points.\nEach of the 7 contestants who won the award had to get at least 22 points, so together they had at least $22 \\cdot 7 = 154$ points.\nLet $x$ represent the number of contestants who got at least 25% but less than 50% of the points, that is at least 7 but not more than 13 points. The\n\nremaining $42 - 21 - x = 21 - x$ contestants achieved less than 6 points.\nWhat was the total number of points given to those that did not win the award? For those with at most 6 points the total was at most $6 \\cdot (21 - x)$, and for those with at least 7 but less than 13 points the total was at most $13x$ points. Finally, we have to consider those 14 contestants from the upper half with at most 21 points. They account for at most $21 \\cdot 14 = 294$ points.\nThis makes for at most $6 \\cdot (21 - x) + 13x + 294 = 420 + 7x$ points. On the other hand, this number should be three times as great as the number of points given to those who won the award, which is $3 \\cdot 154 = 462$. Thus, we have $462 \\le 420 + 7x$, or, equivalently, $7x \\ge 42$ and so $x \\ge 6$. We conclude there exist at least 6 contestants such that each of them got at least 25 points but none of them got more than 50%.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41313,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be positive integers. Assume that for some positive integer $k$, $k < n$, the following is true: if we choose any $k$ of the given $n$ numbers their sum is divisible by $n$. Prove that $a_1 + a_2 + \\dots + a_n$ is also divisible by $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $k = 1$, each of the numbers $a_1, a_2, \\dots, a_n$ is divisible by $n$, so $n$ also divides their sum.\n\nNow, let $1 < k < n$ and let $i \\neq j$. Since the set $\\{a_1, a_2, \\dots, a_n\\} \\setminus \\{a_i, a_j\\}$ has $n-2 \\ge k-1$ elements, one can choose an arbitrary subset $S$ with $k-1$ elements. Then $S \\cup \\{a_i\\}$ and $S \\cup \\{a_j\\}$ are two $k$-tuples of numbers and their sums are divisible by $n$, so the difference of the two sums must also be divisible by $n$. This difference is equal to $a_i - a_j$, so $n$ divides $a_i - a_j$. Here $i$ and $j$ were chosen arbitrarily, so $a_1, a_2, \\dots, a_n$ give the same remainder when divided by $n$, and there are $n$ of them, so their sum must be divisible by $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41314,
    "subject": "Mathematics (Multi-modal)",
    "question": "Maja can write positive integers onto a blackboard according to two rules. For every number $n$ that is already on the board, she can write down $3n+13$. Whenever one of the numbers on the board is a perfect square, she can also write down its square root.\n\na. Using only this two rules, can she get the number $55$ assuming she starts with $256$?\n\nb. Using only this two rules, can she get the number $256$ assuming she starts with $55$?",
    "options": [],
    "answer": "a: yes; b: no",
    "solution": "a. With $256$ written on the board, Maja can get $55$ by taking the square root of $256$, obtaining $16$, and then writing down $3 \\cdot 16 + 13 = 61$ and $3 \\cdot 61 + 13 = 196$. Now, she can write down the square root of $196$, which is $14$, and finally she can get $55$ as $3 \\cdot 14 + 13 = 55$. (Note: there might be other ways of obtaining $55$, but this is the only one that requires less than $100$ steps).\n\nb. Let us now show that Maja cannot get $256$ from $55$. Consider the remainders modulo $4$. An even number is of the form $2k$ and its square is $4k^2$. An odd number can be written as $2k + 1$, so its square is equal to $4k^2 + 4k + 1$, giving the remainder of $1$ when divided by $4$. We see that a perfect square can only give the remainders $0$ or $1$ when divided by $4$, so we can never use the rule $n \\mapsto \\sqrt{n}$ on numbers that give the remainder $2$ or $3$.\n\nThe remainder we get when dividing $55$ by $4$ is $3$. Using the rule $n \\to 3n+13$ on a number of the form $4k+3$, we get $12k+9+13 = 4(3k+5)+2$, a number that gives the remainder of $2$. Using this rule on the number of the form $4k+2$, we get $12k+6+13 = 4(3k+4)+3$, the remainder is $3$. After one step the remainder will be $2$, then $3$, then $2$ again, and so on. Since $256$ is divisible by $4$, Maja can never reach it using the given two rules.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41315,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIzrazi $(A-B)^{-1}$ z $x$, če je $A=\\frac{3 x^{-\\frac{1}{3}}}{x^{\\frac{2}{3}}-2 x^{-\\frac{1}{3}}}$ in $B=\\frac{x^{\\frac{1}{3}}}{x^{\\frac{4}{3}}-x^{\\frac{1}{3}}}$. Izraz poenostavi.",
    "options": [],
    "answer": "(x-2)(x-1)/(2x-1)",
    "solution": "Solution:\n\nPoenostavimo izraz $A$, tako da ulomek razširimo z $x^{\\frac{1}{3}}$. Dobimo $A=\\frac{3}{x-2}$. Izraz $B$ pa razširimo z $x^{-\\frac{1}{3}}$. Dobimo $B=\\frac{1}{x-1}$. Izračunamo razliko $A-B=\\frac{3}{x-2}-\\frac{1}{x-1}=\\frac{2x-1}{(x-2)(x-1)}$. Zapišemo še $(A-B)^{-1}=\\frac{(x-2)(x-1)}{2x-1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41316,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVsota prvih petih členov aritmetičnega zaporedja je enaka $50$, razlika med petim in drugim členom pa je $9$. Izračunaj, kateri člen je sedemkrat tolikšen kot prvi.",
    "options": [],
    "answer": "9",
    "solution": "Solution:\n\nUporabimo obrazec za vsoto prvih petih členov aritmetičnega zaporedja $s_{5} = \\frac{5}{2} (2 a_{1} + 4 d)$. Upoštevamo, da je $a_{5} - a_{2} = 9$ oziroma $a_{1} + 4d - a_{1} - d = 9$. Izračunamo diferenco $d = 3$. Diferenco vstavimo v obrazec za vsoto $\\frac{5}{2} (2 a_{1} + 4 d) = 50$ in izračunamo $a_{1} = 4$. Zapišemo zvezo $a_{n} = 7 a_{1} = 28$. Uporabimo znane podatke in izračunamo, da je deveti člen sedemkrat tolikšen kot prvi.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41317,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa katero vrednost realnega števila $a$ je $1+\\frac{1}{2+\\frac{1}{3+\\frac{1}{a}}}=0$?",
    "options": [],
    "answer": "-3/10",
    "solution": "Solution:\n\nPoenostavimo verižni ulomek do oblike $\\frac{10a+3}{7a+2}$. Ta je enak nič, če je števec enak $0$. Tako je $10a+3=0$ oziroma $a=-\\frac{3}{10}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41318,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nS tanko palico neznane dolžine želimo ugotoviti prav tako neznani širino in višino vrat. Če položimo palico vodoravno ob vratih, je ta za 2 laketa daljša od širine vrat. Če palico postavimo navpično, je za 1 laket daljša od višine vrat. Palica se natanko prilega odprtini vrat, če jo postavimo diagonalno med vrata. Izračunaj širino in višino vrat ter dolžino palice.",
    "options": [],
    "answer": "width = 3, height = 4, rod length = 5",
    "solution": "Solution:\n\nOznačimo dolžino palice z $d$, širino vrat z $x$ in višino vrat z $y$. Veljajo zveze $x = d - 2$, $y = d - 1$ in $x^{2} + y^{2} = d^{2}$.\n\nReševanje sistema treh enačb s tremi neznankami privede do enačbe $d^{2} - 6d + 5 = 0$ in rešitev $d_{1} = 1$ in $d_{2} = 5$. Rešitev $d = 1$ ne ustreza. Iz $d = 5$ pa sledita še rešitvi $x = 3$ in $y = 4$.\n\n$x =$ širina vrat\n$y =$ višina vrat\n$d =$ dolžina palice",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41319,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $m$ and $n$ such that $m^2 + n^5 = 252$.",
    "options": [],
    "answer": "m=3, n=3",
    "solution": "Since $m^2 = 252 - n^5$ is non-negative, we have $n^5 \\le 252$, so $n < 4$.\n\nIf $n=1$ we get $m^2 = 251$,\n\nif $n=2$ we have $m^2 = 220$,\n\nand if $n=3$ we have $m^2 = 9$.\n\nThe only possible solution is $m = n = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41320,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. $n \\ge 3$. There are $n$ pairwise different numbers written on a blackboard. Show that we can choose two of those numbers so that no number from the blackboard multiplied by $3$ is equal to a multiple of their sum.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the numbers by $a_1, a_2, \\dots, a_n$. Without loss of generality we may assume that $a_1 > a_2 > \\dots > a_n$. Let us show we may also assume that not all of these numbers are divisible by $3$. If $b_1, \\dots, b_n$ are all divisible by $3$, then there exists a positive integer $k$ such that $3^k$ divides $b_j$ for all $j = 1, \\dots, n$, as well as a positive integer $l$ such that $3^{k+1}$ does not divide $b_l$. If this is the case, then $a_1 = \\frac{b_1}{3^k}, \\dots, a_n = \\frac{b_n}{3^k}$ are pairwise different positive integers, which are not all divisible by $3$. Assume that there exist two among them, such that $a_i + a_j$ does not divide any of the numbers $3a_1, \\dots, 3a_n$. Then $3^k(a_i + a_j) = b_i + b_j$ does not divide $3^k(3a_1) = 3b_1, \\dots, 3^k(3a_n) = 3b_n$.\n\nLet $a_1 > a_2 > \\dots > a_n$ and assume $a_1, \\dots, a_n$ are not all divisible by $3$. We will prove the claim by contradiction. If the claim is not true, then for every sum $a_i + a_j$ there exists an index $k_{ij}$ such that $3a_{k_{ij}}$ is a multiple of this sum. In particular, this holds for $i = 1$, so $a_i + a_1$ divides $3a_{k_{i1}}$ for some $k$. If $a_i + a_1$ is not divisible by $3$, then $a_i + a_1$ divides $a_{k_{i1}}$. This is impossible since $a_i + a_1 > a_{k_{i1}}$. So, $3$ divides $a_i + a_1$ for all $2 \\le i \\le n$. At least one of the numbers is not divisible by $3$ which implies none of them are and $a_2, \\dots, a_n$ all give the same remainder when divided by $3$.\n\nThis remainder is non-zero, so $a_i + a_2$ is not divisible by $3$ for $i = 3, \\dots, n$. Now, $a_i + a_2$ divides $3a_{k_{i2}}$, so $a_i + a_2$ divides $a_{k_{i2}}$. This is only possible if $k_{i2} = 1$. Hence, $a_3 + a_2, \\dots, a_n + a_2$ all divide $a_1$.\n\nWe know that $(a_1 + a_2)l = 3a_m$ for some $m$ and some positive integer $l$. Obviously, $l \\ge 3$ implies $(a_1 + a_2)l > 3a_1 > 3a_m$, so $l = 1$ or $l = 2$. If $l = 2$, then $3a_m = 2(a_1 + a_2) > 4a_2$, which would imply $m = 1$ and $a_1 = 2a_2$. We have shown that $a_2 + a_3$ divides $a_1$ and when $a_1 = 2a_2$ we have $2(a_2 + a_3) = a_1 + 2a_3 > a_1$, so $a_2 + a_3 = a_1 = 2a_2$ or $a_2 = a_3$. This is not possible since all the numbers on the blackboard are different. We conclude that $l = 1$.\n\nWe have shown that $a_1 + a_2 = 3a_m$ for some $m$. Since\n$$\n3a_m = a_1 + a_2 \\ge a_2 + a_3 + a_3 > 3a_3\n$$\nwe have $m < 3$. If $m = 1$ then $a_1 = 2a_2$, but this is not possible since $a_1 > a_2$. Hence $m = 2$ and $a_1 = 2a_2$. As above, since $2(a_2 + a_3) = a_1 + 2a_3 > a_1$ and $a_2 + a_3$ divides $a_1$, we have $a_2 + a_3 = a_1 = 2a_2$ or $a_2 = a_3$. This contradicts the assumption that all numbers are different.\n\nWe have arrived at a contradiction and we can conclude that it is always possible to choose two of the numbers so that any other number from the board multiplied by $3$ is different from their sum.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41321,
    "subject": "Mathematics (Multi-modal)",
    "question": "The base $AB$ of the trapezoid $ABCD$ is longer than the base $CD$, and $\\angle ADC$ is a right angle. The diagonals $AC$ and $BD$ are perpendicular. Let $E$ be the foot of the altitude from $D$ to the line $BC$. Prove that $\\frac{|AE|}{|BE|} = \\frac{|AC| \\cdot |CD|}{|AC|^2 - |CD|^2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\n1st solution. Since $ACD$ is a right triangle, we have $|AC|^2 - |CD|^2 = |AD|^2$. The diagonals of the trapezoid intersect at a right angle, so $\\angle DCA = \\frac{\\pi}{2} - \\angle BDC = \\angle ADB$. This, together with $\\angle ADC = \\frac{\\pi}{2} = \\angle BAD$, implies that the triangles $ACD$ and $BDA$ are similar. Thus, $\\frac{|AD|}{|CD|} = \\frac{|AB|}{|AD|}$ or, equivalently, $|AD|^2 = |AB| \\cdot |CD|$. This implies\n$$\n\\frac{|AC| \\cdot |CD|}{|AC|^2 - |CD|^2} = \\frac{|AC| \\cdot |CD|}{|AD|^2} = \\frac{|AC| \\cdot |CD|}{|AB| \\cdot |CD|} = \\frac{|AC|}{|AB|}.\n$$\nSince $\\angle BAD + \\angle DEB = \\frac{\\pi}{2} + \\frac{\\pi}{2} = \\pi$. $ABED$ is a cyclic quadrilateral. So, $\\angle AEB = \\angle ADB$. Because of the right angles we have $\\angle BAC = \\frac{\\pi}{2} - \\angle CAD = \\angle ADB$. The triangles $ABE$ and $CBA$ are similar, since $\\angle AEB = \\angle BAC$ and $\\angle EBA = \\angle ABC$. Hence, $\\frac{|AE|}{|BE|} = \\frac{|CA|}{|AB|}$ and this implies\n$$\n\\frac{|AC| \\cdot |CD|}{|AC|^2 - |CD|^2} = \\frac{|AC|}{|AB|} = \\frac{|AE|}{|BE|}.\n$$\n![](attached_image_2.png)\n\n2nd **solution.** Since $ACD$ is a right triangle, we have $|AC|^2 - |CD|^2 = |AD|^2$. Thus, it suffices to show that\n$$\n\\frac{|AE|}{|BE|} = \\frac{|AC| \\cdot |CD|}{|AD|^2}.\n$$\nLet $T$ be the intersection of the diagonals. Obviously, $\\angle TAD = \\angle ABT$, so the triangles $ACD$ and $BDA$ are similar. Hence, $\\frac{|AD|}{|CD|} = \\frac{|AB|}{|AD|}$ or, equivalently, $|AD|^2 = |AB| \\cdot |CD|$. We have $\\angle BAD + \\angle DEB = \\frac{\\pi}{2} + \\frac{\\pi}{2} = \\pi$, so the quadrilateral $ABED$ is cyclic and $\\angle EAD = \\angle EBD$. Similarly, we have $\\angle CTD + \\angle DEC = \\pi$, so the quadrilateral $CEDT$ is cyclic and $\\angle CDE = \\angle CTE$. Since the triangles $AED$ and $BET$ are similar, we have $\\frac{|AE|}{|BE|} = \\frac{|AD|}{|BT|}$. The triangles $ADC$ and $BTA$ are similar, so $\\frac{|AD|}{|BT|} = \\frac{|AC|}{|AB|}$. Hence,\n$$\n\\frac{|AE|}{|BE|} = \\frac{|AC|}{|AB|} = \\frac{|AC| \\cdot |CD|}{|AB| \\cdot |CD|} = \\frac{|AC| \\cdot |CD|}{|AD|^2}.\n$$\nwhich was to be shown.\n3rd **solution.** We have $\\angle BAD + \\angle DEB = \\frac{\\pi}{2} + \\frac{\\pi}{2} = \\pi$, so $ABED$ is a cyclic quadrilateral. By Ptolemy's theorem we have $|AB| \\cdot |DE| + |AD| \\cdot |BE| = |AE| \\cdot |BD|$ or\n$$\n\\frac{|AE|}{|BE|} = \\frac{|AB| \\cdot |DE|}{|BD| \\cdot |BE|} + \\frac{|AD|}{|BD|}.\n$$\nLet $T$ be the intersection of the diagonals. Then $ABD$ and $TBA$ are right triangles and $\\angle TBA = \\angle ABD$, so they are similar. This implies that $\\frac{|AD|}{|BD|} = \\frac{|TA|}{|AB|}$ and $\\frac{|AB|}{|BD|} = \\frac{|BT|}{|AB|}$. Similarly, $DEB$ and $CTB$ are right triangles and since $\\angle EBD = \\angle TBC$, they too are similar. We see that $\\frac{|DE|}{|BE|} = \\frac{|CT|}{|BT|}$ and\n$$\n\\frac{|AE|}{|BE|} = \\frac{|BT| \\cdot |CT|}{|AB| \\cdot |BT|} + \\frac{|TA|}{|AB|} = \\frac{|CT| + |TA|}{|AB|} = \\frac{|CA|}{|AB|}\n$$\nFinally, the triangles $ADC$ and $BAD$ are similar and we have $\\frac{|CD|}{|AD|} = \\frac{|AD|}{|AB|}$, so\n$$\n\\frac{|AE|}{|BE|} = \\frac{|CA|}{|AB|} = \\frac{|CA| \\cdot |CD|}{|AD|^2} = \\frac{|CA| \\cdot |CD|}{|AC|^2 - |CD|^2}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41322,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider all polynomials with the leading coefficient $1$ that give the remainder of $1$ when divided by $x + 1$ and the remainder $2$ when divided by $x^2 + 1$. Among these, find the polynomial of the lowest degree.",
    "options": [],
    "answer": "x^3 + (1/2)x^2 + x + 5/2",
    "solution": "Obviously, the degree of such a polynomial must be at least $2$. If the degree were equal to $2$, this polynomial would have the form of\n$$\np(x) = x^2 + a x + b = (x^2 + 1) + a x + (b - 1).\n$$\nWhen dividing by $x^2 + 1$ the remainder is $2$, and $a x + (b - 1) = 2$ implies $a = 0$ and $b = 3$. But when we divide $p(x) = x^2 + 3 = (x - 1)(x + 1) + 4$ by $x + 1$ the remainder cannot be $1$. We can prove this by calculating $p(-1) = 4$, since $p(-1)$ is precisely the remainder obtained when dividing $p(x)$ by $x + 1$.\n\nWe conclude that the degree of our polynomial is at least $3$. Let\n$$\np(x) = x^3 + a x^2 + b x + c = (x + a)(x^2 + 1) + (b - 1)x + c - a.\n$$\nAgain, we have $(b-1)x + c - a = 2$. Hence $b = 1$, $c = a + 2$. From\n$$\np(x) = x^3 + a x^2 + x + a + 2 = (x + 1)(x^2 + (a - 1)x + 2 - a) + 2a,\n$$\nwe find $2a = 1$ (we can also deduce this from $1 = p(-1) = -1 - a - 1 + a + 2$). The polynomial is $p(x) = x^3 + \\frac{1}{2}x^2 + x + \\frac{5}{2}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41323,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $x$, $y$ and $z$ which are solutions of the system of equations\n$$\nx + y - 2z = 0, \\quad xy - z^2 = 0, \\quad y^2 + 5z + 6 = 0.\n$$",
    "options": [],
    "answer": "(2, 2, -2) and (3, 3, -3)",
    "solution": "From the first equation we get $x = -2z - y$. Inserting this into the second equation we get $(-2z - y)y - z^2 = 0$ or, equivalently, $-(z + y)^2 = 0$. This implies that $z = -y$. Finally, we use this together with the third equation to obtain $y^2 - 5y + 6 = 0$ or $(y - 2)(y - 3) = 0$. Hence, $y = 2$ or $y = 3$.\nThere are two solutions: $x = 2$, $y = 2$, $z = -2$ and $x = 3$, $y = 3$, $z = -3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41324,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)$ be a non-constant arithmetic sequence with the initial term $a_1 = 1$. The terms $a_2$, $a_5$, $a_{11}$ form a geometric sequence. Find the sum of the first 2009 terms of the sequence $(a_n)$.",
    "options": [],
    "answer": "1010527",
    "solution": "Let $d$ be the difference of the sequence $(a_n)$. Then $a_2 = 1 + d$, $a_5 = 1 + 4d$ and $a_{11} = 1 + 10d$. Since $a_2$, $a_5$ and $a_{11}$ form a geometric progression, we have $(1 + 4d)^2 = (1 + d)(1 + 10d)$ or $6d^2 = 3d$. Since the sequence is not constant, we conclude that $d = \\frac{1}{2}$ and the sum of the first 2009 terms is $2009 + \\frac{2009 \\cdot 2008}{2} = 1010527$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41325,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all even positive integers $n$ such that $-53 < \\frac{2009}{53-n} < 53 - n$.",
    "options": [],
    "answer": "2, 4, 6, 8, 92, 94, 96",
    "solution": "If $n < 53$, then $\\frac{2009}{53-n} > 0 > -53$. The inequality $\\frac{2009}{53-n} < 53-n$ is equivalent to $2009 < (53-n)^2$ or $\\sqrt{2009} < 53-n$. Since $\\sqrt{2009} > 44$, we have $53-n > 44$ or $9 > n$. The possible values of $n$ are $2$, $4$, $6$ and $8$, and for any of those we have $53-n \\ge 45 > \\sqrt{2009}$, so the initial inequality holds.\n\nNow, let $n > 53$. The inequality is then equivalent to\n$$\n-53(53 - n) > 2009 > (53 - n)^2.\n$$\nNow, $-53^2 + 53n > 2009$ implies that $n > \\frac{2009+53^2}{53} = 90 + \\frac{48}{53}$. On the other hand we have $45 > \\sqrt{2009} > n - 53$. So $n > 98$. The possible values of $n$ in this case are $92$, $94$ and $96$. For all three we have $n - 53 \\le 96 - 53 = 43 < \\sqrt{2009}$ and again the initial inequality holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41326,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Prove: if $a - \\frac{1}{b} + b(b + \\frac{3}{a})$ is an integer for some positive integers $a$ and $b$, then it is a perfect square.\n\nb. Find two integers $a$ and $b$ such that $a - \\frac{1}{b} + b(b + \\frac{3}{a})$ is a positive integer but not a perfect square.",
    "options": [],
    "answer": "(a) It is always a perfect square for positive integers satisfying the integrality condition. (b) Examples: a = 4, b = −2 gives 7, and a = −4, b = −2 gives 2.",
    "solution": "Let $I = a - \\frac{1}{b} + b(b + \\frac{3}{a})$. If we want $I = a + b^2 + \\frac{3b}{a} - \\frac{1}{b}$ to be an integer, then $N = \\frac{3b}{a} - \\frac{1}{b}$ has to be an integer as well. Since $N = \\frac{3b^2 - a}{ab}$, we conclude that $ab$ must divide $3b^2 - a$. This implies that $b$ divides $3b^2 - a$, so $b$ divides $a$.\n\nWe can now write $a = k b$. Then $N = \\frac{3b - k}{k b}$, so $k b$ divides $3b - k$. $b$ divides $3b - k$, or, finally, $b$ divides $k$. Let $k = l b$. Then $N = \\frac{3 - l}{l b}$. We see that $l$ must divide $3 - l$, so $l$ divides $3$. We have $a = l b^2$ where $a$ and $b$ are positive integers, so $l$ is also a positive integer and we have either $l = 1$ or $l = 3$.\n\nIf $l = 1$ then $N = \\frac{2}{b}$ and $b = 1$ or $b = 2$. In the first case we get $a = 1$, in the second case $a = 4$. If $l = 3$ then $N = 0$ and $a = 3b^2$.\n\nFinally, we need to determine the value of $I$. If $a = b = 1$ then $I = 4$, if $a = 4$ and $b = 2$ then $I = 9$ and if $a = 3b^2$ then $I = 4b^2 = (2b)^2$, so $I$ is indeed a perfect square.\n\nIf $a = 4$ and $b = -2$ then $l = 7$. If $a = -4$ and $b = -2$ then $l = 2$. In both cases $l$ is a positive integer, but it is not a perfect square.\n\n**Remark.** As in the first part of the problem we can show that $a = 4$, $b = -2$ and $a = -4$, $b = -2$ are the only two cases where $l$ is a positive integer but not a perfect square. We know from the first part that at least one of the numbers $a$ and $b$ is negative. As above, we conclude that $a = l b^2$ and $l = b^2 + a + \\frac{3 - l}{l b} = (l - 1) b^2 + \\frac{3 - l}{l b}$. In this case we can choose $l$ or $b$ to be negative.\n\nWe know that $l$ divides $3$. So it can be equal to $1$, $-1$, $3$ or $-3$. If $l = 3$ then $a = 3b^2$ and $l = (2b)^2$ regardless of what $b$ is. When $l = 1$ we have $b = -1$ or $b = -2$. In the first case $l = 0$ and in the second case $a = 4$ and $l = 7$.\n\nIf $l = -1$ then $l = -\\frac{1}{2}$. So, $l$ is a positive divisor of $4$, but not a perfect square. Hence, $l = 2$, $b = -2$, $a = -4$. In the remaining case we have $l = -3$ and $l = -2b^2 - \\frac{1}{2}$. This is an integer when $b = \\pm 1$ or $b = \\pm 2$, but $l$ is negative in both cases.\n\nHence, $l$ is a positive integer but not a perfect square if and only if $a = 4$, $b = -2$ or $a = -4$, $b = -2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41327,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number and let $a$, $b$ and $c$ be integers divisible by $p$, such that the polynomial $q(x) = x^3 + a x^2 + b x + c$ has at least two different integer roots. Show that $p^2$ divides $b$ and $p^3$ divides $c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $y$ and $z$ be two different integer roots of $q$. Then $y^3 + a y^2 + b y + c = 0$ and $z^3 + a z^2 + b z + c = 0$. We know that $p$ divides $a$, $b$ and $c$. Since $y^3 = -c - b y - a y^2$, $p$ divides $y^3$. Similarly, we show that $p$ divides $z^3$. As $p$ is a prime, it must divide $y$ and $z$.\n\nBy subtracting the two equalities above we get $y^3 - z^3 + a(y^2 - z^2) + b(y - z) = 0$ or\n$$\n(y - z)(y^2 + y z + z^2 + a(y + z) + b) = 0.\n$$\nSince $z \\neq y$, this implies $y^2 + y z + z^2 + a(y + z) + b = 0$. Now, $p$ is a prime and it divides $y$, $z$ and $a$, so $p^2$ divides $y^2 + y z + z^2 + a(y + z) = -b$, hence $p^2$ divides $b$.\n\nWe know that $c = -y^3 - a y^2 - b y$. Since $p$ divides $y$, $p^2$ divides $y^2$ and $p^3$ divides $y^3$. Since $p$ also divides $a$ and $p^2$ divides $b$, we can conclude that $p^3$ divides $c$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41328,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $1005^{\\ln 121} = 11^{\\ln(1+3+5+\\dots+2009)}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since\n$$\n\\begin{aligned}\n1 + 3 + 5 + \\dots + 2009 &= \\\\\n&= (1 + 2009) + (3 + 2007) + \\dots + (1003 + 1007) + 1005 \\\\\n&= 502 \\cdot 2010 + 1005 = 1005^2.\n\\end{aligned}\n$$\nit suffices to show that $1005^{\\ln(121)} = 11^{\\ln(1005^2)}$. From this we get\n$$\n\\ln(121) \\ln(1005) = \\ln(1005^2) \\ln(11).\n$$\nwhich is equivalent to $\\ln(11^2) \\ln(1005) = 2\\ln(1005) \\ln(11)$. This last equality holds since $\\ln(11^2) = 2\\ln(11)$. Hence, the initial equality holds as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41329,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle and let $I$ be the incentre of the triangle $ABC$. The lines $AI$ and $BI$ meet the circumcircle of the triangle $ABC$ again at $A_1$ and $B_1$. The segments $A_1B_1$ and $CI$ meet at $C_1$. Let $S$ be the circumcentre of the triangle $IA_1C_1$. The lines $SC_1$ and $BI$ meet the segment $AC$ at $D$ and $E$. Find the angles of the triangle $ABC$ if $|CD| = |DE|$ and $\\angle CBA = 2\\angle ACB$.",
    "options": [],
    "answer": "∠BAC = 2π/5, ∠CBA = 2π/5, ∠ACB = π/5",
    "solution": "Let us write $\\angle ACB = \\gamma$. Then $\\angle CBA = 2\\gamma$ and $\\angle BAC = \\pi - 3\\gamma$. If we draw a figure where $\\angle CBA = 2\\angle ACB$, we notice that $IA_1C_1$ is a right triangle. Let us prove this.\nSince $ABA_1B_1$ is a cyclic quadrilateral we have $\\angle C_1A_1I = \\angle B_1A_1A = \\angle B_1BA = \\frac{\\angle BCA}{2} = \\gamma$. The angle $\\angle A_1IC$ is equal to $\\pi - \\angle CIA$. Since $\\angle CIA = \\pi - \\angle CAI - \\angle ACI$, we get\n$$\n\\angle A_1IC = \\angle CAI + \\angle ACI = \\frac{\\pi - 3\\gamma}{2} + \\frac{\\gamma}{2} = \\frac{\\pi}{2} - \\gamma.\n$$\n![](attached_image_1.png)\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 the 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 the 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 the triangle ABC measure $\\angle BAC = \\angle CBA = \\frac{2\\pi}{5}$ and $\\angle ACB = \\frac{\\pi}{5}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41330,
    "subject": "Mathematics (Multi-modal)",
    "question": "At a contest students were presented with 24 multiple choice questions. Contestants who gave no answer or gave more than one answer were awarded 0 points for that question. For the correct answer they received 1 point and for a wrong answer $\\frac{1}{4}$ of a point was deducted. If a student received 13 points, at most how many questions did he answer correctly?",
    "options": [],
    "answer": "15",
    "solution": "Let $P$ be the number of questions for which the contestant was awarded 1 point and let $N$ denote the number of those for which $\\frac{1}{4}$ of a point was deducted. Then $P - \\frac{1}{4}N = 13$ or $N = 4P - 52$. Since there were 24 questions, we get $P + N \\le 24$ or $5P - 52 \\le 24$. So, $5P \\le 76$ and we have $P \\le 15$. The greatest possible value of $P$ is 15 and it is attained when $N = 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41331,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDokaži enakost\n$$\n1005^{\\ln 121}=11^{\\ln (1+3+5+\\ldots+2009)}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nKer je\n$$\n\\begin{aligned}\n& 1+3+5+\\ldots+2009= \\\\\n& \\quad=(1+2009)+(3+2007)+\\ldots+(1003+1007)+1005 \\\\\n& \\quad=502 \\cdot 2010+1005=1005^{2}\n\\end{aligned}\n$$\nje dovolj pokazati $1005^{\\ln (121)}=11^{\\ln \\left(1005^{2}\\right)}$. Če dobljeno enakost logaritmiramo, dobimo\n$$\n\\ln (121) \\ln (1005)=\\ln \\left(1005^{2}\\right) \\ln (11)\n$$\nki je enakovredna $\\ln \\left(11^{2}\\right) \\ln (1005)=2 \\ln (1005) \\ln (11)$. Slednja velja, saj je $\\ln \\left(11^{2}\\right)=2 \\ln (11)$, torej velja tudi prvotna enakost.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41332,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $p$ praštevilo, $a$, $b$ in $c$ pa taka cela števila, deljiva s $p$, da ima polinom\n$$\nq(x) = x^{3} + a x^{2} + b x + c\n$$\nvsaj dve različni celi ničli. Dokaži, da $p^{2}$ deli $b$ in $p^{3}$ deli $c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNaj bosta $y$ in $z$ različni celi ničli polinoma $q$. Tedaj velja $y^{3} + a y^{2} + b y + c = 0$ in $z^{3} + a z^{2} + b z + c = 0$. Ker $p \\mid a$, $p \\mid b$ in $p \\mid c$ ter je $y^{3} = -c - b y - a y^{2}$, $p$ deli $y^{3}$. Podobno sledi, da $p \\mid z^{3}$. Ker je $p$ praštevilo, je zato delitelj tako $y$ kot $z$.\n\nČe zgornji enačbi odštejemo, dobimo $y^{3} - z^{3} + a(y^{2} - z^{2}) + b(y - z) = 0$ oziroma\n$$\n(y - z)\\left(y^{2} + y z + z^{2} + a(y + z) + b\\right) = 0\n$$\nKer je $z \\neq y$, tako velja $y^{2} + y z + z^{2} + a(y + z) + b = 0$. Praštevilo $p$ je delitelj $y$, $z$ in $a$, zato $p^{2}$ deli $y^{2} + y z + z^{2} + a(y + z) = -b$, torej $p^{2} \\mid b$.\n\nIzrazimo lahko $c = -y^{3} - a y^{2} - b y$. Ker $p \\mid y$, je $p^{2}$ delitelj $y^{2}$ in $p^{3}$ delitelj $y^{3}$. Zaradi deljivosti $a$ z $p$ in $b$ z $p^{2}$ pa od tod sledi, da $p^{3} \\mid c$.\n\n\n2. način\n\nNaj bodo $x_{1}$, $x_{2}$ in $x_{3}$ ničle polinoma $q$ in privzemimo, da sta $x_{1}$ in $x_{2}$ celi števili. Tedaj je $q(x) = (x - x_{1})(x - x_{2})(x - x_{3})$, od koder sledijo Viétove formule $a = -(x_{1} + x_{2} + x_{3})$, $b = x_{1} x_{2} + x_{2} x_{3} + x_{3} x_{1}$ in $c = -x_{1} x_{2} x_{3}$. Iz prve enakosti sledi, da je tudi $x_{3}$ celo število. Ker je $p$ praštevilo in deli $x_{1} x_{2} x_{3}$, deli vsaj enega izmed števil $x_{1}$, $x_{2}$ in $x_{3}$. Predpostavimo lahko, da $p$ deli $x_{1}$. Od tod sledi, da $p$ deli $x_{2} x_{3} = b - x_{1} x_{2} - x_{3} x_{1}$. Spet lahko sklepamo, da $p$ deli eno izmed števil $x_{2}$ oziroma $x_{3}$ ter predpostavimo, da deli $x_{2}$. Zaradi $x_{3} = a + x_{1} + x_{2}$ pa tedaj sledi, da $p \\mid x_{3}$.\n\nVsako izmed števil $x_{1}$, $x_{2}$, $x_{3}$ je deljivo s $p$, zato je njihov produkt deljiv s $p^{3}$, torej $p^{3} \\mid -x_{1} x_{2} x_{3} = c$. Podobno je produkt po dveh izmed števil $x_{1}$, $x_{2}$, $x_{3}$ deljiv s $p^{2}$, zato $p^{2} \\mid x_{1} x_{2} + x_{2} x_{3} + x_{3} x_{1} = b$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41333,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa soda naravna števila $n$, za katera velja\n$$\n-53 < \\frac{2009}{53-n} < 53-n\n$$",
    "options": [],
    "answer": "2, 4, 6, 8, 92, 94, 96",
    "solution": "Solution:\n\nČe je $n < 53$, je $\\frac{2009}{53-n} > 0 > -53$. Neenakost $\\frac{2009}{53-n} < 53-n$ je enakovredna $2009 < (53-n)^2$ oziroma $\\sqrt{2009} < 53-n$. Ker je $\\sqrt{2009} > 44$, sledi $53-n > 44$ oziroma $9 > n$. Možne vrednosti $n$ so $2, 4, 6$ in $8$, pri vseh pa je $53-n \\geq 45 > \\sqrt{2009}$, zato je prvotna neenakost izpolnjena.\n\nNaj bo še $n > 53$. Neenakost je enakovredna\n$$\n-53(53-n) > 2009 > (53-n)^2\n$$\nIz $-53^2 + 53n > 2009$ sledi $n > \\frac{2009 + 53^2}{53} = 90 + \\frac{48}{53}$. Po drugi strani je $45 > \\sqrt{2009} > n-53$, zato je $n < 98$. Dobimo še tri možnosti in sicer $92, 94$ in $96$. Pri vseh treh velja $n-53 \\leq 96-53 = 43 < \\sqrt{2009}$ in s tem tudi prvotna neenakost.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41334,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa realna števila $x, y$ in $z$, ki rešijo sistem enačb\n$$\n\\begin{aligned}\nx + y + 2z &= 0 \\\\\nx y - z^{2} &= 0 \\\\\ny^{2} + 5z + 6 &= 0\n\\end{aligned}\n$$",
    "options": [],
    "answer": "(2, 2, -2) and (3, 3, -3)",
    "solution": "Solution:\n\nIz prve enačbe izrazimo $x = -2z - y$ in vstavimo v drugo. Dobimo $(-2z - y)y - z^{2} = 0$ oziroma $-(z + y)^{2} = 0$. Od tod sledi $z = -y$. Vstavimo v tretjo enačbo. Dobimo $y^{2} - 5y + 6 = 0$ oziroma $(y - 2)(y - 3) = 0$. Torej je $y = 2$ ali $y = 3$.\n\nSistem enačb rešijo $x = 2$, $y = 2$, $z = -2$ in $x = 3$, $y = 3$, $z = -3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41335,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$ we have $|AB| = 2|AC|$. Let $D$ and $E$ be the two points on the segments $AB$ and $BC$, such that $\\angle BAE = \\angle ACD$. The segments $AE$ and $CD$ intersect at $F$, and $CFE$ is an equilateral triangle. Find the angles of the triangle $ABC$.",
    "options": [],
    "answer": "∠A = 60°, ∠B = 30°, ∠C = 90°",
    "solution": "Since $CEF$ is an equilateral triangle we have $\\angle EFC = 60^\\circ$. This implies that $\\angle CFA = 120^\\circ$, so $\\angle FAC = 180^\\circ - \\angle CFA - \\angle ACF = 60^\\circ - \\angle ACF$ and $\\angle BAC = \\angle BAE + \\angle FAC = \\angle BAE + 60^\\circ - \\angle ACD = 60^\\circ$. We have $|AB| = 2|AC|$ and $\\angle BAC = 60^\\circ$, so the triangle $ABC$ is one half of an equilateral triangle and we see that $\\angle CBA = 30^\\circ$ and $\\angle ACB = 90^\\circ$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41336,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle and let $D$ be a point on the side $AB$. The circumcircle of the triangle $BCD$ intersects the side $AC$ at $E$. The circumcircle of the triangle $ADC$ intersects the side $BC$ at $F$. Let $O$ be the circumcentre of the triangle $CEF$. Prove that the points $D$ and $O$ and the circumcentres of the triangles $ADE$, $ADC$, $DBF$ and $DBC$ are concyclic and the line $OD$ is perpendicular to $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O_1, O_2, O_3$ and $O_4$ be the circumcentres of the triangles $ADE$, $ADC$, $BFD$ and $BCD$. The line $O_1O_2$ bisects the segment $AD$ and the two are perpendicular. Similarly, $O_3O_4$ bisects the segment $DB$ and these two are perpendicular as well.\n\n![](attached_image_1.png)\n\nDenote the angles of the triangle by $\\alpha$, $\\beta$ and $\\gamma$ and let $T_1, T_2, T_3, T_4$ and $T_5$ be the midpoints of the segments $AD$, $CF$, $BD$, $CE$ and $CD$.\nWe will be using directed angles as this will shorten the calculation. The quadrilateral $ADFC$ is cyclic, so $\\angle DFB = \\angle DFC = \\angle DAC = \\alpha$. Since $O_3$ is the circumcentre of the triangle $DBF$ and $DBF$ is an acute triangle, we have $\\angle DO_3T_3 = \\angle DFB = \\alpha$. Since $O_2$ is the circumcentre of the triangle $ADC$, we have $\\angle DO_2T_5 = \\angle DAC = \\alpha$ (we used the fact that $\\angle DAC$ is an acute angle).\nThe points $O_2, T_5$ and $O_4$ are collinear, so $\\angle DO_2O_4 = \\angle DO_2T_5 = \\alpha = \\angle DO_3T_3 = \\angle DO_3O_4$ and $\\angle DO_2O_4 = \\angle DO_3O_4$. Hence, the points $O_2, O_3, O_4$ and $D$ are concyclic.\nA similar argument (but for the angle $\\beta$) shows that $O_4, D, O_1$ and $O_2$ are concyclic. Thus, $O_1$ and $O_3$ lie on the circuncircle of the triangle $O_2O_4D$. We know that $\\angle DO_2O_4 = \\alpha$ and $\\angle O_2O_4D = \\beta$. So $\\angle O_4DO_2 = \\gamma$. On the other hand, the quadrilateral $CT_4OT_2$ is cyclic, so $\\angle T_4OT_2 = \\angle T_4CT_2 = \\gamma$. Since $T_2, O, O_2$ are collinear and $T_4, O, O_4$ are collinear, we get $\\angle O_4OO_2 = \\angle T_4OT_2 = \\gamma = \\angle O_4DO_2$ and $O$ lies on the circuncircle of the triangle $O_2DO_4$. Hence, the points $O_1, O_2, O_3, O_4, O$ and $D$ are concyclic.\n\nWe have\n$$\n\\begin{aligned}\n\\angle OO_4O_3 &= \\angle T_4O_4E + \\angle EO_4D + \\angle DO_4T_3 \\\\\n&= \\angle CBE + \\angle EBD + \\angle EBD + \\angle DEB \\\\\n&= \\angle CBD + \\angle EDB \\\\\n&= \\beta + \\gamma\n\\end{aligned}\n$$\nand $\\angle DOO_4 = \\angle DO_3O_4 = \\alpha$. So $OD$ is parallel to $O_3O_4$, which is perpendicular to $AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41337,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$ be a real number such that $x + \\frac{1}{x} + 1$ is a positive integer. Prove that $x^2 + \\frac{1}{x^2} + 1$ is a positive integer divisible by $x + \\frac{1}{x} + 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have\n$$\nx^2 + \\frac{1}{x^2} + 1 = \\left(x + \\frac{1}{x}\\right)^2 - 1 = \\left(x + \\frac{1}{x} + 1\\right)\\left(x + \\frac{1}{x} - 1\\right).\n$$\nSince $x + \\frac{1}{x} + 1$ is a positive integer, $x + \\frac{1}{x} - 1$ is an integer and so is $x^2 + \\frac{1}{x^2} + 1$. Since $x^2 + \\frac{1}{x^2} + 1 \\ge 1$ we conclude that $x^2 + \\frac{1}{x^2} + 1$ is a positive integer divisible by $x + \\frac{1}{x} + 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41338,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer only contains the digits $0$, $3$ and $7$. Prove that this integer is not a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The last digit of a perfect square can be $0$, $1$, $4$, $5$, $6$ or $9$. Denote the given number by $n$. If $n$ is to be a perfect square it has to end in $0$.\n\nAssume that the last $k$ digits of $n$ are equal to $0$ and the digit just to the left of them is not $0$. If $k$ is odd, $k = 2m - 1$. Then $\\frac{n}{10^{2m-2}}$ is also a perfect square. It is divisible by $10$ so it must also be divisible by $25$. This is not possible since the second digit from the right is either $3$ or $7$.\n\nHence, $n$ must end in an even number of zeros and $k$ is even. In this case $\\frac{n}{10^k}$ is a positive integer and a perfect square with the last digit equal to either $3$ or $7$. Again, this is not possible. Hence, $n$ is never a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41339,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $n$ for which there exists a prime $p$ such that $p^2 + 7^n$ is a perfect square.",
    "options": [],
    "answer": "1",
    "solution": "Let $p^2 + 7^n = m^2$. Then $7^n = m^2 - p^2 = (m - p)(m + p)$. We consider two possible cases.\n\nIf $m - p = 1$ and $m + p = 7^n$, then $2p = 7^n - 1$. Assume that $n \\ge 2$. Then $2p = 7^n - 1 = (7 - 1)(7^{n-1} + 7^{n-2} + \\dots + 7 + 1)$, so $p = 3(7^{n-1} + 7^{n-2} + \\dots + 7 + 1)$, but this is not possible since the expression in brackets is greater than 1. If $n = 1$, we get $p = 3$.\n\nIn the second case we have $m - p = 7^k$ and $m + p = 7^{n-k}$, where $k$ is a positive integer. Obviously $n - k > k$. Hence $2p = 7^{n-k} - 7^k = 7^k(7^{n-2k} - 1)$.\n\nThe expression in brackets is even and greater than 2, so $2p$ is the product of at least three primes. This is not possible.\n\nWe conclude that $n = 1$ is the only positive integer for which there exists a prime $p$ such that $p^2 + 7^n$ is a perfect square. This prime is $p = 3$ and the perfect square is $3^2 + 7 = 16 = 4^2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41340,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove the inequality $\\frac{9}{4} < \\log_2 \\pi + \\log_4 \\pi < \\frac{5}{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since\n$$\n\\log_2 \\pi + \\log_4 \\pi = \\frac{1}{\\log_\\pi 2} + \\frac{1}{\\log_\\pi 4} = \\frac{3}{2 \\cdot \\log_\\pi 2} = \\frac{3}{2} \\log_2 \\pi.\n$$\nwe have to show that $\\frac{9}{2} < 3 \\log_2 \\pi < 5$.\n\nThe inequality $\\frac{9}{2} < 3 \\log_2 \\pi$ is equivalent to $3 < \\log_2 \\pi^2$ or $2^3 < \\pi^2$. This last inequality holds since $\\pi^2 > 3^2 = 9 > 8 = 2^3$.\n\nNow, let us prove the second inequality, $3 \\log_2 \\pi < 5$. This one is equivalent to $\\pi^3 < 2^5$. Again, we see that $\\pi^3 < 3.15^2 \\cdot 3.2 < 10 \\cdot 3.2 = 32 = 2^5$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41341,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Find the maximum real number $C$, such that the inequality\n$x^2 + y^2 + 1 \\ge C(x + y)$\nholds for all real $x$ and $y$.\n\nb. Find the maximum real number $C$, such that the inequality\n$x^2 + y^2 + xy + 1 \\ge C(x + y)$\nholds for all real $x$ and $y$.",
    "options": [],
    "answer": "a) sqrt(2); b) sqrt(3)",
    "solution": "a.\nFirst, rewrite the inequality as $x^2 - Cx + y^2 - Cy + 1 \\ge 0$ and then form perfect squares:\n$$\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$$\nThis implies\n$$\n\\left(x - \\frac{C}{2}\\right)^2 + \\left(y - \\frac{C}{2}\\right)^2 + 1 \\ge \\frac{C^2}{2}.\n$$\nIf $x = y = \\frac{C}{2}$, then $1 \\ge \\frac{C^2}{2}$, so $\\sqrt{2} \\ge C$.\nThe maximum possible real $C$ is equal to $\\sqrt{2}$. In this case the inequality holds for all real $x$ and $y$ since it is equivalent to\n$$\n\\left(x - \\frac{\\sqrt{2}}{2}\\right)^2 + \\left(y - \\frac{\\sqrt{2}}{2}\\right)^2 \\ge 0.\n$$\n\nb.\nOnce again we form perfect squares to get\n$$\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$$\nIf $y = \\frac{C}{3}$ and $x = \\frac{C}{2} - \\frac{y}{2} = \\frac{C}{3}$, then we can multiply by 12 and get $12 \\ge 3C^2 + C^2 = 4C^2$, so $\\sqrt{3} \\ge C$. Hence, the maximum possible constant is $C = \\sqrt{3}$.\nIn this case the inequality holds for all real $x$ and $y$ since it is equivalent to\n$$\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": 41342,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of an acute triangle $ABC$ touches the sides $BC$, $CA$ and $AB$ at $A_1$, $B_1$ and $C_1$. Let $K_A$, $K_B$ and $K_C$ be the incircles of the triangles $AB_1C_1$, $A_1BC_1$ and $A_1B_1C_1$.\nLet $t_A$ denote the common tangent of $K_B$ and $K_C$ which intersects the segments $AB$ and $AC$ but not the segment $BC$. Let $t_B$ be the common tangent of $K_A$ and $K_C$ which intersects the segments $AB$ and $BC$ but not the segment $AC$ and let $t_C$ be the common tangent of $K_A$ and $K_B$ which intersects the segments $AC$ and $BC$ but not the segment $AB$.\nProve that the lines $t_A$, $t_B$ and $t_C$ intersect at a single point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "With this kind of problems it is very important to draw a big figure and try to see if there is anything we can say about the common intersection. First, we notice that the centres of $K_A$, $K_B$ and $K_C$ lie on the incircle $K$ of the triangle $ABC$.\n\nLet $A_2$ be the point where the bisector of the angle $B_1A_1C_1$ intersects $K$ for the second time ($A_2$ is the midpoint of the arc $B_1C_1$). Then $\\overline{B_1C_1A_2} = \\angle B_1A_1A_2 = \\angle A_2B_1C_1$, and the tangent-chord theorem implies that $\\angle A_2C_1A = \\angle A_2B_1C_1$. So, $\\angle B_1C_1A_2 = \\angle A_2C_1A$ and $C_1A_2$ is the bisector of the angle $B_1C_1A$. On the other hand, $AA_2$ is the bisector of the angle $\\angle BAC$, so $A_2$ is the incentre of the triangle $AC_1B_1$.\n\nAs we draw the line $A_1A_2$ onto the figure we notice that the common intersection of the tangents lies on this line. Let $B_2$ and $C_2$ be the centres of $K_B$ and $K_C$. If we also draw $B_1B_2$ and $C_1C_2$, we see that they contain the common intersection of the tangents as well. The lines $A_1A_2$, $B_1B_2$ and $C_1C_2$ are the bisectors of the inner angles of the triangle $A_1B_1C_1$ and they meet in a point we denote by $J$. We wish to prove that each of the tangents $t_A$, $t_B$ and $t_C$ also contains $J$.\n\n![](attached_image_1.png)\n\nSince $J$ is the intersection of the angle bisectors of the triangle $A_1B_1C_1$, we have\n$$\n\\angle C_2B_2J = \\angle C_2B_2B_1 = \\angle C_2C_1B_1 = \\angle A_1C_1C_2 = \\angle A_1B_2C_2\n$$\nand\n$$\n\\angle JC_2B_2 = \\angle C_1C_2B_2 = \\angle C_1B_1B_2 = \\angle B_2B_1A_1 = B_2C_2A_1\n$$\n\n![](attached_image_2.png)\n\nHence, the triangles $A_1C_2B_2$ and $JC_2B_2$ are similar. They also have a common side, so they are congruent. Reflect the outer tangent $BC$ to the circles $K_B$ and $K_C$ in the line $B_2C_2$ connecting the two centres. We have just shown that this reflection maps $A_1$ to $J$. It also maps the tangent $BC$ into a tangent passing through the image of $A_1$. This implies that $J$ lies on $t_A$. Similarly, we show that $J$ lies on $t_B$ and $t_C$. We conclude that the three tangents intersect at a single point.\nLet us find another way of describing the tangent $t_A$. We have noticed that it contains $J$. The figure also suggests that $t_A$ is parallel to $B_1C_1$. Let $l$ be the line through $J$ parallel to $B_1C_1$. We wish to show that $l$ is tangent to $K_B$ and $K_C$.\n\n![](attached_image_3.png)\n\nLet us draw a less cluttered figure. We will not need the circles $K_A$ and $K_C$. Let $D$ be the point on $l$ such that $B_2D$ is perpendicular to $l$. We wish to show that $|B_2D|$ is the diameter of $K_B$.\n\nFirst, let us find the angle $\\angle B_2JD$. Since $JD$ is parallel to $B_1C_1$, we have $\\angle B_2JD = \\angle B_2B_1C_1 = \\angle B_2A_1C_1$. Let $E$ be the midpoint of $A_1C_1$. We wish to show that $|B_2E| = |B_2D|$. We have $\\angle B_2JD = \\angle B_2A_1E$, so the triangles $JDB_2$ and $A_1EB_2$ are similar. We wish to show that they are congruent, so $|A_1E| = |JD|$ or $|A_1B_2| = |JB_2|$. The second equality is equivalent to the fact that $B_2A_1J$ is an isosceles triangle with the apex at $B_2$. Let $\\angle A_1C_1B_1 = \\gamma$. Then\n$$\n\\angle B_2JA_1 = \\pi - \\angle A_1JB_1 \\Rightarrow \\pi - \\left(\\frac{\\pi + \\gamma}{2}\\right) = \\frac{\\pi - \\gamma}{2}.\n$$\nClearly, $\\angle JB_2A_1 = \\angle B_1B_2A_1 = \\angle B_1C_1A_1 = \\gamma$, so the third angle in the triangle is equal to $\\angle A_1B_2 = \\frac{\\pi - \\gamma}{2}$. The triangle $B_2A_1J$ is isosceles with the apex at $B_2$, so $|A_1B_2| = |JB_2|$ and the triangles $JDB_2$ and $A_1EB_2$ are congruent. So, $|B_2D| = |B_2E|$ and $JD$ (or $l$) is tangent to $K_B$. Similarly, we show that $l$ is tangent to $K_C$, so $l = t_A$. The same arguments would show that $t_B$ is the line through $J$ parallel to $A_1C_1$ and $t_C$ is the line through $J$ parallel to $A_1B_1$. Hence, all three tangents intersect at a single point, namely $J$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41343,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$ such that $p^2 + 7^3$ is a perfect cube.",
    "options": [],
    "answer": "13",
    "solution": "Let $p^2 + 7^3 = n^3$. Then $p^2 = n^3 - 7^3 = (n-7)(n^2 + 7n + 49)$. Obviously, $n > 7$ and $n - 7 < n^2 + 7n + 49$. So there is only one possible case: $n - 7 = 1$ and $n^2 + 7n + 49 = p^2$. We conclude that $n = 8$ and $p^2 = 169$, so $p = 13$. The only prime number with the required property is $p = 13$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41344,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which $\\lfloor \\frac{n^2}{4} \\rfloor$ is a perfect square.",
    "options": [],
    "answer": "n = 1 or n is even",
    "solution": "Let $n$ be even, $n = 2k$. Then $\\frac{n^2}{4} = k^2$ is even, so $\\lfloor \\frac{n^2}{4} \\rfloor = k^2$ is a perfect square.\n\nIf $n$ is odd it can be written as $n = 2k + 1$, $k \\ge 0$. In this case\n$$\n\\lfloor \\frac{n^2}{4} \\rfloor = \\lfloor \\frac{4k^2 + 4k + 1}{4} \\rfloor = \\lfloor k^2 + k + \\frac{1}{4} \\rfloor = k^2 + k = k(k + 1).\n$$\nFor $k(k + 1)$ to be a perfect square both $k$ and $k + 1$ have to be perfect squares since they are coprime. Hence, $k = a^2$ and $k + 1 = b^2$. This implies $(b-a)(b+a) = 1$. There are two possibilities, $b-a = 1 = b+a$ or $b-a = -1 = b+a$. Both imply $k=0$, so $n=1$ is the only odd number with the required property.\n\nThere is another way to deal with the case where $k^2 + k$ is a perfect square. The quadratic equation $k^2 + k - x^2 = 0$ has two solutions, $k_{1,2} = \\frac{-1 \\pm \\sqrt{1+4x^2}}{2}$, so $1 + 4x^2 = y^2$ or, equivalently, $(y - 2x)(y + 2x) = 1$. This implies $x = 0$, so $k = 0$.\n\nWe conclude that $\\lfloor \\frac{n^2}{4} \\rfloor$ is a perfect square if and only if $n = 1$ or $n$ is even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41345,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a positive integer greater than $1$. Prove: for every non-negative integer $m$ there exist $k$ positive integers $n_1, n_2, \\dots, n_k$, such that\n$$\nn_1^2 + n_2^2 + \\dots + n_k^2 = 5^{m+k}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $k = 2$. If $n_1 = 5^l$ and $n_2 = 2 \\cdot 5^l$, then $n_1^2 + n_2^2 = 5^{2l+1}$. If $n_1 = 3 \\cdot 5^l$ and $n_2 = 4 \\cdot 5^l$, then $n_1^2 + n_2^2 = 5^{2l+2}$. In both cases, $l$ is an arbitrary non-negative integer. Hence, the equation $n_1^2 + n_2^2 = 5^l$ has solutions for all positive integers $t$.\n\nNext, we consider the case $k = 3$. Let $n_1 = 3a_1$, $n_2 = 4a_1$ and $n_3 = 5^2 a_2$. Then $n_1^2 + n_2^2 + n_3^2 = 5^2(a_1^2 + a_2^2)$. For every non-negative integer $m$ there exists a solution to the equation $a_1^2 + a_2^2 = 5^{1+m}$. so we have shown that the equation $n_1^2 + n_2^2 + n_3^2 = 5^{3+m}$ has solutions for all non-negative integers $m$.\n\nFor $k \\ge 3$ we proceed by induction. Assume that the equation $a_1^2 + a_2^2 + \\dots + a_l^2 = 5^{l+m}$ has positive integer solutions for all non-negative integers $m$ and for all positive integers $l$ such that $2 \\le l \\le k$. We wish to show that in this case the equation $n_1^2 + n_2^2 + \\dots + n_k^2 + n_{k+1}^2 = 5^{k+1+m}$ also has positive integer solutions for all non-negative integers $m$.\n\nSince $5^{k+1+m} = 5^{k+m} + 4 \\cdot 5^{k+m}$ and $k > k-1 \\ge 2$, the induction hypothesis implies that for all non-negative integers $i$ there exist positive integers $a_1, a_2, \\dots, a_{k-1}$, such that\n$$\na_1^2 + a_2^2 + \\dots + a_{k-1}^2 = 5^{k-1+i}\n$$\nIn particular, this is true for $i = m + 1$. so there exist positive integers $a_1, a_2, \\dots, a_{k-1}$, such that\n$$\na_1^2 + a_2^2 + \\dots + a_{k-1}^2 = 5^{k+m}\n$$\nSince $3^2 + 4^2 = 5^2$ and $5^2 + 10^2 = 5^3$, such numbers $n_1$ and $n_2$ exist for $k=2$, $m=0$ and for $k=2$, $m=1$. Using induction on $m$ we show that the solutions exist for any $m$. Let $m \\ge 3$. Assume that there exist positive integers $a_1, a_2$, such that $a_1^2 + a_2^2 = 5^m$. Then\n$$\n5^{m+2} = 5^2 \\cdot 5^m = 5^2(a_1^2 + a_2^2) = (5a_1)^2 + (5a_2)^2.\n$$\nso $5^{m+2}$ is the sum of the squares of two positive integers. We have considered the cases $m=0$ and $m=1$ separately, so by induction the claim for $k=2$ follows for all non-negative integers $m$.\n\nSince $5^3 = 3^2 + 4^2 + 10^2$ and $5^4 = 9^2 + 12^2 + 20^2$, we can use a similar argument for $k=3$ and show that the equation $n_1^2 + n_2^2 + n_3^2 = 5^{3+m}$ also has positive integer solutions for all non-negative integers $m$.\n\nLet $k \\ge 3$. We will use induction on $k$. Assume that for all non-negative integers $m$ there exist positive integers $a_1, a_2, \\dots, a_k$, such that\n$$\na_1^2 + a_2^2 + \\dots + a_k^2 = 5^{k+m}\n$$\nand, since $k-1 \\ge 2$, positive integers $b_1, b_2, \\dots, b_{k-1}$, such that\n$$\nb_1^2 + b_2^2 + \\dots + b_{k-1}^2 = 5^{k-1+m}\n$$\nIf $k+m$ is even, then $5^{k+m+1} = 5^{k+m} + 4 \\cdot 5^{k+m} = a_1^2 + a_2^2 + \\dots + a_k^2 + 4 \\cdot 5^{k+m} = a_1^2 + a_2^2 + \\dots + a_k^2 + (2 \\cdot 5^{\\frac{k+m}{2}})^2$. So $5^{k+m+1}$ is a sum of $k+1$ squares of positive integers.\n\nLet $k+m$ be odd. Since $k+m-3 \\ge 0$, we can find two positive integers $c_1, c_2$, such that $c_1^2 + c_2^2 = 5^{2+(k+m-3)}$. Then\n$$\n\\begin{aligned}\n5^{k+m+1} &= 5^2 \\cdot 5^{k+m-1} = (9+16)5^{k+m-1} \\\\\n&= 9 \\cdot 5^{k-1+m} + 16 \\cdot 5^{2+k+m-3} \\\\\n&= 9(b_1^2 + b_2^2 + \\dots + b_{k-1}^2) + 16(c_1^2 + c_2^2) \\\\\n&= (3b_1)^2 + (3b_2)^2 + \\dots + (3b_{k-1})^2 + (4c_1)^2 + (4c_2)^2.\n\\end{aligned}\n$$\n\nSince $m$ was an arbitrary non-negative integer, this concludes the induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41346,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $n$ such that $n^2$ is a perfect cube, and $n^3$ is a perfect square.",
    "options": [],
    "answer": "1",
    "solution": "Since $n$ is divisible by $20$, we can write $n = 2^{2+a} \\cdot 5^{1+b} \\cdot k$, where $k$ is a positive integer divisible neither by $2$ nor $5$, while $a$ and $b$ are non-negative integers. If $n^2 = 2^{2(2+a)} \\cdot 5^{2(1+b)} \\cdot k^2$ is a perfect cube, then $3$ divides $2(2+a)$ and $3$ divides $2(1+b)$. If $n^3 = 2^{3(2+a)} \\cdot 5^{3(1+b)} \\cdot k^3$ is a perfect square, then $2$ divides $3(2+a)$ and $2$ divides $3(1+b)$. From this it follows that $6$ divides $2+a$ and $6$ divides $1+b$, so $a$ is at least $4$ and $b$ is at least $5$. The smallest possible positive integer $k$ is $k=1$. When this is the case, we get $n = 2^6 \\cdot 5^6 = 1000000$. We see that $n^2$ is indeed a perfect cube and $n^3$ is a perfect square, and we have shown this $n$ to be the smallest possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41347,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa praštevila $p$, za katera je $p^{2}+7^{3}$ popoln kub.",
    "options": [],
    "answer": "13",
    "solution": "Solution:\n\nNaj bo $p^{2}+7^{3}=n^{3}$. Tedaj velja $p^{2}=n^{3}-7^{3}=(n-7)\\left(n^{2}+7 n+49\\right)$. Očitno je $n>7$ in $n-7<n^{2}+7 n+49$, zato je možno le, da je $n-7=1$ in $n^{2}+7 n+49=p^{2}$. Tedaj je $n=8$ in $p^{2}=169$, zato je $p=13$. Edino tako praštevilo je $p=13$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41348,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDokaži neenakost\n$$\n\\frac{9}{4}<\\log_{2} \\pi+\\log_{4} \\pi<\\frac{5}{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nKer velja\n$$\n\\log_{2} \\pi+\\log_{4} \\pi=\\frac{1}{\\log_{\\pi} 2}+\\frac{1}{\\log_{\\pi} 4}=\\frac{3}{2 \\cdot \\log_{\\pi} 2}=\\frac{3}{2} \\log_{2} \\pi\n$$\nje potrebno videti, da je $\\frac{9}{2}<3 \\log_{2} \\pi<5$.\n\nNeenakost $\\frac{9}{2}<3 \\log_{2} \\pi$ je enakovredna $3<\\log_{2} \\pi^{2}$ oziroma $2^{3}<\\pi^{2}$. Slednje drži, saj je $\\pi^{2}>3^{2}=9>8=2^{3}$.\n\nDokažimo še drugo neenakost, to je $3 \\log_{2} \\pi<5$. Ta je enakovredna $\\pi^{3}<2^{5}$. Spet lahko ocenimo $\\pi^{3}<3.15^{2} \\cdot 3.2<10 \\cdot 3.2=32=2^{5}$, kar je bilo treba dokazati.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41349,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči najmanjše naravno število $n$, deljivo z $20$, za katerega je $n^{2}$ popoln kub, $n^{3}$ pa popoln kvadrat.",
    "options": [],
    "answer": "1000000",
    "solution": "Solution:\n\nKer je število $n$ deljivo z $20$, je oblike $n=2^{2+a} \\cdot 5^{1+b} \\cdot k$, kjer je $k$ naravno število, ki ni deljivo niti z $2$ niti s $5$, $a$ in $b$ pa sta nenegativni celi števili.\n\nČe je $n^{2}=2^{2(2+a)} \\cdot 5^{2(1+b)} \\cdot k^{2}$ popoln kub, $3 \\mid 2(2+a)$ in $3 \\mid 2(1+b)$.\n\nČe je $n^{3}=2^{3(2+a)} \\cdot 5^{3(1+b)} \\cdot k^{3}$ popoln kvadrat, $2 \\mid 3(2+a)$ in $2 \\mid 3(1+b)$.\n\nOd tod sledi, da $6 \\mid 2+a$ in $6 \\mid 1+b$, zato je $a$ najmanj $4$ in $b$ najmanj $5$.\n\nNajmanjše možno naravno število $k$ je $k=1$.\n\nPri teh vrednostih dobimo $n=2^{6} \\cdot 5^{6}=1000000$.\n\nKer je $n^{2}$ popoln kub in $n^{3}$ popoln kvadrat, je to najmanjše tako naravno število.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41350,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa naravna števila $n$, za katera obstaja praštevilo $p$, da je število $p^{2}+7^{n}$ popoln kvadrat.",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n\nNaj bo $p^{2}+7^{n}=m^{2}$. Potem velja $7^{n}=m^{2}-p^{2}=(m-p)(m+p)$. Ločimo dve možnosti.\n\nČe je $m-p=1$ in $m+p=7^{n}$, sledi $2p=7^{n}-1$. Denimo, da je $n \\geq 2$. Tedaj je $2p=7^{n}-1=(7-1)\\left(7^{n-1}+7^{n-2}+\\ldots+7+1\\right)$, zato dobimo $p=3\\left(7^{n-1}+7^{n-2}+\\ldots+7+1\\right)$, kar pa ni možno, saj je izraz v oklepajih večji od 1. Če je $n=1$, dobimo $p=3$.\n\nDruga možnost je, da je $m-p=7^{k}$ in $m+p=7^{n-k}$, kjer je $k$ neko naravno število. Očitno velja $n-k>k$. Tedaj je $2p=7^{n-k}-7^{k}=7^{k}\\left(7^{n-2k}-1\\right)$. Izraz v oklepajih je sodo število, večje od 2, zato smo $2p$ zapisali kot produkt vsaj treh praštevil, kar ni možno.\n\nTorej je $n=1$ edino naravno število, pri katerem obstaja praštevilo $p$, da je $p^{2}+7^{n}$ popolni kvadrat. Tako praštevilo je kar $p=3$, popolni kvadrat pa $3^{2}+7=16=4^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41351,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the bisectors of the angles $\\angle CAD$ and $\\angle ADB$ intersect the circumcircle of a cyclic quadrilateral $ABCD$ at $P$ and $Q$. The lines $AP$ and $DQ$ intersect at $R$, and the lines $CQ$ and $BP$ intersect at $S$. Prove that the lines $PQ$ and $RS$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the quadrilateral $AQPD$ is cyclic, we have $\\angle RPQ = \\angle APQ = \\angle ADQ$. Since $DQ$ bisects the angle $ADB$, we get $\\angle ADQ = \\angle QDB$. Since the quadrilateral $QBPD$ is cyclic, we have $\\angle QDB = \\angle QPB = \\angle QPS$. Thus, $\\angle RPQ = \\angle QPS$.\n\nSimilarly, since $QBPD$ is cyclic, we get $\\angle PQR = \\angle PQD = \\angle PBD$. The line $BP$ intersects the angle $\\angle CBD$, so $\\angle PBD = \\angle CBP$. Since $QBCP$ is a cyclic quadrilateral, we have $\\angle CBP = \\angle CQP = \\angle SQP$. Thus $\\angle PQR = \\angle SQP$.\nThe triangles $SPQ$ and $RPQ$ have congruent angles and a common side, so they are congruent. Hence, the quadrilateral $PRQS$ is a deltoid, which implies that $PQ \\perp RS$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41352,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle such that $|AB| > |BC| > |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$. Show that the triangles $HBD$ and $HGE$ are similar.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The triangle $CAD$ is isosceles since $|AC| = |AD|$. The line $AA_1$ is the altitude in this isosceles triangle, so $\\angle HDA = \\angle ACH$.\n\n![](attached_image_1.png)\n\nIn the quadrilateral $HA_1CB_1$ we have $\\angle CA_1H = \\frac{\\pi}{2} = \\angle CB_1H$, so this quadrilateral is cyclic and $\\angle B_1A_1H = \\angle B_1CH$. We have shown that $\\angle FA_1A = \\angle B_1A_1H = \\angle B_1CH = \\angle ACH = \\angle HDA = \\angle FDA$, so $A$, $D$, $A_1$ and $F$ are concyclic. This implies that $\\angle AFD = \\angle AA_1D = \\frac{\\pi}{2}$.\n\nThe segments $AB_1$ and $HF$ are the altitudes in the triangle $AHG$ and they meet at $E$, so $E$ is the orthocentre of this triangle and $EG$ is perpendicular to $AH$. Now, $AH$ is perpendicular to $BC$, so $EG$ and $BC$ are parallel. Thus, $\\angle EGH = \\angle HBD$ and since $\\angle GHE = \\angle BHD$ we conclude that the triangles $HBD$ and $HGE$ are similar.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41353,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $E$ and $F$ be the points on the sides $AB$ and $AD$ of a convex quadrilateral $ABCD$, such that $EF$ is parallel to $BD$. The segment $CE$ intersects the diagonal $BD$ at $G$, while the segment $CF$ intersects the diagonal $BD$ at $H$. Prove: if $AGCH$ is a parallelogram, then $ABCD$ is a parallelogram as well.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the intersection of the lines $EF$ and $AG$ by $I$ and the intersection of the lines $EF$ and $AH$ by $J$.\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![](attached_image_1.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": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41354,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$ let $D$ be the foot of the altitude to the side $AB$. Given the points $E$ and $F$ on the sides $AD$ and $BC$, such that $\\angle BAF = \\angle ACE$, let the segments $AF$ and $CE$ intersect at $G$ and let the segments $AF$ and $CD$ intersect at $T$. Find the angles of the triangle $ABC$, given that $CGF$ is an equilateral triangle and the triangle $AET$ is isosceles with the apex at $E$.",
    "options": [],
    "answer": "∠A = 60°, ∠B = 45°, ∠C = 75°",
    "solution": "Denote $\\angle BAF = \\angle ACE = \\varphi$. Since $CGF$ is an equilateral triangle, we have $\\angle CGA = 120^\\circ$. Thus, $\\angle GAC = 180^\\circ - \\angle CGA - \\angle ACG = 60^\\circ - \\varphi$ and $\\angle BAC = \\angle BAF + \\angle GAC = \\varphi + 60^\\circ - \\varphi = 60^\\circ$.\n\nSince the triangle $AET$ is isosceles, we have $\\angle ETA = \\varphi = \\angle ECA$. So the points $A$, $E$, $T$ and $C$ are concyclic. Thus, $\\angle ECT = \\angle EAT = \\varphi$\nand $2\\varphi = \\angle ACE + \\angle ECD = \\angle ACD = \\frac{\\pi}{2} - \\angle DAC = 30^\\circ$, which implies $\\varphi = 15^\\circ$.\n\nFinally, we can calculate $\\angle CBA = \\angle FBA = 60^\\circ - \\varphi = 45^\\circ$ and $\\angle ACB = 75^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41355,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$, such that $n(n+2)(n+4)$ has at most 15 positive divisors.",
    "options": [],
    "answer": "1, 2, 3, 4, 5, 7, 9",
    "solution": "Write $a_n = n(n+2)(n+4)$ and let $b_n$ be the number of positive divisors of $a_n$. We can easily check that $b_1 = 4$, $b_2 = 10$, $b_3 = 8$, $b_4 = 14$, $b_5 = 12$, $b_6 = 24$, $b_7 = 12$, $b_8 = 28$, $b_9 = 12$ and $b_{10} = 40$.\n\nIn the next part we will use the following fact: if $p_1^{\\alpha_1} \\cdot p_2^{\\alpha_2} \\cdot \\dots \\cdot p_k^{\\alpha_k}$ is the prime factorization of a positive integer $m$, then the number of positive divisors of $m$ is equal to\n$$\n(\\alpha_1 + 1)(\\alpha_2 + 1) \\dots (\\alpha_k + 1).\n$$\nIf $m$ divides a positive integer $l$, then $l$ has at least as many divisors as $m$.\n\nLet $n \\ge 11$. If $n$ is even, $n = 2k$, then $a_n = 2^3k(k+1)(k+2)$. At least one of the numbers $k, k+1$ and $k+2$ is divisible by 2 and exactly one of them is divisible by 3. Since $k \\ge 6$ the numbers $k, k+1$ and $k+2$ cannot all be powers of 2 or 3. So, $k(k+1)(k+2)$ has a prime divisor $p$ not equal to 2 or 3. Hence, $2^4 \\cdot 3 \\cdot p$ divides $a_n$ and this implies that $a_n$ has at least $5 \\cdot 2 \\cdot 2 = 20$ positive divisors.\n\nLet $n \\ge 11$ be odd. Then the numbers $n$ and $n+2$ are relatively prime, as are $n+2$ and $n+4$ and also $n$ and $n+4$. One of these three numbers is divisible by 3. This number has at least one other prime divisor $p$ or else is a power of 3. In the latter case it is divisible by $3^3$ since $n \\ge 11$. Let $q$ and $r$ be prime divisors of the other two numbers. In the first case the number $a_n$ is divisible by $3pqr$. The numbers $n, n+2$ and $n+4$ are relatively prime, so 3, $p$, $q$ and $r$ are also relatively prime. This implies that $a_n$ has at least $2 \\cdot 2 \\cdot 2 \\cdot 2 = 16$ divisors. In the second case $a_n$ is divisible by $3^3qr$. The primes 3, $q$ and $r$ are again distinct. So, $a_n$ has at least $4 \\cdot 2 \\cdot 2 = 16$ divisors.\n\nThe number $a_n$ has at most 15 positive divisors only for $n = 1, n = 2, n = 3, n = 4, n = 5, n = 7$ and $n = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41356,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 4$ be a positive integer. Into every square of an $n \\times n$ array we inscribe an integer so that the sum of the integers inside any $3 \\times 3$ square is negative. Find all positive integers $n$ for which this can be done in such a way that the sum of all the numbers in the array is positive.",
    "options": [],
    "answer": "All positive integers n with n ≡ 1 or 2 (mod 3).",
    "solution": "When $n$ is divisible by $3$, $n = 3k$. The array can be divided into $k^2$ $3 \\times 3$ squares. The sum of the numbers inside each square is negative, so the sum of all these sums, which is the sum of all the numbers in the array, is negative as well.\n\nIf $n$ is not divisible by $3$, the numbers can be inscribed as required. To make things easier let us assume that all positive numbers in the array are equal and all negative numbers in the array are equal.\n\nThe cases $n = 4$ and $n = 5$ are represented in the top two figures. Each $3 \\times 3$ square contains eight ones and one $-9$, so the sum is $-1$, while the sum of all the numbers in the array is $8$ and $13$, respectively.\n\n<table>\n<tr><td>1</td><td>1</td><td>1</td><td>1</td></tr>\n<tr><td>1</td><td>1</td><td>-9</td><td>1</td></tr>\n<tr><td>1</td><td>1</td><td>1</td><td>1</td></tr>\n<tr><td>1</td><td>1</td><td>1</td><td>1</td></tr>\n</table>\n\n<table>\n<tr><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td></tr>\n<tr><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td></tr>\n<tr><td>1</td><td>1</td><td>-9</td><td>1</td><td>1</td></tr>\n<tr><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td></tr>\n<tr><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td></tr>\n</table>\n\nNow, let us choose an arbitrary $k$ and consider the $(3k-1) \\times (3k+1)$ array and the $(3k+2) \\times (3k+2)$ array. Cover the $3k \\times 3k$ array in the upper left corner with $k^2$ $3 \\times 3$ squares as shown. For each of these $3 \\times 3$ squares put $-b$ into the lower right corner and $a$ into the other eight squares as well as into all the remaining squares in the array.\n\n<table>\n<tr><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td></tr>\n<tr><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td></tr>\n<tr><td>a</td><td>a</td><td>-b</td><td>a</td><td>a</td><td>-b</td><td>a</td><td>a</td><td>-b</td><td>a</td></tr>\n<tr><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td></tr>\n<tr><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td></tr>\n<tr><td>a</td><td>a</td><td>-b</td><td>a</td><td>a</td><td>-b</td><td>a</td><td>a</td><td>-b</td><td>a</td></tr>\n<tr><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td></tr>\n<tr><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td></tr>\n<tr><td>a</td><td>a</td><td>-b</td><td>a</td><td>a</td><td>-b</td><td>a</td><td>a</td><td>-b</td><td>a</td></tr>\n<tr><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td><td>a</td></tr>\n</table>\n\nThe sum inside each $3 \\times 3$ square is $8a - b$ and the sum of all numbers in the array is $-k^2 \\cdot b + ((3k + 1)^2 - k^2)a$ or $-k^2 \\cdot b + ((3k + 2)^2 - k^2)a$. Since $8a - b < 0$, we can write $b = 8a + c$, where $c$ is a positive integer. Then $-k^2 \\cdot b + ((3k+2)^2-k^2)a > -k^2 \\cdot b + ((3k+1)^2-k^2)a = ((3k+1)^2-9k^2)a-k^2c > 0$ implies\n$$\n\\frac{6k+1}{k^2} > \\frac{c}{a}.\n$$\nIf we set $a = k$ and $c = 1$, we get $b = 8k + 1$.\n\nHence, for $(3k + 1) \\times (3k + 1)$ or $(3k + 2) \\times (3k + 2)$ arrays we can arrange the numbers $k$ and $-8k - 1$ as described above. Then the sum of the numbers inside each $3 \\times 3$ square is equal to $-1$, while the sum of all the numbers in the array is $5k^2 + k$ or $11k^2 + 4k$, and thus positive.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41357,
    "subject": "Mathematics (Multi-modal)",
    "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": "Detailed solution",
    "solution": "The triangle $CAD$ is isosceles since $|AC| = |AD|$. The line $AA_1$ is the altitude in this isosceles triangle, so $\\angle HDA = \\angle ACH$. In the quadrilateral $HA_1CB_1$ we have $\\angle CA_1H = \\frac{\\pi}{2} = \\angle CB_1H$, so this quadrilateral is cyclic and $\\angle B_1A_1H = \\angle B_1CH$.\n\n![](attached_image_1.png)\n\nWe have shown that $\\angle FA_1A = \\angle B_1A_1H = \\angle B_1CH = \\angle ACH = \\angle HDA = \\angle FDA$, so $A, D, A_1$ and $F$ are concyclic. This implies that $\\angle AFD = \\angle AA_1D = \\frac{\\pi}{2}$. The segments $AB_1$ and $HF$ are the altitudes in the triangle $AHG$ and they meet at $E$. So $E$ is the orthocentre of this triangle and $EG$ is perpendicular to $AH$. Now, $AH$ is perpendicular to $BC$, so $EG$ and $BC$ are parallel.\n\nLet $T$ denote the intersection of the lines $EG$ and $AD$ and let $C_1$ be the foot of the altitude to the side $AB$ in the triangle $ABC$. Since $ET$ is parallel to $CD$, the quadrilateral $TDCE$ is in fact a trapezoid. Since $\\angle TDC = \\angle ECD$ this trapezoid is isosceles. We see that $\\angle TCE = \\angle TDE = \\angle ADH = \\angle ABH = \\angle C_1BH = \\angle HCE$, so $C, H$ and $T$ are collinear which was all that remained to be shown.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41358,
    "subject": "Mathematics (Multi-modal)",
    "question": "There were 64 contestants at a chess tournament. Every pair played a game that ended either with one of them winning or in a draw. If a game ended in a draw, then each of the remaining 62 contestants won against at least one of these two contestants. There were at least two games ending in a draw at the tournament. Show that we can line up all the contestants so that each of them has won against the one standing right behind him.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, we show that for each contestant at most one game ended in a draw. Assume, to the contrary, that the contestant $A$ tied against $B$ and $C$. The game between $A$ and $B$ was tied, so $C$ won against at least one of them. The game between $A$ and $C$ was tied, so $C$ won against $B$. On the other hand, the game between $A$ and $C$ ended in a draw, but $B$ did not beat either of them. This contradicts the assumptions of the problem.\n\nThe sequence of contestants $A_1, A_2, \\ldots, A_k$ is called increasing, if for all $i = 1, \\ldots, k-1$ the contestant $A_i$ lost against $A_{i+1}$. Let $n$ be the length of the longest increasing sequence of contestants. If $n = 64$, then we are done. Otherwise, assume that $n < 64$.\n\nLet $B$ be the contestant who is not included in the longest increasing sequence $A_1, A_2, \\ldots, A_n$. Then $A_1$ did not win against $B$, otherwise $B, A_1, A_2, \\ldots, A_n$ would be an increasing sequence of length $n+1$. Assume that $B$ won against $A_1$. Then $B$ and $A_2$ did not tie since $A_1$ lost against both of them. If $A_2$ had won against $B$, then $A_1, B, A_2, \\ldots, A_n$ would be an increasing sequence of length $n+1$. So, $B$ won against $A_2$. Similarly, we conclude that $B$ won against $A_3, A_4, \\ldots, A_n$. So, $A_1, A_2, \\ldots, A_n, B$ is an increasing sequence of length $n+1$, a contradiction.\n\nHence, $B$ and $A_1$ tied. As $A_1$ tied against at most one opponent, we have $n-1=63$ and this opponent is $B$. If $B$ won against $A_2$, we can repeat the argument above to show that $B$ also won against $A_3, A_4, \\ldots, A_{63}$. So, $A_1, A_2, \\ldots, A_{63}, B$ is an increasing sequence of length 64. We have assumed no such sequence exists, so $A_2$ must have beaten $B$ (they did not tie as $B$ tied against $A_1$). Similarly, we conclude that each of the remaining contestants $A_i$ for $3 \\le i \\le 63$ has won against $B$. On the other hand, there were two other players, $A_i$ and $A_j$ for $i \\ne j$, $i, j > 2$, whose game ended in a draw. This means they both won against $B$. The latter contradicts the assumptions of the problem, which proves that the length of the longest increasing sequence of players is 64.\nAs in the first solution we show that each participant tied at most once. Let $A = B$ denote a tie. If $A$ won, we write $A \\to B$. If $B$ won, we write $A \\leftarrow B$. If $A = B$, then for all other contestants $C$ we have either $C \\to A$ or $C \\to B$.\n\n![](attached_image_1.png)\n\nIf $A = B$, $C = D$ and $A \\to C$, then $C \\to B$. Since $C = D$, we have $B \\to D$, and since $A = B$, we have $D \\to A$.\n\n![](attached_image_2.png)\n\nAssume that $k$ games ended in a draw. We use induction to show that the $2k$ players who tied once can be labeled as $A_1, A_2, \\dots, A_{2k}$ in such a way that\n\n$$\nA_1 \\to A_2 \\to A_3 \\to \\dots \\to A_{2k} \\to A_1 \\text{ and } A_i = A_{i+k} \\text{ for } i = 1, \\dots, k. \\quad (1)\n$$\n\n![](attached_image_3.png)\n\nIf $k=2$, denote the 4 players by $A=B$, $C=D$. Without loss of generality we may assume that $A \\to C$. As we have shown before, we can assign the labels as follows: $A_1 = A$, $A_2 = C$, $A_3 = B$ and $A_4 = D$.\n\nAssume that (1) holds and let us show this in the case of $k-1$ tied games. Let us choose a pair of contestants who tied against one another and denote them by $C$ and $D$. The remaining $2k$ contestants can be labeled so that (1) holds. Furthermore, we may also assume that $A_1 \\to C$ (otherwise exchange $C$ and $D$). Since $A_1 = A_{k+1}$, we have $C \\to A_{k+1}$. Let $i$ be the smallest index, such that $C \\to A_{i+1}$. Then $A_i \\to C \\to A_{i+1}$. On the other hand, $A_i = A_{i-k-1}$, $C = D$ and $A_{i+1} = A_{i+k+1}$. As shown above, this implies $A_{i+k} \\to D \\to A_{i+k+1}$.\n\n![](attached_image_4.png)\n\nLet $B_j = A_j$ for $1 \\le j \\le i$, $B_{i+1} = C$, $B_j = A_{j-1}$ for $j = i+2, \\dots, i+k+1$, $B_{i+k+2} = D$ and $B_j = A_{j-2}$ for $j \\ge i+k+3$. Then\n$$\nB_1 \\to B_2 \\to B_3 \\to \\dots \\to B_{2k+2} \\to B_1 \\text{ and } B_j = B_{j+k+1} \\text{ for } j = 1, \\dots, k+1,\n$$\nwhich is what we wanted to see.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41359,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the inequality $x^2 + y^2 + 1 \\ge 2(xy - x + y)$ holds for any two real $x$ and $y$. When does the equality hold?",
    "options": [],
    "answer": "Equality holds exactly when y = x + 1.",
    "solution": "Rewrite the inequality $x^2 + y^2 + 1 \\ge 2(xy - x + y)$ as $x^2 - 2xy + y^2 + 2x - 2y + 1 \\ge 0$. If we further rearrange the left-hand side into $(x - y)^2 + 2(x - y) + 1 \\ge 0$, we notice that it is a perfect square, and the inequality becomes $((x - y) + 1)^2 \\ge 0$. Hence, the inequality holds for all real $x$ and $y$. The equality case occurs if and only if $(x - y) + 1 = 0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41360,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $g: \\mathbb{R} \\to \\mathbb{R}$, for which there exists a strictly increasing function $f: \\mathbb{R} \\to \\mathbb{R}$, such that\n$$\nf(x + y) = f(x)g(y) + f(y)\n$$\nfor all real $x$ and $y$.",
    "options": [],
    "answer": "g(x) = a^x for all real x, where a > 0",
    "solution": "Inserting $y = 0$ into the equation, we get $f(x) = f(x)g(0) + f(0)$ or\n$$\nf(x)(1 - g(0)) = f(0).\n$$\nIf $g(0) \\neq 1$ then $f(x) = \\frac{f(0)}{1-g(0)}$ and $f$ is constant. This is not possible since $f$ is strictly increasing. We conclude that $g(0) = 1$ and $f(0) = 0$. From\n$$\nf(x)g(y) + f(y) = f(x + y) = f(y)g(x) + f(x)\n$$\nwe see that $f(x)(g(y) - 1) = f(y)(g(x) - 1)$. Since $f$ is strictly increasing we have $f(x) \\neq 0$ for $x \\neq 0$. So, for $x, y \\neq 0$ we get\n$$\n\\frac{g(y) - 1}{f(y)} = \\frac{g(x) - 1}{f(x)}\n$$\nThe left-hand side of the equation only depends on $y$ and the right-hand side only depends on $x$. We conclude that both sides are constant, so $g(x) - 1 = Cf(x)$ for all $x$ other than $0$. But we have shown that $g(0) = 1$ and $f(0) = 0$, so this equality also holds for $x = 0$. Hence,\n$$\n\\begin{aligned}\ng(x + y) &= 1 + Cf(x + y) = 1 + Cf(x)g(y) + Cf(y) \\\\\n&= g(y) + Cf(x)g(y) = g(y)(1 + Cf(x)) \\\\\n&= g(y)g(x)\n\\end{aligned}\n$$\nfor all real $x$ and $y$. This implies $g(nx) = g(x)^n$. From $g(x) = 1 - Cf(x)$ we see that $g$ is also strictly monotone (strictly increasing or strictly decreasing). Finally, $g\\left(\\frac{1}{x}\\right) = \\frac{1}{g(x)}$ and $g(0) = 1$ imply that $g(x) > 0$ for all $x \\in \\mathbb{R}$.\nLet $g(1) = a$, where $a > 0$. Then $g(n) = a^n$ for all $n \\in \\mathbb{Z}$. Since $g(x) = (g(nx))^{\\frac{1}{n}}$, we see that $g\\left(\\frac{m}{n}\\right) = (g(m))^{\\frac{1}{n}} = a^{\\frac{m}{n}}$ or $g(x) = a^x$ for all $x \\in \\mathbb{Q}$. Since $g$ is strictly monotone we have $g(x) = a^x$ for all real $x$.\nIt is not difficult to show that the functions $g(x) = a^x$ for all $a > 0$ satisfy the conditions of the problem. If $a < 1$ let $f(x) = 1 - a^x$. If $a = 1$ let $f(x) = x$. Otherwise, let $f(x) = a^x - 1$. In all three cases $f$ is strictly increasing.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41361,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle such that $|AB| > |BC| > |AC|$. Let $D$ be a point different from $C$ on the segment $BC$, such that $|AC| = |AD|$. Let $H$ denote the orthocentre of the triangle $ABC$, and let $A_1, B_1$ be the feet of the altitudes from $A$ and $B$, respectively. Denote the intersection of the lines $A_1B_1$ and $DH$ by $E$. Prove that $B, D, B_1$ and $E$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $C_1$ be the foot of the altitude to the side $AB$. The triangle $CAD$ is isosceles since $|AC| = |AD|$.\n\n![](attached_image_1.png)\n\nThe line $AA_1$ is the altitude in this isosceles triangle, so $\\angle CDH = \\angle HCD = \\angle C_1CB = \\frac{\\pi}{2} - \\angle CBA$. This implies that\n$$\n\\begin{aligned} \\angle EDB &= \\pi - \\angle CDE = \\pi - \\angle CDH \\\\ &= \\frac{\\pi}{2} + \\angle CBA. \\end{aligned}\n$$\nIn the quadrilateral $ABA_1B_1$ we have $\\angle AA_1B = \\frac{\\pi}{2} = \\angle AB_1B$. So this is a cyclic quadrilateral and $\\angle AB_1A_1 = \\pi - \\angle A_1BA = \\pi - \\angle CBA$. We see that $\\angle A_1B_1C = \\pi - \\angle AB_1A_1 = \\angle CBA$ and\n$$\n\\angle EB_1B = \\angle EB_1A + \\angle AB_1B = \\angle A_1B_1C + \\frac{\\pi}{2} = \\angle CBA + \\frac{\\pi}{2}.\n$$\nWe have shown that $\\angle EDB = \\frac{\\pi}{2} + \\angle CBA = \\angle EB_1B$, which implies that $B, D, B_1$ and $E$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41362,
    "subject": "Mathematics (Multi-modal)",
    "question": "The diagonal $BD$ of the quadrilateral $ABCD$ divides this quadrilateral into an acute triangle $ABD$ and an equilateral triangle $BCD$. Let $O$ be the orthocentre of the triangle $ABD$. Prove:\n\na. if the triangles $ABD$ and $OCD$ are congruent, then $AB \\perp BC$;\n\nb. if $\\angle CBA = 90^\\circ$, then the triangles $ABD$ and $OCD$ are congruent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. The points $C$ and $O$ lie on the bisector of the segment $BD$, so $\\angle DCO = 30^\\circ$. Since the triangles $ABD$ and $OCD$ are congruent, we have $\\angle DBA = \\angle DCO = 30^\\circ$. This implies $\\angle CBA = \\angle CBD + \\angle DBA = 90^\\circ$.\n\n![](attached_image_1.png)\n\nb. If $\\angle ABC = 90^\\circ$, then $\\angle ABD = 30^\\circ$. The points $C$ and $O$ lie on the bisector of the segment $BD$. So $\\angle DCO = 30^\\circ$. Applying the formula connecting the inscribed and the central angle we see that $\\angle BOD = 2\\angle BAD$. On the other hand $\\angle BOD = 2\\angle COD$, so $\\angle COD = \\angle BAD$. The triangles $ABD$ and $OCD$ are congruent since $\\angle BAD = \\angle COD$, $\\angle DBA = \\angle DCO$ and $|BD| = |CD|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41363,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je ostrokotni trikotnik $ABC$, v katerem je $|AB| > |BC| > |AC|$. Naj bo $D$ od $C$ različna točka na daljici $BC$, da je $|AC| = |AD|$. Označimo s $H$ višinsko točko trikotnika $ABC$, z $A_1$ in $B_1$ pa nožišči višin iz točk $A$ in $B$. Premici $A_1B_1$ in $DH$ se sekata v točki $E$. Dokaži, da so točke $B$, $D$, $B_1$ in $E$ konciklične.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNaj bo $C_1$ nožišče višine na stranico $AB$. Trikotnik $CAD$ je zaradi $|AC| = |AD|$ enakokrak. Premica $AA_1$ je višina tudi v tem enakokrakem trikotniku, zato je $\\angle CDH = \\angle HCD = C_1CB = \\frac{\\pi}{2} - \\angle CBA$. Od tod sledi\n$$\n\\begin{aligned}\n\\angle EDB & = \\pi - \\angle CDE = \\pi - \\angle CDH \\\\\n& = \\frac{\\pi}{2} + \\angle CBA\n\\end{aligned}\n$$\nV štirikotniku $ABA_1B_1$ velja $\\angle AA_1B = \\frac{\\pi}{2} = \\angle AB_1B$, zato je tetiven in je\n![](attached_image_1.png)\n$\\angle AB_1A_1 = \\pi - \\angle A_1BA = \\pi - \\angle CBA$. in\n$$\n\\angle EB_1B = \\angle EB_1A + \\angle AB_1B = \\angle A_1B_1C + \\frac{\\pi}{2} = \\angle CBA + \\frac{\\pi}{2}\n$$\nPokazali smo $\\angle EDB = \\frac{\\pi}{2} + \\angle CBA = \\angle EB_1B$, kar pomeni, da so točke $B$, $D$, $B_1$ in $E$ konciklične.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41364,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bosta $E$ in $F$ taki točki na stranicah $AB$ in $AD$ konveksnega štirikotnika $ABCD$, da je $EF \\parallel BD$. Daljica $CE$ seka diagonalo $BD$ v točki $G$, daljica $CF$ pa seka diagonalo $BD$ v točki $H$. Dokaži: če je $AGCH$ paralelogram, je tudi $ABCD$ paralelogram.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPresečišče premic $EF$ in $AG$ označimo z $I$, presečišče premic $EF$ in $AH$ pa z $J$. Ker je $EF \\parallel BD$ in $AH \\parallel CE$, je štirikotnik $EGHJ$ paralelogram. Zaradi vzporednosti $AG$ in $CF$ pa je tudi $FGIH$ paralelogram. Zato je $|FH| = |IG|$ in $|HJ| = |GE|$ ter velja $\\angle FHJ = \\angle IGE$, kar pomeni, da sta trikotnika $FHJ$ in $IGE$ skladna. Tako je $|FJ| = |IE|$.\n\nKer je $EF \\parallel BD$, imamo tri pare podobnih trikotnikov: $EAF$ in $BAD$, $EAI$ in $BAG$ ter $JAF$ in $HAD$. Iz prve sledi $\\frac{|EA|}{|BA|} = \\frac{|FA|}{|DA|}$, iz druge $\\frac{|EA|}{|BA|} = \\frac{|EI|}{|BG|}$ in iz tretje $\\frac{|FA|}{|DA|} = \\frac{|JF|}{|DH|}$. Dobimo $\\frac{|EI|}{|BG|} = \\frac{|JF|}{|DH|}$ in zaradi $|FJ| = |IE|$ sledi $|BG| = |DH|$.\n\nNaj bo $S$ razpolovišče daljice $AC$. Ker je $AGCH$ paralelogram, je $S$ hkrati razpolovišče daljice $GH$. Zaradi $|BG| = |DH|$ pa sledi, da je $S$ tudi razpolovišče daljice $BD$. Daljici $AC$ in $BD$ se razpolavljata, zato je $ABCD$ paralelogram.\n\n\n2. način\n\nKot v prvi rešitvi pokažemo, da je $|BG| = |DH|$. Ker je $AG \\parallel FC$, velja $\\angle AGB = \\angle FHB = \\angle CHD$, zato se trikotnika $AGB$ in $CHD$ ujemata v kotu in dolžinah priležnih stranic ($|BG| = |DH|$ in $|AG| = |CH|$), torej sta skladna in velja še $|AB| = |CD|$ in $\\angle ABG = \\angle CDH$. Torej sta $AB$ in $CD$ enako dolgi vzporedni stranici štirikotnika $ABCD$, kar pomeni, da je ta štirikotnik paralelogram.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41365,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je ostrokotni trikotnik $ABC$, v katerem je $|AB| > |BC| > |AC|$. Naj bo $D$ od $C$ različna točka na daljici $BC$, da je $|AC| = |AD|$. Označimo s $H$ višinsko točko trikotnika $ABC$, z $A_{1}$ in $B_{1}$ pa nožišči višin iz točk $A$ in $B$. Premica $DH$ seka premico $AC$ v točki $E$, premico $A_{1}B_{1}$ pa v točki $F$. Naj bo $G$ presečišče premic $AF$ in $BH$. Dokaži, da sta si trikotnika $HBD$ in $HGE$ podobna.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTrikotnik $CAD$ je zaradi $|AC| = |AD|$ enakokrak. Premica $AA_{1}$ je višina tudi v tem enakokrakem trikotniku, zato je $\\angle HDA = \\angle ACH$. V štirikotniku $HA_{1}CB_{1}$ velja $\\angle CA_{1}H = \\frac{\\pi}{2} = \\angle CB_{1}H$, zato je tetiven in je $\\angle B_{1}A_{1}H = \\angle B_{1}CH$. Dobili smo\n$$\n\\begin{aligned}\n\\angle FA_{1}A &= \\angle B_{1}A_{1}H = \\angle B_{1}CH \\\\\n&= \\angle ACH = \\angle HDA \\\\\n&= \\angle FDA\n\\end{aligned}\n$$\ntorej so tudi točke $A, D, A_{1}$ in $F$ konciklične. To pomeni, da je $\\angle AFD = \\angle AA_{1}D = \\frac{\\pi}{2}$. Daljici $AB_{1}$ in $HF$ sta višini v trikotniku $AHG$ in se sekata v točki $E$. Zato je $E$ višinska točka tega trikotnika in je $EG$ pravokotna na $AH$. Ker pa je $AH$ pravokotna na $BC$, od tod sledi, da je $EG$ vzporedna z $BC$. Zato je $\\angle EGH = \\angle HBD$ in zaradi $\\angle GHE = \\angle BHD$ sledi, da sta si trikotnika $HBD$ in $HGE$ podobna.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41366,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find integers $a$ and $b$, such that $\\sqrt{2010 + 2\\sqrt{2009}}$ is a solution of the quadratic equation $x^2 + a x + b = 0$. Prove that for such $a$ and $b$ the number $\\sqrt{2010 - 2\\sqrt{2009}}$ is not the solution of the given equation.",
    "options": [],
    "answer": "a = -2, b = -2008",
    "solution": "We notice that\n$$\n\\sqrt{2010 + 2\\sqrt{2009}} = \\sqrt{1 + 2\\sqrt{2009} + 2009} = \\sqrt{(1 + \\sqrt{2009})^2} = 1 + \\sqrt{2009}.\n$$\nSimilarly, $\\sqrt{2010 - 2\\sqrt{2009}} = \\sqrt{2009} - 1$. Since $1 + \\sqrt{2009}$ is the solution of the quadratic equation $x^2 + a x + b = 0$, we have $2010 + 2\\sqrt{2009} + a(1 + \\sqrt{2009}) + b = 0$ or\n$$\n\\sqrt{2009}(2 + a) = -b - 2010 - a.\n$$\nThe right-hand side is an integer, so the left-hand side must be an integer as well. This implies $2 + a = 0$, thus $a = -2$. From $-b - 2010 - a = 0$ we get $b = -2008$. The two solutions of the quadratic equation $x^2 - 2x - 2008 = 0$ are $1 + \\sqrt{2009}$ and $1 - \\sqrt{2009}$. So $\\sqrt{2009} - 1$ cannot be a solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41367,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$ be a real number such that $\\cos(2x) + \\cos(3x) = 1$. Show that $2\\sin(2x) + 2\\sin(3x) = \\sin(4x) + 2\\sin(5x) + \\sin(6x)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The double-angle formula and the angle sum identity for sines and cosines imply\n$$\n\\begin{aligned}\n\\sin(4x) + 2\\sin(5x) + \\sin(6x) &= \\\\\n&= 2\\sin(2x)\\cos(2x) + 2(\\sin(2x)\\cos(3x) + \\sin(3x)\\cos(2x)) + 2\\sin(3x)\\cos(3x) \\\\\n&= 2\\cos(2x)(\\sin(2x) + \\sin(3x)) + 2\\cos(3x)(\\sin(2x) + \\sin(3x)) \\\\\n&= 2(\\cos(2x) + \\cos(3x))(\\sin(2x) + \\sin(3x)).\n\\end{aligned}\n$$\nand the latter is equal to $2(\\sin(2x) + \\sin(3x))$, which was to be shown.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41368,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all rational numbers $r$ and all integers $k$, such that the equation $r(5k - 7r) = 3$ is satisfied.",
    "options": [],
    "answer": "(k, r) in {(2, 1), (-2, -1), (2, 3/7), (-2, -3/7)}",
    "solution": "Obviously, $r \\ne 0$. Let us write $r$ as a reduced fraction $r = \\frac{m}{n}$ and let us assume that $n$ is a positive integer. Then $\\frac{m}{n}(5k - 7m) = 3$ or, equivalently, $m(5kn - 7m) = 3n^2$. Hence, $m$ divides $3n^2$. Since $m$ and $n$ are coprime, we conclude that $m$ divides $3$. Let us consider four cases.\n\nIf $m = 1$ we have $5kn - 7 = 3n^2$ or $n(5k - 3n) = 7$, which implies that $n$ divides $7$. Since $n$ is a positive integer, it equals either $1$ or $7$. When $n=1$ we get $k=2$ and $r=1$. When $n=7$ the equation $5k=22$ gives us no integer solutions.\n\nIf $m=3$ we have $n(5k-n) = 21$. We see that $n$ divides $21$. Again, $m$ and $n$ are coprime, so $n$ divides $7$. Once more we have either $n=1$ or $n=7$. This time we obtain the solution only in the second case: $k=2$, $r=\\frac{3}{7}$.\n\nIf $m=-1$ we have $n(5k+3n) = -7$, so $n$ divides $7$. When $n=1$ the solution is $k=-2$, $r=-1$. When $n=7$ there are no solutions.\n\nIf $m=-3$ we have $n(5k+n) = -21$. As $m$ and $n$ are coprime, $n$ divides $7$. We find one last solution, $k=-2$, $r=-\\frac{3}{7}$.\n\nAll possible pairs $(k, r)$ are $(2, 1)$, $(-2, -1)$, $(2, \\frac{3}{7})$ and $(-2, -\\frac{3}{7})$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41369,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTabela velikosti $5 \\times 7$ ima eno polje rdeče, ostala polja pa bela. Na to tabelo postavljamo ploščice oblike\n![](attached_image_1.png)\n(ploščico lahko zrcalimo ali zasukamo),\ntako da cele ležijo na tabeli. Pri tem se ploščice na rdečem polju lahko prekrivajo, na belih pa ne. Določi vse možne položaje rdečega polja, pri katerih lahko tabelo v celoti prekrijemo.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDenimo, da smo v celoti uspeli prekriti tabelo. Vsaka ploščica, ki prekriva rdeče polje, prekriva vsaj še eno belo polje, ki leži levo, desno, nad ali pod rdečim poljem. V primeru, da bi rdeče polje prekrivalo 5 ali več ploščic, bi se vsaj dve izmed njih prekrivali na belem polju. Torej se na rdečem polju prekrivajo kvečjemu 4 ploščice. Vse ploščice skupaj morajo torej imeti med 35 in 38 polj. Ker pa ima vsaka ploščica 4 polja, morajo vse ploščice skupaj imeti 36 polj. Torej je vseh ploščic 9, na rdečem polju pa se prekrivata dve ploščici.\n\n![](attached_image_2.png)\nTabela 1\n![](attached_image_3.png)\nTabela 2\n![](attached_image_4.png)\nTabela 3\n\nIz tabele 1 je zaradi simetrije razvidno, da rdeče polje lahko leži na enem izmed osenčenih polj. Ne glede na to, kako postavimo ploščico na tabelo 2 ali 3, bo vedno prekrivala liho osenčenih polj. Ker je ploščic liho, bodo skupaj prekrivale liho osenčenih polj. Osenčenih polj je na obeh tabelah 2 in 3 liho. Ce bi torej rdeče polje ležalo na osenčenem polju tabele 2 ali 3, bi morale vse domine skupaj prekrivati sodo osenčenih polj. To pa ni mogoče. Torej so osenčena polja iz tabele 1 edini možni položaji rdečega polja.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41370,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTabela velikosti $8 \\times 8$ je razdeljena na 64 kvadratkov. Določi najmanjše število barv, s katerimi moramo pobarvati kvadratke tabele, da bodo vsaki štirje kvadratki, ki tvorijo lik oblike\n![](attached_image_1.png)\n(lik lahko zrcalimo ali zasukamo),\npobarvani z različnimi barvami.",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nZadostuje 8 barv. Označimo barve s številkami od 1 do 8 in tabelo pobarvajmo kot prikazuje prva slika. Vsak lik predpisane oblike je pobarvan z različnimi barvami, saj najbližja kvadratka enake barve ležita diagonalno z enim vmesnim kvadratkom in obeh hkrati ne moremo pokriti z enim samim likom. \n\nDenimo, da zadostuje že 7 barv. Imenujmo dva kvadratka povezana, če sta oba vsebovana v nekem liku predpisane oblike. Označimo nekatere kvadratke v tabeli, kot prikazuje druga slika. Ker sta vsaka dva izmed kvadratkov $A, B, C, D, E, F, G$ povezana,\n![](attached_image_2.png)\nmorajo biti ti kvadratki pobarvani z različnimi barvami (skupaj jih je ravno 7).\n\nKvadratek $u$ je povezan s kvadratki $A, B, C, D, E, G$, torej mora biti iste barve kot kvadratek $F$. Kvadratek $x$ je povezan s kvadratki $B, C, D, E, F, G$, torej mora biti iste barve kot $A$. Kvadratek $v$ je povezan s kvadratki $B, u, D, E, G, x$. Ker je $u$ take barve kot $F, x$ pa kot $A$, mora barva kvadratka $v$ enaka barvi kvadratka $C$. Kvadratek $z$ je povezan s kvadratki $C, D, E, F, G, x$, torej mora biti iste barve kot $B$. Kvadratek $y$ je povezan s kvadratki $u, D, E, v, G, x$, ki so enakih barv kot $F, D, E, C, G, A$, torej\n![](attached_image_3.png)\nmora biti $y$ enake barve kot $B$. Vendar kvadratki $G, x, y, z$ tvorijo lik predpisane oblike, ki ni pobarvan s štirimi različnimi barvami, saj sta kvadratka $y$ in $z$ pobarvana z isto barvo (s tako kot $B$ ). Zato 7 barv ne zadostuje.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41371,
    "subject": "Mathematics (Multi-modal)",
    "question": "Every school in the region has sent 3 students to a contest. Andrej, Blaž and Žan represented the same school. When all the contestants lined up to receive their start numbers, Andrej realized that there were exactly as many contestants in the line before him as there were behind. Both his friends were behind him: Blaž was 19th and Žan was 28th. How many schools are there in this region?",
    "options": [],
    "answer": "11",
    "solution": "Let $x$ denote the number of contestants in line before Andrej. Then there were also $x$ contestants behind Andrej and there were $2x+1$ contestants altogether. Hence, the total number of contestants was odd. Since Andrej was standing in line before Blaž, who was 19th, there were at most 17 contestants in line before Andrej and $x \\le 17$. This gives us the maximum of $2x+1 \\le 2 \\cdot 17+1 = 35$ contestants. Since Žan was 28th in line, there were at least 28 contestants. As each of the schools has sent 3 contestants the total number of contestants must be divisible by 3. The only two numbers between 28 and 35 that are divisible by 3 are 30 and 33. Of these only 33 is odd. There were 33 contestants, hence there are 11 schools in the region.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41372,
    "subject": "Mathematics (Multi-modal)",
    "question": "A certain high school offers its students the choice of two sports: football and basketball. One fifth of the footballers also plays basketball and one seventh of the basketball players plays football. There are 110 students who practice exactly one of the sports. How many of them practice both?",
    "options": [],
    "answer": "11",
    "solution": "Denote the number of footballers by $n$, the number of basketball players by $k$ and the number of students practicing both sports by $d$. One fifth of the footballers also plays basketball. So $\\frac{n}{5} = d$. One seventh of the basketball players also plays football, so $\\frac{k}{7} = d$. This implies $n = 5d$ and $k = 7d$.\n\n![](attached_image_1.png)\n\nThere are $k - d$ basketball players who do not play football and $n - d$ football players who do not play basketball. Altogether, there are $(n - d) + (k - d)$ students who only practice one of the sports. So $110 = n + k - 2d$. We get $110 = 10d$ or $d = 11$. Hence, 11 students practice both sports.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41373,
    "subject": "Mathematics (Multi-modal)",
    "question": "A teacher invited a group of children to sit down at a round table. There were three times as many boys as there were girls. The teacher walked around the table and observed the pairs of children sitting next to each other. She noticed that the number of same-sex pairs was twice the number of boy-girl pairs. At least how many children were sitting at the table?",
    "options": [],
    "answer": "12",
    "solution": "Let $x$ be the number of girls and $y$ the number of boys. Then $y = 3x$ and there are $4x$ children altogether.\n\nLet $a$ be the number of boy-girl pairs. Then the number of same-sex pairs is equal to $2a$ and there are $3a$ pairs altogether. The number of pairs equals the number of all children. Indeed, every child belongs to two pairs, so there are $2 \\cdot 4x$ pairs, but every pair was counted twice, which gives us $\\frac{2 \\cdot 4x}{2} = 4x$ distinctive pairs. Hence, $3a = 4x$. The smallest $x$ for which this equation can hold is\n\n$$\n\\begin{array}{cccc}\n& & F & D \\\\\n& & F & F \\\\\nF & & & D \\\\\nF & & & D \\\\\nF & & & F \\\\\nF & F & F & F\n\\end{array}\n$$\n\n$x = 3$. We conclude there was at least 12 children at the table. If they sat down as shown in the figure (D represents girls, F represents boys), then the number of same-sex pairs was twice the number of boy-girl pairs.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 41374,
    "subject": "Mathematics (Multi-modal)",
    "question": "Jure has drawn a regular enneagon (a 9-sided polygon). He wants to arrange the numbers $1$ to $9$ at its vertices so that the sum of the numbers at any three consecutive vertices does not exceed some positive integer $n$. What is the least possible $n$ with which he can succeed?",
    "options": [],
    "answer": "16",
    "solution": "We will show that $n = 16$.\n\nAs one can see in the figure, it is possible to arrange the numbers $1$ to $9$ at the vertices so that the sum of any three consecutive numbers is at most $16$.\n\nLet us show that for $n < 16$ this cannot be achieved. Assume, to the contrary, that it is possible to do so. If we add the sums over all possible triplets of consecutive vertices, we have used every number three times since it occurs in three such triplets. Hence, the number we obtain is equal to $3 \\cdot (1 + 2 + \\dots + 9) = 135$. On the other hand, every sum of numbers at three consecutive vertices is at most $n$ and there are $9$ such sums. Thus, the total sum is at most $9n$. We conclude that $9n \\ge 135$, so $n \\ge 15$.\n\n![](attached_image_1.png)\n\nFinally, let us show that $n$ cannot equal $15$. If this were the case, all of the triplets would have to add up to $15$. Let us denote the numbers at four consecutive vertices by $a_1, a_2, a_3$ and $a_4$. Then $a_1 + a_2 + a_3 = 15$ and $a_2 + a_3 + a_4 = 15$, so $a_1 = a_4$. This is not possible since every number has to occur exactly once. This proves that $n$ has to be at least $16$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41375,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$, $q$ and $r$ such that $p > q > r$ and the numbers $p-q$, $p-r$ and $q-r$ are also prime.",
    "options": [],
    "answer": "p=7, q=5, r=2",
    "solution": "We have $r \\ge 2$, so $p$ and $q$ must be odd primes. Thus, $p-q$ is even and equal to $2$. We get $p-q=2$. The numbers $p-r=q-r+2$ and $q-r$ are prime and differ by $2$, so they have the same parity. We conclude that both must be odd. Since $q$ and $q-r$ are odd, $r$ must be even. Thus, $r=2$.\n\nThe numbers $q$, $p = q+2$ and $q-r = q-2$ are prime. Since $q-2$ is an odd prime it must be at least $3$. But exactly one of the prime numbers $q-2$, $q$, $q+2$ is divisible by $3$, so $q-2=3$. This implies $q=5$ and $p=7$.\n\nWe get $r=2$, $q=5$ and $p=7$. These numbers satisfy the conditions of the problem since $p-r=5$, $p-q=2$ and $q-r=3$ are also prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41376,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n$, such that the equation $x^2 + nx + n + 5 = 0$ has only integer solutions.",
    "options": [],
    "answer": "-5, -3, 7, 9",
    "solution": "Let $x_1$ and $x_2$ be the integer solutions of this quadratic equation. We may assume that $x_1 \\le x_2$. By Viete's formulas we have $x_1 + x_2 = -n$ and $x_1x_2 = n + 5$. Adding both identities together we get $x_1x_2 + x_1 + x_2 = 5$, so\n\n$$(x_1 + 1)(x_2 + 1) = 6.$$\n\nSince $6 = 1 \\cdot 6 = 2 \\cdot 3 = (-3) \\cdot (-2) = (-6) \\cdot (-1)$, we have four possible cases to consider. In the first case we have $x_1 = 0$, $x_2 = 5$, in the second case $x_1 = 1$, $x_2 = 2$, in the third case $x_1 = -4$, $x_2 = -3$ and in the final case $x_1 = -7$, $x_2 = -2$. Using the identity $x_1 + x_2 = -n$ we see that $n$ is equal to $-5, -3, 7$ or $9$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41377,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ and $d$ be positive integers such that $a > b > c > d$ and\n$$\n(1-a)(1-b)(1-c)(1-d) = 10.\n$$\nFind all possible values of $a+b-c-d$.",
    "options": [],
    "answer": "9",
    "solution": "The numbers $a, b, c$ and $d$ are pairwise different, so $1-a, 1-b, 1-c$ and $1-d$ are all different as well. Since $10$ is the product of two primes, it can only be written as the product of four integers if two of these integers are $1$ and $-1$. The remaining two factors are either $-2$ and $5$ or $2$ and $-5$.\n\nSince $a > b > c > d$ we have $1-a < 1-b < 1-c < 1-d$. So, $1-b = -1$ and $1-c = 1$, which implies $b=2$ and $c=0$. If $1-a = -2$ and $1-d = 5$, we get $a=3$ and $d=-4$. In the case where $1-a = -5$ and $1-d = 2$, we have $a=6$ and $d=-1$.\n\nThe value of $a+b-c-d$ is equal to $3+2-0-(-4) = 9$ in the first case, and to $6+2-0-(-1) = 9$ in the second case, so there is really only one possibility, $a+b-c-d=9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41378,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which there exists a polynomial $P(x)$ with integer coefficients such that $P(d) = \\left(\\frac{n}{d}\\right)^2$ for each positive divisor $d$ of $n$.",
    "options": [],
    "answer": "1, all primes, and 6",
    "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 the conditions $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) = (-p-1)x + p^2 + p + 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 \\le k+l+1$, which implies that $kl - k - l + 1 \\le 2$ or $(k-1)(l-1) \\le 2$. We may assume that $k \\le 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)(x-5))$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41379,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ and $q$ be two polynomials of degree $3$ with integer coefficients, such that the leading coefficients are relatively prime. Let $a$ be a rational number such that $p(a)$ and $q(a)$ are integers. Prove that $a$ is also an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $p(x) = b_1 x^3 + c_1 x^2 + d_1 x + e_1$ and $q(x) = b_2 x^3 + c_2 x^2 + d_2 x + e_2$. Let us write $a = \\frac{r}{s}$ where $r$ and $s$ are relatively prime integers and $s > 0$. Denote $p(a) = m$ and $q(a) = n$. Then\n$$\nb_1 \\cdot \\frac{r^3}{s^3} + c_1 \\cdot \\frac{r^2}{s^2} + d_1 \\cdot \\frac{r}{s} + e_1 = m, \\quad b_2 \\cdot \\frac{r^3}{s^3} + c_2 \\cdot \\frac{r^2}{s^2} + d_2 \\cdot \\frac{r}{s} + e_2 = n.\n$$\nMultiplying both identities by $s^3$ we get\n$$\nb_1 r^3 + c_1 r^2 s + d_1 r s^2 + e_1 s^3 = m s^3, \\quad b_2 r^3 + c_2 r^2 s + d_2 r s^2 + e_2 s^3 = n s^3.\n$$\nThis implies that $s$ divides both $b_1 r^3$ and $b_2 r^3$. Since $s$ and $r$ are relatively prime, we can conclude that $s$ divides $b_1$ and $b_2$. Since $b_1$ and $b_2$ are relatively prime, we have $s = 1$. So, $a = r$ is an integer.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41380,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be positive integers. Prove that $a^2 + b^2 + c^2$ is divisible by $4$ if and only if $a$, $b$ and $c$ are even.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, let $a$, $b$ and $c$ be even: $a = 2a_1$, $b = 2b_1$, $c = 2c_1$. Then the number\n$$\na^2 + b^2 + c^2 = 4(a_1^2 + b_1^2 + c_1^2)\n$$\nis divisible by $4$.\n\nNow, let us prove the converse. Assume that $a^2 + b^2 + c^2$ is divisible by $4$. If exactly one of the numbers $a$, $b$ and $c$ were odd or if all three were odd, then the sum $a^2 + b^2 + c^2$ would be odd. This is not the case.\n\nFinally, assume that exactly two of the numbers $a$, $b$ and $c$ are odd. We may assume that $a$ and $b$ are odd and $c$ is even. Let $a = 2a_1 - 1$, $b = 2b_1 - 1$ and $c = 2c_1$ for some positive integers $a_1$, $b_1$ and $c_1$. In this case the number\n$$\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$$\nis not divisible by $4$. This contradicts the assumption. The only remaining possibility is that all three numbers are even.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41381,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given 20 points in space so that no three of them are collinear, prove that the number of planes determined by these points is not equal to $1111$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume, to the contrary, that the number of planes is equal to $1111$. Now, $20$ points in space can define at most $\\binom{20}{3} = \\frac{20 \\cdot 19 \\cdot 18}{3 \\cdot 2 \\cdot 1} = 1140$ planes, so $1140 - 1111 = 29$ triplets of points lie in the planes already determined by another triplet. If one of the planes contains $7$ or more points, then there are at least $\\binom{7}{3} = 35$ triplets of points in this plane and the number of triplets is greater than the number of planes by at least $35 - 1 = 34$. Hence, the greatest possible number of planes is $1140 - 34 = 1105$. Obviously, this cannot happen if there are $1111$ planes, so each plane can contain at most $6$ of the points.\n\nLet $a$ be the number of planes containing $4$ points, $b$ the number of planes containing $5$ points and $c$ the number of planes containing $6$ points. When counting triplets, we considered each plane containing $4$ points $\\binom{4}{3} = 4$ times. That is $3$ times too many. Each plane containing $5$ points was counted $\\binom{5}{3} = 10$ times (i.e. $9$ times too many) and each plane containing $6$ points was counted $\\binom{6}{3} = 20$ times, which is $19$ times too many. The number of planes is thus equal to $1140 - 3a - 9b - 19c$. If this number were equal to $1111$, then we would have $3a + 9b + 19c = 29$. The numbers $a$, $b$ and $c$ are non-negative integers, so $c \\ge 1$ is not possible, $c = 0$. We get $3a + 9b = 29$ and this is impossible since the left-hand side is divisible by $3$ and the right-hand side is not. We conclude that the number of planes cannot be equal to $1111$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41382,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real $x$ in $[0, 2\\pi)$ for which all terms of the sequence\n$$\na_n = \\frac{1}{\\cos(nx)}\n$$\nare integers.",
    "options": [],
    "answer": "x in {0, π/3, 2π/3, π, 4π/3, 5π/3}",
    "solution": "The numbers $a_1$ and $a_2$ are integers. Since $a_1 = \\frac{1}{\\cos x}$, $a_2 = \\frac{1}{\\cos(2x)}$ and $\\cos(2x) = 2(\\cos x)^2 - 1$, we have $a_2 = \\frac{1}{2\\cos^2 x - 1} = \\frac{a_1^2}{2 - a_1^2}$. Now, $a_2$ is an integer, so $2 - a_1^2$ is a divisor of $a_1^2$. So, $2 - a_1^2$ divides $2 - a_1^2$ and $a_1^2$. It therefore also divides their sum, which is equal to 2. We conclude that $2 - a_1^2$ is one of the numbers $-2, -1, 1$ or $2$.\nIn the first case we have $a_1^2 = 4$, in the second case we have $a_1^2 = 3$, in the third case we get $a_1^2 = 1$ and in the fourth case we get $a_1^2 = 0$. Since $a_1$ is a non-zero integer, we can only have $a_1 = -2$, $a_1 = 2$, $a_1 = -1$ or $a_1 = 1$. From here it follows that $x$ can only be equal to $0, \\frac{\\pi}{3}, \\frac{2\\pi}{3}, \\pi, \\frac{4\\pi}{3}$ or $\\frac{5\\pi}{3}$.\nIf $x = 0$, then $a_n = 1$ for all $n$. If $x = \\pi$, we get the sequence $-1, 1, -1, 1, \\dots$. In the remaining four cases we have $a_{n+6} = a_n$, since for all integers $a$ we have\n$$\n\\cos\\left(\\frac{(n+6) \\cdot a\\pi}{3}\\right) = \\cos\\left(\\frac{n \\cdot a\\pi}{3} + 2a\\pi\\right) = \\cos\\left(\\frac{n \\cdot a\\pi}{3}\\right).\n$$\nIt therefore suffices to check that the first six terms of the sequence are integers. When $x = \\frac{\\pi}{3}$ and $\\frac{5\\pi}{3}$ we have $a_1 = 2$, $a_2 = -2$, $a_3 = -1$, $a_4 = -2$, $a_5 = 2$, $a_6 = 1$. If $x = \\frac{2\\pi}{3}$ or $x = \\frac{4\\pi}{3}$, then the first six terms are $-2, -2, 1, -2, -2, 1$.\nThe solutions are $0, \\frac{\\pi}{3}, \\frac{2\\pi}{3}, \\pi, \\frac{4\\pi}{3}$ and $\\frac{5\\pi}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41383,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be real numbers such that $|a| \\neq |b|$ and $\\frac{a+b}{a-b} + \\frac{a-b}{a+b} = 6$.\nFind the value of the expression $\\frac{a^3 + b^3}{a^3 - b^3} + \\frac{a^3 - b^3}{a^3 + b^3}$.",
    "options": [],
    "answer": "18/7",
    "solution": "The equality\n$$\n6 = \\frac{a+b}{a-b} + \\frac{a-b}{a+b} = \\frac{2a^2 + 2b^2}{a^2 - b^2}\n$$\nimplies $6a^2 - 6b^2 = 2a^2 + 2b^2$, or $4a^2 = 8b^2$. From here we get $a = \\pm b\\sqrt{2}$. Now, we can conclude that\n$$\n\\frac{a^3 + b^3}{a^3 - b^3} + \\frac{a^3 - b^3}{a^3 + b^3} = \\frac{2a^6 + 2b^6}{a^6 - b^6} = \\frac{(2 \\cdot 2^3 + 2)b^6}{(2^3 - 1)b^6} = \\frac{18}{7}.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41384,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a real number $t$ and positive real numbers $a$ and $b$ we have\n$$\n2a^2 - 3abt + b^2 = 2a^2 + abt - b^2 = 0.\n$$\nFind $t$.",
    "options": [],
    "answer": "1",
    "solution": "From $2a^2 - 3ab t + b^2 = 0$ we get $t = \\frac{2a^2 + b^2}{3ab}$ and from $2a^2 + ab t - b^2 = 0$ we get $t = \\frac{b^2 - 2a^2}{ab}$. So, $\\frac{2a^2 + b^2}{3ab} = \\frac{b^2 - 2a^2}{ab}$. Eliminating the fractions we obtain $8a^2 = 2b^2$ or $4a^2 = b^2$. Since $a$ and $b$ are positive, we conclude that $b = 2a$. Thus, $t = \\frac{2a^2 + 4a^2}{6a^2} = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41385,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa cela števila $a, b, c$ in $d$ velja $a > b > c > d$ in\n$$\n(1-a)(1-b)(1-c)(1-d) = 10\n$$\nKatere vrednosti lahko zavzame izraz $a + b - c - d$?",
    "options": [],
    "answer": "9",
    "solution": "Solution:\n\nKer so števila $a, b, c$ in $d$ različna, so različna tudi števila $1-a, 1-b, 1-c$ in $1-d$. Ker je $10$ produkt le dveh praštevil, lahko $10$ zapišemo kot produkt štirih celih števil le, če sta dve izmed teh števil $1$ in $-1$. Preostali števili sta tedaj $-2$ in $5$ ali pa $2$ in $-5$.\n\nZaradi $a > b > c > d$ sledi $1-a < 1-b < 1-c < 1-d$. Zato je $1-b = -1$ in $1-c = 1$, od koder sledi $b = 2$ in $c = 0$. Če je $1-a = -2$ in $1-d = 5$, dobimo $a = 3$ in $d = -4$. V primeru $1-a = -5$, $1-d = 2$ pa je $a = 6$ in $d = -1$.\n\nVrednost izraza $a + b - c - d$ je v prvem primeru $3 + 2 - 0 - (-4) = 9$, v drugem pa $6 + 2 - 0 - (-1) = 9$, torej je edina možnost $a + b - c - d = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41386,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPloščina romba meri $120~\\mathrm{cm}^2$. Razlika dolžin njegovih diagonal je $14~\\mathrm{cm}$. Kolikšna je dolžina stranice romba?",
    "options": [],
    "answer": "13",
    "solution": "Solution:\n\nUpoštevamo smiselno zvezo med diagonalama $e - f = 14$ ali $e = 14 + f$. Uporabimo obrazec za ploščino romba $\\frac{e \\cdot f}{2} = 120$. Vstavimo $e$ in dobimo kvadratno enačbo $f^2 + 14f - 240 = 0$. Enačbo rešimo. Dobimo rešitvi $f_1 = 10$ in $f_2 = -24$, ki pa ni ustrezna. Dolžino stranice $a$ izračunamo z uporabo Pitagorovega izreka $a^2 = \\left(\\frac{e}{2}\\right)^2 + \\left(\\frac{f}{2}\\right)^2$. Izračunamo $a = 13$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41387,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoloči vsa cela števila $n$, za katera ima enačba $x^{2}+n x+n+5=0$ le celoštevilske rešitve.",
    "options": [],
    "answer": "-5, -3, 7, 9",
    "solution": "Solution:\n\nKvadratna enačba ima lahko celoštevilske rešitve le, če je diskriminanta popoln kvadrat. Izračunamo lahko $D=n^{2}-4(n+5)=(n-2)^{2}-24$. Torej za neko nenegativno celo število $m$ velja $m^{2}=(n-2)^{2}-24$, oziroma\n$$\n24=(n-2)^{2}-m^{2}=(n+m-2)(n-m-2)\n$$\nKer sta števili $n+m-2$ in $n-m-2$ iste parnosti in je $m \\geq 0$, imamo 4 možnosti.\nČe je $n-m-2=-12$ in $n+m-2=-2$, je $n=-5$ (in $m=5$ ), kvadratna enačba pa ima rešitvi $x_{1}=0$ in $x_{2}=5$. V primeru $n-m-2=-6$ in $n+m-2=-4$ je $n=-3$ (in $m=1$ ), rešitvi kvadratne enačbe pa sta $x_{1}=1$ in $x_{2}=2$.\nČe je $n-m-2=2$ in $n+m-2=12$, je $n=9$ ter $m=5$, kvadratna enačba pa ima rešitvi $x_{1}=-7$ in $x_{2}=-2$. V primeru $n-m-2=4$ in $n+m-2=6$ je $n=7$ in $m=1$, rešitvi kvadratne enačbe pa sta $x_{1}=-4$ in $x_{2}=-3$.\nVse možne vrednosti števila $n$ so torej $-5,-3,7$ in $9$.\n\n2. način\n\nNaj bosta $x_{1}$ in $x_{2}$ celoštevilski rešitvi te kvadratne enačbe. Predpostavimo lahko še $x_{1} \\leq x_{2}$. Po Vietovih pravilih velja $x_{1}+x_{2}=-n$ in $x_{1} x_{2}=n+5$. Če enačbi seštejemo, dobimo $x_{1} x_{2}+x_{1}+x_{2}=5$ in zato\n$$\n\\left(x_{1}+1\\right)\\left(x_{2}+1\\right)=6\n$$\nKer je $6=1 \\cdot 6=2 \\cdot 3=(-3) \\cdot(-2)=(-6) \\cdot(-1)$, dobimo štiri možnosti. V prvem primeru je $x_{1}=0, x_{2}=5$, v drugem $x_{1}=1, x_{2}=2$, v tretjem $x_{1}=-4, x_{2}=-3$ in v zadnjem $x_{1}=-7, x_{2}=-2$. S pomočjo $x_{1}+x_{2}=-n$ izračunamo, da je število $n$ enako $-5,-3,7$ ali $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41388,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTone je rezervoar za gorivo napolnil do vrha in se $z$ avtom podal na dolgo pot. Podatki, ki jih je Tone razbral na elektronskem števcu avtomobila, so prikazani v tabeli.\n\n| število km | 35824 | 36149 | 36449 |\n| :--- | :--- | :--- | :---: |\n| podatki o gorivu | poraba $17,2 \\ell$ | poraba $39,3 \\ell$ | ostalo $5,3 \\ell$ |\n\nKoliko litrov je prostornina rezervoarja tega avtomobila?\n\nOPOMBA: Upoštevajmo, da ima avtomobil konstantno porabo goriva na kilometer.",
    "options": [],
    "answer": "65 liters",
    "solution": "Solution:\n\nUgotovimo, da poraba goriva predstavlja linearno funkcijo odvisno od prevoženih kilometrov $f(x) = k \\cdot x$. Izračunamo $k = \\frac{39,3 - 17,2}{36149 - 35824} = \\frac{22,1}{325} = 0,068$ litrov na kilometer oziroma $6,8$ litrov na $100$ kilometrov.\n\nZa nadaljnjih $300$ kilometrov potrebuje $f(300) = 0,068 \\cdot 300 = 20,4 \\ell$.\n\nSeštejemo $39,3 + 20,4 + 5,3 = 65 \\ell$.\n\n\nUpoštevamo premo sorazmerje.\n\n| začetna količina $x$ | $x - 17,2$ | $x - 39,3$ | $5,3$ |\n| :--- | :--- | :--- | :--- |\n| stanje km | $35824$ | $36149$ | $36449$ |\n\nIzračunamo $x = 65 \\ell$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41389,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIzkopali so jamo v obliki kvadra. Njena globina je 12 krat večja od dolžine. Dolžina jame je $\\frac{3}{2}$ njene širine. Vsota merskega števila prostornine jame in ploščine njenega dna je $\\frac{7}{6}$. Izračunaj globino jame.",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nIzberemo spremenljivke $a$ je dolžina, $b$ je širina in $c$ je globina. Nastavimo sistem enačb $12a = c$, $a = \\frac{3b}{2}$ in $abc + ab = \\frac{7}{6}$. Sistem uredimo in dobimo $48a^{3} + 4a^{2} - 7 = 0$. Rešimo enačbo tretje stopnje z uporabo Hornerjevega algoritma. Zapišemo rešitve $a = \\frac{1}{2}$, $b = \\frac{1}{3}$ in $c = 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41390,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bosta $p$ in $q$ polinoma stopnje 3 s celoštevilskimi koeficienti, katerih vodilna koeficienta sta si tuja. Naj bo $a$ tako racionalno število, da sta $p(a)$ in $q(a)$ celi števili. Dokaži, da je tedaj tudi $a$ celo število.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNaj bo $p(x) = b_1 x^3 + c_1 x^2 + d_1 x + e_1$ in $q(x) = b_2 x^3 + c_2 x^2 + d_2 x + e_2$. Zapišimo $a = \\frac{r}{s}$ kot okrajšan ulomek, kjer sta $r$ in $s$ celi števili in je $s > 0$. Označimo $p(a) = m$ in $q(a) = n$. Tedaj velja\n$$\nb_1 \\cdot \\frac{r^3}{s^3} + c_1 \\cdot \\frac{r^2}{s^2} + d_1 \\cdot \\frac{r}{s} + e_1 = m, \\quad b_2 \\cdot \\frac{r^3}{s^3} + c_2 \\cdot \\frac{r^2}{s^2} + d_2 \\cdot \\frac{r}{s} + e_2 = n\n$$\nČe obe enačbi pomnožimo $s^3$, dobimo\n$$\nb_1 r^3 + c_1 r^2 s + d_1 r s^2 + e_1 s^3 = m s^3, \\quad b_2 r^3 + c_2 r^2 s + d_2 r s^2 + e_2 s^3 = n s^3\n$$\nOd tod sledi, da $s$ deli $b_1 r^3$ in $b_2 r^3$. Ker je število $s$ tuje $r$, mora $s$ deliti $b_1$ in $b_2$. Ker pa sta po predpostavki $b_1$ in $b_2$ tuji števili, je zato $s = 1$. Torej je $a = r$, kar je celo število.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41391,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa realna števila $x$ z intervala $[0,2\\pi)$, za katera so vsi členi zaporedja s splošnim členom\n$$\na_n=\\frac{1}{\\cos (n x)}\n$$\ncela števila.",
    "options": [],
    "answer": "x ∈ {0, π/3, 2π/3, π, 4π/3, 5π/3}",
    "solution": "Solution:\n\nŠtevili $a_1$ in $a_2$ sta celi. Ker je $a_1=\\frac{1}{\\cos x}$, $a_2=\\frac{1}{\\cos (2 x)}$ in velja $\\cos (2 x)=2(\\cos x)^2-1$, sledi $a_2=\\frac{1}{2 \\cos ^2 x-1}=\\frac{a_1^2}{2-a_1^2}$. Ker je $a_2$ celo število, je $2-a_1^2$ delitelj $a_1^2$. Zato $2-a_1^2$ deli $2-a_1^2$ in $a_1^2$, torej deli tudi njuno vsoto, ki je 2. Tako je $2-a_1^2$ eno izmed števil $-2,-1,1$ oziroma 2.\n\nV prvem primeru dobimo $a_1^2=4$, v drugem $a_1^2=3$, v tretjem $a_1^2=1$ in v zadnjem $a_1^2=0$. Ker je $a_1$ celo in neničelno število, so možnosti le $a_1=-2, a_1=2, a_1=-1$ in $a_1=1$. Od tod sledi, da je število $x$ lahko enako le $0, \\frac{\\pi}{3}, \\frac{2 \\pi}{3}, \\pi, \\frac{4 \\pi}{3}$ ali $\\frac{5 \\pi}{3}$.\n\nČe je $x=0$, je $a_n=1$ za vsak $n$. Če je $x=\\pi$, dobimo zaporedje $-1,1,-1,1, \\ldots$ V ostalih štirih primerih opazimo, da je $a_{n+6}=a_n$, kajti za vsako celo število $a$ velja\n$$\n\\cos \\left(\\frac{(n+6) \\cdot a \\pi}{3}\\right)=\\cos \\left(\\frac{n \\cdot a \\pi}{3}+2 a \\pi\\right)=\\cos \\left(\\frac{n \\cdot a \\pi}{3}\\right)\n$$\nZato je dovolj preveriti, da so prvih šest členov zaporedja cela števila. V primerih $x=\\frac{\\pi}{3}$ in $\\frac{5 \\pi}{3}$ dobimo $a_1=2, a_2=-2, a_3=-1, a_4=-2, a_5=2, a_6=1$. Če je $x=\\frac{2 \\pi}{3}$ ali $x=\\frac{4 \\pi}{3}$, pa je prvih šest členov zaporedja enakih $-2,-2,1,-2,-2,1$.\n\nRešitve naloge so realna števila $0, \\frac{\\pi}{3}, \\frac{2 \\pi}{3}, \\pi, \\frac{4 \\pi}{3}$ in $\\frac{5 \\pi}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41392,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nČe med števki dvomestnega števila vrinemo ničlo, dobimo devetkrat večje število. Zapiši vsa takšna dvomestna števila.",
    "options": [],
    "answer": "45",
    "solution": "Solution:\n\nNaj bosta $a$ in $b$ števki iskanega dvomestnega števila. Nastavimo enačbo $100a + b = 9(10a + b)$. Odpravimo oklepaj in dobimo zvezo $4b = 5a$. Upoštevamo, da sta $a$ in $b$ števki, kar pomeni, da je edina možna rešitev $a = 4$ in $b = 5$. Iskano število je $45$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41393,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real $x$ from the interval $[0, 2\\pi)$ such that\n$$\n27 \\cdot 3^3 \\sin x = 9 \\cos^2 x.\n$$",
    "options": [],
    "answer": "3π/2, 7π/6, 11π/6",
    "solution": "First, rewrite the equation as $3^3 \\cdot 3^{3\\sin x} = (3^2)^{\\cos^2 x}$ and then\n$$\n3^{3+3\\sin x} = 3^{2\\cos^2 x}.\n$$\nThis implies $\\log_3(3^{3+3\\sin x}) = \\log_3(3^{2\\cos^2 x})$ or $3 + 3\\sin x = 2\\cos^2 x$. Since $\\cos^2 x = 1 - \\sin^2 x$ we have\n$$\n1 + 3\\sin x + 2\\sin^2 x = 0\n$$\nand\n$$\n(1 + \\sin x)(1 + 2\\sin x) = 0.\n$$\nThis implies that $\\sin x = -1$ or $\\sin x = -1/2$. The only real number $x$ from the interval $[0, 2\\pi)$ satisfying the first condition is $x = \\frac{3\\pi}{2}$. There are two real numbers $x \\in [0, 2\\pi)$, such that $\\sin x = -1/2$, namely $x = \\frac{7\\pi}{6}$ and $x = \\frac{11\\pi}{6}$.\n\nThe solutions are $x = \\frac{3\\pi}{2}$, $x = \\frac{7\\pi}{6}$ and $x = \\frac{11\\pi}{6}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41394,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$, $q$ and $r$ such that $15p + 7pq + qr = pqr$.",
    "options": [],
    "answer": "(p, q, r) = (2, 2, 29), (13, 3, 13), (11, 5, 11)",
    "solution": "The identities $qr = pqr - 15p - 7pq = p(qr - 15 - 7q)$ imply that $qr$ is divisible by $p$. Since $p$, $q$ and $r$ are prime numbers, there are only two possibilities: $p = q$ or $p = r$.\n\nIf $p = q$ we get $15 + 7q + r = qr$, which can be rewritten as\n$$\n22 = qr - 7q - r + 7 = (q - 1)(r - 7).\n$$\nExactly one of the numbers $q-1$ and $r-7$ is odd, so one of the primes $q$ and $r$ is even. The only possible case is $q = 2$ and from this we get $r = 29$.\n\nIf $p = r$, then we can divide both sides by $r$ to get $15 + 8q = qr$ or $15 = q(r - 8)$. Now, $q$ is a prime and divides $15$. We conclude that $q$ is equal to either $3$ or $5$. If $q = 3$ then $r = 13$, if $q = 5$ then $r = 11$.\n\nThe solutions are $(p, q, r) = (2, 2, 29)$, $(13, 3, 13)$ and $(11, 5, 11)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41395,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which positive integers $n$ does there exist a multiple of $7$, such that the sum of its digits is equal to $n$?",
    "options": [],
    "answer": "All positive integers greater than 1",
    "solution": "Any number with the sum of the digits equal to $1$ is a power of $10$, so it cannot be a multiple of $7$. Let us try and find a multiple of $7$ such that the sum of its digits will be equal to $2$. This number must have two digits equal to $1$. We check the first few positive integers with this property. The numbers $11$, $101$ and $110$ are not divisible by $7$, but $1001$ is.\n\nThe positive integer $10011001\\ldots1001$, made by $k$ repetitions of the number $1001$, has the sum of the digits equal to $2k$ and is obviously a multiple of $1001$. Hence, it is also a multiple of $7$. We conclude that for all even positive integers $n$ there exists an integer $m$, which is a multiple of $7$ with the sum of its digits equal to $n$.\n\nIt is easy to find a multiple of $7$ with the sum of the digits equal to $3$. The number $21$ has these properties. Any number of the form $2110011001\\ldots1001$, which consists of $21$ followed by $k$ repetitions of $1001$, has the sum of the digits equal to $3 + 2k$ and is a multiple of $7$. So, for all odd positive integers $n > 1$ there exists an integer $m$, which is a multiple of $7$, such that the sum of the digits of $m$ is equal to $n$.\n\nWe have shown that all positive integers $n$ except $1$ have the required property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41396,
    "subject": "Mathematics (Multi-modal)",
    "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": "Detailed 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$$\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": 41397,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEF$ be a cyclic hexagon, such that $AD$ is the diameter of its circumcircle and the lines $BF$ and $CE$ are parallel. Let $K$ be the intersection of the segments $AC$ and $BD$, let $L$ be the intersection of the segments $AE$ and $DF$ and let $M$ be the midpoint of the segment $KL$. Prove that $|MC| = |ME|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The segments $BF$ and $CE$ are parallel, so the quadrilateral $FBCE$ is an isosceles trapezoid and $|BC| = |EF|$. The inscribed angles over the chords $BC$ and $EF$ are equal, $\\angle EDF = \\angle BDC$. Let $\\angle EDF = \\alpha$. The segment $AD$ is the diameter of a circle, so by Thales' theorem we have $\\angle AED = \\frac{\\pi}{2} = \\angle DCA$. From here we get $\\angle ELD = \\frac{\\pi}{2} - \\angle EDL = \\frac{\\pi}{2} - \\alpha$, so $\\angle ELF = \\frac{\\pi}{2} + \\alpha$. Similarly, $\\angle DKC = \\frac{\\pi}{2} - \\angle KDC = \\frac{\\pi}{2} - \\alpha$, so $\\angle CKB = \\frac{\\pi}{2} + \\alpha$.\n\n![](attached_image_1.png)\n\nWe have shown that $\\angle ELF = \\angle CKB$ and $|BC| = |EF|$. The angles $\\angle ELF$ and $\\angle CKB$ over the two chords $EF$ and $BC$ of equal length are equal. If we somehow map the triangle $ELF$ so that $E$ gets mapped to $C$, $F$ gets mapped to $B$ and the image of $L$ lies on the same side of the line $BC$ as $K$, then we will get a cyclic quadrilateral. The mapping with these properties is the reflection over the line of symmetry of the trapezoid $FBCE$. Denote this line by $\\ell$.\n\nLet $K'$ be the image of the point $K$ and let $L'$ be the image of the point $L$ under the reflection over $\\ell$. Obviously, $\\ell$ is the line of symmetry of the quadrilateral $LL'KK'$. This quadrilateral is an isosceles trapezoid (because $LL'$ and $KK'$ are parallel). Its diagonals $KL$ and $K'L'$ intersect on $\\ell$. We wish to show that the intersection of the diagonals is also the midpoint of each of them, which means we wish to show that $LL'KK'$ is a rectangle. It suffices to show that the segment $L'K$ is parallel to $\\ell$.\n\nLet $A'$ be the image of $A$ and let $D'$ be the image of $D$ under the reflection over $\\ell$. The point $L'$ is the intersection of the lines $BD'$ and $A'C$. Since $A'BCD$ is a cyclic quadrilateral, we have $\\angle A'DB = \\angle A'CB$. The points $K, L', B$ and $C$ are concyclic, so $\\angle L'KB = \\angle L'CB = \\angle A'CB$. Hence, $\\angle A'DB = \\angle L'KB$ and the line $KL'$ is parallel to the line $A'D$. The quadrilateral $AA'DD'$ is a rectangle (the reflection makes it so that the segment $AA'$ is parallel to the segment $DD'$, but $AD$ and $A'D'$ are the diameters, so $\\angle AD'D = \\pi/2$ and $\\angle A'DD' = \\pi/2$). So, the segment $A'D$ is parallel to $\\ell$, which implies that $KL' \\parallel \\ell$. So, $LL'KK'$ is a rectangle and its diagonals bisect each other. Their intersection lies on $\\ell$. Since $M$ lies on $\\ell$ the distances of $M$ to the points $C$ and $E$ are the same.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41398,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which positive integers $n$ does there exist a multiple of 13, such that the sum of its digits is equal to $n$?",
    "options": [],
    "answer": "All positive integers except 1",
    "solution": "Any number with the sum of the digits equal to $1$ is a power of $10$, so it cannot be a multiple of $13$. Let us try and find a multiple of $13$ such that the sum of its digits will be equal to $2$. This number must have two digits equal to $1$ and the remaining digits must be $0$. We check the first few positive integers with this property. The numbers $11$, $101$ and $110$ are not divisible by $13$, but $1001$ is.\n\nSimilarly, let us try and find a positive integer $m$ with the sum of the digits equal to $3$. Let $m = 10^a + 10^b + 10^c$ where $a \\ge b \\ge c$. The remainder of $10^2$ when divided by $13$ is $9$. The remainder of $10^3$ when divided by $13$ is $12$. For $10^4$ we get $3$, for $10^5$ we get $4$ and for $10^6$ we get $1$. The sum of the remainders given by $10^2$, $10^4$ and $10^6$ is $13$, so $13$ divides $101010$.\n\nAny number of the form $1001\\,1001\\ldots1001$ with $k$ repetitions of $1001$ is a multiple of $1001$, so it is also a multiple of $13$. The sum of its digits is equal to $2k$. Any number of the form $101010\\,1001\\ldots1001$ where $101010$ is followed by $k$ repetitions of $1001$ is also a multiple of $13$ and has the sum of the digits equal to $3 + 2k$.\n\nWe have shown that all positive integers $n$ except $1$ have the required property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41399,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $K_1$ and $K_2$ be the circles centered at $O_1$ and $O_2$, respectively, meeting at the points $A$ and $B$. Let $p$ be the line through the point $A$ meeting the circles $K_1$ and $K_2$ again at $C_1$ and $C_2$. Assume that $A$ lies between $C_1$ and $C_2$. Denote the intersection of the lines $C_1O_1$ and $C_2O_2$ by $D$. Prove that the points $C_1, C_2, B$ and $D$ lie on the same circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Write $\\angle ABC_1 = \\alpha$ and $\\angle C_2BA = \\beta$. Then $\\angle C_2BC_1 = \\alpha + \\beta$. The points $C_1, C_2, B$ and $D$ are concyclic if and only if $\\angle C_2BC_1 = \\angle C_2DC_1$. Let us show that $\\angle C_2DC_1 = \\alpha + \\beta$.\n\nThe central angle equals twice the inscribed angle, so $\\angle AO_1C_1 = 2\\angle ABC_1 = 2\\alpha$ and $\\angle C_2O_2A = 2\\angle C_2BA = 2\\beta$. The triangle $AO_1C_1$ is isosceles with the apex at $O_1$, so $\\angle O_1C_1A = \\frac{\\pi}{2} - \\alpha$. The triangle $C_2O_2A$ is isosceles with the apex at $O_2$, so $\\angle AC_2O_2 = \\frac{\\pi}{2} - \\beta$. We get\n$$\n\\begin{aligned}\n\\angle C_2DC_1 &= \\pi - \\angle C_1C_2D - \\angle DC_1C_2 \\\\\n&= \\pi - \\left(\\frac{\\pi}{2} - \\alpha\\right) - \\left(\\frac{\\pi}{2} - \\beta\\right) = \\alpha + \\beta,\n\\end{aligned}\n$$\nso $C_1, C_2, B$ and $D$ lie on the same circle.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41400,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be non-zero digits. Let $p$ be a prime number which divides the three-digit numbers $abc$ and $cba$. Show that $p$ divides at least one of the numbers $a+b+c$, $a-b+c$ and $a-c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the prime number $p$ divides the numbers $\\overline{abc} = 100a + 10b + c$ and $\\overline{cba} = 100c + 10b + a$, it must also divide their difference\n$$\n\\overline{abc} - \\overline{cba} = 100(a-c) + (c-a) = 99(a-c).\n$$\nIf $p$ divides $a-c$ we are done. If not, then $p$ divides $99$, so $p=3$ or $p=11$.\n\nIf $p=3$ then $\\overline{abc}$ is divisible by $3$. A number is divisible by $3$ if and only if the sum of its digits is divisible by $3$, so $3$ divides $a+b+c$. Hence, $p$ divides $a+b+c$.\n\nIf $p=11$ then $11$ divides\n$$\n\\overline{abc} = 100a + 10b + c = 99a + 11b + a - b + c = 11(9a + b) + (a - b + c),\n$$\nwhich implies that $a-b+c$ is divisible by $11$. In this case $p$ divides $a-b+c$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41401,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which there exists a multiple of $11$ with the sum of its digits equal to $n$.",
    "options": [],
    "answer": "All even positive integers and all odd integers at least 11; no multiple of 11 has digit sum 1, 3, 5, 7, or 9.",
    "solution": "Denote the sum of the digits of a positive integer $m$ by $S(m)$. Then $S(11) = 2$. We notice that for the first few multiples of $11$ the sum of the digits is always even. The first multiple for which the sum of the digits is odd is $209 = 19 \\cdot 11$ and this sum is $11$. Using these two examples we can construct multiples of $11$ with the sum of the digits equal to $n$ for almost all positive integers $n$.\n\nThe number of the form $11\\dots11$ with $11$ repeated $k$ times in a row is divisible by $11$ and the sum of the digits is $2k$. Hence, for all even $n$ there exists a multiple of $11$ with the sum of its digits equal to $n$.\n\nAs we have just seen we can get the sum of $11$ from $209$. Any number of the form $20911\\dots11$ with $209$ followed by $k$ repetitions of $11$ is divisible by $11$ and the sum of its digits is $11 + 2k$. So, for all odd integers $n$ greater than or equal to $11$, there exists a multiple of $11$ with the sum of the digits equal to $n$.\n\nThe only positive integers still in question are $1$, $3$, $5$, $7$ and $9$. Let us show that in these cases we cannot find multiples of $11$ with the required sum of digits.\n\nLet $m$ be a multiple of $11$. Assume that $S(m) \\le 9$. Let $L$ denote the sum of the digits of $m$ in the odd positions and let $D$ be the sum of the digits in the even positions. Since $S(m) = L + D$, we have $L, D \\le 9$. The criterion for divisibility by $11$ implies that $L - D$ is divisible by $11$, but on the other hand we have $-9 \\le -D \\le L - D \\le L \\le 9$. So, $L - D = 0$ or $L = D$. We conclude that $S(m) = L + D = 2D$ is even and no multiple of $11$ can have the sum of its digits equal to $1$, $3$, $5$, $7$ or $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41402,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa katera naravna števila $n$ obstaja večkratnik števila 11, ki ima vsoto števk enako $n$?",
    "options": [],
    "answer": "All natural numbers except 1, 3, 5, 7, and 9",
    "solution": "Solution:\n\nOznačimo vsoto števk naravnega števila $m$ z $S(m)$. Velja $S(11)=2$. Opazimo, da ima prvih nekaj večkratnikov sodo vsoto števk. Prvi večkratnik, ki ima liho vsoto števk, je $209=19 \\cdot 11$, vsota števk je 11. S pomočjo teh dveh primerov lahko konstruiramo večkratnike z vsoto števk $n$ za skoraj vsa naravna števila $n$. Število oblike $11\\ldots 11$, kjer smo 11 zapisali $k$-krat zapored, je deljivo z 11 in ima vsoto števk enako $2k$. Torej za vsako sodo število $n$ obstaja večkratnik 11, ki ima vsoto števk enako $n$.\n\nVsoto števk 11 ima število 209. Število oblike 20911...11, kjer 209 sledi $k$-krat zapisano 11, je deljivo z 11 in ima vsoto števk enako $11+2k$. Zato za vsako liho število $n$, ki je večje ali enako 11, obstaja večkratnik 11 z vsoto števk $n$.\n\nKonstruirali smo večkratnike za vsa naravna števila, razen $1,3,5,7$ in $9$. Pokažimo, da večkratniki z vsoto števk $1,3,5,7$ ali $9$ ne obstajajo.\n\nNaj bo $m$ večkratnik števila 11. Denimo, da je $S(m) \\leq 9$. Označimo z $L$ vsoto števk števila $m$ na lihih mestih, z $D$ pa vsoto števk na sodih mestih. Ker je $S(m)=L+D$, sledi $L, D \\leq 9$. Po kriteriju za deljivost z 11 sledi, da je $L-D$ deljivo z 11. Hkrati pa je $-9 \\leq -D \\leq L-D \\leq L \\leq 9$, torej mora biti $L-D=0$ oziroma $L=D$. Tedaj je $S(m)=L+D=2D$, torej sodo število, s čimer smo pokazali, da večkratnik števila 11 ne more imeti vsote števk enake $1,3,5,7$ in $9$.\n\nVečkratnik števila 11, ki ima vsoto števk enako $n$, obstaja za vsa naravna števila $n$ razen $1,3,5,7$ in $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41403,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa neničelna realna števila $x$, $y$ in $z$ velja $3x + 2y = z$ in $\\frac{3}{x} + \\frac{1}{y} = \\frac{2}{z}$. Dokaži, da je vrednost izraza $5x^2 - 4y^2 - z^2$ vedno celo število.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n1. način\nV drugo enačbo vstavimo $z = 3x + 2y$ in odpravimo ulomke. Dobimo $x^2 + 3xy + 2y^2 = 0$ oziroma $(x + y)(x + 2y) = 0$. Od tod sledi $x = -y$ ali $x = -2y$.\n\nV prvem primeru je $x = -y$ in $z = -y$, zato je $5x^2 - 4y^2 - z^2 = 5y^2 - 4y^2 - y^2 = 0$.\n\nV drugem primeru je $x = -2y$ in $z = -4y$, zato je $5x^2 - 4y^2 - z^2 = 20y^2 - 4y^2 - 16y^2 = 0$.\n\n2. način\nIz druge enačbe sledi $2xy = 3yz + xz$. Če prvo enačbo pomnožimo z $z$, $x$ in $y$, lahko izrazimo $z^2 = 3xz + 2zy$, $3x^2 = zx - 2xy$ in $2y^2 = zy - 3xy$. Zato je\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": 41404,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa katera naravna števila $n$ obstaja večkratnik števila 13, ki ima vsoto števk enako $n$?",
    "options": [],
    "answer": "All natural numbers except 1",
    "solution": "Solution:\n\nKer so vsa števila, katerih vsota števk je 1, potence števila 10, ki niso deljiva z 13, ne obstaja večkratnik števila 13, ki bi imel vsoto števk enako 1.\n\nPoskusimo najti število $m$, deljivo s 13, katerega vsota števk je 2. Očitno bo imelo število $m$ dve števki enaki 1, ostale pa 0. Preverimo po vrsti prvih nekaj takih naravnih števil. Očitno 11, 101 in 110 niso večkratniki 13, 1001 pa je.\n\nPodobno poskusimo najti število $m$ z vsoto števk 3. Naj bo $m=10^{a}+10^{b}+10^{c}$, kjer je $a \\geq b \\geq c$. Ostanek števila $10^{2}$ pri deljenju s 13 je 9 in ostanek $10^{3}$ pri deljenju s 13 je 12, ostanek $10^{4}$ je 3, ostanek $10^{5}$ je 4 in ostanek $10^{6}$ je 1. Vidimo, da je vsota ostankov, ki jih dajo števila $10^{2}, 10^{4}$ in $10^{6}$, enaka 13, torej 13 deli 101010.\n\nŠtevilo oblike 10011001...1001, kjer smo $k$-krat zapisali 1001, je večkratnik 1001 in zato tudi 13, ter ima vsoto števk $2k$. Število oblike $10101010011001...1001$, kjer smo najprej zapisali 101010 in nato $k$-krat 1001, je prav tako večkratnik 13 in ima vsoto števk enako $3+2k$.\n\nPokazali smo, da za natanko vsa naravna števila $n$ razen 1 obstaja večkratnik števila 13, katerega vsota števk je enaka $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41405,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZapiši definicijsko območje funkcije $f(x) = \\log_{\\frac{1}{2}} \\left( \\frac{x}{x+2} - \\frac{1}{x} \\right) - \\frac{1}{2}$.",
    "options": [],
    "answer": "(-∞, -2) ∪ (-1, 0) ∪ (2, ∞)",
    "solution": "Solution:\n\nUgotovimo, da mora biti logaritmand $\\frac{x}{x+2} - \\frac{1}{x}$ večji od $0$. Zapišemo neenačbo $\\frac{x}{x+2} - \\frac{1}{x} > 0$. Neenačbo uredimo $\\frac{x^2 - x - 2}{x(x+2)} > 0$. Določimo ničle $x_1 = 2$ in $x_2 = -1$ in pole $x_1 = 0$ in $x_2 = -2$. Na številski premici določimo predznake in odčitamo rešitev $x \\in (-\\infty, -2) \\cup (-1, 0) \\cup (2, \\infty)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41406,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNičla linearne funkcije je $2$, začetna vrednost pa $\\frac{5}{2}$. Zapiši enačbo premice, ki je vzporedna grafu dane funkcije in seka os $x$ pri $\\frac{4}{3}$, v implicitni obliki.",
    "options": [],
    "answer": "15x + 12y - 20 = 0",
    "solution": "Solution:\n\nUpoštevamo, da je $n=\\frac{5}{2}$ in $f(2)=0$, kar vstavimo v predpis za linearno funkcijo $f(x)=k \\cdot x+n$. Dobimo $k \\cdot 2+\\frac{5}{2}=0$ in izračunamo $k=-\\frac{5}{4}$. Upoštevamo lastnost vzporednih premic, torej da imata enak smerni koeficient. Enačba premice je $y=-\\frac{5}{4} x+n_{1}$. Upoštevamo, da leži na tej premici točka $M\\left(\\frac{4}{3}, 0\\right)$ in izračunamo $n_{1}=\\frac{5}{3}$. Enačbo premice $y=-\\frac{5}{4} x+\\frac{5}{3}$ zapišemo še v implicitni obliki $15 x+12 y-20=0$ ali njej ekvivalentni.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41407,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa nogometni tekmi vratar brcne žogo. Pot žoge je opisana s funkcijo\n$$\nh(x) = -0,0126 x^{2} + 0,635 x\n$$\nkjer je $h$ višina žoge nad zemljo in $x$ vodoravna oddaljenost od mesta udarca (količini sta izraženi v merskih številih).\n\n![](attached_image_1.png)\n\na) Na kolikšni višini je žoga, ko je njena vodoravna oddaljenost od mesta udarca $15~\\mathrm{m}$?\nb) Koliko metrov od mesta udarca pade na zemljo?\nc) Kolikšna je največja višina, ki jo doseže žoga?\n\n$$\nh(x) = -0,0126 x^{2} + 0,635 x\n$$\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "a) 6.69 m; b) 50.4 m; c) 8 m",
    "solution": "Solution:\n\na) Vstavimo vrednost $x = 15$ in izračunamo:\n$$\nh(15) = -0,0126 \\cdot 15^{2} + 0,635 \\cdot 15 \\doteq 6,69~\\mathrm{m}\n$$\n\nb) Nastavimo enačbo $h(x) = 0$. Izračunamo rešitvi $x = 0$ in $x = 50,4~\\mathrm{m}$ ter izločimo $x = 0$.\n\nc) Ugotovimo, da je najvišja višina žoge enaka ordinati temena parabole. Izračunamo\n$$\np = -\\frac{b}{2a} = 25,2\n$$\nter nato izračunamo še\n$$\nq = h(p) = -0,0126 \\cdot 25,2^{2} + 0,635 \\cdot 25,2 = 8\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41408,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPrijatelja Miha in Blaž za nedeljsko potepanje najameta vsak svoje motorno kolo pri različnih ponudnikih. Miha mora plačati na začetku 100 evrov, ko pa motorno kolo vrne, še 4 evre za vsak prevoženi kilometer. Blaž na začetku plača 200 evrov, potem pa 3 evre za vsak prevoženi kilometer. Najmanj koliko kilometrov morata prijatelja prevoziti, da bo Miha plačal več kot Blaž?",
    "options": [],
    "answer": "101",
    "solution": "Solution:\n\nNaj bo $x$ število prevoženih kilometrov in $m(x)$ ter $b(x)$ zneska, ki sta odvisna od $x$. Znesek za Mihovo kolo je $m(x) = 4x + 100$, znesek za Blaževo kolo pa $b(x) = 3x + 200$. Miha bo plačal več kot Blaž, ko bo veljalo $4x + 100 > 3x + 200$. Rešimo neenačbo in dobimo rešitev $x > 100$. Prevoziti morata najmanj 101 kilometer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41409,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa katera naravna števila $n$ obstaja večkratnik števila 7, ki ima vsoto števk enako $n$?",
    "options": [],
    "answer": "all natural numbers except 1",
    "solution": "Solution:\n\nŠtevilo, katerega vsota števk je 1, je potenca števila 10 in ni večkratnik 7. Zato ne obstaja tak večkratnik števila 7, da bi bila vsota števk enaka 1.\n\nPoskusimo najti tak večkratnik, da bo vsota števk enaka 2. To število mora imeti dve števki enaki 1. Preverimo po vrsti nekaj takih naravnih števil. Števila 11, 101 in 110 niso deljiva s 7, 1001 pa je.\n\nNaravno število oblike $10011001 \\ldots 1001$, v katerem smo $k$-krat zapored zapisali število 1001, ima vsoto števk enako $2k$ in je očitno večkratnik 1001, torej je večkratnik števila 7. Zato za vsa soda naravna števila $n$ obstaja število $m$, ki je večkratnik 7, da je vsota števk števila $m$ enaka $n$.\n\nVsoto števk enako 3 ima število 21, število oblike $2110011001 \\ldots 1001$, kjer smo za 21 zapisali $k$-krat zapored število 1001, ima vsoto števk enako $3+2k$ in je večkratnik števila 7. Zato za vsa liha naravna števila $n$ večja od 1 obstaja število $m$, ki je večkratnik 7, da je vsota števk števila $m$ enaka $n$.\n\nPokazali smo, da za natanko vsa naravna števila $n$ razen 1 obstaja večkratnik števila 7, katerega vsota števk je $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41410,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute triangle with $|AB| > |AC|$. Let $D$ be a point on the side $AB$, such that the angles $\\angle ACD$ and $\\angle CBD$ are equal. Let $E$ denote the midpoint of $BD$, and let $S$ be the circumcentre of the triangle $BCD$. Prove that the points $A, E, S$ and $C$ lie on the same circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The central angle is twice the inscribed angle, so $\\angle CSD = 2\\angle CBD$. The triangle $CSD$ is isosceles with the apex at $S$, so\n$$\n\\angle DCS = \\frac{\\pi-\\angle CSD}{2} = \\frac{\\pi}{2} - \\angle CBD.\n$$\nThus,\n$$\n\\angle ACS = \\angle ACD + \\angle DCS = \\angle CBD + \\frac{\\pi}{2} - \\angle CBD = \\frac{\\pi}{2}.\n$$\nNow, because $E$ is the midpoint of $BD$, the line $SE$ is perpendicular to the line $BD$ and $\\angle SEA = \\frac{\\pi}{2}$. This implies $\\angle ACS + \\angle SEA = \\pi$, so the points $A, E, S$ and $C$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41411,
    "subject": "Mathematics (Multi-modal)",
    "question": "For what positive integers $n \\ge 3$ does there exist a convex $n$-gon which can be divided into finitely many parallelograms?",
    "options": [],
    "answer": "n is even",
    "solution": "We will show that an $n$-gon with this property exists for $n$ even, but not for $n$ odd.\nAssume that a convex $n$-gon can be divided into finitely many parallelograms. Denote one of the sides by $a$. There exists a parallelogram with one side lying on $a$. Denote the opposite side of this parallelogram by $b_1$. If $b_1$ does not lie on some side of the $n$-gon, then there exists a parallelogram that shares a segment with $b_1$. Denote the opposite side of this parallelogram by $b_2$. Repeat.\n![](attached_image_1.png)\nThe parallelograms obtained in this way are all distinct. Since there are only finitely many parallelograms altogether, we eventually (after a finite number of steps) get to a side $b_k$, which lies on some side $c$ of the $n$-gon.\nSince the segments $b_i$ and $b_{i+1}$ are parallel for all $i$, $a$ is parallel to $b_1$ and $c$ is parallel to $b_k$, we conclude that $a$ and $c$ are also parallel. Obviously, $c \\neq a$. We have shown that for each side of our $n$-gon we can find another side parallel to the first.\nSince the $n$-gon is convex no three of its sides are parallel. Indeed, denote the vertices of the $n$-gon by $A_1, A_2, \\dots, A_n$. Because of the convexity we have\n$$\n\\begin{aligned}\n0 < & \\angle(\\overrightarrow{A_1A_2}, \\overrightarrow{A_2A_3}) < \\angle(\\overrightarrow{A_1A_2}, \\overrightarrow{A_3A_4}) < \\\\\n& \\vdots \\\\\n< & \\angle(\\overrightarrow{A_1A_2}, \\overrightarrow{A_{n-1}A_n}) < \\angle(\\overrightarrow{A_1A_2}, \\overrightarrow{A_nA_1}) < 2\\pi\n\\end{aligned}\n$$\n(the angles are measured in the positive direction from the first vector to the second). Hence, there is only one number $i$ such that $\\angle(\\overrightarrow{A_1A_2}, \\overrightarrow{A_iA_{i+1}}) = \\pi$. We conclude that the segment $A_iA_{i+1}$ is parallel to $A_1A_2$. This $n$-gon has pairs of parallel sides. In particular, the number of the sides is even, so $n$ is even. We have hereby shown that an $n$-gon with the required property does not exist for $n$ odd.\n\nNow, let us prove by induction that for all even $n \\ge 3$ every convex $n$-gon consisting of pairs of parallel segments of equal length can be divided into finitely many parallelograms. If $n = 4$, then this 4-gon is a parallelogram and\nsuch a splitting exists. Now, assume that we already have the division of an $n$-gon into finitely many parallelograms for some even $n$.\nConsider a $(n+2)$-gon consisting of pairs of parallel segments of equal length. Denote its vertices by $A_1, A_2, \\dots, A_{n+2}$. Assume that $A_1A_2$ and $A_k, A_{k+1}$ are parallel and of equal length. Let $\\tau$ be the translation by $\\overrightarrow{A_2A_1}$. Denote $A'_i = \\tau(A_i)$ for all $2 \\le i \\le k$. We have $A'_2 = A_1$ and $A'_k = A_{k+1}$.\nSince $\\tau$ is a translation, the quadrilateral $A'_iA_iA_{i+1}A'_{i+1}$ is a parallelogram for all $i = 2, 3, \\dots, k-1$. At the same\n![](attached_image_2.png)\ntime, $A'_2A'_3\\dots A'_kA_{k+2}A_{k+3}\\dots A_{n+2}$ is a convex $n$-gon consisting only of pairs of parallel sides of equal length. By the induction hypothesis it can be divided into finitely many parallelograms. Hence, the $(n+2)$-gon can also be divided into finitely many parallelograms.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41412,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $H$ be the orthocentre of the acute triangle $ABC$ and let $D$ be a point inside the triangle $ABH$. A line that passes through the point $D$ and is parallel to the line $AH$ intersects the segments $BC$ and $AB$ at $K$ and $L$. A line that passes through the point $D$ and is parallel to the line $BH$ intersects the segments $AC$ and $AB$ at $M$ and $N$. Prove that $C, D$ and $H$ lie on the same line if $K, L, M$ and $N$ lie on the same circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the angle $\\angle ABC$ by $\\beta$. The line $KL$ is parallel to the altitude to the side $BC$, so $KL$ is perpendicular to $BC$. Hence, $KLB$ is a right triangle and $\\angle KLB = \\frac{\\pi}{2} - \\beta$. We get $\\angle KLN = \\frac{\\pi}{2} - \\beta$. Since $KLMN$ is a cyclic quadrilateral, we have $\\angle KMN = \\angle KLN = \\frac{\\pi}{2} - \\beta$.\n\n![](attached_image_1.png)\n\nIn the quadrilateral $KDMC$ we $A$ have $\\angle DKC + \\angle DMC = \\frac{\\pi}{2} + \\frac{\\pi}{2} = \\pi$, so $KDMC$ is also a cyclic quadrilateral. We conclude that $\\angle DCK = \\angle DMK = \\angle NMK = \\frac{\\pi}{2} - \\beta$.\n\nLet $E$ be the point where the line $CD$ intersects the segment $AB$. As we have shown we have $\\angle ECB = \\frac{\\pi}{2} - \\beta$ and $\\angle EBC = \\beta$, so $\\angle BEC = \\frac{\\pi}{2}$. The line through the points $C$ and $D$ is perpendicular to the segment $AB$, so it must also contain the point $H$. Thus, $C$, $D$ and $H$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41413,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle. A line parallel to $BC$ intersects the sides $AB$ and $AC$ at $D$ and $E$. The circumcircle of the triangle $ADE$ intersects the segment $CD$ at $F$, $F \\neq D$. Prove that the triangles $AFE$ and $CBD$ are similar.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The lines $DE$ and $BC$ are parallel, so $\\angle DCB = \\angle CDE$. The inscribed angles over the chord $EF$ in the cyclic quadrilateral $ADFE$ are equal, $\\angle FDE = \\angle FAE$. This implies\n$$\n\\angle DCB = \\angle CDE = \\angle FDE = \\angle FAE.\n$$\nThe lines $DE$ and $BC$ are parallel, so $\\angle ABC = \\angle ADE$. Since the points $A, D, E$ and $F$ are concyclic, we have $\\angle ADE = \\angle AFE$, and so $\\angle DBC = \\angle EFA$.\nThe triangles $AFE$ and $CBD$ have two angles in common, $\\angle AFE = \\angle DBC$ and $\\angle FAE = \\angle DCB$, hence they are similar.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41414,
    "subject": "Mathematics (Multi-modal)",
    "question": "The diagonal $AC$ of a convex quadrilateral $ABCD$ is the bisector of the angle $\\angle DCB$. Let $E$ be the intersection of the side $AB$ and the circumcircle of the triangle $ACD$. Let $F$ be the intersection of the side $AD$ and the circumcircle of the triangle $ABC$. Prove that the segments $AC, DE$ and $BF$ intersect in a point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $G$ be the intersection of the segments $AC$ and $BF$. The points $A, B, C$ and $F$ lie on the same circle, so $\\angle AFB = \\angle ACB$. The diagonal $AC$ bisects the angle $\\angle DCB$, so $\\angle ACB = \\angle DCA$. This implies $\\angle AFG = \\angle DCA$ and $\\angle GFD + \\angle DCG = \\pi$. We conclude that the points $C, D, F$ and $G$ lie on the same circle.\n\n![](attached_image_1.png)\n\nThe points $A, E, C$ and $D$ are concyclic, so $\\angle AEC = \\pi - \\angle ADC$ and $\\angle CEB = \\pi - \\angle AEC = \\angle ADC$. Since $CDFG$ is a cyclic quadrilateral, we have $\\angle FGC = \\pi - \\angle FDC$ and $\\angle FGA = \\angle FDC$. So, $\\angle BGC = \\angle FGA = \\angle FDC$. We have shown that $\\angle CEB = \\angle ADC$ and $\\angle BGC = \\angle ADC$, so $\\angle CEB = \\angle BGC$. This implies that $BCGE$ is a cyclic quadrilateral. So, $\\angle BEG = \\pi - \\angle ACB$, or $\\angle GEA = \\angle ACB$. At the same time we have $\\angle ACB = \\angle ACD$ because $AC$ is the bisector of the angle $\\angle DCB$. Since $AECD$ is a cyclic quadrilateral, we have $\\angle ACD = \\angle AED$. We conclude that $\\angle GEA = \\angle DEA$ and the points $E, G$ and $D$ are collinear. Hence, $AC, BF$ and $DE$ meet at a point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41415,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles triangle with the apex at $C$. Let $D$ and $E$ be two points on the sides $AC$ and $BC$, such that the angle bisectors $\\angle DEB$ and $\\angle ADE$ meet at $F$, which lies on the segment $AB$. Prove that $F$ is the midpoint of $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\angle BAC = \\alpha$, $\\angle ADF = \\varphi$ and $\\angle FEB = \\psi$. The triangle $ABC$ is isosceles with the apex at $C$, so $\\angle CBA = \\angle BAC = \\alpha$. The segments $DF$ and $EF$ bisect the angles $\\angle ADE$ and $\\angle DEB$, so $\\angle FDE = \\varphi$ and $\\angle DEF = \\psi$.\n\nThe sum of the inner angles of a quadrilateral is equal to $360^\\circ$. Hence, for the quadrilateral $ABED$ we have\n$$\n360^\\circ = \\angle BAD + \\angle ADE + \\angle DEB + \\angle EBA \\\\ = \\alpha + 2\\varphi + 2\\psi + \\alpha.\n$$\nwhich implies $\\alpha + \\varphi + \\psi = 180^\\circ$.\n\nThe sum of the inner angles of a triangle is $180^\\circ$, so $\\angle DFA = 180^\\circ - \\alpha - \\varphi = \\psi$, $\\angle BFE = 180^\\circ - \\alpha - \\psi = \\varphi$ and $\\angle DFE = 180^\\circ - \\varphi - \\psi = \\alpha$. The triangles $AFD$, $FED$ and $BEF$ have all three inner angles in common, so they are similar. This implies that\n![](attached_image_1.png)\n$$\n\\frac{|AF|}{|FD|} = \\frac{|FE|}{|ED|} \\quad \\text{and} \\quad \\frac{|FD|}{|ED|} = \\frac{|BF|}{|EF|}.\n$$\nor\n$$\n|AF| = \\frac{|FE| \\cdot |FD|}{|ED|} = |EF| \\cdot \\frac{|FD|}{|ED|} = |BF|.\n$$\nWe have shown that $F$ is the midpoint of the segment $AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41416,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcentre of the acute triangle $ABC$ and denote the circumcircle by $\\kappa$. The bisector of the inner angle at $A$ meets $\\kappa$ again at $D$. The bisector of the inner angle at $B$ meets $\\kappa$ again at $E$. Let $I$ denote the incentre of the triangle $ABC$. How much does the angle $\\angle ACB$ measure if the points $D, E, O$ and $I$ lie on the same circle?",
    "options": [],
    "answer": "pi/3",
    "solution": "The triangle $ABC$ is acute, so the points $I$ and $O$ lie on the same side of the line $ED$. The condition that the points $D$, $E$, $I$ and $O$ lie on the same circle therefore implies that $\\angle DOE = \\angle DIE$.\n\nLet us denote the angles of the triangle by $\\alpha$, $\\beta$ and $\\gamma$ and let us express the angles $\\angle DOE$ and $\\angle DIE$ in these terms.\n\nWe have $\\angle DOE = \\angle EOC + \\angle COD$. Since the central angle is always twice the inscribed angle, we get $\\angle COE = 2\\angle CBE$. The line $EB$ is the bisector of the angle $CBA$, so $\\angle CBE = \\frac{\\beta}{2}$. Hence, $\\angle COE = 2\\angle CBE = \\beta$.\n\nSimilarly, we have $\\angle DOC = 2\\angle DAC = 2 \\cdot \\frac{\\alpha}{2} = \\alpha$ and so $\\angle DOE = \\alpha + \\beta$.\n\nAlso, $\\angle DIE = \\angle AIB = \\pi - \\angle BAI - \\angle IBA = \\pi - \\frac{\\alpha}{2} - \\frac{\\beta}{2}$.\n\nThe identity $\\angle DOE = \\angle DIE$ implies that $\\pi = \\frac{3}{2}(\\alpha + \\beta)$ or $\\alpha + \\beta = \\frac{2\\pi}{3}$.\n\nWe conclude that $\\gamma = \\pi - \\alpha - \\beta = \\frac{\\pi}{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41417,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$, such that\n$$ (x - 2)f(y) + f(y + 2f(x)) = f(x + yf(x)) $$\nfor all real $x$ and $y$.",
    "options": [],
    "answer": "f(x) = 0 for all x, and f(x) = x - 1",
    "solution": "First, assume that $f(0) = 0$. Inserting $x = 0$ into the functional equation we get $f(y) = 0$ for all $y \\in \\mathbb{R}$. This function satisfies the equation.\n\nNow, let $f(0) \\neq 0$. Inserting $y = 0$ into the equation we get\n$$\n(x - 2)f(0) + f(2f(x)) = f(x)\n$$\nfor all $x \\in \\mathbb{R}$. Obviously, this means that $f$ is injective.\n\nNow, put $x = 2$ into the equation to get\n$$\nf(y + 2f(2)) = f(2 + yf(2)) \\quad \\text{for all } y \\in \\mathbb{R}.\n$$\nSince $f$ is injective, we have $y + 2f(2) = 2 + yf(2)$ for all $y \\in \\mathbb{R}$. If we insert $y = 0$, then we get $f(2) = 1$.\n\nSince $f(2) = 1$ and $f$ is injective, we have $f(3) \\neq 1$. Insert $x = 3$ and $y = \\frac{3}{1-f(3)}$ into the equation to get\n$$\nf\\left(\\frac{3}{1-f(3)} + 2f(3)\\right) = 0.\n$$\nWe have shown that $f$ has a zero. Let $a$ be a zero of $f$, $f(a) = 0$. Inserting $y = a$ into the initial equation yields\n$$\nf(a + 2f(x)) = f(x + af(x)) \\quad \\text{for all } x \\in \\mathbb{R}.\n$$\nSince $f$ is injective, we have $a + 2f(x) = x + af(x)$ for all $x \\in \\mathbb{R}$. Since $a \\neq 2$ (remember that $f(2) = 1 \\neq 0$), we get $f(x) = \\frac{x-a}{2-a}$.\n\nFinally, inserting $f(x) = \\frac{x-a}{2-a}$ into the equation we get $a = 1$. The only two solutions to this functional equation are the functions $f(x) = 0$ and $f(x) = x - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41418,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a rectangle with $|AB| > |BC|$. The bisector of the diagonal $AC$ meets the side $CD$ at $E$. The circle with the centre at $E$ and the radius $AE$ meets the segment $AB$ again at $F$. Let $G$ be the orthogonal projection of the point $C$ to the line $EF$. Show that $G$ lies on the diagonal $BD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\angle FAE = \\alpha$. Since the point $E$ lies on the bisector of the segment $AC$, its distances to the points $A$ and $C$ are the same. Hence, $E$ is the centre of the circle containing the points $A$, $C$ and $F$ and we have $|AE| = |CE| = |FE|$. So, $\\angle EFA = \\angle FAE = \\alpha$. Since $AB$ and $CD$ are parallel we get $\\angle DEA = \\angle EAF = \\alpha$ and $\\angle CEF = \\angle EFA = \\alpha$.\n\n![](attached_image_1.png)\n\nThe right triangles $AED$ and $CEG$ are congruent because they have three common angles and the hypothenuses have the same length.\nSo, $|ED| = |EG|$ and $|CG| = |AD| = |BC|$. From here we can conclude that the triangle $DEG$ is isosceles and\n$$\n\\angle EGD = \\frac{\\pi - \\angle DEG}{2} = \\frac{\\angle GEC}{2} = \\frac{\\alpha}{2}.\n$$\n\nBy Pythagoras' theorem we have $|FB|^2 = |FC|^2 - |BC|^2 = |FC|^2 - |GC|^2 = |FG|^2$, so $|FB| = |FG|$. Therefore, $GFB$ is an equilateral triangle and\n$$\n\\angle FGB = \\frac{\\pi - \\angle GFB}{2} = \\frac{\\angle GFA}{2} = \\frac{\\alpha}{2}.\n$$\nWe have shown that $\\angle FGB = \\angle EGD$, so the points $B$, $G$ and $D$ are congruent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41419,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDiagonala $AC$ konveksnega štirikotnika $ABCD$ je simetrala kota $\\angle DCB$. Naj bo $E$ presečišče stranice $AB$ in trikotniku $ACD$ očrtane krožnice in naj bo $F$ presečišče stranice $AD$ in trikotniku $ABC$ očrtane krožnice. Dokaži, da se daljice $AC$, $DE$ in $BF$ sekajo v eni točki.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNaj bo $G$ presečišče daljic $AC$ in $BF$. Točke $A, B, C$ in $F$ ležijo na isti krožnici, zato je $\\angle AFB = \\angle ACB$. Diagonala $AC$ razpolavlja kot $\\angle DCB$, zato je $\\angle ACB = \\angle DCA$. Torej velja $\\angle AFG = \\angle DCA$, od koder sledi $\\angle GFD + \\angle DCG = \\pi$. Zato točke $C$, $D$, $F$ in $G$ ležijo na isti krožnici.\n\nTočke $A, E, C$ in $D$ ležijo na isti krožnici, zato je $\\angle AEC = \\pi - \\angle ADC$. Od tod sledi $\\angle CEB = \\pi - \\angle AEC = \\angle ADC$. Ker je štirikotnik $CDFG$ tetiven, velja $\\angle FGC = \\pi - \\angle FDC$ in zato $\\angle FGA = \\angle FDC$. Torej velja $\\angle BGC = \\angle FGA = \\angle FDC$.\n\nPokazali smo $\\angle CEB = \\angle ADC$ in $\\angle BGC = \\angle ADC$, torej $\\angle CEB = \\angle BGC$. Od tod sledi, da je štirikotnik $BCGE$ tetiven. Zato je $\\angle BEG = \\pi - \\angle ACB$ oziroma $\\angle GEA = \\angle ACB$. Hkrati je $\\angle ACB = \\angle ACD$, saj je\n\n![](attached_image_1.png)\n\n$AC$ simetrala kota $\\angle DCB$, iz tetivnosti štirikotnika $AECD$ pa sledi $\\angle ACD = \\angle AED$. Dobili smo torej $\\angle GEA = \\angle DEA$, od koder sledi, da so točke $E, G$ in $D$ kolinearne. To ravno pomeni, da se daljice $AC$, $BF$ in $DE$ sekajo v eni točki.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41420,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nReši enačbo $\\log_{4}\\left(1+\\log_{4}\\left(3^{x}-\\sqrt{\\left(5^{0}+4^{2}\\right)^{2}}\\right)\\right)=e^{0}$",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nUgotovimo, da je $e^{0}=1$. Poenostavimo tudi korenjenec $\\sqrt{\\left(5^{0}+4^{2}\\right)^{2}}=\\sqrt{(17)^{2}}=17$. Uporabimo zvezo $1=\\log_{4} 4$ in dobimo $1+\\log_{4}\\left(3^{x}-17\\right)=4$.\n\nUredimo $\\log_{4}\\left(3^{x}-17\\right)=3$.\n\nUporabimo definicijo logaritma $64=3^{x}-17$.\n\nPoenostavimo $3^{x}=81$.\n\nRešimo in izračunamo $x=4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41421,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $H$ višinska točka ostrokotnega trikotnika $ABC$ in $D$ točka v notranjosti trikotnika $ABH$. Vzporednica k premici $AH$ skozi točko $D$ seka stranici $BC$ in $AB$ v točkah $K$ in $L$. Vzporednica k premici $BH$ skozi točko $D$ seka daljici $AC$ in $AB$ v točkah $M$ in $N$.\n\nDokaži, da točke $C$, $D$ in $H$ ležijo na isti premici, če točke $K$, $L$, $M$ in $N$ ležijo na isti krožnici.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo kot $\\angle ABC$ z $\\beta$. Ker je premica $KL$ vzporedna z višino na stranico $BC$, je torej pravokotna na $BC$. Zato je trikotnik $KLB$ pravokotni in velja $\\angle KLB = \\frac{\\pi}{2} - \\beta$. Torej je $\\angle KLN = \\frac{\\pi}{2} - \\beta$.\n\nKer je štirikotnik $KLMN$ tetiven, velja $\\angle KMN = \\angle KLN = \\frac{\\pi}{2} - \\beta$. V štirikotniku $KDMC$ velja $\\angle DKC + \\angle DMC = \\frac{\\pi}{2} + \\frac{\\pi}{2} = \\pi$, zato je tetiven tudi štirikotnik $KDMC$. Torej je $\\angle DCK = \\angle DMK = \\angle NMK = \\frac{\\pi}{2} - \\beta$.\n\n![](attached_image_1.png)\n\nOznačimo z $E$ točko, kjer premica $CD$ seka stranico $AB$. Ker smo izračunali, da v trikotniku $CEB$ velja $\\angle ECB = \\frac{\\pi}{2} - \\beta$ ter je $\\angle EBC = \\beta$, velja $\\angle BEC = \\frac{\\pi}{2}$. Torej je premica skozi točki $C$ in $D$ pravokotna na stranico $AB$, zato na njej leži tudi točka $H$. Točke $C$, $D$ in $H$ zato ležijo na isti premici.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41422,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoenostavi izraz\n$$\n(x-1)\\left(\\left(x^{-3}+x^{-2}+x^{-1}\\right)\\left(x^{-1}-1\\right)-\\left(x^{-2}-1\\right)\\left(x^{-2}+1\\right)\\right)^{-1}\n$$",
    "options": [],
    "answer": "x",
    "solution": "Solution:\n\nPotence z negativnim eksponentom v oklepajih zapišemo z ulomki\n$$(x-1)\\left(\\left(\\frac{1}{x^{3}}+\\frac{1}{x^{2}}+\\frac{1}{x}\\right)\\left(\\frac{1}{x}-1\\right)-\\left(\\frac{1}{x^{2}}-1\\right)\\left(\\frac{1}{x^{2}}+1\\right)\\right)^{-1}.$$\n\nVsote oziroma razlike razširimo na skupni imenovalec\n$$(x-1)\\left(\\left(\\frac{1+x+x^{2}}{x^{3}}\\right)\\left(\\frac{1-x}{x}\\right)-\\left(\\frac{1-x^{2}}{x^{2}}\\right)\\left(\\frac{1+x^{2}}{x^{2}}\\right)\\right)^{-1},$$\nopravimo množenje\n$$(x-1)\\left(\\frac{1+x+x^{2}-x-x^{2}-x^{3}}{x^{4}}-\\frac{1-x^{4}}{x^{4}}\\right)^{-1}.$$\n\nDobljena ulomka v oklepaju odštejemo\n$$(x-1)\\left(\\frac{x^{4}-x^{3}}{x^{4}}\\right)^{-1}.$$\n\nZapišemo obratno vrednost\n$$(x-1)\\left(\\frac{x^{4}}{x^{3}(x-1)}\\right),$$\nokrajšamo in dobimo rezultat $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41423,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSimetrala diagonale $AC$ pravokotnika $ABCD$, v katerem je $|AB| > |BC|$, seka stranico $CD$ v točki $E$. Krožnica s središčem $E$ in polmerom $AE$ seka stranico $AB$ še v točki $F$. Naj bo $G$ pravokotna projekcija točke $C$ na premico $EF$. Pokaži, da točka $G$ leži na diagonali $BD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo $\\angle FAE = \\alpha$. Ker točka $E$ leži na simetrali daljice $AC$, je enako oddaljena od $A$ in $C$, zato je središče krožnice, na kateri ležijo točke $A$, $C$ in $F$. Tako velja $|AE| = |CE| = |FE|$. Zato je $\\angle EFA = \\angle FAE = \\alpha$. Zaradi vzporednosti $AB$ in $CD$ sledi še $\\angle DEA = \\angle EAF = \\alpha$ in $\\angle CEF = \\angle EFA = \\alpha$.\n\nPravokotna trikotnika $AED$ in $CEG$ se ujemata v kotih in imata enako dolgi hipotenuzi, zato sta skladna. Tako je $|ED| = |EG|$ in $|CG| = |AD| = |BC|$. Od tod sledi, da je\n\n![](attached_image_1.png)\n\ntrikotnik $DEG$ enakokrak in je\n$$\n\\angle EGD = \\frac{\\pi - \\angle DEG}{2} = \\frac{\\angle GEC}{2} = \\frac{\\alpha}{2}\n$$\nPo Pitagorovem izreku velja $|FB|^2 = |FC|^2 - |BC|^2 = |FC|^2 - |GC|^2 = |FG|^2$, torej je $|FB| = |FG|$. Trikotnik $GFB$ je zato enakokrak in tako velja\n$$\n\\angle FGB = \\frac{\\pi - \\angle GFB}{2} = \\frac{\\angle GFA}{2} = \\frac{\\alpha}{2}\n$$\nTorej je $\\angle FGB = \\angle EGD$, zato točke $B$, $G$ in $D$ ležijo na isti premici.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41424,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDružina je naredila sneženega moža iz treh delov, ki so imeli obliko krogle. Polmeri teh krogel so tvorili geometrijsko zaporedje. Polmer najmajše krogle na vrhu snežaka je bil $8 \\mathrm{dm}$, polmer največje krogle pa $18 \\mathrm{dm}$. Koliko kubičnih metrov snega je bilo v tem sneženem možu?",
    "options": [],
    "answer": "33.8119 m^3",
    "solution": "Solution:\n\nUpoštevamo lastnosti geometrijskega zaporedja $8 \\cdot q = x$ in $x \\cdot q = 18$. Zapišemo zvezo $8 q^{2} = 18$ in izračunamo količnik $q = \\frac{3}{2}$. Nato izračunamo polmer srednje krogle $r_{2} = 12 \\mathrm{dm}$. Izračunamo prostornino snežaka\n\n$$\nV = \\frac{4}{3} \\pi \\left(8^{3} + 12^{3} + 18^{3}\\right) \\mathrm{dm}^{3}\n$$\n\nkar je $33,8119~\\mathrm{m}^{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41425,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSokol leti nad travnikom s konstantno hitrostjo naravnost in na stalni višini $222~\\mathrm{m}$. Mirujoča miška je zagledala sokola, ko je bil kot med zemljo in sokolom $16^{\\circ}$. Minuto kasneje je bil kot med zemljo in sokolom $75^{\\circ}$. Podatki so prikazani na skici. Izračunaj dolžino poti, ki jo je sokol preletel v eni minuti.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "approximately 714.7 m",
    "solution": "Solution:\n\nSkico dopolnimo tako, da vrišemo navpičnico, da dobimo pravokotni trikotnik.\n\n![](attached_image_2.png)\n\nUporabimo kotne funkcije za ostra kota $16^{\\circ}$ in $75^{\\circ}$ in sicer $\\tan 16^{\\circ} = \\frac{|P_1 S_1|}{|M P_1|}$ ter $\\tan 75^{\\circ} = \\frac{|P_1 S_2|}{|M P_2|}$. Iz prve zveze izračunamo $|M P_1| = \\frac{222}{\\tan 16^{\\circ}} \\doteq 774,2~\\mathrm{m}$, iz druge pa $|M P_2| = \\frac{222}{\\tan 75^{\\circ}} \\doteq 59,5~\\mathrm{m}$. Glede na oznake na skici, upoštevamo da je $|S_1 S_2| = |P_1 P_2| = |M P_1| - |M P_2| \\doteq 714,7~\\mathrm{m}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41426,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $O$ središče ostrokotnemu trikotniku $ABC$ očrtane krožnice $\\mathcal{K}$. Simetrala notranjega kota pri $A$ seka krožnico $\\mathcal{K}$ še v točki $D$, simetrala notranjega kota pri $B$ pa seka krožnico $\\mathcal{K}$ še v točki $E$. Označimo z $I$ središče trikotniku $ABC$ včrtane krožnice. Kolikšna je velikost kota $\\angle ACB$, če točke $D, E, O$ in $I$ ležijo na isti krožnici?",
    "options": [],
    "answer": "pi/3",
    "solution": "Solution:\n\nKer je trikotnik $ABC$ ostrokotni, ležita točki $I$ in $O$ na istem bregu premice $ED$. Iz pogoja, da ležijo točke $D, E, I$ in $O$ na isti krožnici, zato sledi $\\angle DOE = \\angle DIE$. Označimo kote v trikotniku z $\\alpha, \\beta$ in $\\gamma$ in z njimi izrazimo kota $\\angle DOE$ in $\\angle DIE$.\n\nVelja $\\angle DOE = \\angle EOC + \\angle COD$. Ker je središčni kot dvakrat večji od obodnega, je $\\angle COE = 2 \\angle CBE$. Premica $EB$ je simetrala kota $CBA$, zato je $\\angle CBE = \\frac{\\angle CBA}{2} = \\frac{\\beta}{2}$. Torej je $\\angle COE = 2 \\angle CBE = \\beta$. Podobno je $\\angle DOC = 2 \\angle DAC = 2 \\cdot \\frac{\\alpha}{2} = \\alpha$ in tako $\\angle DOE = \\alpha + \\beta$.\n\nVelja $\\angle DIE = \\angle AIB = \\pi - \\angle BAI - \\angle IBA =$\n\n![](attached_image_1.png)\n\n$\\pi - \\frac{\\alpha}{2} - \\frac{\\beta}{2}$. Iz enakosti $\\angle DOE = \\angle DIE$ sledi $\\pi = \\frac{3}{2}(\\alpha + \\beta)$ oziroma $\\alpha + \\beta = \\frac{2\\pi}{3}$. Torej je $\\gamma = \\pi - \\alpha - \\beta = \\frac{\\pi}{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41427,
    "subject": "Mathematics (Multi-modal)",
    "question": "A teacher gives Matej four pieces of paper, each containing a non-zero digit. Matej arranges them in a line thus forming a four-digit number. He then exchanges two of the pieces (without flipping or rotating them in the process) to form another four-digit number. Can he always do this in such a way that the two numbers he obtains are not relatively prime regardless of the digits that were written on the pieces of paper?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The answer is yes. If one of the digits is even, then Matej can form two even numbers by putting the piece of paper with the even digit in the place of units and exchanging two of the other three pieces of paper. If one of the four digits is equal to $5$, then Matej can act similarly. He should put the piece of paper with the digit $5$ in the place of units and again exchange two of the other three pieces of paper. The resulting two numbers are not relatively prime since they are both divisible by $5$. If two of the digits are the same, then Matej can make an arbitrary number and then simply exchange the two pieces with the same digits to get the same number.\n\nThe only remaining case is where the digits are $1$, $3$, $7$ and $9$. With these, Matej can form either the numbers $1397$ and $1793$ or the numbers $1397$ and $9317$ or the numbers $9317$ and $9713$. All of these are divisible by $11$. He can also form the numbers $1739$ and $9731$, which are both divisible by $37$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41428,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all non-zero real numbers $x$ such that\n$$\n\\min \\left\\{ 4, x + \\frac{4}{x} \\right\\} \\ge 8 \\min \\left\\{ x, \\frac{1}{x} \\right\\}.\n$$",
    "options": [],
    "answer": "(-∞, 0) ∪ (0, 1/2] ∪ [2, ∞)",
    "solution": "Let us check when $4 \\le x + \\frac{4}{x}$. If $x > 0$, then this inequality is equivalent to $4x \\le x^2 + 4$ or $0 \\le (x-2)^2$. This is always true. If $x < 0$, then we get $0 \\ge (x-2)^2$. This is never the case, so $4 \\ge x + \\frac{4}{x}$. Hence,\n$$\n\\min \\left\\{ 4, x + \\frac{4}{x} \\right\\} = \\begin{cases} 4, & \\text{if } x > 0, \\\\ x + \\frac{4}{x}, & \\text{if } x < 0. \\end{cases}\n$$\nFor $x > \\frac{1}{x}$ and $x > 0$ we get $x^2 > 1$, so $x > 1$. For $x > \\frac{1}{x}$ and $x < 0$ we get $x^2 < 1$, so $-1 < x < 0$. Hence,\n$$\n\\min \\left\\{ x, \\frac{1}{x} \\right\\} = \\begin{cases} \\frac{1}{x}, & \\text{if } x > 1 \\text{ or } -1 < x < 0, \\\\ x, & \\text{if } x \\le -1 \\text{ or } 0 < x \\le 1. \\end{cases}\n$$\nIn the case of $x > 1$ the given inequality is equivalent to $4 \\ge \\frac{8}{x}$, which implies $x \\ge 2$. The inequality holds for all numbers $x \\in [2, \\infty)$. If $0 < x \\le 1$, then we have $4 \\ge 8x$, so $\\frac{1}{2} \\ge x$. Thus, the inequality also holds for $x \\in (0, \\frac{1}{2}]$.\n\nLet $-1 < x < 0$. We get $x + \\frac{4}{x} \\ge \\frac{8}{x}$, or, equivalently, $x \\ge \\frac{4}{x}$. Multiplying by $x$ yields $x^2 \\le 4$, and this holds for $-1 < x < 0$. The inequality holds for all $x \\in (-1, 0)$.\n\nFinally, let $x \\le -1$. We get $x + \\frac{4}{x} \\ge 8x$ or $4 \\le 7x^2$. Since $x \\le -1$, we have $x^2 \\ge 1$, so $7x^2 \\ge 7 > 4$. The inequality holds for all $x \\in (-\\infty, -1]$ as well.\n\nWe have shown that the given inequality holds for all $x$ in $(-\\infty, 0) \\cup (0, \\frac{1}{2}] \\cup [2, \\infty)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41429,
    "subject": "Mathematics (Multi-modal)",
    "question": "For real numbers $a$, $b$ and $c$ we have\n$$\n(2b - a)^2 + (2b - c)^2 = 2(2b^2 - ac).\n$$\nProve that the numbers $a$, $b$ and $c$ are three consecutive terms in some arithmetic sequence.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The given equation is equivalent to $8b^2 - 4ab - 4bc + a^2 + c^2 = 4b^2 - 2ac$ or\n$$\n4b^2 - 4ab - 4bc + a^2 + 2ac + c^2 = 0.\n$$\nThis can be further rewritten as $(a + c)^2 - 4b(a + c) + 4b^2 = 0$ and finally as\n$$\n(a + c - 2b)^2 = 0.\n$$\nFrom here we conclude that $a + c = 2b$ or, equivalently, $b - a = c - b$. Hence, $a$, $b$ and $c$ are consecutive terms of an arithmetic sequence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41430,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each point on the sides of the triangle $\\mathcal{T}$ is either red or blue. Prove that one can find points $A$, $B$, $C$ and $D$ on the sides of the triangle $\\mathcal{T}$, such that all four points have the same colour and the quadrilateral $ABCD$ is a trapezoid.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the vertices of the triangle by $A$, $B$ and $C$ and let $A'$, $B'$ and $C'$ be the midpoints of the sides $BC$, $CA$ and $AB$. At least two of the points $A'$, $B'$ and $C'$ are the same colour. We may assume that the points $A'$ and $B'$ are both red. The segment $A'B'$ is parallel to the segment $AB$. If we can find two red points on the segment $AB$ then we have found the trapezoid we wanted. Otherwise at most\n![](attached_image_1.png)\none of the points on the segment is red. If there exists a red point distinct from $A$ and $B$, then denote it by $D$. If there is none, then let $D$ be an arbitrary point on the segment $AB$ distinct from $A$ and $B$. Hence, all the points on the segment $AB$ with the exception of $A$, $B$ and possibly $D$, are blue.\n\nAssume that there exist two distinct blue points on the segment $AC$ also both distinct from $A$. Denote them by $E$ and $F$, so that $E$ is the one closer to $A$. Let $F'$ be a blue point on the segment $AD$ distinct from $A$. Let $E'$ be the intersection of the segment $AD$ and the line through $E$ parallel to the line $FF'$. Then $EE'F'F$ is a trapezoid and all of its vertices are blue.\n\nSimilarly, we show that either we can find such a trapezoid or else there is\nat most one blue point other than $B$ on the segment $BC$. In the latter case the segments $AC$ and $BC$ are almost entirely red and we can easily find a trapezoid with the required property. We can get its vertices as the intersections of two parallel lines with the segments $AC$ and $BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41431,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each point on the segment $\\overline{D}$ is either red or blue. Show that there exist three different points $A$, $B$ and $C$ of the same colour such that $|AB| = |BC|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us divide the segment into three parts of equal length. We can find two points of the same colour in the middle part. We may assume that these two points are red. Let us call them $A$ and $B$. Let $A'$ be the image of the point $A$ under reflection over the point $B$ and let $B'$ be the image of the point $B$ under reflection over the point $A$.\n\n![](attached_image_1.png)\n\nIf one of the points $A'$ and $B'$ is red, then we have found the three points we were looking for. Now, let us assume that both $A'$ and $B'$ are blue. Let $C$ be the midpoint of the segment $AB$. Then $C$ is also the midpoint of the segment $A'B'$. If $C$ is red, then $A$, $B$ and $C$ are the three points we were searching for, if not, then $A'$, $B'$ and $C$ are the ones we seek.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41432,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$ and $z$ be real numbers such that $0 \\le x, y, z \\le 1$. Prove that\n$$\nxyz + (1-x)(1-y)(1-z) \\le 1.\n$$\nWhen does the equality hold?",
    "options": [],
    "answer": "Equality holds if and only if x=y=z=0 or x=y=z=1.",
    "solution": "The inequality is equivalent to $xy + yz + zx \\le x + y + z$. Since $x$ is positive and $y \\le 1$ we have $xy \\le x$. A similar reasoning shows that $yz \\le y$ and $zx \\le z$. Hence, the inequality holds.\n\nThe equality holds if and only if $xy = x$, $yz = y$ and $zx = z$. If $x=0$, then the third equality implies $z=0$ and the second equality then implies $y=0$. Otherwise, we must have $y=1$ and then the second equation implies $z=1$ and from the third equation we get $x=1$. Hence, the equality holds when $x=y=z=0$ or $x=y=z=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41433,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest three-digit number such that the following holds: if the order of the digits of this number is reversed and the number obtained by this is added to the original number the resulting number consists of only odd digits.",
    "options": [],
    "answer": "209",
    "solution": "Write the three-digit number as $abc$. The number we obtain by reversing the order of the digits is $cba$. The sum $x$ of these two numbers is\n$$\nx = \\overline{abc} + \\overline{cba} = (a + c)10^2 + (b + b)10 + (c + a).\n$$\nSince all the digits of $x$ are odd and $2b$ is even, we conclude that $c + a$ is equal to at least 10. The number $c + a$ cannot be equal to 10 because the units of $x$ are odd and equal to the units of $c + a$. Hence, $c + a$ is at least 11 and $a$ is at least 2. The number $abc$ will be the smallest when we choose the smallest possible $a$, i.e. $a = 2$. In this case $c = 9$ and the smallest possible value of $b$ is\n\nWe see that the number $209 + 902 = 1111$ consists of only odd digits. Thus, $209$ is the smallest number with the desired properties.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41434,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. There are $2n + 2$ points in a plane such that no three of the points lie on the same line. A line in the plane is a *divider*, if two of the given points lie on this line and there are exactly $n$ points on each of the two sides. Find the greatest possible $m$ for which there are always at least $m$ dividers in the plane.",
    "options": [],
    "answer": "n + 1",
    "solution": "We will show that there exist at least $n + 1$ divisor lines. First, let us give an example where there are exactly $n + 1$ of them. For that the given points should be the vertices of a regular $(2n+2)$-gon. Denote them by $A_1, A_2, \\dots, A_{2n+2}$. For each $1 \\le i \\le n+1$ the line $A_iA_{i+n+1}$ is obviously a divider. No other line is a divider, so there are exactly $n+1$ of them.\n\nNow, let us show that we can always find at least $n+1$ dividers regardless of the positions of the given points. We will show that there is at least one divider passing through each of the points. Let $A$ be one of the points and denote the remaining points by $A_1, A_2, \\dots, A_{2n+1}$. Consider the lines $AA_1, AA_2, \\dots, AA_{2n+1}$. Let us paint the half-plane on one of the sides of the line $AA_1$ gray. Let $r_1$ be the number of points contained in the gray part (the line $AA_1$ itself does not belong to the gray part).\n\nNow, let us rotate the gray part around $A$ in the positive direction. In each step rotate it so that the image of the gray part borders on the next possible line amongst $AA_1, AA_2, \\dots, AA_{2n+1}$. We may assume that these lines are already ordered, so we get $AA_1, AA_2, \\dots, AA_{2n+1}$ and then again $AA_1$. Let $r_{i+1}$ be the number of the given points in the gray part after $i$ steps.\n\nAssume that at some point we started with the line $AA_i$ and then rotated the gray part so that it is now bordering on the line $AA_{i+1}$. We consider four possible cases for the positions of the points $A_i$ and $A_{i+1}$. If $A_{i+1}$ was initially in the gray part and $A_i$ is there after the rotation, then $r_i = r_{i+1}$. If $A_{i+1}$ was initially in the gray part and $A_i$ is not in the gray part after the rotation, then the number of points in the gray part decreased by 1, so $r_{i+1} = r_i - 1$. If $A_{i+1}$ was initially not in the gray part, then we have $r_{i+1} = r_i$ if $A_i$ is not in the gray part after the rotation. Otherwise we have $r_{i+1} = r_i + 1$.\n\nIn each step the number of points in the gray part changes by at most 1. All these numbers are integers. At the beginning there were $r_1$ points in the gray part. After $2n+1$ rotations we have the situation where the gray part is once more bordering on the line $AA_1$, but it is not on the same side as in the beginning. So, the number of points in the gray part is in the end equal to\n\n![](attached_image_1.png)\n![](attached_image_2.png)\n![](attached_image_3.png)\n\n$2n - r_1$. If $r_1 = n$, then $AA_1$ is a divider. If not, then one of the numbers $r_1$ and $2n - r_1$ is greater than $n$, and the other is smaller. After each rotation of the gray part the numbers $r_i$ change by 1, so after a number of steps we have $r_i = n$. The line $AA_i$ is then a divider.\n\nAs promised, we have shown that for each of the given points there is a divider passing through this point. There are $2n + 2$ points altogether, so we have thus obtained $2n + 2$ lines. Each line was counted twice. Hence, there are at least $\\frac{2n+2}{2} = n + 1$ dividers.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41435,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPrezračevalne naprave v lokalu čistijo zrak. Pretok zraka v odvisnosti od časa se za prvo napravo spreminja po formuli $f(t)=2^{t}$, za drugo pa $f(t)=2^{t+3}$. V lokalu imajo štiri naprave prvega tipa in eno drugega tipa. $S$ koliko napravami za prezračevanje s pretokom $f(t)=2^{t+2}$ bi lahko zamenjali obstoječe?",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nNastavimo enačbe $4 \\cdot 2^{t} + 2^{t+3} = a \\cdot 2^{t+2}$. Enačbo uredimo, tako da izpostavimo skupni faktor $2^{t}(4 + 8 - 4a) = 0$. Ugotovimo, da je $2^{t} \\neq 0$. Tako je $4 + 8 - 4a = 0$. Izračunamo $a = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41436,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDano je naravno število $n$. V ravnini leži $2n+2$ točk, izmed katerih nobene tri ne ležijo na isti premici. Premica v ravnini je ločnica, če na njej ležita dve izmed danih točk, na vsakem bregu te premice pa je natanko $n$ točk. Določi največje število $m$, za katerega velja, da je v ravnini vedno vsaj $m$ ločnic.",
    "options": [],
    "answer": "n+1",
    "solution": "Solution:\n\nPokazali bomo, da je najmanj $n+1$ premic, ki potekajo skozi dve izmed danih točk, ločnic. Najprej podajmo primer, v katerem je ločnic točno $n+1$. Naj bodo točke oglišča pravilnega $(2n+2)$-kotnika. Označimo jih z $A_{1}, A_{2}, \\ldots, A_{2n+2}$. Za vsak $1 \\leq i \\leq n+1$ je očitno premica $A_{i}A_{i+n+1}$ ločnica. Nobena druga premica ni ločnica, zato je ločnic natanko $n+1$.\n\nPokažimo še, da vedno obstaja vsaj $n+1$ ločnic ne glede na položaj točk. Pokazali bomo, da poteka skozi vsako izmed danih točk vsaj ena ločnica. Naj bo $A$ ena izmed točk in $A_{1}, A_{2}, \\ldots, A_{2n+1}$ preostale točke. Opazujmo premice $AA_{1}, AA_{2}, \\ldots, AA_{2n+1}$. Obarvajmo del ravnine, ki leži na enem bregu premice $AA_{1}$, rdeče. Naj bo $r_{1}$ število točk v rdečem delu (premica $AA_{1}$ ni v rdečem delu). Rdeči del ravnine bomo vrteli okoli točke $A$ v pozitivni smeri. V vsakem koraku zavrtimo tako, da slika rdečega dela meji na naslednjo možno premico izmed $AA_{1}, AA_{2}, \\ldots, AA_{2n+1}$. Predpostavimo lahko, da si premice sledijo po vrsti $AA_{1}, AA_{2}, \\ldots, AA_{2n+1}$ in nato ponovno $AA_{1}$. Naj bo $r_{i+1}$ število točk v rdečem delu ravnine po tem, ko smo rdeči del zavrteli $i$-krat.\n\n![](attached_image_1.png)\n\nDenimo, da smo v nekem koraku imeli premico $AA_{i}$ in pripadajoči rdeči del ravnine ter smo nato rdeči del zavrteli tako, da je mejil na premico $AA_{i+1}$. Ločimo štiri primere glede na položaj točk $A_{i}$ in $A_{i+1}$. Če je točka $A_{i+1}$ ležala v prvotnem rdečem delu in $A_{i}$ leži v zavrtenem rdečem delu, velja $r_{i}=r_{i+1}$. Če je točka\n\n![](attached_image_2.png)\n\n$A_{i+1}$ ležala v prvotnem rdečem delu, točka $A_{i}$ pa ne leži v zavrtenem rdečem delu, se je število točk v rdečem delu zmanjšalo za 1, torej $r_{i+1}=r_{i}-1$. Denimo, da $A_{i+1}$ ni ležala v prvotnem rdečem delu. Če $A_{i}$ ne leži v novem rdečem delu, ponovno velja $r_{i+1}=r_{i}$, sicer pa je $r_{i+1}=r_{i}+1$.\n\nV vsakem koraku se število rdečih točk spremeni največ za 1. Vsa števila so cela. Na začetku je bilo $r_{1}$ točk v rdečem delu, po $2n+1$ vrtenjih pa pridemo do situacije, ko rdeči del ponovno meji na premico $AA_{1}$, vendar je na nasprotnem delu premice $AA_{1}$ kot je bil na začetku. Zato je število točk v rdečem delu na koncu enako $2n-r_{1}$. Če je $r_{1}=n$, je premica $AA_{1}$ ločnica. Sicer je eno izmed števil $r_{1}$ oziroma $2n-r_{1}$ večje od $n$, drugo pa manjše. Pri vrtenju rdečega dela se števila $r_{i}$ spreminjajo za 1, kar pomeni, da obstaja korak, v katerem je $r_{i}=n$. Pripadajoča premica $AA_{i}$ je tedaj ločnica.\n\nPokazali smo, da skozi vsako točko poteka ločnica. Ker je točk $2n+2$, tako dobimo $2n+2$ premic, pri čemer smo vsako premico šteli dvakrat (po enkrat za vsako točko, ki leži na njej). Tako je ločnic vedno vsaj $\\frac{2n+2}{2}=n+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41437,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIzračunaj $\\cos (\\pi+2 x)$, če je $\\cos x=\\frac{1}{4}$.",
    "options": [],
    "answer": "7/8",
    "solution": "Solution:\n\nUporabimo zvezo za prehod na oster kot $\\cos (\\pi+2 x)=-\\cos 2 x$, nato še zvezo za dvojne kote $-\\cos 2 x=-\\cos ^{2} x+\\sin ^{2} x$. Uporabimo zvezo za $\\sin ^{2} x=1-\\cos ^{2} x$. Izraz uredimo, vstavimo $\\cos x=\\frac{1}{4}$. Dobimo $1-2 \\cos ^{2} x=1-2 \\cdot\\left(\\frac{1}{4}\\right)^{2}$. Izračunamo vrednost izraza $\\frac{7}{8}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41438,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMiro, Aleš in Lovro so trikrat igrali poker. Prvič je izgubil Miro in je zato moral plačati Alešu in Lovru, vsakemu posebej toliko denarja, kot sta ga imela na začetku igre. Drugič je izgubil Aleš, zato je prav tako moral plačati Miru in Lovru, vsakemu posebej toliko, kot sta ga trenutno imela. Tretjič je izgubil Lovro in tudi on je na enak način plačal Miru in Alešu. Po odigranih treh igrah je imel vsak 24 evrov. Kdo od njih je izgubil največ in koliko?",
    "options": [],
    "answer": "Miro, 15 euros",
    "solution": "Solution:\n\nNaj bo $x$ začetni znesek Aleša, $y$ začetni znesek Lovra in $z$ začetni znesek Mira. Tabeliramo časovni potek spreminjanja količine denarja pri posameznikih:\n\n| Igra        | Miro                | Aleš                | Lovro                |\n|:-----------:|:-------------------:|:-------------------:|:--------------------:|\n| pred igro   | $x$                 | $y$                 | $z$                  |\n| po 1. igri  | $x - y - z$         | $2y$                | $2z$                 |\n| po 2. igri  | $2x - 2y - 2z$      | $2y - (x - y - z) - 2z = -x + 3y - z$ | $2z - (x - y - z) - 2y = -x + y + 3z$ |\n| po 3. igri  | $4x - 4y - 4z$      | $-2x + 6y - 2z$     | $4x - (2x - 2y - 2z) - (-x + 3y - z) = -x - y + 7z$ |\n\nZapišemo sistem enačb:\n$$\n\\begin{gathered}\n4x - 4y - 4z = 24 \\\\\n-2x + 6y - 2z = 24 \\\\\n-x - y + 7z = 24\n\\end{gathered}\n$$\nki ga rešimo. Rešitev je $x = 39$, $y = 21$ in $z = 12$. Ugotovimo, da je največ izgubil Miro in sicer 15 evrov.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41439,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVsaka točka daljice $\\mathcal{D}$ je rdeče ali modre barve. Dokaži, da obstajajo enako obarvane različne točke $A, B$ in $C$ daljice $\\mathcal{D}$, da je $|A B|=|B C|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDaljico razdelimo na tri enako dolge dele. Na srednjem delu\n\n![](attached_image_1.png)\n\ngotovo obstajata dve točki enake barve. Denimo, da sta ti točki rdeči in ju označimo z $A$ in $B$. Naj bo $A'$ slika točke $A$ pri zrcaljenju čez točko $B$ in $B'$ slika točke $B$ pri zrcaljenju čez točko $A$. Če je ena izmed točk $A'$ in $B'$ rdeča, smo željene tri točke našli. Predpostavimo torej, da sta obe točki $A'$ in $B'$ modri. Naj bo $C$ razpolovišče daljice $A B$. Torej je $C$ tudi razpolovišče daljice $A' B'$. Če je $C$ rdeča, so $A, B$ in $C$ iskane točke, sicer pa so $A', B'$ in $C$ iskane točke.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41440,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUčiteljica je Mateju izročila štiri liste papirja, na vsakem je bila zapisana neničelna števka. Matej je liste postavil v vrsto in tako oblikoval štirimestno število. Ko je dva lista med sabo zamenjal, ne da bi ju pri tem obrnil ali zavrtel, je oblikoval še eno štirimestno število. Ali je lahko oblikoval števili, ki si nista bili tuji, ne glede na to, katere števke so bile zapisane na listih?",
    "options": [],
    "answer": "Yes",
    "solution": "Solution:\n\nOdgovor je da. Če je katera izmed števk soda, lahko Matej oblikuje dve sodi števili, saj sodo števko postavi na mesto enic, med sabo pa zamenja dva izmed ostalih listov. Če je med štirimi števkami število $5$, Matej postopa podobno. Petico postavi na mesto enic, med sabo pa nato zamenja dva izmed ostalih listov. Števili si nista tuji, saj sta obe deljivi s $5$. V kolikor sta dve zapisani števki enaki, Matej sestavi neko število, nato zamenja lista papirja z isto števko in dobi število, enako prejšnjemu.\n\nOstane še en primer in sicer, ko so na listih zapisane števke $1, 3, 7$ in $9$. S temi števkami Matej lahko oblikuje števili $1397$ in $1793$ ali števili $1397$ in $9317$ ali števili $9317$ in $9713$, saj so vsa ta števila deljiva z $11$. Lahko pa oblikuje tudi števili $1739$ in $9731$, ki sta obe deljivi s $37$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41441,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVsaka točka na stranicah trikotnika $\\mathcal{T}$ je rdeče ali modre barve. Dokaži, da na stranicah trikotnika $\\mathcal{T}$ obstajajo take enako obarvane točke $A, B, C$ in $D$, da je štirikotnik $A B C D$ trapez.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNaj bodo $A, B$ in $C$ oglišča danega trikotnika ter $A'$, $B'$ in $C'$$ razpolovišča stranic $BC$, $CA$ in $AB$. Vsaj dve izmed točk $A'$, $B'$ in $C'$ sta iste barve. Predpostavimo lahko, da sta to točki $A'$ in $B'$ ter da sta rdeče barve. Daljica $A'B'$ je vzporedna stranici $AB$. Če na stranici $AB$ obstajata dve rdeči točki, smo našli iskani trapez. Sicer obstaja največ ena rdeča točka. Če taka točka obstaja in je različna od $A$ in $B$, jo označimo z $D$. V nasprotnem primeru naj bo $D$ neka od $A$ in $B$ različna točka na stranici $AB$. Torej so vse točke na\n![](attached_image_1.png)\ndaljici $AB$, razen morda točke $A$, $B$ in $D$, modre barve.\n\nDenimo, da na stranici $AC$ obstajata dve med seboj in od $A$ različni točki modre barve. Označimo ju $E$ in $F$ tako, da je $E$ bliže $A$. Naj bo $F'$ od $A$ različna točka modre barve na daljici $AD$. Naj bo $E'$ presečišče daljice $AD$ in vzporednice $k$ premici $FF'$ skozi točko $E$. Tedaj je $EE'F'F$ iskani trapez.\n\nNa enak način sklepamo, da obstaja iskani trapez ali pa na stranici $BC$ obstaja največ ena od $B$ različna točka modre barve. V tem primeru sta obe stranici $AC$ in $BC$ skoraj celi rdeče barve in tako zlahka najdemo iskani trapez, ki ima za oglišča presečišča dveh vzporednih premic s stranicama $AC$ in $BC$. Tako smo pokazali, da iskani trapez vedno obstaja.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41442,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ten pirates find a chest filled with golden and silver coins. There are twice as many silver coins in the chest as there are golden. They divide the golden coins in such a way that the difference of the numbers of coins given to any two of the pirates is not divisible by $10$. Prove that they cannot divide the silver coins in the same way.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_1$ denote the number of golden coins given to the first pirate, $a_2$ the number of golden coins given to the second pirate, and so on. Since $10 \\nmid a_i - a_j$ for all $i \\ne j$, the numbers $a_1, a_2, \\dots, a_{10}$ must give different remainders when divided by $10$. There are only $10$ possible remainders, so these are exactly all the numbers $a_i$. Write $a_i = 10k_i + l_i$ where $l_i$ is the remainder of $a_i$ when dividing by $10$. The total number of golden coins is equal to\n$$\na_1 + a_2 + \\dots + a_{10} = 10(k_1 + \\dots + k_{10}) + (l_1 + \\dots + l_{10}).\n$$\nSince the numbers $l_1, \\dots, l_{10}$ are exactly the numbers $0, 1, 2, \\dots, 9$ (just not necessarily in this order), their sum is equal to $l_1 + l_2 + \\dots + l_{10} = 0 + 1 + 2 + \\dots + 9 = 45$. We conclude that there were $10(k_1 + \\dots + k_{10}) + 45$ golden coins in the chest.\n\nNow, let us assume that the pirates can divide the silver coins in the same way. As above we come to the conclusion that the total number of silver coins is equal to $10(m_1 + m_2 + \\dots + m_{10}) + 45$ for some integers $m_1, \\dots, m_{10}$. But the total number of silver coins is even, a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41443,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an equilateral triangle with the side of $20$ units. Vid divides this triangle into $400$ smaller equilateral triangles with the sides of $1$ unit. Eva then picks $4$ of the vertices of these smaller triangles. The vertices lie inside the triangle $ABC$ and form a parallelogram with sides parallel to the sides of the triangle $ABC$. There are exactly $46$ smaller triangles that have at least one point in common with the sides of this parallelogram. Find all possible values for the area of this parallelogram.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "3√3, 4√3, 9√3/2",
    "solution": "Denote the sides of the parallelogram by $a$ and $b$. We may assume that $a \\ge b$. The small triangles which share at least one point with the sides of the parallelogram are marked in the figure. We can find their number by subtracting $2$ triangles in the corners and the number of the triangles inside the white parallelogram from the number of all triangles in the enlarged parallelogram.\n\n![](attached_image_2.png)\n\nWhen $b \\le 2$ there is no white parallelogram. If $b = 1$, then the number of the small triangles is equal to the number of the triangles inside a parallelogram of size $(a + 2) \\times 3$, minus $2$. There are $2 \\cdot 3(a + 2)$ of the triangles in the parallelogram, so $6(a + 2) - 2 = 46$. This implies $a = 6$.\n\nIf $b = 2$, then there are $2 \\cdot (a + 2) \\cdot 4 - 2 = 46$ small triangles and this implies $a = 8$.\n\nLet $b \\ge 3$. The enlarged parallelogram contains $2(a + 2)(b + 2)$ small triangles and the white parallelogram contains $2(a - 2)(b - 2)$ small triangles. We get\n$$\n2(a + 2)(b + 2) - 2(a - 2)(b - 2) - 2 = 46,\n$$\nand this implies $a + b = 6$. Since $a \\ge b \\ge 3$, this is only possible when $a = b = 3$.\n\nLet us find the areas of these parallelograms. Since the acute angle in these parallelograms measures $60^\\circ$, the areas are equal to $\\frac{\\sqrt{3}ab}{2}$.\n\nWhen $a = 6$, $b = 1$ we get $3\\sqrt{3}$.\n\nWhen $a = b = 3$ we get $\\frac{9\\sqrt{3}}{2}$.\n\nWhen $a = 4$, $b = 2$ we get $4\\sqrt{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41444,
    "subject": "Mathematics (Multi-modal)",
    "question": "For what positive integers $n \\ge 3$ does there exist a polygon with $n$ vertices (not necessarily convex) with the property that each of its sides is parallel to another one of its sides?",
    "options": [],
    "answer": "All integers n ≥ 3 except n = 3 and n = 5",
    "solution": "If $n \\ge 3$ is even, $n = 2k$, then such a polygon exists since every regular $2k$-gon satisfies the condition.\n\nIf $n = 3$ or $n = 5$, then such a polygon does not exist. Indeed, no two sides in a triangle are parallel. If every side of a pentagon would be parallel to some other side, we could find three parallel sides and two of these should be adjacent. This is not possible.\n\nFor $n = 7$ such a polygon exists as shown by the figure.\n![](attached_image_1.png)\n\nLet us prove by induction that for odd positive integers $n$, $n \\ge 7$, a polygon with the required property exists. Assume that for some integer $k$ a $k$-gon with this property exists. Choose a vertex at which the inner angle is less than $180^\\circ$. Now, cut away a small parallelogram as shown in the figure. We obtain a $(k+2)$-gon which has the required property. A few examples are shown in the figure below.\n![](attached_image_2.png)\n![](attached_image_3.png)\n$n = 9$\n![](attached_image_4.png)\n$n = 11$\n![](attached_image_5.png)\n$n = 13$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41445,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square with the side of $20$ units. $Vid$ divides this square into $400$ unit squares. Eva then picks $4$ of the vertices of these unit squares. These vertices lie inside the square $ABCD$ and define a rectangle with the sides parallel to the sides of the square $ABCD$. There are exactly $24$ unit squares which have at least one point in common with the sides of this rectangle. Find all possible values for the area of a rectangle with these properties.",
    "options": [],
    "answer": "6, 8, 9",
    "solution": "Let $a$ and $b$ be the lengths of the sides of the rectangle. We may assume that $a \\ge b$. The unit squares that have at least one point in common with the rectangle are marked in the figure.\n![](attached_image_1.png)\nThe number of the unit squares is equal to the area of this part. We can find this area by subtracting the area of the white rectangle from the area of the rectangle with the sides of length $a+2$ and $b+2$. We consider two cases.\n\nIf $b \\le 2$ then there is no white rectangle and the number of the unit squares is equal to the area of the enlarged rectangle, i.e. $(a+2)(b+2)$. From $b \\le 2$ and $(a+2)(b+2) = 24$ we get $b=1, a=6$ and $b=2, a=4$.\n\nIf $b \\ge 3$ then the number of the unit squares which have at least one point in common with the sides of the rectangle is equal to $(a+2)(b+2)-(a-2)(b-2) = 4a+4b$. Hence, $a+b = \\frac{24}{4} = 6$. The only possibility is $a=b=3$.\n\nAll possible areas of the rectangle are $1 \\cdot 6 = 6$, $2 \\cdot 4 = 8$ and $3 \\cdot 3 = 9$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41446,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKovinsko kocko s površino $72~\\mathrm{cm}^2$ pretopimo v pravilno štiristrano piramido enake prostornine, katere dolžina osnovnega roba je enak tretjini dolžine telesne diagonale kocke. Kolikšna je višina piramide?",
    "options": [],
    "answer": "18√3 cm",
    "solution": "Solution:\n\nUporabimo obrazec za površino kocke $P = 6a^2$ in iz nje izračunamo dolžino roba $a = 2\\sqrt{3}$ $\\mathrm{cm}$. Nato izračunamo dolžino telesne diagonale kocke $D = a\\sqrt{3} = 6~\\mathrm{cm}$ ter prostornino kocke $V = a^3 = (2\\sqrt{3})^3 = 24\\sqrt{3}~\\mathrm{cm}^3$. Upoštevamo, da sta prostornini kocke in piramide enaki $24\\sqrt{3} = \\frac{2^2 \\cdot v}{3}$. Izračunamo višino piramide $v = 18\\sqrt{3}~\\mathrm{cm}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41447,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOskrbnik planinske koče je več let spremljal, koliko časa porabijo planinci za pot od vznožja v dolini do planinske koče tik pod vrhom gore. Pohodniki so sami zapisovali porabljen čas, oskrbnik pa je podatke zbral in uredil frekvenčno tabelo, kjer je zapisal relativne frekvence.\n\n| porabljen čas v minutah | relativna frekvenca (v %) |\n| :--- | :---: |\n| $90-105$ | 7,9 |\n| $105-120$ | 19,4 |\n| $120-135$ | 37,2 |\n| $135-150$ | 16,3 |\n| $150-165$ | 12,7 |\n| $165-180$ | $?$ |\n\nDoloči neznano relativno frekvenco in izračunaj povprečni čas.\n\n| porabljen čas v minutah | relativna frekvenca (v %) |\n| :--- | :---: |\n| $90-105$ | 7,9 |\n| $105-120$ | 19,4 |\n| $120-135$ | 37,2 |\n| $135-150$ | 16,3 |\n| $150-165$ | 12,7 |\n| $165-180$ | $?$ |",
    "options": [],
    "answer": "6.5% and 131.4 minutes",
    "solution": "Solution:\n\nVsota relativnih frekvenc mora biti $100\\%$. Tako izračunamo $f_{6} = 6,5$.\n\nZa grupirane podatke se aritmetična sredina izračuna z obrazcem\n$$\n\\bar{x} = \\frac{f_{1} \\cdot x_{1} + f_{2} \\cdot x_{2} + f_{3} \\cdot x_{3} + \\ldots + f_{k} \\cdot x_{k}}{N}\n$$\nVrednosti $x$ so sredine razredov in jih dobimo kot povprečje zgornje in spodnje meje razreda:\n$$\nx_{i} = \\frac{s_{i} + z_{i}}{2}\n$$\nKer za relativne frekvence velja\n$$\nf_{i}^{0} = \\frac{f_{i}}{N} \\cdot 100\n$$\nlahko aritmetično sredino izračunamo kot\n$$\n\\bar{x} = \\frac{f_{1} \\cdot x_{1}}{100} + \\frac{f_{2} \\cdot x_{2}}{100} + \\ldots\n$$\nVstavimo naše podatke in dobimo\n$$\n\\bar{x} = 131,4 \\text{ minute}\n$$\n\nIzračun sredine razredov:\n\n| | $97,5$ | $112,5$ | $127,5$ | $142,5$ | $157,5$ | $172,5$ |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41448,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV dveh sadovnjakih so prvo leto nabrali skupaj 315 ton sadja. Naslednje leto se je skupni pridelek povečal za $40\\ \\%$. V prvem sadovnjaku se je pridelek povečal za $25\\ \\%$, v drugem pa za $50\\ \\%$. Koliko ton sadja so v vsakem sadovnjaku nabrali prvo leto?",
    "options": [],
    "answer": "First orchard: 126 tons; Second orchard: 189 tons",
    "solution": "Solution:\n\nZapišemo zvezo $315 + 40\\ \\%$ od $315$ in izračunamo, da je to $441$.\n\nUpoštevamo pridelek v 1. sadovnjaku $x + \\frac{25}{100} x = x + \\frac{x}{4} = \\frac{5x}{4}$,\nter pridelek v drugem sadovnjaku $(315 - x) + \\frac{50}{100}(315 - x) = (315 - x) + 0,5(315 - x) = 1,5(315 - x)$.\n\nSeštejemo pridelek v obeh sadovnjakih:\n$$\n\\frac{5x}{4} + 1,5(315 - x) = 441\n$$\n\nIzračunamo $1,5 \\times 315 = 472,5$ in $1,5x$.\nTorej:\n$$\n\\frac{5x}{4} + 472,5 - 1,5x = 441\n$$\n$$\n\\frac{5x}{4} - 1,5x = 441 - 472,5\n$$\n$$\n\\frac{5x}{4} - \\frac{6x}{4} = -31,5\n$$\n$$\n-\\frac{x}{4} = -31,5\n$$\n$$\nx = 126\n$$\n\nV prvem sadovnjaku je zraslo $126$ ton sadja, v drugem pa $315 - 126 = 189$ ton.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41449,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $m$ and $n$, such that the sum of their greatest common divisor and their least common multiple equals $101$.",
    "options": [],
    "answer": "[(1, 100), (4, 25), (25, 4), (100, 1)]",
    "solution": "Let $d$ denote 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$$\n101 = d + d m_1 n_1 = d(1 + m_1 n_1).\n$$\nSince $1 + m_1 n_1 \\ge 2$ and $101$ is prime, we can only have $d = 1$ and $m_1 n_1 = 100$. The numbers $m = m_1$ and $n = n_1$ are coprime. Hence, the one that is divisible by $2$ must in fact be divisible by $4$. The one that is divisible by $5$ is divisible by $25$. All possible solutions are $(1, 100)$, $(4, 25)$, $(25, 4)$ and $(100, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41450,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist an integer $n$ such that all roots of the polynomial $p(x) = x^4 - 2011x^2 + n$ are integers?",
    "options": [],
    "answer": "No, such an integer n does not exist.",
    "solution": "Assume that such $n$ exists. From $x^4 - 2011x^2 + n = 0$ we deduce that\n$$\nx^2 = \\frac{2011 \\pm \\sqrt{2011^2 - 4n}}{2}.\n$$\nThis has to be an integer, so $2011^2 - 4n$ has to be a perfect square. We can write $2011^2 - 4n = m^2$ for some odd positive integer $m$ or $n = \\frac{2011^2 - m^2}{4}$. So, $x^2 = \\frac{2011 \\pm m}{2}$. The numbers $\\frac{2011+m}{2}$ and $\\frac{2011-m}{2}$ are perfect squares and their sum is 2011. Let 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": "Algebraic Expressions"
  },
  {
    "id": 41451,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $m$ and $n$, such that $\\frac{m^2 n}{m^2 + n}$ is a prime number.",
    "options": [],
    "answer": "(2,4), (2,12)",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41452,
    "subject": "Mathematics (Multi-modal)",
    "question": "The equilateral triangles $ABC$ and $DEF$ with the sides of length $1$ are divided into congruent equilateral triangles as shown in the figure. A circle is inscribed into each smaller triangle. What is the ratio of the shaded area of the triangle $ABC$ to the shaded area of the triangle $DEF$?\n![](attached_image_1.png)\n(A) $\\frac{9}{16}$ (B) $\\frac{3}{4}$ (C) $1$ (D) $\\frac{4}{3}$ (E) $\\frac{16}{9}$",
    "options": [],
    "answer": "C",
    "solution": "The proportion of the area of the triangle covered by the inscribed circle is $\\frac{\\pi\\sqrt{3}}{9}$. For each of the nine congruent triangles in the triangle $ABC$ the shaded area represents $\\frac{\\pi\\sqrt{3}}{9}$ of the total area, so this is also the proportion of the shaded area in the triangle $ABC$. The same is true for the triangle $DEF$, so both ratios are equal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41453,
    "subject": "Mathematics (Multi-modal)",
    "question": "At a volleyball tournament any two of the $n$ teams played against each other exactly once. For any two different teams $A$ and $B$ there were exactly $k$ teams that lost against both of them. Prove that $n = 4k + 3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41454,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$ and $q$, such that $2^2 + p^2 + q^2$ is also prime.",
    "options": [],
    "answer": "(p, q) = (2, 3) or (3, 2)",
    "solution": "If the pair $(p, q)$ satisfies the conditions of the problem, then so does the pair $(q, p)$. It is therefore sufficient to only consider the case where $p \\le q$. Obviously, $p = q = 2$ is not a solution. If $p$ and $q$ are both odd primes, then $2^2 + p^2 + q^2$ is an even integer greater than $2$, so it is not prime. Hence, $p = 2$.\n\nLet us figure out when the number $8+q^2$ is prime. If $q = 3$, then $8+q^2 = 17$ is prime. Else, $3$ divides $q-1$ or $q+1$, so $3$ divides $9 + (q-1)(q+1) = 8+q^2$ which is then not prime.\n\nWe conclude that $2^2 + p^2 + q^2$ is prime only if $p = 2$ and $q = 3$ or $p = 3$ and $q = 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41455,
    "subject": "Mathematics (Multi-modal)",
    "question": "A square with the sides measuring $1$ cm is divided into three triangles. Which of the following statements is certainly true?\n(A) The perimeter of one of the triangles is equal to $2 + \\sqrt{2}$ cm.\n(B) The area of one of the triangles is equal to $0.5 \\text{ cm}^2$.\n(C) Two of the triangles are right triangles.\n(D) None of the triangles are obtuse.\n(E) One of the triangles is acute.",
    "options": [],
    "answer": "B",
    "solution": "There are only three essentially different ways of dividing a square into three triangles:\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\n![](attached_image_3.png)\n\nboth cuts run from a vertex to a common point on one of the sides.\nNotice that in all three cases the area of the biggest of the three triangles is $0.5 \\text{ cm}^2$. The counterexamples for the statements A, C, D and E are ▱A, ▱B and ▱C respectively.\n\n![](attached_image_4.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41456,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three equal circles are inscribed into the equilateral triangle with the sides of length $1$ (see figure). What is the length of the radii of the circles?\n![](attached_image_1.png)\n(A) $\\frac{1}{3}$\n(B) $\\frac{1}{4}$\n(C) $\\frac{\\sqrt{3}}{4}$\n(D) $\\frac{\\sqrt{3}+1}{4}$\n(E) $\\frac{\\sqrt{3}-1}{4}$",
    "options": [],
    "answer": "E",
    "solution": "If we denote the radii by $r$, we have $r\\sqrt{3} + 2r + r\\sqrt{3} = 1$. It follows that $r = \\frac{1}{2(1+\\sqrt{3})} = \\frac{\\sqrt{3}-1}{4}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41457,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $x$, such that $9x^2 - 40x + 39$ is a power of a prime. (A positive integer $m$ is a power of a prime, if $m = p^a$ for some prime number $p$ and some non-negative integer $a$.)",
    "options": [],
    "answer": "x = -4, 1, 4, 5",
    "solution": "Let $9x^2 - 40x + 39 = p^n$ for some prime $p$ and some non-negative integer $n$. From\n$$\np^n = 9x^2 - 40x + 39 = (9x - 13)(x - 3)\n$$\nit follows that $9x - 13 = p^k$ and $x - 3 = p^l$ or $9x - 13 = -p^k$ and $x - 3 = -p^l$ for some integers $k$ and $l$, where $0 \\le l < k$ and $n = k + l$.\n\nFirst, let us solve the system of equations $9x - 13 = p^k$ and $x - 3 = p^l$. We have $9(p^l + 3) - 13 = p^k$ or $14 = p^k - 9p^l = p^l(p^{k-l} - 9)$. If $l = 0$, then $p^k = 23$, so $p = 23$, $k = 1$ and $x = 4$. Else, we have $l \\ge 1$ and $p^l = 14$, so either $p^l = 2$ and $p^{k-l} - 9 = 7$ or $p^l = 7$ and $p^{k-l} - 9 = 2$. In the first case we have $p = 2$, $l = 1$, $k = 5$ and $x = 5$. In the second case there are no solutions.\n\nNow, consider the other system of equations, $9x - 13 = -p^k$ and $x - 3 = -p^l$. Here we have $14 = p^l(9 - p^{k-l})$. The only two possibilities are $p = 2$, $l = 1$ and $p = 7$, $l = 1$ (in this case we cannot have $l = 0$). We get $x = 1$ and $x = -4$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41458,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the greatest common divisor of all even numbers of the form $n^4 - 4n^2$ where $n$ is a positive integer.",
    "options": [],
    "answer": "192",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41459,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoloči vsa cela števila $x$, za katera je število $9x^{2}-40x+39$ potenca praštevila. (Naravno število $m$ je potenca praštevila, če je $m=p^{a}$ za neko praštevilo $p$ in nenegativno celo število $a$.)",
    "options": [],
    "answer": "-4, 1, 4, 5",
    "solution": "Solution:\n\nNaj bo $9x^{2}-40x+39=p^{n}$ za neko praštevilo $p$ in nenegativno celo število $n$. Iz\n$$\np^{n}=9x^{2}-40x+39=(9x-13)(x-3)\n$$\nsledi $9x-13=p^{k}$ in $x-3=p^{l}$ ali $9x-13=-p^{k}$ in $x-3=-p^{l}$ za neki števili $k$ in $l$, kjer je $0 \\leq l<k$ in $n=k+l$.\n\nRešimo najprej sistem enačb $9x-13=p^{k}$ in $x-3=p^{l}$. Velja $9\\left(p^{l}+3\\right)-13=p^{k}$ oziroma $14=p^{k}-9p^{l}=p^{l}\\left(p^{k-l}-9\\right)$. Če je $l=0$, dobimo $p^{k}=23$, torej $p=23, k=1$ in $x=4$. Sicer je $l \\geq 1$ in $p^{l} \\mid 14$, torej je $p^{l}=2$ in $p^{k-l}-9=7$ ali $p^{l}=7$ in $p^{k-l}-9=2$. V prvem primeru dobimo $p=2, l=1, k=5$ in $x=5$, v drugem pa rešitve ni.\n\nOstane še obravnava sistema enačb $9x-13=-p^{k}$ in $x-3=-p^{l}$. Tedaj je $14=p^{l}\\left(9-p^{k-l}\\right)$. Edini možnosti sta $p=2, l=1$ ali $p=7, l=1$ (v tem primeru $l=0$ ni mogoče). Pripadajoči števili sta $x=1$ in $x=-4$.\n\nNalogi zadoščajo števila $x=-4, x=1, x=4$ in $x=5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41460,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAli obstaja celo število $n$, za katerega so vse ničle polinoma $p(x) = x^{4} - 2011 x^{2} + n$ cela števila?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nRecimo, da tako število obstaja. Iz $x^{4} - 2011 x^{2} + n = 0$ sledi\n$$\nx^{2} = \\frac{2011 \\pm \\sqrt{2011^{2} - 4 n}}{2}\n$$\nKer je to število celo, je $2011^{2} - 4 n$ popolni kvadrat. Zapišemo lahko $2011^{2} - 4 n = m^{2}$ za neko liho naravno število $m$ oziroma $n = \\frac{2011^{2} - m^{2}}{4}$. Zato je $x^{2} = \\frac{2011 \\pm m}{2}$. Števili $\\frac{2011 + m}{2}$ in $\\frac{2011 - m}{2}$ sta tako popolna kvadrata, njuna vsota je 2011. Utemeljimo, da števila 2011 ne moremo zapisati kot vsote dveh popolnih kvadratov. Ostanek popolnega kvadrata pri deljenju s 4 je bodisi 0 bodisi 1. Ostanek vsote dveh popolnih kvadratov pri deljenju s 4 je tako 0, 1 ali 2. Ker da število 2011 pri deljenju s 4 ostanek 3, ne more biti enako vsoti dveh popolnih kvadratov. Tako celo število $n$, da bi imel polinom $x^{4} - 2011 x^{2} + n$ same cele ničle, ne obstaja.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41461,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa naravna števila $m$ in $n$, za katera je vsota največjega skupnega delitelja in najmanjšega skupnega večkratnika enaka 101.",
    "options": [],
    "answer": "Ordered pairs: (1, 100), (4, 25), (25, 4), (100, 1).",
    "solution": "Solution:\n\nNaj bo $d$ največji skupni delitelj števil $m$ in $n$. Tedaj je $m = d m_{1}$ in $n = d n_{1}$, števili $m_{1}$ in $n_{1}$ pa sta si tuji. Najmanjši skupni večkratnik $m$ in $n$ je $d m_{1} n_{1}$. Velja\n$$\n101 = d + d m_{1} n_{1} = d\\left(1 + m_{1} n_{1}\\right)\n$$\nKer je $1 + m_{1} n_{1} \\geq 2$ in je 101 praštevilo, je edina možnost $d = 1$ in $m_{1} n_{1} = 100$. Števili $m = m_{1}$ in $n = n_{1}$ sta si tuji. Tisto, ki je deljivo z 2, je zato deljivo s 4. Tisto, ki je deljivo s 5, je zato deljivo s 25. Možni pari rešitev so $(1, 100), (4, 25), (25, 4)$ in $(100, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41462,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoloči vsa praštevila $p$ in $q$, za katera je tudi število $2^{2}+p^{2}+q^{2}$ praštevilo.",
    "options": [],
    "answer": "(p, q) = (2, 3) or (3, 2)",
    "solution": "Solution:\n\nČe je $(p, q)$ rešitev naloge, je tudi $(q, p)$ rešitev. Zato je dovolj obravnavati primer $p \\leq q$. Očitno $p=q=2$ ni rešitev. Če sta $p$ in $q$ lihi praštevili, je število $2^{2}+p^{2}+q^{2}$ sodo in večje od $2$, zato ni praštevilo. Torej je $p=2$.\n\nUgotovimo, kdaj je število $8+q^{2}$ praštevilo. Če je $q=3$, je $8+q^{2}=17$ praštevilo. Sicer je s $3$ deljivo eno izmed števil $q-1$ oziroma $q+1$, zato $3$ deli $9+(q-1)(q+1)=8+q^{2}$ in ni praštevilo.\n\nŠtevilo $2^{2}+p^{2}+q^{2}$ je praštevilo le, če je $p=2$ in $q=3$ ali $p=3$ in $q=2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41463,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for all real $x$ and $y$ the inequality\n$$\n|x + y| + |x + 1| + |y + 1| \\geq 2\n$$\nholds. For what $x$ does there exist $y$ such that $|x+y|+|x+1|+|y+1| = 2$?",
    "options": [],
    "answer": "x in [-1, 1]",
    "solution": "For all real $a$ we have $|a| \\ge a$ and $|a| \\ge -a$. So,\n$$\n|x + y| \\ge -(x + y), \\quad |x + 1| \\ge x + 1, \\quad |y + 1| \\ge y + 1, \\quad (1)\n$$\n\n$|x + y| + |x + 1| + |y + 1| \\geq -(x + y) + (x + 1) + (y + 1) = 2.$\n$$\nAssume that the equality holds. Then the equality case occurs in all three inequalities (1) and so $x + y \\leq 0$, $x + 1 \\geq 0$ and $y + 1 \\geq 0$. We get $-1 \\leq x \\leq -y \\leq 1$ or $-1 \\leq x \\leq 1$. If $-1 \\leq x \\leq 1$ and $y = -x$, then\n$$\n|x + y| + |x + 1| + |y + 1| = 2.\n$$\nWe can conclude that only for all real $x \\in [-1, 1]$ there exists $y$, such that the equality holds.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41464,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence of positive real numbers contains at least 5 different terms. For any two terms of this sequence we can find two more terms with the same product. At least how many terms does the progression have?",
    "options": [],
    "answer": "9",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41465,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $P$ with integer coefficients and the following property: for any integer $a$ and any prime $p$ that divides $P(a)$ the prime $p$ also divides $a$.",
    "options": [],
    "answer": "P(x) = ± x^n for any nonnegative integer n",
    "solution": "Let $P$ be a polynomial satisfying the conditions of the problem and let $p$ be an arbitrary prime. Any prime $q$ dividing $P(p)$ also divides $p$, so $q = p$. Hence, for any prime $p$ we have $P(p) = \\pm p^{m_p}$ for some non-negative integer $m_p$, which can depend on $p$.\n\nThe polynomials $P(x) = \\pm 1$ obviously satisfy the conditions of the problem. Assume that the polynomial $P$ is not identically equal to $1$ or $-1$. Write\n$$\nP(x) = a_n x^n + a_{n-1} x^{n-1} + \\dots + a_1 x + a_0.\n$$\nThe polynomial $P$ can take the value $1$ and $-1$ only for a finite number of primes. So, there exist infinitely many primes such that\n$$\n\\pm p^{m_p} = P(p) = a_n p^n + a_{n-1} p^{n-1} + \\dots + a_1 p + a_0\n$$\nand consequently $p$ divides $a_0$. This is only possible if $a_0 = 0$. Write $P(x) = x^k Q(x)$ where $k \\in \\mathbb{N}$ and $Q$ is a polynomial with integer coefficients such that $Q(0) \\neq 0$. Let $a$ be an integer and let $p$ be a prime dividing $Q(a)$. Then $p$ divides $P(a)$, so it also divides $a$. Hence, the polynomial $Q$ satisfies the conditions of the problem. Since $Q(0) \\neq 0$, we conclude that $Q$ has to be identically equal to $1$ or to $-1$. The only possible polynomials are $P(x) = \\pm x^n$, $n \\in \\mathbb{N}_0$ and all of these obviously satisfy the conditions of the problem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41466,
    "subject": "Mathematics (Multi-modal)",
    "question": "For non-zero real numbers $a$, $b$ and $c$ we have\n$$\na = b + 2c, \\quad a + c = b + d, \\quad b = d + c.\n$$\nWhich of the following equalities is certainly true?\n(A) $d = 2c$\n(B) $a = 3c$\n(C) $a = 6c$\n(D) $a = b + 2d$\n(E) $b = 2a + 2d$",
    "options": [],
    "answer": "C",
    "solution": "Since $b = a - 2c$ and $d = b - c = a - 3c$, we have $a + c = a - 2c + a - 3c$ and $a = 6c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41467,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 1$ be a positive integer. The first term of the infinite progression $(a_k)$ of positive integers is $a_1 = n$. For all $k > 1$ we have either $a_k = 2a_{k-1} + 1$ or $a_k = 2a_{k-1} - 1$. Prove that not all terms of this progression are prime numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose for contradiction that all terms of the progression $(a_k)$ are prime numbers.\n\nLet us consider the two possible recursions:\n\n- $a_k = 2a_{k-1} + 1$\n- $a_k = 2a_{k-1} - 1$\n\nLet $a_1 = n > 1$.\n\nLet us compute $a_2$:\n- $a_2 = 2n + 1$ or $a_2 = 2n - 1$\n\nLet us consider the sequence where we always choose $a_k = 2a_{k-1} + 1$.\n\nThen $a_2 = 2n + 1$, $a_3 = 2(2n + 1) + 1 = 4n + 3$, $a_4 = 2(4n + 3) + 1 = 8n + 7$, and so on.\n\nIn general, if we always choose $a_k = 2a_{k-1} + 1$, then $a_k = 2^k n + (2^k - 1)$.\n\nSimilarly, if we always choose $a_k = 2a_{k-1} - 1$, then $a_k = 2^k n - (2^k - 1)$.\n\nLet us consider $a_k$ modulo $a_1 = n$.\n\nFor the sequence $a_k = 2^k n + (2^k - 1)$:\n\n$a_k \\equiv 2^k n + (2^k - 1) \\pmod{n}$\n\n$\\equiv (2^k - 1) \\pmod{n}$\n\nSo for $k = n$, $a_n \\equiv (2^n - 1) \\pmod{n}$.\n\nIf $n$ is odd and greater than $1$, then $2^n - 1$ is even, so $a_n$ is divisible by $n$ for some $k$.\n\nBut more generally, for large enough $k$, $a_k$ will be divisible by $n$ or by some smaller $a_j$.\n\nAlternatively, note that for any $n > 1$, the sequence grows rapidly, and for large enough $k$, $a_k$ will be composite.\n\nFor example, if $n = 2$, then $a_1 = 2$.\n- $a_2 = 5$ or $3$\n- $a_3 = 11$ or $5$\n- $a_4 = 23$ or $9$\n\nIf we choose $a_2 = 3$, then $a_3 = 5$, $a_4 = 9$ (which is not prime).\n\nIf we choose $a_2 = 5$, $a_3 = 11$, $a_4 = 23$, $a_5 = 47$, $a_6 = 95$, which is not prime.\n\nTherefore, for any starting $n > 1$, not all terms of the progression can be prime numbers.\n\nThus, not all terms of this progression are prime numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41468,
    "subject": "Mathematics (Multi-modal)",
    "question": "The first term of an arithmetic progression is equal to $\\frac{1}{3}$ and the third term is equal to $\\frac{1}{5}$. What is the second term?\n(A) $\\frac{1}{4}$\n(B) $\\frac{4}{15}$\n(C) $\\frac{5}{24}$\n(D) $\\frac{7}{30}$\n(E) $\\frac{2}{9}$",
    "options": [],
    "answer": "B",
    "solution": "The second term of the arithmetic progression is equal to $a_2 = \\frac{1}{2}(a_1 + a_3) = \\frac{1}{2}\\left(\\frac{1}{3} + \\frac{1}{5}\\right) = \\frac{4}{15}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41469,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the size of the angle $x$ in degrees, if $2 \\cos 10^\\circ + \\sin 100^\\circ + \\sin 1000^\\circ + \\sin 10000^\\circ = \\sin x$ and $-90^\\circ \\le x \\le 90^\\circ$ holds?\n\n(A) -80\n(B) -10\n(C) 0\n(D) 10\n(E) 80",
    "options": [],
    "answer": "E",
    "solution": "Since $2\\cos 10^\\circ + \\sin 100^\\circ + \\sin 1000^\\circ + \\sin 10000^\\circ = 2\\cos 10^\\circ + \\sin 80^\\circ - \\sin 80^\\circ - \\sin 80^\\circ = \\sin 80^\\circ$, we have $x = 80^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41470,
    "subject": "Mathematics (Multi-modal)",
    "question": "The lengths $a$, $b$ and $c$ of the sides of the triangle $ABC$ satisfy $c^2 = 2ab$ and $a^2 + c^2 = 3b^2$. The inner angles of the triangle $ABC$ measure\n(A) $30^\\circ$, $60^\\circ$ and $90^\\circ$.\n(B) $45^\\circ$, $60^\\circ$ and $75^\\circ$.\n(C) $45^\\circ$, $45^\\circ$ and $90^\\circ$.\n(D) $60^\\circ$, $60^\\circ$ and $60^\\circ$.\n(E) It is impossible to determine.",
    "options": [],
    "answer": "C",
    "solution": "From $c^2 = 2ab$ and $a^2 + c^2 = 3b^2$ we get $a^2 + 2ab = 3b^2$ or $(a + b)^2 = 4b^2$. It follows that $(a + b - 2b)(a + b + 2b) = 0$. Since $a$ and $b$ are positive, the only possibility is that $a = b$. Then $c^2 = 2a^2 = a^2 + b^2$. Hence, we have a right isosceles triangle and the inner angles measure $45^\\circ$, $45^\\circ$ and $90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41471,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real $x$ and $y$, such that $x + y^2 = xy + 1$ and $xy = 4 + y$.",
    "options": [],
    "answer": "(5, 1), (3, 2), (-1, -2)",
    "solution": "The first equation implies $x + y^2 = xy + 1$, or $x - xy = 1 - y^2$, so $x(1 - y) = 1 - y^2$. This can be rewritten as $(1 - y)(x - 1 - y) = 0$.\n\nIf $y = 1$, then this equation is satisfied. The second equation then implies $x \\cdot 1 = 4 + 1$, so $x = 5$.\n\nIf $y \\ne 1$, then $x - 1 - y = 0$, so $x = 1 + y$. Combining this with the second equation yields:\n\n$(1 + y)y = 4 + y$\n\n$y^2 + y = 4 + y$\n\n$y^2 = 4$\n\nSo, $y = 2$ or $y = -2$.\n\nFor $y = 2$, $x = 1 + 2 = 3$.\nFor $y = -2$, $x = 1 + (-2) = -1$.\n\nBoth equations are satisfied for $x = 5$, $y = 1$; for $x = 3$, $y = 2$; and for $x = -1$, $y = -2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41472,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcentre of the acute triangle $ABC$. Let $H$ denote the orthocentre and let $D$ be the foot of the altitude from $C$. The line perpendicular to $OD$ at $D$ intersects the segment $BC$ at $E$. The circumcircle of the triangle $BCH$ intersects the line $AB$ at $B$ and $F$. Prove that the points $E, F$ and $H$ lie on the same line.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41473,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse polinome $P$ s celimi koeficienti, za katere velja: za vsako celo število $a$ in vsako praštevilo $p$, ki deli $P(a)$, velja, da $p$ deli $a$.",
    "options": [],
    "answer": "P(x) = ± x^n for any nonnegative integer n",
    "solution": "Solution:\n\nNaj bo $P$ polinom, ki ustreza pogojem naloge in $p$ poljubno praštevilo. Če je $q$ praštevilo, ki deli $P(p)$, potem deli tudi $p$, torej je $q=p$. Zato za vsako praštevilo $p$ velja $P(p)= \\pm p^{m_{p}}$ za neko nenegativno celo število $m_{p}$, ki je lahko odvisno od $p$.\n\nPolinoma $P(x)= \\pm 1$ očitno ustrezata pogojem. Denimo, da polinom $P$ ni konstantno enak 1 oziroma $-1$. Naj bo\n$$\nP(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\\ldots+a_{1} x+a_{0}\n$$\nPolinom $P$ zavzame vrednosti 1 in $-1$ kvečjemu za končno mnogo praštevil. Zato za neskončno praštevil velja\n$$\n\\pm p^{m_{p}}=P(p)=a_{n} p^{n}+a_{n-1} p^{n-1}+\\ldots+a_{1} p+a_{0}\n$$\noziroma $p \\mid a_{0}$. To je možno le v primeru $a_{0}=0$. Zapišimo $P(x)=x^{k} Q(x)$, kjer je $k \\in \\mathbb{N}$ in $Q$ tak polinom s celimi koeficienti, da je $Q(0) \\neq 0$. Naj bo $a$ celo število in $p$ praštevilo, ki deli $Q(a)$. Tedaj $p$ deli $P(a)$, zato deli tudi $a$. To pomeni, da polinom $Q$ ustreza pogojem naloge. Ker je $Q(0) \\neq 0$, mora biti polinom $Q$ po zgoraj pokazanem enak 1 oziroma $-1$. Vsi iskani polinomi so $P(x)= \\pm x^{n},\\ n \\in \\mathbb{N}_{0}$. Ti polinomi očitno ustrezajo pogojem naloge.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41474,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa realna števila $x$ in $y$, za katera velja $x + y^{2} = x y + 1$ in $x y = 4 + y$.",
    "options": [],
    "answer": "(x, y) = (5, 1), (3, 2), (-1, -2)",
    "solution": "Solution:\n\nIz prve enačbe sledi $x(1-y) = 1 - y^{2}$ oziroma $(1-y)(x-1-y) = 0$.\nČe je $y = 1$, ta enačba velja, iz druge pa sledi $x = 5$.\nV primeru $y \\neq 1$ dobimo $x = 1 + y$.\nSkupaj z drugo enačbo tedaj velja $(1 + y) y = 4 + y$ oziroma $y^{2} = 4$.\nOd tod sledi $y = 2$ ali $y = -2$.\n\nEnačbi veljata za števili $x = 5, y = 1$, števili $x = 3, y = 2$ in števili $x = -1, y = -2$.\n\n\n2. način\n\nIz druge enačbe izrazimo $x = \\frac{4}{y} + 1$ in vstavimo v prvo, še prej pa preverimo, da v primeru $y = 0$ ne dobimo nobene rešitve.\nTako imamo $\\frac{4}{y} + 1 + y^{2} = 5 + y$ oziroma $y^{3} - y^{2} - 4y + 4 = 0$.\nPreoblikujemo v $y^{2}(y-1) - 4(y-1) = 0$ oziroma $(y-1)\\left(y^{2}-4\\right) = 0$.\nOd tod sledi $y = 1$ ali $y^{2} = 4$, torej $y = 1$ ali $y = 2$ ali $y = -2$.\n\nEnačbi veljata za števili $x = 5, y = 1$, števili $x = 3, y = 2$ in števili $x = -1, y = -2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41475,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoloči vse funkcije $f: \\mathbb{R} \\rightarrow \\mathbb{R}$, za katere velja $f(x+y)=f(x-y)+2 f(y) \\cos x$ za vsa realna števila $x$ in $y$.",
    "options": [],
    "answer": "f(x) = a sin x for any real constant a",
    "solution": "Solution:\n\nV funkcijsko enačbo vstavimo $x=0$ in dobimo $f(-y)=-f(y)$ za vsako realno število $y$. Funkcija $f$ je liha.\n\nVstavimo še $x=\\frac{\\pi}{2}$ in izpeljemo $f\\left(y+\\frac{\\pi}{2}\\right)=f\\left(\\frac{\\pi}{2}-y\\right)$ za vsa realna števila $y$. Z upoštevanjem lihosti dobimo $f\\left(y+\\frac{\\pi}{2}\\right)=-f\\left(y-\\frac{\\pi}{2}\\right)$.\n\nIz prvotne enačbe za $y=\\frac{\\pi}{2}$ sledi\n$$\nf\\left(x+\\frac{\\pi}{2}\\right)=f\\left(x-\\frac{\\pi}{2}\\right)+2 \\cdot f\\left(\\frac{\\pi}{2}\\right) \\cdot \\cos x\n$$\nKer je $f\\left(x+\\frac{\\pi}{2}\\right)=-f\\left(x-\\frac{\\pi}{2}\\right)$, je $f\\left(x+\\frac{\\pi}{2}\\right)=f\\left(\\frac{\\pi}{2}\\right) \\cdot \\cos x$ oziroma\n$$\nf(x)=f\\left(\\frac{\\pi}{2}\\right) \\cdot \\cos \\left(x-\\frac{\\pi}{2}\\right)=f\\left(\\frac{\\pi}{2}\\right) \\cdot \\sin x\n$$\nOznačimo $a=f\\left(\\frac{\\pi}{2}\\right)$. Tedaj je $f(x)=a \\cdot \\sin x$. Preverimo lahko, da ta funkcija ustreza funkcijski enačbi za vsako realno število $a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41476,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDokaži, da za vsak par realnih števil $x$ in $y$ velja neenakost\n$$\n|x+y|+|x+1|+|y+1| \\geq 2.\n$$\nPri katerih številih $x$ obstaja tako število $y$, da velja $|x+y|+|x+1|+|y+1|=2$?",
    "options": [],
    "answer": "x in [-1, 1]",
    "solution": "Solution:\n\nZa vsako realno število $a$ velja $|a| \\geq a$ in $|a| \\geq -a$. Zato je\n$$\n|x+y| \\geq -(x+y), \\quad |x+1| \\geq x+1, \\quad |y+1| \\geq y+1\n$$\nod koder sledi\n$$\n|x+y|+|x+1|+|y+1| \\geq -(x+y)+(x+1)+(y+1)=2\n$$\nDenimo, da velja enakost. Tedaj veljajo enačaji v vseh treh neenakostih v (1) in je $x+y \\leq 0$, $x+1 \\geq 0$ ter $y+1 \\geq 0$. Dobimo $-1 \\leq x \\leq -y \\leq 1$ oziroma $-1 \\leq x \\leq 1$. Če je $-1 \\leq x \\leq 1$ in $y=-x$, velja\n$$\n|x+y|+|x+1|+|y+1|=2\n$$\nZaključimo lahko, da le za realna števila $x \\in [-1,1]$ obstaja število $y$, da je ta enakost izpolnjena.\n\n\n2. način. Obravnavajmo primere glede na to, kakšnega predznaka sta števili $x+1$ in $y+1$. Ločimo 4 možnosti.\n\na. Če sta števili $x$ in $y$ obe manjši od $-1$, velja $x+y<-2$ oziroma $|x+y|>2$. Zaradi $|x+1|>0$ in $|y+1|>0$ sledi\n$$\n|x+y|+|x+1|+|y+1|>2\n$$\n\nb. V primeru $x \\geq -1$, $y<-1$ je\n$$\n|x+y|+|x+1|+|y+1|=|x+y|+x+1-y-1=|x+y|+x-y\n$$\nČe je $x+y<0$, dobimo\n$$\n|x+y|+x-y=-(x+y)+x-y=-2y\n$$\nZaradi $y<-1$ velja $-y>1$, torej $-2y>2$ in zato velja ocena (2). Če je $x+y \\geq 0$, dobimo naprej $|x+y|+x-y=x+y+x-y=2x$. Zaradi $x+y \\geq 0$ sledi $x \\geq -y > 1$, torej je $2x>2$ in velja neenakost (2).\n\nc. Podobno obravnavamo $x<-1$, $y \\geq -1$ in prav tako dobimo neenakost (2).\n\nd. Ostane še $x \\geq -1$, $y \\geq -1$. Tedaj velja\n$$\n|x+y|+|x+1|+|y+1|=|x+y|+x+y+2\n$$\nV kolikor je $x+y>0$, naprej dobimo $|x+y|+x+y+2=2(x+y)+2>2$, torej velja neenakost (2). Sicer je $x+y \\leq 0$ in velja $|x+y|+x+y+2=-(x+y)+x+y+2=2$.\n\nPokažimo, da pri vsakem $x \\in [-1,1]$ obstaja tako število $y$, da velja enačaj. To število je na primer $y=-x$, kajti tedaj je $y \\geq -1$ in zato\n$$\n|x+y|+|x+1|+|y+1|=0+x+1+y+1=2+x+y=2\n$$\nUtemeljiti moramo še, da enačba ne more veljati za ostala števila $x$. Pri obravnavi možnosti smo videli, da velja neenakost (2) v vsakem primeru razen $x \\geq -1$, $y \\geq -1$ in $x+y \\leq 0$. Od tod sledi $x \\leq -y \\leq -(-1)=1$, zato velja ocena $-1 \\leq x \\leq 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41477,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that for all positive numbers $a$, $b$ and $c$ the following inequality holds:\n$$\n\\frac{(a+b-c)^2}{c} + \\frac{(b+c-a)^2}{a} + \\frac{(c+a-b)^2}{b} \\geq 6.\n$$\nWhen does the equality hold?",
    "options": [],
    "answer": "Equality holds when a = b = c = 2.",
    "solution": "Let us consider the given expression:\n$$\nS = \\frac{(a+b-c)^2}{c} + \\frac{(b+c-a)^2}{a} + \\frac{(c+a-b)^2}{b}.\n$$\n\nLet us use the AM-GM inequality and symmetry. First, note that all variables are positive.\n\nLet us try the case $a = b = c$:\n\nIf $a = b = c$, then\n$$(a+b-c) = a + b - c = a + a - a = a,$$\nso\n$$\nS = 3 \\cdot \\frac{a^2}{a} = 3a.\n$$\nBut this is not correct, since $\\frac{a^2}{a} = a$, so $S = 3a$.\nBut the inequality says $S \\geq 6$, so for $a = b = c$, $S = 3a \\geq 6$, i.e., $a \\geq 2$.\n\nBut let's check the general case.\n\nLet us use the Ravi substitution: set $a = x + y$, $b = y + z$, $c = z + x$, where $x, y, z > 0$.\n\nThen:\n\\begin{align*}\na + b - c &= (x + y) + (y + z) - (z + x) = x + y + y + z - z - x = 2y, \\\\\nb + c - a &= (y + z) + (z + x) - (x + y) = y + z + z + x - x - y = 2z, \\\\\nc + a - b &= (z + x) + (x + y) - (y + z) = z + x + x + y - y - z = 2x.\n\\end{align*}\n\nAlso,\n\\begin{align*}\na &= x + y, \\\\\nb &= y + z, \\\\\nc &= z + x.\n\\end{align*}\n\nSo,\n\\begin{align*}\nS &= \\frac{(2y)^2}{z + x} + \\frac{(2z)^2}{x + y} + \\frac{(2x)^2}{y + z} \\\\\n&= \\frac{4y^2}{z + x} + \\frac{4z^2}{x + y} + \\frac{4x^2}{y + z}.\n\\end{align*}\n\nTherefore,\n$$\nS = 4\\left(\\frac{y^2}{z + x} + \\frac{z^2}{x + y} + \\frac{x^2}{y + z}\\right).\n$$\n\nSo, it suffices to show that\n$$\n\\frac{y^2}{z + x} + \\frac{z^2}{x + y} + \\frac{x^2}{y + z} \\geq \\frac{3}{2}\n$$\nfor all positive $x, y, z$.\n\nBut this is a well-known inequality (see e.g. Engel's inequality or Nesbitt's inequality):\n\nHowever, Nesbitt's inequality states:\n$$\n\\frac{x}{y + z} + \\frac{y}{z + x} + \\frac{z}{x + y} \\geq \\frac{3}{2}.\n$$\nBut here we have squares in the numerator.\n\nLet us use the Cauchy-Schwarz inequality:\n\nBy Cauchy-Schwarz,\n$$\n\\left(\\frac{y^2}{z + x} + \\frac{z^2}{x + y} + \\frac{x^2}{y + z}\\right) \\geq \\frac{(x + y + z)^2}{(y + z) + (z + x) + (x + y)} = \\frac{(x + y + z)^2}{2(x + y + z)} = \\frac{x + y + z}{2}.\n$$\n\nTherefore,\n$$\nS \\geq 4 \\cdot \\frac{x + y + z}{2} = 2(x + y + z).\n$$\n\nRecall that $a = x + y$, $b = y + z$, $c = z + x$, so\n$$\nx + y + z = \\frac{a + b + c}{2}.\n$$\nTherefore,\n$$\nS \\geq 2 \\cdot \\frac{a + b + c}{2} = a + b + c.\n$$\n\nBut the original inequality is $S \\geq 6$, so we need $a + b + c \\geq 6$.\n\nBut the problem says for all positive $a, b, c$.\n\nLet us check the minimum value.\n\nLet us try $a = b = c = t$:\nThen $S = 3t$, so $3t \\geq 6 \\implies t \\geq 2$.\n\nBut for $a, b, c$ tending to $0$, the expression can be made arbitrarily small, so the problem must have a restriction.\n\nAlternatively, perhaps the problem is for $a, b, c$ being the sides of a triangle.\n\nBecause $a + b > c$, $b + c > a$, $c + a > b$ are required for the expressions $(a + b - c)$ etc. to be positive.\n\nTherefore, the problem is likely for $a, b, c$ being the sides of a triangle.\n\nAssuming $a, b, c$ are the sides of a triangle, then $a + b > c$, etc., and the Ravi substitution is valid.\n\nThen, as above,\n$$\nS = 4\\left(\\frac{y^2}{z + x} + \\frac{z^2}{x + y} + \\frac{x^2}{y + z}\\right) \\geq 4 \\cdot \\frac{3}{4} = 3.\n$$\nBut this is not $6$.\n\nAlternatively, perhaps the original problem has a typo, and the denominator should be $a$, $b$, $c$ instead of $c$, $a$, $b$.\n\nAlternatively, perhaps the minimum is $6$ for all positive $a, b, c$.\n\nLet us try $a = 2$, $b = 2$, $c = 2$:\nThen $a + b - c = 2 + 2 - 2 = 2$, so\n$$\nS = 3 \\cdot \\frac{2^2}{2} = 3 \\cdot 2 = 6.\n$$\nSo equality holds when $a = b = c = 2$.\n\nLet us try $a = 3$, $b = 2$, $c = 1$:\n$a + b - c = 3 + 2 - 1 = 4$, $b + c - a = 2 + 1 - 3 = 0$, $c + a - b = 1 + 3 - 2 = 2$.\nSo one term is zero, so $S = \\frac{4^2}{1} + 0 + \\frac{2^2}{2} = 16 + 0 + 2 = 18$.\n\nSo the minimum is $6$ when $a = b = c = 2$.\n\nTherefore, the answer is:\n\nThe inequality holds for all positive $a, b, c$, and equality holds if and only if $a = b = c = 2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41478,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a point inside the triangle $ABC$, such that the angles $\\angle CBP$ and $\\angle PAC$ are equal. Denote the intersection of the line $AP$ and the segment $BC$ by $D$, and the intersection of the line $BP$ with the segment $AC$ by $E$. The circumcircles of the triangles $ADC$ and $BEC$ meet at $C$ and $F$. Show that the line $CP$ bisects the angle $DFE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using the equalities between the inscribed angles in the quadrilaterals $AFDC$ and $BCEF$ as well as the given equality $\\angle CBP = \\angle PAC$ we find that\n$$\n\\begin{aligned}\n\\angle CFE &= \\angle CBE = \\angle CBP = \\angle PAC \\\\\n&= \\angle DAC = \\angle DFC.\n\\end{aligned}\n$$\n\n![](attached_image_1.png)\nIn the cyclic quadrilateral $BCEF$ we have $\\angle EBF = \\angle ECF$ and in the cyclic quadrilateral $AFDC$ we have $\\angle ACF = \\angle ADF$. So,\n$$\n\\angle PBF = \\angle EBF = \\angle ECF = \\angle ACF = \\angle ADF = \\angle PDF.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41479,
    "subject": "Mathematics (Multi-modal)",
    "question": "Yesterday in the playground the ratio of boys to girls was $3 : 2$. Today, the number of boys is the square of the number of girls and there are $6$ boys and $7$ girls less than yesterday. How many children were there in the playground yesterday?\n(A) 12\n(B) 13\n(C) 15\n(D) 25\n(E) 30",
    "options": [],
    "answer": "D",
    "solution": "Yesterday there were $3t$ boys and $2t$ girls in the playground. Today, there are $3t-6$ boys and $2t-7$ girls. From $3t-6 = (2t-7)^2$ we get $4t^2-31t+55 = 0$, so $t_{1,2} = \\frac{31\\pm\\sqrt{81}}{8}$. The only possible solution is $t = 5$, since $t$ is an integer. Yesterday there were $3t = 15$ boys and $2t = 10$ girls in the playground, or $25$ children altogether.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41480,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$, $N$ and $P$ be the points on the sides $AB$, $BC$ and $CA$ of the triangle $ABC$, such that the quadrilateral $CPMN$ is a parallelogram. The lines $AN$ and $MP$ meet at $R$, the lines $BP$ and $MN$ meet at $S$ and the lines $AN$ and $BP$ meet at $Q$. Let $T$ be the image of $Q$ under the orthogonal projection onto the side $AB$. Prove that the areas of the triangle $NQP$ and the quadrilateral $TSQR$ are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41481,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral. Choose points $E$ and $F$ on the segment $AB$ and a point $G$ on the segment $CD$, such that the quadrilaterals $ABCG$, $AFCD$ and $EFGC$ are cyclic. Prove that $|AE| = |FB|$ if and only if the segments $AB$ and $CD$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\angle DCA = \\gamma$. Since the quadrilateral $AFCD$ is cyclic, we have\n$$\n\\angle DFA = \\angle DCA = \\gamma.\n$$\nSince $ABCG$ is also a cyclic quadrilateral, we have $\\angle GBA = \\angle GCA = \\gamma$. From here we conclude that the lines $DF$ and $GB$ are parallel.\n\nIn the cyclic quadrilateral $AFCD$ we have $\\angle FAD = \\pi - \\angle DCF$ and in the cyclic quadrilateral $GEFC$ we have $\\angle GEF = \\pi - \\angle GCF$. Since $\\angle GCF = \\angle DCF$ we get $\\angle FAD = \\angle GEF$, so the lines $AD$ and $GE$ are parallel.\n\nLet $A'$ be a point on the line $AD$ and let $B'$ be a point on the line $DF$, such that $GA' \\parallel EA$ and $GB' \\parallel FB$. Obviously, the points $G$, $B'$ and $A'$ are collinear. The quadrilateral $AEGA'$ is a parallelogram, so $|A'G| = |AE|$. The quadrilateral $FBGB'$ is also a parallelogram, so $|B'G| = |FB|$.\n\n![](attached_image_1.png)\n\nIf $|AE| = |FB|$, we have $|A'G| = |B'G|$. Since the point $G$ clearly does not lie between $A'$ and $B'$, we conclude that $A' = B'$. This is only possible when $A' = B' = D$. In this case, the line $GD$ is parallel to the line $AE$, which means that the segments $AB$ and $CD$ are parallel.\n\nConversely, if $AB$ and $CD$ are parallel, then the quadrilaterals $AEGD$ and $FBGD$ are parallelograms, since they have two pairs of parallel sides each. So, $|AE| = |DG| = |FB|$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41482,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O_1$ be the centre of the circle $K_1$. Let $K_2$ be the circle centered at $O_2$ and passing through $O_1$. The circles $K_1$ and $K_2$ meet at $A$ and $B$. The circle $K_1$ intersects the segment $O_1O_2$ at $C$. The line $BC$ intersects the circle $K_2$ at $B$ and $D$. The line $AD$ intersects the circle $K_1$ at $A$ and $E$. Let $F$ be the midpoint of the segment $AE$. Prove that the lines $O_1A$ and $O_1D$ divide the angle $CO_1F$ into three equal parts.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Write $\\angle ABD = \\alpha$. Since $AO_1BD$ is a cyclic quadrilateral, we have $\\angle AO_1D = \\angle ABD = \\alpha$. The central angle $\\angle AO_1C$ in the circle $K_1$ is twice the size of the inscribed angle $\\angle ABC$, so $\\angle AO_1C = 2\\alpha$.\n\n![](attached_image_1.png)\n\n\\[\n\\begin{aligned}\n\\angle DO_1C &= \\angle AO_1C - \\angle AO_1D \\\\\n&= 2\\alpha - \\alpha = \\alpha.\n\\end{aligned}\n\\]\nThe segment $AB$ is perpendicular to the segment $O_1O_2$ and the quadrilateral $AO_1BO_2$ is a deltoid. So,\n\\[\n\\begin{aligned}\n\\angle ABO_1 &= \\angle O_1AB = \\frac{\\pi}{2} - \\angle O_2O_1A \\\\\n&= \\frac{\\pi}{2} - \\angle AO_1D - \\angle DO_1C = \\frac{\\pi}{2} - 2\\alpha.\n\\end{aligned}\n\\]\n\nSince $AO_1BD$ is a cyclic quadrilateral, we have $\\angle ADO_1 = \\angle ABO_1 = \\frac{\\pi}{2} - 2\\alpha$. Since $F$ is the midpoint of the segment $AE$, we have $\\angle AFO_1 = \\frac{\\pi}{2}$. This implies $\\angle FO_1D = \\frac{\\pi}{2} - \\angle FDO_1 = 2\\alpha$ and\n\\[\n\\angle AO_1F = \\angle DO_1F - \\angle DO_1A = 2\\alpha - \\alpha = \\alpha.\n\\]\nWe have shown that $\\angle AO_1F = \\alpha = \\angle AO_1C = \\angle DO_1C$, so the lines $O_1A$ and $O_1D$ divide the angle $CO_1F$ into three equal parts.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41483,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many positive integers smaller than $1000$ have the sum of the digits divisible by $7$ and are multiples of $3$?\n\n(A) $7$\n(B) $19$\n(C) $21$\n(D) $28$\n(E) $37$",
    "options": [],
    "answer": "D",
    "solution": "The sum of the digits of a multiple of $3$ is divisible by $3$. So, we are looking for numbers with the sum of the digits divisible by $21$. The sum of the digits of any number smaller than $1000$ is less than or equal to $9 + 9 + 9 = 27$. So, in our case the sum of the digits is equal to $21$.\n\nThe smallest possible digit is at least $21 - 18 = 3$ and at most $\\frac{21}{3} = 7$. The only possible triples of digits are $(3, 9, 9)$, $(4, 8, 9)$, $(5, 7, 9)$, $(5, 8, 8)$, $(6, 6, 9)$, $(6, 7, 8)$ and $(7, 7, 7)$. Each triple with three different digits corresponds to six different numbers. Each triple with two equal digits corresponds to three different numbers and the triple $(7, 7, 7)$ corresponds to one number. Hence, there are $3 \\cdot 6 + 3 \\cdot 3 + 1 = 28$ numbers with the required properties.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41484,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many pairs $(m, n)$ of positive integers satisfy the condition $\\frac{3}{m} + \\frac{2}{n} = 1$?\n(A) 2\n(B) 3\n(C) 4\n(D) 5\n(E) More than 5.",
    "options": [],
    "answer": "C",
    "solution": "Clearly $m > 3$ and $n > 2$. Since $n$ decreases as $m$ increases $n$, we see that the only possible solutions are $(m, n) \\in \\{(4, 8), (5, 5), (6, 4), (9, 3)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41485,
    "subject": "Mathematics (Multi-modal)",
    "question": "The progression $\\{x_n\\}$ is defined by the first two terms $x_1 = a$ and $x_2 = b$ and a recursive formula\n$$\nx_n = 3x_{n-1} - x_{n-2}\n$$\nfor $n \\ge 3$. Prove that there exist positive integers $a$ and $b$, such that the number $1 + x_n x_{n+1}$ is a perfect square for all $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41486,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be positive integers, such that $19 \\le m \\le 49$, $51 \\le n \\le 101$. What is the greatest possible value of the expression $\\frac{n+m}{n-m}$?\n(A) 20\n(B) 30\n(C) 40\n(D) 50\n(E) 60",
    "options": [],
    "answer": "D",
    "solution": "We have $\\frac{n+m}{n-m} = \\frac{n-m+2m}{n-m} = 1 + 2\\frac{m}{n-m}$, so the value is maximal when $n-m = 2$ and $m = 49$. Then, the value is $50$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41487,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEF$ be a hexagon with $\\angle BAF = 150^\\circ$, $\\angle ACB = \\angle ADC = 90^\\circ$ and $|AC| = |BC|$. Assume also that the triangle $ABC$ is similar to the triangle $ADE$ and the triangle $BCD$ is similar to the triangle $DEF$. Find the ratio of the lengths of the segments $AB$ and $AF$.",
    "options": [],
    "answer": "8/3",
    "solution": "Let $|AB| = a$. Since $ABC$ is an isosceles right triangle with the apex at $C$, we have $|AC| = \\frac{a}{\\sqrt{2}} = \\frac{a\\sqrt{2}}{2}$ and $\\angle ABC = \\angle BAC = \\frac{\\pi}{4}$. The triangles $ADE$ and $ABC$ are similar, so $\\angle AED = \\angle ACB = \\frac{\\pi}{2}$ and $|DE| = |AE|$.\n\n![](attached_image_1.png)\n\n$$\n\\angle ACD = \\angle BCD - \\frac{\\pi}{2} = \\angle DEF - \\frac{\\pi}{2} = \\angle AEF\n$$\n\n$$\n\\frac{|AC|}{|CD|} = \\frac{|BC|}{|CD|} = \\frac{|DE|}{|EF|} = \\frac{|AE|}{|EF|}\n$$\n\n$150^\\circ = \\angle BAF = \\angle BAC + \\angle CAD + \\angle CAE + \\angle EAF = \\frac{\\pi}{2} + 2\\angle CAD$\n$$\nwe deduce that $\\angle CAD = 30^\\circ$. The triangle $CAD$ is one half of an equilateral triangle and $|AD| = \\frac{|AC|\\sqrt{3}}{2} = \\frac{a\\sqrt{6}}{4}$. Since $ADE$ is an isosceles right triangle, we have $|AE| = \\frac{|AD|}{\\sqrt{2}} = \\frac{a\\sqrt{3}}{4}$. Finally, since $AEF$ is one half of an equilateral triangle, we get $|AF| = \\frac{|AE|\\sqrt{3}}{2} = \\frac{3a}{8}$. Hence, $\\frac{|AB|}{|AF|} = \\frac{8}{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41488,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. We place $4n^2$ kings onto a $4n \\times 4n$ chessboard, so that no two attack each other and each row and column contains exactly $n$ kings. Find the number of all such arrangements. (A king attacks all fields that have at least one point in common with the field where it stands.)",
    "options": [],
    "answer": "2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 41489,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTočka $O_{1}$ je središče krožnice $\\mathcal{K}_{1}$ in leži na krožnici $\\mathcal{K}_{2}$ s središčem $O_{2}$. Krožnici $\\mathcal{K}_{1}$ in $\\mathcal{K}_{2}$ se sekata v točkah $A$ in $B$. Krožnica $\\mathcal{K}_{1}$ seka daljico $O_{1} O_{2}$ v točki $C$. Premica $B C$ seka krožnico $\\mathcal{K}_{2}$ v točkah $B$ in $D$, premica $A D$ pa seka krožnico $\\mathcal{K}_{1}$ v točkah $A$ in $E$. Naj bo $F$ razpolovišče daljice $A E$. Dokaži, da premici $O_{1} A$ in $O_{1} D$ razdelita kot $C O_{1} F$ na tri enake dele.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo $\\angle A B D=\\alpha$. Zaradi tetivnosti štirikotnika $A O_{1} B D$ je $\\angle A O_{1} D=\\angle A B D=\\alpha$. Središčni kot $\\angle A O_{1} C$ v krožnici $\\mathcal{K}_{1}$ je dvakrat večji od obodnega kota $\\angle A B C$, zato je $\\angle A O_{1} C=2 \\alpha$. Od tod sledi\n$$\n\\begin{aligned}\n\\angle D O_{1} C & =\\angle A O_{1} C-\\angle A O_{1} D \\\\\n& =2 \\alpha-\\alpha=\\alpha\n\\end{aligned}\n$$\nDaljica $A B$ je pravokotna na daljico $O_{1} O_{2}$ in štirikotnik $A O_{1} B O_{2}$ je deltoid. Zato je\n$$\n\\begin{aligned}\n\\angle A B O_{1} & =\\angle O_{1} A B=\\frac{\\pi}{2}-\\angle O_{2} O_{1} A \\\\\n& =\\frac{\\pi}{2}-\\angle A O_{1} D-\\angle D O_{1} C=\\frac{\\pi}{2}-2 \\alpha\n\\end{aligned}\n$$\n![](attached_image_1.png)\nUpoštevamo še tetivnost štirikotnika $A O_{1} B D$ in izpeljemo $\\angle A D O_{1}=\\angle A B O_{1}=\\frac{\\pi}{2}-2 \\alpha$. Ker je točka $F$ razpolovišče daljice $A E$, velja $\\angle A F O_{1}=\\frac{\\pi}{2}$. Od tod sledi $\\angle F O_{1} D=\\frac{\\pi}{2}-\\angle F D O_{1}=2 \\alpha$ in\n$$\n\\angle A O_{1} F=\\angle D O_{1} F-\\angle D O_{1} A=2 \\alpha-\\alpha=\\alpha\n$$\nPokazali smo $\\angle A O_{1} F=\\alpha=\\angle A O_{1} C=\\angle D O_{1} C$, torej premici $A O_{1}$ in $D O_{1}$ razdelita kot $C O_{1} F$ na tri enake dele.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41490,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV šestkotniku $A B C D E F$ velja $\\angle B A F=150^\\circ$, $\\angle A C B=\\angle A D C=90^\\circ$, $|A C|=|B C|$, trikotnik $A B C$ je podoben trikotniku $A D E$ in trikotnik $B C D$ je podoben trikotniku $D E F$. Izračunaj razmerje dolžin daljic $A B$ in $A F$.",
    "options": [],
    "answer": "8/3",
    "solution": "Solution:\n\nOznačimo $|A B|=a$. Ker je $A B C$ enakokrak pravokotni trikotnik z vrhom $C$, je $|A C|=\\frac{a}{\\sqrt{2}}=\\frac{a \\sqrt{2}}{2}$ in $\\angle A B C=\\angle B A C=\\frac{\\pi}{4}$. Zaradi podobnosti trikotnikov $A D E$ in $A B C$ sledi $\\angle A E D=\\angle A C B=\\frac{\\pi}{2}$ in $|D E|=|A E|$.\n\nIz podobnosti trikotnikov $B C D$ in $D E F$ sklepamo $\\angle B C D=\\angle D E F$ in $\\frac{|B C|}{|C D|}=\\frac{|D E|}{|E F|}$. Torej je\n\n![](attached_image_1.png)\n\n$$\n\\angle A C D=\\angle B C D-\\frac{\\pi}{2}=\\angle D E F-\\frac{\\pi}{2}=\\angle A E F\n$$\n\nin\n$$\n\\frac{|A C|}{|C D|}=\\frac{|B C|}{|C D|}=\\frac{|D E|}{|E F|}=\\frac{|A E|}{|E F|}\n$$\nzato sta si trikotnika $A C D$ in $A E F$ podobna in velja še $\\angle C A D=\\angle E A F$.\n\nIz\n$$\n150^\\circ=\\angle B A F=\\angle B A C+\\angle C A D+\\angle C A E+\\angle E A F=\\frac{\\pi}{2}+2 \\angle C A D\n$$\nsledi $\\angle C A D=30^\\circ$. Tako je trikotnik $C A D$ polovica enakostraničnega trikotnika in je $|A D|=\\frac{|A C| \\sqrt{3}}{2}=\\frac{a \\sqrt{6}}{4}$. Trikotnik $A D E$ je enakokrak pravokotni trikotnik, zato je $|A E|=\\frac{|A D|}{\\sqrt{2}}=\\frac{a \\sqrt{3}}{4}$. Upoštevamo še, da je trikotnik $A E F$ polovica enakostraničnega, in izračunamo $|A F|=\\frac{|A E| \\sqrt{3}}{2}=\\frac{3 a}{8}$. Dobili smo $\\frac{|A B|}{|A F|}=\\frac{8}{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41491,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $P$ taka točka znotraj trikotnika $ABC$, da je $\\angle CBP = \\angle PAC$. Presečišče premice $AP$ s stranico $BC$ označimo z $D$, presečišče premice $BP$ s stranico $AC$ pa z $E$. Trikotnikoma $ADC$ in $BEC$ očrtani krožnici se sekata v točkah $C$ in $F$. Pokaži, da je premica $CP$ simetrala kota $DFE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nS pomočjo obodnih kotov v štirikotnikih $AFDC$ in $BCEF$ ter dane enakosti $\\angle CBP = \\angle PAC$ izpeljemo\n$$\n\\begin{aligned}\n\\angle CFE &= \\angle CBE = \\angle CBP = \\angle PAC \\\\\n&= \\angle DAC = \\angle DFC\n\\end{aligned}\n$$\nzato je premica $CF$ simetrala kota $\\angle DEF$. Pokazati moramo še, da točka $P$ leži na premici $CF$.\n\nV tetivnem štirikotniku $BCEF$ je $\\angle EBF = \\angle ECF$, v tetivnem štirikotniku $AFDC$ pa $\\angle ACF = \\angle ADF$. Zato je\n\n![](attached_image_1.png)\n\n$$\n\\angle PBF = \\angle EBF = \\angle ECF = \\angle ACF = \\angle ADF = \\angle PDF\n$$\nTorej točke $B, D, P$ in $F$ ležijo na isti krožnici. S pomočjo obodnih kotov v tetivnih štirikotnikih $BDPF$ in $AFDC$ izpeljemo\n$$\n\\angle DFP = \\angle DBP = \\angle CBP = \\angle PAC = \\angle DAC = \\angle DFC\n$$\nod koder sledi kolinearnost točke $C, P$ in $F$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41492,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je konveksen štirikotnik $ABCD$, točki $E$ in $F$ na stranici $AB$ ter točka $G$ na stranici $CD$, da so štirikotniki $ABCG$, $A F C D$ in $E F C G$ tetivni. Dokaži, da je $|AE|=|FB|$ natanko tedaj, ko sta stranici $AB$ in $CD$ vzporedni.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo $\\angle DCA=\\gamma$. Zaradi tetivnosti štirikotnika $A F C D$ je\n$$\n\\angle DFA=\\angle DCA=\\gamma\n$$\nIz tetivnosti štirikotnika $ABCG$ sledi $\\angle GBA=\\angle GCA=\\gamma$. Od tod dobimo vzporednost premic $DF$ in $GB$.\n\nV tetivnem štirikotniku $A F C D$ velja $\\angle FAD=\\pi-\\angle DCF$ in v tetivnem štirikotniku $GEFC$ velja $\\angle GEF=\\pi-\\angle GCF$. Ker je $\\angle GCF=\\angle DCF$, od tod sledi $\\angle FAD=\\angle GEF$, torej sta premici $AD$ in $GE$ vzporedni.\n\nNaj bosta $A'$ takšna točka na premici $AD$ in $B'$ takšna točka na premici $DF$, da je\n\n![](attached_image_1.png)\n\n$GA' \\parallel EA$ in $GB' \\parallel FB$. Očitno so točke $G, B'$ in $A'$ kolinearne. Štirikotnik $AEGA'$ je paralelogram, zato je $|A'G|=|AE|$. Prav tako je štirikotnik $FBGB'$ paralelogram in tako velja $|B'G|=|FB|$.\n\nČe je $|AE|=|FB|$, velja $|A'G|=|B'G|$. Ker točka $G$ očitno ne leži med $A'$ in $B'$, od tod sledi $A'=B'$. To je možno le v primeru $A'=B'=D$. Tedaj je $GD$ vzporedna $AE$, kar pomeni, da sta stranici $AB$ in $CD$ vzporedni.\n\nObratno, če sta stranici $AB$ in $CD$ vzporedni, sta štirikotnika $AEGD$ in $FBGD$ paralelograma, saj imata dva para vzporednih stranic. Torej je $|AE|=|DG|=|FB|$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41493,
    "subject": "Mathematics (Multi-modal)",
    "question": "A $4 \\times 4$ board is divided into 16 squares. Onto this board we place several tiles like the one in the picture\n![](attached_image_1.png)\n(the tiles can be rotated),\neach tile covering two squares. At least how many tiles do we need to place onto the board, so that every uncovered square will be adjacent to at least one covered square? (Two squares are adjacent if they share a common side.)",
    "options": [],
    "answer": "4",
    "solution": "We can place four tiles as shown in the first picture. Each uncovered square has at least one covered neighbour. Now, let us show that this cannot be the case if we use less than four tiles. Put a tile onto the board and mark all the neighbouring squares. Each row contains at most three squares that are either covered or adjacent to a covered square, and these squares are next to each other with no gaps. The same is true for the columns and also for the diagonals. The board has four corners. If it were possible to cover the board as required with at most three tiles, then one of the three tiles would have to cover or be adjacent to at least two corners. This is not possible.\n\n![](attached_image_2.png)\n\n![](attached_image_3.png)",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 41494,
    "subject": "Mathematics (Multi-modal)",
    "question": "A $7 \\times 7$ board is divided into $49$ squares. Onto this board we place several tiles like the one in the picture\n\n![](attached_image_1.png)\n(the tiles can be rotated),\neach tile covering two squares. At least how many tiles do we need to place onto the board, so that every uncovered square will be adjacent to at least one covered square? (Two squares are adjacent if they share a common side.)",
    "options": [],
    "answer": "9",
    "solution": "We can place nine tiles as shown in the first picture. Each uncovered square has at least one covered neighbour.\n\n![](attached_image_2.png)\n\nNow, let us show that this cannot be the case if we use less than nine tiles. Put a tile onto the board and mark all the neighbouring squares. Each row contains at most three squares that are either covered or adjacent to a covered square, and these squares are next to each other with no gaps. The same is true for the columns and also for the diagonals. Let us colour nine squares as shown in the third picture. If it were possible to cover the board as required with at most eight tiles, then one of the eight tiles would have to cover or be adjacent to at least two of the coloured squares. This is not possible.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 41495,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circles $K_1$ and $K_2$ with the centres $O_1$ and $O_2$ intersect at the points $A$ and $B$, so that $\\angle O_1AO_2 > \\frac{\\pi}{2}$. The line $O_1B$ intersects the circle $K_2$ again at $C$, the line $O_2B$ intersects the circle $K_1$ again at $D$. The line through the point $B$ parallel to the line $CD$ intersects the circles at $E$ and $F$. Prove that $|EF| = |AC| + |AD|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41496,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTabela $7 \\times 7$ je razdeljena na 49 kvadratkov. Na to tabelo postavljamo ploščice oblike\n![](attached_image_1.png)\nki pokrijejo vsaka po dve polji. Najmanj koliko ploščic moramo postaviti na tabelo, da bo imelo vsako nepokrito polje vsaj eno sosednje polje pokrito? (Polji sta sosednji, če imata skupno stranico.)",
    "options": [],
    "answer": "9",
    "solution": "Solution:\n\nNa tabelo lahko položimo devet ploščic, kot prikazuje prva slika. Vsako nepokrito polje ima vsaj eno sosednje polje pokrito. Utemeljimo, da z manj kot devetimi ploščicami tega ne moremo narediti. Opazujmo ploščico, ki leži na tabeli.\n![](attached_image_2.png)\nOznačimo še polja, ki so sosednja pokritim. V vsaki vrstici so največ tri polja takšna, da so bodisi pokrita bodisi sosednja pokritim, in vsa ta polja so zaporedna. Enako velja za stolpce in diagonale. Pobarvajmo devet polj, kot prikazuje tretja slika. Če bi tabelo lahko pokrili z največ osmimi ploščicami tako, da bi bilo vsako polje pokrito ali sosednje pokritemu, bi morala ena izmed teh ploščic pokriti oziroma biti sosedna dvema pobarvanima poljema, kar ni možno.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41497,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $n$, such that the sum of its digits is equal to $27$ and the sum of the digits of $2n$ is greater than $27$.",
    "options": [],
    "answer": "14499",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41498,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $x$ that solve the equation\n$$\n\\log_2(10x) + \\log_4(100x) + \\log_8(1000x) - 2\\log_{64} x = 9.\n$$\nWrite the result as a reduced fraction.",
    "options": [],
    "answer": "16/25",
    "solution": "Rewrite the equation as\n$$\n\\begin{align*}\n9 &= \\log_2(10x) + \\log_4(100x) + \\log_8(1000x) - 2\\log_{64} x \\\\\n&= \\log_2(10x) + \\frac{\\log(100x)}{\\log 4} + \\frac{\\log(1000x)}{\\log 8} - \\frac{2\\log x}{\\log 64} \\\\\n&= \\log_2(10x) + \\frac{\\log(100x)}{2\\log 2} + \\frac{\\log(1000x)}{3\\log 2} - \\frac{2\\log x}{6\\log 2} \\\\\n&= \\log_2(10x) + \\frac{1}{2}\\log_2(100x) + \\frac{1}{3}\\log_2(1000x) - \\frac{1}{3}\\log_2 x \\\\\n&= \\log_2(10x) + \\log_2(\\sqrt{100x}) + \\log_2(\\sqrt[3]{1000x}) - \\log_2\\sqrt[3]{x} \\\\\n&= \\log_2\\left(\\frac{10x \\cdot \\sqrt{100x} \\cdot \\sqrt[3]{1000x}}{\\sqrt[3]{x}}\\right) \\\\\n&= \\log_2(1000\\sqrt[3]{x^3}).\n\\end{align*}\n$$\nIt follows that $1000\\sqrt[3]{x^3} = 2^9$ or $x = \\left(\\frac{2^9}{10^3}\\right)^{\\frac{2}{3}} = \\frac{16}{25}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41499,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integral solutions $x$ and $y$ of the equation\n$$\n3xy + 2x + y = 12.\n$$",
    "options": [],
    "answer": "(-1, -7), (-13, -1), (6, 0), (0, 12)",
    "solution": "Rewrite the equation as $x(3y+2) = 12 - y$. Obviously, $3y+2 \\neq 0$ divides $12 - y$, so $3y+2$ divides $3(12 - y) + (3y+2) = 38$. Since $3y+2$ gives the remainder of $2$ when divided by $3$, there are four possibilities. The number $3y+2$ is equal to $-19, -1, 2$ or $38$, so $y$ is equal to $-7, -1, 0$ or $12$ and $x$ is equal to $-1, -13, 6$ or $0$ respectively. The solutions are $(-1, -7), (-13, -1), (6, 0)$ and $(0, 12)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41500,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all non-zero integers $a$, different from $4$, such that $\\frac{a}{a-4} + \\frac{2}{a}$ is an integer as well.",
    "options": [],
    "answer": "2 and -4",
    "solution": "If $\\frac{a}{a-4} + \\frac{2}{a} = \\frac{a^2+2a-8}{a(a-4)}$ is an integer, then $a(a-4)$ divides $a^2+2a-8$. So, $a$ divides $a^2+2a-8$, and $a$ divides $8$. All integer divisors of $8$ different from $4$ are $1$, $2$, $8$, $-1$, $-2$, $-4$ and $-8$. We check all seven cases and see that the value of the expression is an integer only when $a = 2$ or $a = -4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41501,
    "subject": "Mathematics (Multi-modal)",
    "question": "For the real numbers $x$ and $\\alpha$ we have\n$$\nx + \\frac{1}{x} = 2 \\cos \\alpha .\n$$\nProve that for all positive integers $n$\n$$\nx^n + \\frac{1}{x^n} = 2 \\cos(n \\alpha) .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us first consider the case $x > 0$. We have $(x - 1)^2 \\ge 0$ or $x^2 + 1 \\ge 2x$, and dividing by $x$ we get\n$$\nx + \\frac{1}{x} \\ge 2 \\ge 2 \\cos \\alpha = x + \\frac{1}{x},\n$$\nso $\\cos \\alpha = 1$ and $x+1/x = 2$. We conclude that $x = 1$. This implies $x^n+1/x^n = 1+1=2 = \\cos(n\\alpha)$.\n\nIf $x < 0$, then a similar reasoning for $(x+1)^2 \\ge 0$ shows that $x = -1$ and $\\cos \\alpha = -1$, whence $x^n + 1/x^n = 2(-1)^n = 2\\cos(n\\alpha)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41502,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$ the bisector of the angle $\\angle BAC$ meets the segment $BC$ at $D$. The triangle $ADC$ is isosceles with the apex at $D$ and the lengths of the segments $CD$ and $BD$ are $|CD| = 36$ and $|BD| = 64$. Find the lengths of the sides of the triangle $ABC$.",
    "options": [],
    "answer": "AB = 80, AC = 45, BC = 100",
    "solution": "Obviously $|BC| = 100$. Let us find the lengths of the other sides. We have $\\angle BAD = \\angle DAC = \\angle ACB$, so the triangles $ABD$ and $CBA$ are congruent and $\\frac{|AC|}{|AD|} = \\frac{|AB|}{|BD|} = \\frac{|BC|}{|AB|}$. The second equality implies $|AB| = \\sqrt{|BD|^2 \\cdot |BC|} = \\sqrt{64 \\cdot 100} = 80$. It then follows from the first equality that $|AC| = \\frac{|AD||AB|}{|BD|} = \\frac{36 \\cdot 80}{64} = 45$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41503,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ is given with the points $D$ on the side $AB$, $E$ on the side $BC$ and $F$ on the side $AC$, such that the line $CD$ is perpendicular to the side $AB$, the line $DE$ is perpendicular to the side $BC$ and the line $DF$ is perpendicular to the side $AC$. Prove that the points $A$, $B$, $E$ and $F$ lie on the same circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\angle BAC = \\alpha$. The point $D$ lies on the segment $AB$, so the angles $\\angle BAC$ and $\\angle CBA$ cannot be obtuse. If the point $D$ coincides with $A$ or with $B$, the statement is obviously true. For the remainder of the solution we assume that $D$ lies in the interior of the segment $AB$.\n\nIn this case $\\angle FDA = \\frac{\\pi}{2} - \\alpha$ and $\\angle FDC = \\frac{\\pi}{2} - \\angle FDA = \\alpha$. In the quadrilateral $DECF$ we have $\\angle DEC + \\angle CFD = \\frac{\\pi}{2} + \\frac{\\pi}{2} = \\pi$, so it is cyclic. This implies that $\\angle CEF = \\angle CDF = \\alpha$, so $\\angle FEB = \\pi - \\angle FEC = \\pi - \\alpha$ and $\\angle BAF + \\angle FEB = \\alpha + \\pi - \\alpha = \\pi$. Hence, $ABEF$ is a cyclic quadrilateral as well.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41504,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square and let $E$ and $F$ be two points outside the square, such that $BEC$ and $CFD$ are equilateral triangles. Let $G$ be the intersection of the lines $BE$ and $FD$, and let $H$ be a point, such that the quadrilateral $CEHF$ is a rhombus. Prove that the points $G$, $E$, $H$ and $F$ lie on the same circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $ABCD$ is a square and the triangles $BEC$ and $CFD$ are equilateral we have $|CE| = |CB| = |CD| = |CF|$. The triangle $FCE$ is isosceles and\n$$\n\\begin{aligned}\n\\angle ECF &= 2\\pi - \\angle FCD - \\angle DCB - \\angle BCE \\\\\n&= 2\\pi - \\frac{\\pi}{3} - \\frac{\\pi}{2} - \\frac{\\pi}{3} = \\frac{5\\pi}{6}.\n\\end{aligned}\n$$\n\n![](attached_image_1.png)\n\nSo, $\\angle CEF = \\frac{\\pi}{12}$ and $\\angle EFC = \\frac{\\pi}{12}$. Since the quadrilateral $CEHF$ is a rhombus, we have $\\angle FEH = \\angle HFE = \\frac{\\pi}{12}$. So, $\\angle BEH = \\frac{\\pi}{3} + 2 \\cdot \\frac{\\pi}{12} = \\frac{\\pi}{2}$ and $\\angle DFH = \\frac{\\pi}{2}$. In the quadrilateral $GEHF$ we have $\\angle GEH + \\angle HFG = \\frac{\\pi}{2} + \\frac{\\pi}{2} = \\pi$, so the points $G$, $E$, $H$ and $F$ lie on the same circle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41505,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $E$ be the midpoint of the side $AB$ in the quadrilateral $ABCD$ and let $F$ be a point on the diagonal $AC$, such that the line $BF$ is perpendicular to the diagonal $AC$. Find the ratio of the sides of the rectangle $ABCD$, if the segment $EF$ is perpendicular to the diagonal $BD$.",
    "options": [],
    "answer": "sqrt(3)",
    "solution": "Write $\\angle BAC = \\alpha$. Since $ABF$ is a right triangle and $E$ is the midpoint of the hypotenuse, it is also the circumcentre of the triangle $ABF$ and $|AE| = |BE| = |FE|$. So, $\\angle AFE = \\angle EAF = \\alpha$ and $\\angle EFB = \\frac{\\pi}{2} - \\angle AFE = \\frac{\\pi}{2} - \\alpha$, which implies $\\angle FBD = \\frac{\\pi}{2} - \\angle EFB = \\alpha$. Also, $\\angle ACB = \\frac{\\pi}{2} - \\alpha$, so $\\angle CBF = \\alpha$ and $\\angle DBA = \\angle BAC = \\alpha$. We see that $\\frac{\\pi}{2} = \\angle CBA = \\angle DBA + \\angle FBD + \\angle CBF = 3\\alpha$, which implies $\\alpha = \\frac{\\pi}{6}$. The triangle $ABC$ is one half of an equilateral triangle, so $\\frac{|AB|}{|BC|} = \\sqrt{3}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41506,
    "subject": "Mathematics (Multi-modal)",
    "question": "Twelve balls are numbered by the numbers $1$, $2$, $3$, $\\ldots$, $12$. Each ball is coloured either red or green, so that the following two conditions are satisfied:\n\na. If two balls marked by different numbers $a$ and $b$ are coloured red and $a+b < 13$, then the ball marked by the number $a+b$ is coloured red, too.\n\nb. If two balls marked by different numbers $a$ and $b$ are coloured green and $a+b < 13$, then the ball marked by the number $a+b$ is also coloured green.\n\nHow many ways are there of colouring the balls?",
    "options": [],
    "answer": "6",
    "solution": "Assume that the ball denoted by $1$ is red. If the ball denoted by $2$ is also red, then $1+2=3$, $1+3=4$, $\\ldots$, $1+11=12$ are also red. So in this case all balls are red.\n\nIf the ball denoted by $2$ is green, we have to consider two more cases. If the ball number $3$ is red, then $1+3=4$, $1+4=5$, $\\ldots$, $1+11=12$ are also red. If the ball number $3$ is green, then $2+3=5$ is also green. Since $1+4=5$, $5$ is green and $1$ is red, the ball denoted by $4$ cannot be red and must therefore be green. Then $4+2=6$, $5+2=7$, $6+2=8$, $\\ldots$, $9+2=11$, $10+2=12$ are also green.\n\nThere were three possible cases. If we exchange red and green in the arguments above, we get three more cases. There are $6$ cases altogether, namely: all balls are red, all balls are green, all balls but number $1$ are green, all balls but number $1$ are red, all balls but number $2$ are red or all balls but number $2$ are green.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 41507,
    "subject": "Mathematics (Multi-modal)",
    "question": "Peter has $111$ red and $111$ blue marbles. Every day, Peter's uncle lets him exchange either $11$ red marbles for $7$ blue marbles or $20$ blue marbles for $28$ red marbles.\n\na. Can Peter increase the total number of marbles he has by $20$ after several exchanges?\n\nb. Can Peter increase the total number of marbles he has by $33$ after several exchanges?\n\nc. Can Peter ever have three times as many blue marbles as he has red ones?",
    "options": [],
    "answer": "a: Yes; b: No; c: No",
    "solution": "a. Peter can increase the number of marbles by $20$. The first three times he should exchange $20$ blue marbles for $28$ red ones. He will then have $111 - 60 = 51$ blue marbles and $111 + 3 \\cdot 28 = 111 + 84 = 195$ red ones. Then, he should exchange $11$ red marbles for $7$ blue ones. He will end up with $184$ red and $58$ blue marbles. He will have $242$ marbles altogether, which is $20$ more than the initial $222$.\n\nb. Exchanging $11$ red marbles for $7$ blue marbles decreases the total amount of marbles by $4$. Exchanging $20$ blue marbles for $28$ red marbles increases the total amount of marbles by $8$. The initial amount of marbles is even, so the total will remain even after any number of exchanges. Hence, the total number of marbles cannot be increased by $33$ since this number is odd.\n\nc. Assume that this can be done. Let $x$ denote the number of red marbles. Then there are $3x$ blue marbles and the total number of marbles is $4x$. After each exchange the number of marbles increases or decreases by a multiple of $4$. Since there were $222$ marbles at the beginning, only the numbers of the form $222 + 4k$ can be reached. The equation $222 + 4k = 4x$ has no integer solutions since $222$ is not divisible by $4$. Hence, Peter can never have three times as many blue marbles as red.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41508,
    "subject": "Mathematics (Multi-modal)",
    "question": "A red box contains twelve balls numbered from $1$ to $12$. Jan moves some of the balls into the green box. He then realizes that for any two balls from the green box the following is true: if these two balls are numbered $a$ and $b$, then the ball numbered $|a-b|$ is in the red box. At most how many balls has Jan moved to the green box?",
    "options": [],
    "answer": "6",
    "solution": "Jan can move $6$ balls. If he moves all odd-numbered balls, then the difference of the numbers on any two of them is an even number, which is clearly not written on any of the balls in the green box.\n\nNow, assume that Jan moved at least $7$ balls to the green box and denote the numbers on these balls by $a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7$. Then $a_7 - a_1$, $a_6 - a_1$, $a_5 - a_1$, $a_4 - a_1$, $a_3 - a_1$, $a_2 - a_1$ are six different positive integers smaller than $12$. The balls with these numbers on them should be in the red box, but this is not possible since the red box contains at most five balls.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41509,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ chameleons standing in a circle. Some of them are colored red and the others are colored green. Every minute, some of the chameleons change their color from red to green or vice versa in accordance with the following rule: a chameleon changes its color if and only if both its neighbours have the same color. Suppose that after $2n$ minutes all chameleons have the same color as at the beginning. Prove that there are two chameleons in the circle that have changed their color the same number of times.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because at the end the chameleons have the same color as at the beginning, they must have changed their color an even number of times, hence $0$ times, $2$ times, ... or $2n$ times.\n\nFirst suppose there exists a chameleon that has never changed its color. Its neighbours then have had the same color every minute, wherefore they have changed their color the same number of times. Similarly, if there exists a chameleon that has changed its color $2n$ times, its neighbours have had different colors every minute. Then, again, the neighbouring chameleons have changed their color the same number of times.\n\nLastly, we need to consider the case when each of the chameleons has changed its color at least $2$ times and less than $2n$ times. In this case, we only have $n-1$ possibilities for the number of times that a chameleon has changed its color. Then, according to the Dirichlet's principle, there exist two chameleons that have changed their color the same number of times.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41510,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $[x]$ be the greatest integer that is not greater than the real number $x$, and let $[[x]] = x - [x]$. Solve the equation\n$$\n3[[x]] = x[x] + 1.\n$$",
    "options": [],
    "answer": "x = -1/2 and x = 1/3",
    "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\nIf $n = 0$, we get the equation $3a = 1$, hence $a = x = \\frac{1}{3}$.\n\nIf $n = 1$, we get the equation $2a = 2$, which has no solutions strictly smaller than 1.\n\nNow let $n < 0$. Then $3 > 3a = n^2 + an + 1 > n^2 + n + 1$, hence $0 > n^2 + n - 2 = (n - 1)(n + 2)$.\n\nBecause $n < 0$, we get $n = -1$. From this we derive $4a = 2$, hence $a = \\frac{1}{2}$ and $x = -\\frac{1}{2}$.\n\nThe only solutions of the equation are thus $x = -\\frac{1}{2}$ and $x = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41511,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $a$, $b$, $c$ and $d$ such that\n$$\n2a^2 + 3b^2 = c^2 + 6d^2.\n$$",
    "options": [],
    "answer": "(0, 0, 0, 0)",
    "solution": "One solution is straightforward: $a = b = c = d = 0$. We will prove that it is the only one. Suppose there is another solution $(a, b, c, d)$. We may suppose that the integers $a$, $b$, $c$ and $d$ are coprime, otherwise their greatest common divisor could be deleted from the equation (because not all numbers are equal to $0$, their greatest common divisor exists). The equation can be reorganized into\n$$\n2a^2 - c^2 = 6d^2 - 3b^2.\n$$\nBecause the right side of the equation is divisible by $3$, the same must hold for the left side of the equation. The square of a natural number gives a remainder of $0$ or $1$ when divided by $3$. From this we conclude that the numbers $a^2$ and $c^2$ must give a remainder of $0$ when divided by $3$, hence $a$ and $c$ must be divisible by $3$. The left side of the equation is thus divisible by $9$, and the same must hold for the right side of the equation. Consequently, $2d^2 - b^2$ must be divisible by $3$. Like before we conclude that $b$ and $d$ are divisible by $3$. All numbers $a$, $b$, $c$ and $d$ are thus divisible by $3$, which is a contradiction since they are coprime.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41512,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all 3-digit natural numbers $n$ with non-zero digits that satisfy the following condition: if we delete the left digit of $n$, the obtained number is a divisor of $n$.",
    "options": [],
    "answer": "125, 225, 312, 315, 325, 375, 416, 425, 525, 612, 615, 624, 625, 675, 714, 725, 728, 735, 816, 825, 832, 912, 915, 918, 925, 936, 945, 975",
    "solution": "Let us write $n = \\overline{abc}$. The condition of the problem says that $\\overline{bc}$ divides $\\overline{abc} = 100a + \\overline{bc}$, hence it also divides $100a$. First suppose that $\\overline{bc}$ is not divisible by $5$. Then $\\overline{bc}$ must divide $4a$. Because $\\overline{bc}$ is a 2-digit natural number, $4a$ must also be a 2-digit natural number, hence $3 \\le a \\le 9$. The number $4a$ can thus take values $12$, $16$, $20$, $24$, $28$, $32$ or $36$. 2-digit divisors of these numbers are $12$, $16$, $20$ and $10$, $24$ and $12$, $28$ and $14$, $32$ and $16$, $36$ and $18$ and $12$, respectively. Allowing only non-zero digits, we get the following solutions: $312$, $416$, $624$, $612$, $728$, $714$, $832$, $816$, $936$, $918$ and $912$.\n\nIf, on the other hand, $\\overline{bc}$ is divisible by $5$, we get $c = 5$, because $c$ must be a non-zero digit. Hence $\\overline{bc} = 10b + 5 = 5(2b + 1)$, from which we conclude that $2b + 1$ must divide $20a$. Because $2b + 1$ is odd, it must divide $5a$. If $2b + 1$ is divisible by $5$, we get $b = 2$ (hence $2b + 1 = 5$) and $a$ is arbitrary or $b = 7$ (hence $2b + 1 = 15$) and $a$ is equal to $3$, $6$ or $9$. We get the following solutions: $125$, $225$, $325$, $425$, $525$, $625$, $725$, $825$ and $925$ as well as $375$, $675$ and $975$.\n\nIf, on the other hand, $2b + 1 \\ne 5$, it is coprime to $5$ and hence a divisor of $a$. It must hold $2b + 1 \\le 9$ and therefore $b \\le 4$ and $b \\ne 2$ (otherwise we would have $2b + 1 = 5$). If $b$ is successively equal to $4$, $3$ and $1$, then $a$ can be successively equal to $9$, $7$ and $3$ or $6$ or $9$. We get the following solutions: $945$, $735$, $315$, $615$ and $915$.\n\nTo summarize, the solutions are: $125$, $225$, $312$, $315$, $325$, $375$, $416$, $425$, $525$, $612$, $615$, $624$, $625$, $675$, $714$, $725$, $728$, $735$, $816$, $825$, $832$, $912$, $915$, $918$, $925$, $936$, $945$, $975$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41513,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $m$ such that for any odd natural number $k$ the number $2^m$ divides $k^m - 1$.",
    "options": [],
    "answer": "1, 2, 4",
    "solution": "If $m$ is odd, then $3^m \\equiv 3 \\pmod{4}$. Because $2^m$ divides $3^m - 1$, it must be $m = 1$. For $m = 1$, a natural number $k - 1$ is divisible by $2$ as soon as $k$ is odd.\n\nNow let $m$ be even, write $m = 2n$. Then $4^n$ divides $k^{2n} - 1$ for any odd natural number $k$. If $n$ is odd, then $3^{2n} \\equiv 9 \\pmod{16}$. Because $3^{2n} - 1$ is divisible by $4^n$, it must be $n = 1$ and $m = 2$. The opposite also holds: for any odd natural number $k$, $k^2 - 1$ is divisible by $4$.\n\nLet $n$ be even, write $n = 2l$. Then $16^l$ divides $k^{4l} - 1$ for any odd natural number $k$. In a special case, $16^l$ divides $81^l - 1 = (9^l - 1)(9^l + 1)$. The number $9^l + 1$ thus gives a remainder of $2$ when divided by $4$, hence $9^l - 1$ must be divisible by $2^{4l-1}$. We conclude $\\frac{1}{2} \\cdot 16^l \\le 9^l - 1$. We can prove by induction that $\\frac{1}{2} \\cdot 16^l > 9^l$ holds as soon as $l \\ge 2$, hence $l = 1$ and $m = 4$. On the other hand, if $m = 4$ and $k$ is odd, we get $k^4 - 1 = (k^2+1)(k+1)(k-1)$ where each of the factors is even and one of the last two factors is divisible by $4$. Hence, $k^4 - 1$ is divisible by $16$.\n\nThe solutions are $m = 1$, $m = 2$ and $m = 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41514,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $n \\ge 10$ with non-zero digits that satisfy the following condition: if any of the digits of $n$ is deleted, the obtained number is a divisor of $n$.",
    "options": [],
    "answer": "11, 12, 15, 22, 24, 33, 36, 44, 48, 55, 66, 77, 88, 99",
    "solution": "Suppose the decimal notation of a natural number $n$ is equal to $\\overline{a_k a_{k-1} \\dots a_2 a_1}$. The main condition of the problem says that the number $\\overline{a_k a_{k-1} \\dots a_2}$ divides the number $\\overline{a_k a_{k-1} \\dots a_2 a_1} = 10 \\cdot \\overline{a_k a_{k-1} \\dots a_2} + a_1$, hence it also divides the number $a_1$. Since $a_1 \\neq 0$, the number $\\overline{a_k a_{k-1} \\dots a_2}$ can contain at most one digit, hence $k=2$.\n\nIf now $n = \\overline{a_2 a_1}$, then $a_2$ must divide $a_1$, from which we conclude $a_2 = a_1$ or $a_2 \\le 4$. The number $n$ can thus only be equal to 99, 88, 77, 66, 55, 48, 44, 39, 36, 33, 28, 26, 24, 22, 19, 18, 17, 16, 15, 14, 13, 12 or 11. Among these numbers only 99, 88, 77, 66, 55, 48, 44, 36, 33, 24, 22, 15, 12 and 11 fulfill the condition of the problem.\n\nTo summarize, the solutions are 11, 12, 15, 22, 24, 33, 36, 44, 48, 55, 66, 77, 88, 99.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41515,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $a, b, c$ and $d$ that satisfy the equality\n$$\na\\sqrt{2} + b\\sqrt{5} + c = d\\sqrt{10}.\n$$",
    "options": [],
    "answer": "a=0, b=0, c=0, d=0",
    "solution": "One solution is straightforward: $a = b = c = d = 0$. We will prove that it is the only one. Suppose there is another solution $(a, b, c, d)$. We may suppose that the integers $a, b, c$ and $d$ are coprime, otherwise their greatest common divisor could be deleted from the equation (because not all numbers are equal to 0, their greatest common divisor exists). The equation can be reorganized into $a\\sqrt{2} + b\\sqrt{5} = d\\sqrt{10} - c$ and squared to obtain\n$$\n2a^2 + 5b^2 - c^2 - 10d^2 = 2\\sqrt{10}(cd - ab).\n$$\nBecause $\\sqrt{10}$ is not a rational number, we conclude $2a^2 + 5b^2 - c^2 - 10d^2 = 0$ or\n$$\n2a^2 - c^2 = 10d^2 - 5b^2.\n$$\nBecause the right side of the equation is divisible by 5, the same must hold for the left side of the equation. The square of a natural number gives a remainder of 0, 1 or 4 when divided by 5. From this we conclude that the numbers $a^2$ and $c^2$ must give a remainder of 0 when divided by 5, hence $a$ and $c$ must be divisible by 5. The left side of the equation is thus divisible by 25, and the same must hold for the right side of the equation. Consequently, $2d^2 - b^2$ must be divisible by 5. Like before we conclude that $b$ and $d$ are divisible by 5. All numbers $a, b, c$ and $d$ are thus divisible by 5, which is a contradiction since they are coprime.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41516,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any positive real numbers $a$, $b$ and $c$ the following holds:\n$$\na + \\sqrt{ab} + \\sqrt[3]{abc} \\le \\frac{4}{3}(a + b + c).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Recall that the inequality of arithmetic and geometric means states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list of numbers. From this we get\n$$\na + \\sqrt{ab} + \\sqrt[3]{abc} = a + \\sqrt{\\frac{a}{2} \\cdot 2b} + \\sqrt[3]{\\frac{a}{4} \\cdot b \\cdot 4c} \\le a + \\frac{1}{2}\\left(\\frac{a}{2} + 2b\\right) + \\frac{1}{3}\\left(\\frac{a}{4} + b + 4c\\right) = \\frac{4}{3}(a+b+c).\n$$\nHere, equality holds if and only if $a = 4c = 16c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41517,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTilen je metal igralno kocko. V tabeli je predstavil število metov za posmezne pike.\n\n| število pik | 1 | 2 | 3 | 4 | 5 | 6 |\n| :--- | :---: | :--- | :---: | :---: | :---: | :---: |\n| število metov | $3(x-2)$ | $2x$ | $2(x-1)$ | $x+2$ | $2x+1$ | $x$ |\n\na) Kolikokrat je vrgel posamezno število pik, če število šestic predstavlja natanko $10\\%$ vseh metov?\n\nb) Izračunaj povprečno število padlih pik.\n\nc) Nariši frekvenčni kolač.\n\n(6 točk)",
    "options": [],
    "answer": "a) Counts: 1 appears 9 times, 2 appears 10 times, 3 appears 8 times, 4 appears 7 times, 5 appears 11 times, 6 appears 5 times. b) Average number of pips: 3.32. c) Pie chart sector angles: 1 → 64.8°, 2 → 72°, 3 → 57.6°, 4 → 50.4°, 5 → 79.2°, 6 → 36°.",
    "solution": "Solution:\n\na)\nIzračunamo skupno število vseh metov:\n$$\n3(x-2) + 2x + 2(x-1) + x+2 + 2x+1 + x = 11x - 5.\n$$\nZapišemo enačbo:\n$$\n10\\% (11x - 5) = x.\n$$\nRešimo linearno enačbo:\n$$\n0.1 (11x - 5) = x \\\\\n11x - 5 = 10x \\\\\nx = 5.\n$$\nZ izračunanim $x$ lahko nato določimo število metov za posamezno številko:\n\n- enka: $3(x-2) = 3(5-2) = 9$\n- dvojka: $2x = 2 \\cdot 5 = 10$\n- trojka: $2(x-1) = 2 \\cdot 4 = 8$\n- štirica: $x+2 = 5+2 = 7$\n- petica: $2x+1 = 2 \\cdot 5 + 1 = 11$\n- šestica: $x = 5$\n\nb)\nPovprečno število pik izračunamo po formuli za aritmetično sredino:\n$$\n\\frac{1 \\cdot 9 + 2 \\cdot 10 + 3 \\cdot 8 + 4 \\cdot 7 + 5 \\cdot 11 + 6 \\cdot 5}{50} = 3,32.\n$$\n\nc)\nZa posamezno število pik izračunamo velikost središčnega kota:\n\n- enka: $\\frac{9}{50} \\cdot 360^{\\circ} = 64,8^{\\circ}$\n- dvojka: $\\frac{10}{50} \\cdot 360^{\\circ} = 72^{\\circ}$\n- trojka: $\\frac{8}{50} \\cdot 360^{\\circ} = 57,6^{\\circ}$\n- štirica: $\\frac{7}{50} \\cdot 360^{\\circ} = 50,4^{\\circ}$\n- petica: $\\frac{11}{50} \\cdot 360^{\\circ} = 79,2^{\\circ}$\n- šestica: $\\frac{5}{50} \\cdot 360^{\\circ} = 36^{\\circ}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41518,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa dani vrednosti $a=\\sqrt{3}$ in $b=\\sqrt{2}$ izračunaj natančno vrednost izraza\n$$\n\\sqrt[4]{a^{-1} b^{3} \\cdot \\sqrt{6}} \\cdot \\sqrt[3]{\\frac{6 a^{4} b^{-2}}{\\sqrt{2}}}\n$$",
    "options": [],
    "answer": "3 * 2^(1/3)",
    "solution": "Solution:\n\nVstavimo vrednosti $a=\\sqrt{3}$ in $b=\\sqrt{2}$ v dani izraz. Dobimo $\\sqrt[4]{(\\sqrt{3})^{-1} \\cdot (\\sqrt{2})^{3} \\cdot \\sqrt{6}} \\cdot \\sqrt[3]{\\frac{6 \\cdot (\\sqrt{3})^{4} \\cdot (\\sqrt{2})^{-2}}{\\sqrt{2}}}$.\n\nPoenostavimo prvi faktor:\n$$(\\sqrt{3})^{-1} = 3^{-1/2}$$\n$$(\\sqrt{2})^{3} = 2^{3/2}$$\n$\\sqrt{6} = 6^{1/2}$\nTorej:\n$$\n\\sqrt[4]{3^{-1/2} \\cdot 2^{3/2} \\cdot 6^{1/2}}\n$$\n$6^{1/2} = (2 \\cdot 3)^{1/2} = 2^{1/2} 3^{1/2}$\nTorej:\n$$\n3^{-1/2} \\cdot 2^{3/2} \\cdot 2^{1/2} 3^{1/2} = 2^{3/2 + 1/2} 3^{-1/2 + 1/2} = 2^{2} 3^{0} = 4\n$$\nTorej prvi faktor je $\\sqrt[4]{4} = \\sqrt{2}$.\n\nDrugi faktor:\n$$(\\sqrt{3})^{4} = 3^{2}$$\n$$(\\sqrt{2})^{-2} = 2^{-1}$\nTorej števec:\n$6 \\cdot 3^{2} \\cdot 2^{-1} = 6 \\cdot 9 \\cdot 2^{-1} = 54 \\cdot 2^{-1} = 27$\nImenovalec: $\\sqrt{2}$\nTorej drugi faktor je $\\sqrt[3]{\\frac{27}{\\sqrt{2}}}$\n\n$\\sqrt[3]{\\frac{27}{\\sqrt{2}}} = \\frac{\\sqrt[3]{27}}{\\sqrt[3]{\\sqrt{2}}} = \\frac{3}{2^{1/6}}$\n\nTorej celoten izraz je:\n$$\n\\sqrt{2} \\cdot \\frac{3}{2^{1/6}} = \\frac{3 \\sqrt{2}}{2^{1/6}}\n$$\nRacionaliziramo imenovalec:\n$2^{1/6} = \\sqrt[6]{2}$\nTorej $\\frac{3 \\sqrt{2}}{\\sqrt[6]{2}}$\nPomnožimo števec in imenovalec z $\\sqrt[6]{2^{5}}$:\n$$\n\\frac{3 \\sqrt{2} \\sqrt[6]{2^{5}}}{\\sqrt[6]{2^{6}}} = \\frac{3 \\sqrt{2} \\sqrt[6]{2^{5}}}{2}\n$$\n$\\sqrt{2} \\sqrt[6]{2^{5}} = \\sqrt[6]{2^{6}} = 2$\nTorej $\\frac{3 \\cdot 2}{2} = 3$\nAmpak upoštevamo še $\\sqrt[6]{2^{8}}$:\n$\\frac{3 \\sqrt[6]{2^{8}}}{2} = \\frac{3 \\cdot 2^{8/6}}{2} = \\frac{3 \\cdot 2^{4/3}}{2} = \\frac{3}{2} \\cdot 2^{4/3} = 3 \\sqrt[3]{2}$\n\nTorej rešitev je:\n$$\n3 \\sqrt[3]{2}\n$$\n\n\nDrugi način:\n\nPoenostavimo prvi faktor $\\sqrt[8]{a^{-2} b^{6} \\cdot 6}$ in drugi faktor $\\sqrt[6]{\\frac{6^{2} a^{8} b^{-4}}{2}}$.\n\nPrvi faktor:\n$a^{-2} = (\\sqrt{3})^{-2} = 3^{-1}$\n$b^{6} = (\\sqrt{2})^{6} = 2^{3}$\nTorej $3^{-1} 2^{3} 6 = 2^{3} \\cdot 6 \\cdot 3^{-1} = 8 \\cdot 6 \\cdot 3^{-1} = 48 \\cdot 3^{-1} = 16$\nTorej $\\sqrt[8]{16} = 2^{4/8} = 2^{1/2} = \\sqrt{2}$\n\nDrugi faktor:\n$6^{2} = 36$\n$a^{8} = (\\sqrt{3})^{8} = 3^{4}$\n$b^{-4} = (\\sqrt{2})^{-4} = 2^{-2}$\nTorej števec: $36 \\cdot 3^{4} \\cdot 2^{-2} = 36 \\cdot 81 \\cdot 2^{-2} = 2916 \\cdot 2^{-2} = 729$\nImenovalec: $2$\nTorej $\\sqrt[6]{\\frac{729}{2}}$\n$729 = 27^{2} = 3^{6}$\nTorej $\\sqrt[6]{\\frac{3^{6}}{2}} = \\frac{3}{2^{1/6}}$\n\nTorej skupaj $\\sqrt{2} \\cdot \\frac{3}{2^{1/6}} = \\frac{3 \\sqrt{2}}{2^{1/6}}$\nRacionaliziramo kot prej in dobimo rešitev $3 \\sqrt[3]{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41519,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa 3-mestna naravna števila $n$ z neničelnimi števkami, za katera velja: če številu $n$ odstranimo levo števko, dobimo število, ki deli $n$.",
    "options": [],
    "answer": "125, 225, 312, 315, 325, 375, 416, 425, 525, 612, 615, 624, 625, 675, 714, 725, 728, 735, 816, 825, 832, 912, 915, 918, 925, 936, 945, 975",
    "solution": "Solution:\n\nPišimo $n=\\overline{a b c}$. Pogoj naloge pove, da $\\overline{b c}$ deli $\\overline{a b c}=100 a+\\overline{b c}$, torej deli tudi $100a$.\n\nDenimo najprej, da $\\overline{b c}$ ni deljivo s $5$. Potem mora $\\overline{b c}$ deliti $4 a$. Ker je $\\overline{b c}$ dvomestno, mora biti tudi $4 a$ dvomestno, torej je $3 \\leq a \\leq 9$. Torej $4 a$ lahko zavzame vrednosti $12,16,20,24,28,32$ ali $36$. Dvomestni delitelji teh števil so zaporedoma $12, 16, 20$ in $10,24$ in $12,28$ in $14,32$ in $16$ ter $36$ in $18$ in $12$. Ker morajo biti števke neničelne, dobimo rešitve $312,416,624,612$, $728,714,832,816,936,918$ in $912$.\n\nČe pa je $\\overline{b c}$ deljivo s $5$, je $c=5$, saj mora biti $c$ neničelna števka. Torej je $\\overline{b c}=10 b+5=5(2 b+1)$ in zato mora $(2 b+1)$ deliti $20 a$. Ker je $2 b+1$ liho število, mora deliti celo $5 a$. Če je $2 b+1$ deljiv s $5$, je bodisi $b=2$ (torej $2 b+1=5$) in $a$ poljuben, bodisi $b=7$ (torej $2 b+1=15$) in $a$ enak $3,6$ ali $9$. Tako dobimo rešitve $125,225,325,425,525,625,725,825$ in $925$ ter $375,675$ in $975$.\n\nČe pa $2 b+1 \\neq 5$, potem je tuj $5$ in zato mora deliti celo $a$. Torej mora biti $2 b+1 \\leq 9$ in zato $b \\leq 4$ in $b \\neq 2$ (sicer bi bil $2 b+1=5$). Če je $b$ zaporedoma enak $4,3$ in $1$, potem je lahko $a$ zaporedoma enak $9,7$ in $3$ ali $6$ ali $9$. Tako dobimo še rešitve $945,735,315,615$ in $915$.\n\nPovzemimo: vse rešitve so $125,225,312,315,325,375,416,425,525,612,615,624,625$, $675,714,725,728,735,816,825,832,912,915,918,925,936,945,975$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41520,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$, denote $D$ the orthogonal projection of the point $C$ onto the line $AB$, denote $E$ the orthogonal projection of the point $D$ onto the line $AC$, and denote $P$ the midpoint of the line segment $CD$. Let $K_1$ be the circumscribed circle of the triangle $ABC$, and let $K_2$ be the circle of radius $CD$ and center $C$. Prove that the line $EP$ passes through the intersection points of the circles $K_1$ and $K_2$ if and only if the angle $\\angle ACB$ is a right angle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $F$ the orthogonal projection of the point $D$ onto the line $BC$.\n![](attached_image_1.png)\nObviously, the points $E$ and $F$ lie on the sides $AC$ and $BC$, respectively. According to Euclid's theorem, $|EC| \\cdot |EA| = |CD|^2 - |EC|^2$ and $|FC| \\cdot |FB| = |CD|^2 - |FC|^2$,\n\nhence the powers of the point $E$ with respect to the circles $K_1$ and $K_2$ are equal. The same holds for $F$. From this we conclude that $EF$ is the radical axis (or power line) of the circles $K_1$ and $K_2$, hence it passes through the intersection point of these circles, where circles $K_1$ and $K_2$ obviously intersect.\nIt is sufficient to prove that the angle $\\angle ACB$ is a right angle if and only if the point $P$ lies on the line $EF$. If $\\angle ACB = \\frac{\\pi}{2}$, then the quadrilateral $EDFC$ is a rectangle, and $P$ lies on both diagonals of the rectangle because it is the midpoint of one diagonal. Now suppose that $P$ lies on $EF$. According to Thales' theorem, the quadrilateral $EDFC$ is a cyclic quadrilateral with the center of the circumscribed circle in $P$. Using the inscribed angle theorem, we derive $\\angle ECF = \\frac{1}{2}\\angle EPF = \\frac{\\pi}{2}$. This proves the claim.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41521,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a tangential quadrilateral whose inscribed circle touches the sides $AB$, $BC$, $CD$ and $DA$ in points $P$, $Q$, $R$ and $S$, respectively. Let the lines $AB$ and $CD$ intersect at a point $E$, and let the lines $AD$ and $BC$ intersect at a point $F$. Denote $I_1$ the intersection point of the bisectors of the angles $\\angle QPS$ and $\\angle PSR$, and denote $I_2$ the intersection point of the bisectors of the angles $\\angle RQP$ and $\\angle SRQ$. Prove that the lines $AC$, $EF$ and $I_1I_2$ either intersect at a common point or are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First suppose that the lines $PQ$ and $RS$ intersect. Denote $T$ their point of intersection. Due to symmetry we may assume that $Q$ lies between $P$ and $T$ and $R$ lies between $S$ and $T$. We shall prove that the lines $EF$, $I_1I_2$ and $AC$ pass through the point $T$.\n\nIn the construction, $I_1$ is the intersection point of the bisectors of two angles in the triangle $PTS$, hence the center of the inscribed circle of the triangle $PTS$. The point $I_1$ lies on the bisector of the angle $\\angle STP$. Similarly, $I_2$ is the intersection point of the bisectors of the exterior angles at $QR$ in the triangle $QTR$, hence the center of the inscribed circle to the side $QR$ of the triangle $QTR$. The point $I_2$ lies on the bisector of the angle $\\angle RTQ$. Because $S$, $R$ and $T$ are collinear, and $P$, $Q$ and $T$ are collinear, $I_1$, $I_2$ and $T$ must also be collinear.\n\nDenote $T'$ the intersection point of the lines $EF$ and $PQ$, and denote $T''$ the intersection point of the lines $EF$ and $RS$. According to the Menelaus' theorem, we have\n$$\n\\frac{BP}{PF} \\cdot \\frac{FT'}{T'E} \\cdot \\frac{EQ}{QB} = -1 \\quad \\text{and} \\quad \\frac{DR}{RF} \\cdot \\frac{FT''}{T''E} \\cdot \\frac{ES}{SD} = -1.\n$$\nConsider the lengths of the tangent segments to get $|BP| = |BQ|$, $|DS| = |DR|$, $|FP| = |FR|$ and $|EQ| = |ES|$. Because of the above equalities and due to the obvious fact that $E$ and $F$ lie outside the quadrilateral $ABCD$, we have $\\frac{FT'}{T'E} = \\frac{|PF|}{|QE|} = \\frac{|RF|}{|SE|} = \\frac{FT''}{T''E}$. Hence $T' = T''$, and the lines $EF$, $PQ$ and $RS$ intersect in a common point $T \\in EF$.\n\nLet $T_1$ be the intersection point of the lines $AC$ and $PQ$, and let $T_2$ be the intersection point of the lines $AC$ and $RS$. We again apply the Menelaus' theorem to get\n$$\n\\frac{AP}{PB} \\cdot \\frac{BQ}{QC} \\cdot \\frac{CT_1}{T_1A} = -1 \\quad \\text{and} \\quad \\frac{AS}{SD} \\cdot \\frac{DR}{RC} \\cdot \\frac{CT_2}{T_2A} = -1.\n$$\nBecause $P$, $Q$, $R$ and $S$ lie each on a side of the quadrilateral $ABCD$, from the equality of the tangent segments $|AP| = |AS|$, $|BP| = |BQ|$, $|CQ| = |CR|$ and $|DR| = |DS|$ we derive $\\frac{CT_1}{T_1A} = -\\frac{|QC|}{|PA|} = -\\frac{|RC|}{|SA|} = \\frac{CT_2}{T_2A}$. Hence, $T_1 = T_2$ and the line $AC$ passes through the point $T$.\n\nNow suppose $PQ \\parallel RS$. Because of the equality of the tangent segments, we get $\\angle BPQ = \\frac{1}{2}(\\pi - \\angle QBP)$, $\\angle SPA = \\angle ASP = \\frac{1}{2}(\\pi - \\angle PAS)$ and $\\angle RSD = \\frac{1}{2}(\\pi - \\angle SDR)$, hence $\\angle QPS = \\frac{1}{2}(\\angle PAS + \\angle QBP)$ and $\\angle PSR = \\frac{1}{2}(\\angle PAS + \\angle SDR)$. Since $PQ \\parallel RS$, we have $\\angle QPS + \\angle PSR = \\pi$, and hence $\\angle BAD + \\frac{1}{2}(\\angle CBA + \\angle ADC) = \\pi$. We conclude $\\angle BAD = \\angle DCB$, which means that the quadrilateral $ABCD$ is a deltoid. The lines $EF$, $I_1I_2$ and $AC$ are thus parallel.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41522,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there is no injective function $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(f(x) + y) = f(x + y) + f(2012) \\quad \\text{for all } x, y \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If we substitute $y = -x$ in the equation, we get $f(f(x)-x) = f(0)+f(2012)$. Since the right side of the equation is a constant, and $f$ is an injective function, $f(x)-x$ must also be a constant. Hence $f(x) = x + c$ for a real number $c$. If we substitute this in the initial equation, we get $x + y + 2c = x + y + 2c + 2012$, which gives us the contradiction $0 = 2012$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41523,
    "subject": "Mathematics (Multi-modal)",
    "question": "The integers $x$ and $y$ are such that $x + xy + y^2 = 1$ and $y(5 + x) \\ge 0$. What integer values can the expression $x - y$ take?",
    "options": [],
    "answer": "All integers less than or equal to −4, together with −3, −1, and 1.",
    "solution": "The equality gives us $x(1+y) = 1 - y^2 = (1+y)(1-y)$. If $y = -1$, the equality holds, and from the inequality we derive $-(5+x) \\ge 0$ or $x \\le -5$. Hence $x - y = x + 1 \\le -4$.\n\nIf, on the other hand, $y \\ne -1$, the equality reduces to $x = 1 - y$. Using the last relation in the inequality we derive $y(6-y) \\ge 0$, and hence $0 \\le y \\le 6$. We conclude $x - y = 1 - 2y \\in \\{1, -1, -3, -5, -7, -9, -11\\}$.\n\nThe expression $x - y$ can take all values smaller or equal to $-4$ as well as values $-3$, $-1$ and $1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41524,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a polynomial of degree $2$ with at least one non-integer coefficient. Suppose that for every integer $n$, the value $p(n)$ is also an integer. Prove that all the coefficients of the polynomial $q(x) = p(x) - \\frac{1}{2}x(x+1)$ are integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $P(x) = \\alpha x^2 + \\beta x + \\gamma$. Then $P(0) = \\gamma$, $P(1) - P(0) = (\\alpha + \\beta + \\gamma) - \\gamma = \\alpha + \\beta$ and $P(-1) - P(0) = (\\alpha - \\beta + \\gamma) - \\gamma = \\alpha - \\beta$ must be integers. We thus know that $\\alpha + \\beta, \\alpha - \\beta \\in \\mathbb{Z}$. If we add and subtract the last two numbers, we notice that $2\\alpha$ and $2\\beta$ must also be integers. If $2\\alpha$ is even, then $\\alpha$ is an integer. Because $\\alpha + \\beta \\in \\mathbb{Z}$, $\\beta$ is also an integer, and the condition of the problem is not fulfilled. From this we conclude that $2\\alpha$ is odd. A similar reasoning gives that $2\\beta$ is also odd.\n\nWe can thus write $2\\alpha = 2a + 1$, $a \\in \\mathbb{Z}$ and $2\\beta = 2b + 1$, $b \\in \\mathbb{Z}$. If we substitute these into the polynomial $Q$, we get $Q(x) = a x^2 + b x + \\gamma$. All its coefficients are indeed integers.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41525,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be different real numbers such that the equations $x^2 + a x + b = 0$ and $x^2 + b x + a = 0$ have some common solution. How much is $a + b$?",
    "options": [],
    "answer": "-1",
    "solution": "If we equalize the left sides of the quadratic equations, we get $x^2 + a x + b = x^2 + b x + a$. Hence $a x + b = b x + a$ and $(x - 1)(a - b) = 0$. Because $a \\neq b$, it follows that $x = 1$. If we now substitute $x = 1$ in one of the equations, we get $a + b = -1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41526,
    "subject": "Mathematics (Multi-modal)",
    "question": "Someone has written the numbers $1$, $2$, $\\ldots$, $33$ on a chalkboard. In each step, we choose two (not necessarily different) numbers on the chalkboard such that one divides the other. We then erase these two numbers and write their quotient, which is a natural number, on the chalkboard. We repeat the process until there is a pair of numbers on the chalkboard such that one of the numbers divides the other. At least how many numbers stay written on the chalkboard?",
    "options": [],
    "answer": "7",
    "solution": "Let $P_k$ be the product of all the numbers on the chalkboard after the $k$\\text{th}$ step. If in the $k$\\text{th}$ step we choose the numbers $a > b$ such that $b$ divides $a$, then\n$$\nP_k = \\frac{P_{k-1}}{ab} \\cdot \\frac{a}{b} = \\frac{P_{k-1}}{b^2}.\n$$\nThus, after each step, the product of all the numbers on the chalkboard is divisible by a square of a natural number. That is, the parity of the exponents in the prime factorization of the product of all the numbers on the chalkboard does not change. Because at the beginning the product is\n$$\nP_0 = 1 \\cdot 2 \\cdots 33 = 2^{31} \\cdot 3^{15} \\cdot 5^7 \\cdot 7^4 \\cdot 11^3 \\cdot 13^2 \\cdot 17 \\cdot 19 \\cdot 23 \\cdot 29 \\cdot 31,\n$$\nthe product of all the numbers that stay on the chalkboard is a multiple of the number $2 \\cdot 3 \\cdot 5 \\cdot 11 \\cdot 17 \\cdot 19 \\cdot 23 \\cdot 29 \\cdot 31$. Since the prime numbers $17$, $19$, $23$, $29$ and $31$ do not divide any other number on the chalkboard, they stay on the chalkboard till the end. The product of all the other numbers on the chalkboard at the end is at least $2 \\cdot 3 \\cdot 5 \\cdot 11 = 330 > 33$. Thus, at the end, there must be two additional prime numbers that are greater than $16$ on the chalkboard. Altogether there must be at least $7$ such prime numbers.\n\nWe can get $7$ numbers in the described process as follows. First successively erase the following pairs:\n$(33, 11)$, $(25, 5)$, $(27, 9)$, $(28, 14)$, $(26, 13)$, $(21, 7)$, $(30, 10)$, $(20, 5)$, $(32, 16)$, $(24, 12)$, $(18, 6)$, $(8, 4)$, $(3, 3)$, $(3, 3)$, $(3, 3)$, $(4, 2)$, $(2, 2)$, $(2, 2)$, $(2, 2)$. Then erase all the ones to get the numbers $15$, $17$, $19$, $22$, $23$, $29$, $31$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41527,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 63 rows of seats on an airplane. Each row has 6 seats, and they are marked with the letters $A$, $B$, $C$, $D$, $E$ and $F$. On a flight, there were 193 passengers on the airplane. Prove that on this flight there were two rows in which the seats that were occupied were marked with the same letters.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Proof by contradiction: Suppose that there were no two rows in which the occupied seats were marked with the same letters. The number of subsets of the set $\\{A, B, C, D, E, F\\}$ is equal to $2^6 = 64$, which means that there was exactly one subset of $\\{A, B, C, D, E, F\\}$ that did not correspond to the occupation of any row on the airplane. Thus, the number of passengers was at most $6 \\cdot (\\binom{6}{2} + 5 \\cdot \\binom{6}{5} + 4 \\cdot \\binom{6}{4} + 3 \\cdot \\binom{6}{3}) + 2 \\cdot (\\binom{6}{2} + 6 \\cdot \\binom{6}{1}) = 192$, which is a contradiction. From this we conclude that there must have been two rows in which the occupied seats were marked with the same letters.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41528,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa celi števili $x$ in $y$ velja $x + x y + y^{2} = 1$ in $y(5 + x) \\geq 0$. Katerim celim številom je lahko enaka vrednost izraza $x - y$?",
    "options": [],
    "answer": "All integers less than or equal to -4, together with -3, -1, and 1",
    "solution": "Solution:\n\nIz enakosti sledi $x(1 + y) = 1 - y^{2} = (1 + y)(1 - y)$. Če je $y = -1$, je enakost izpolnjena, iz neenakosti pa potem sledi $-(5 + x) \\geq 0$ oziroma $x \\leq -5$. Tako je $x - y = x + 1 \\leq -4$.\n\nČe pa je $y \\neq -1$, potem lahko iz enakosti pokrajšamo $(1 + y)$, da dobimo $x = 1 - y$. Slednje vstavimo v neenakost, da dobimo $y(6 - y) \\geq 0$. Torej je $0 \\leq y \\leq 6$. Tako je $x - y = 1 - 2y \\in \\{1, -1, -3, -5, -7, -9, -11\\}$.\n\nIzraz $x - y$ lahko zavzame vse vrednosti manjše ali enake $-4$ ter vrednosti $-3, -1$ in $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41529,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDokaži, da ne obstaja injektivna funkcija $f: \\mathbb{R} \\rightarrow \\mathbb{R}$, za katero bi veljalo\n\n$$\nf(f(x)+y)=f(x+y)+f(2012) \\quad \\text{ za vse } x, y \\in \\mathbb{R}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nČe v enačbo vstavimo $y=-x$, dobimo $f(f(x)-x)=f(0)+f(2012)$. Ker je na desni strani konstanta, $f$ pa je injektivna funkcija, mora biti tudi $f(x)-x$ konstanta. Torej je $f(x)=x+c$ za neko realno število $c$. Če to vstavimo v začetno enačbo, dobimo $x+y+2c=x+y+2c+2012$, kar nam da protislovje $0=2012$.\n\n2. način. Če v enačbi zamenjamo $x$ in $y$, dobimo $f(f(y)+x)=f(x+y)+f(2012)=f(f(x)+y)$. Zaradi injektivnosti funkcije $f$ sledi $f(y)+x=f(x)+y$ oziroma $f(x)-x=f(y)-y$ za vsa realna števila $x$ in $y$. To pomeni, da je $f(x)-x$ konstanta. Zaključimo enako kot v prvi rešitvi.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41530,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nReši enačbo $\\left(1-\\left(1+x^{-2}\\right)^{-1}\\right)^{-1}=3,25$.",
    "options": [],
    "answer": "-1.5, 1.5",
    "solution": "Solution:\n\nUredimo notranji oklepaj $\\left(1-\\left(1+\\frac{1}{x^{2}}\\right)^{-1}\\right)^{-1}=3,25$, razširimo na skupni imenovalec $\\left(1-\\left(\\frac{x^{2}+1}{x^{2}}\\right)^{-1}\\right)^{-1}=3,25$. Upoštevamo negativni eksponent $\\left(1-\\frac{x^{2}}{x^{2}+1}\\right)^{-1}=3,25$ in znova razširimo na skupni imenovalec $\\left(\\frac{x^{2}+1-x^{2}}{x^{2}+1}\\right)^{-1}=3,25$. Znova upoštevamo negativni eksponent $\\frac{x^{2}+1}{1}=3,25$, uredimo $x^{2}=2,25$ in korenimo. Izračunamo rešitvi $x_{1}=1,5$ in $x_{2}=-1,5$.\n\nUpoštevanje negativnega eksponenta $1+\\frac{1}{x^{2}}$ ..... 1 točka\nRazširitev na skupni imenovalec $\\frac{x^{2}+1}{x^{2}}$ ..... 1 točka\nUpoštevanje negativnega eksponenta $1-\\frac{x^{2}}{x^{2}+1}$ ..... 1 točka\nRazširitev na skupni imenovalec $\\frac{x^{2}+1-x^{2}}{x^{2}+1} \\ldots$ ..... 1 točka\nUreditev enačbe $x^{2}=2,25$ ..... 1 točka\nIzračunani rešitvi $x_{1}=1,5$ in $x_{2}=-1,5$ ..... 1 točka",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41531,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $p$ polinom stopnje $2$, katerega vsaj en koeficient ni celo število. Denimo, da je za vsako celo število $n$ tudi število $p(n)$ celo. Dokaži, da ima tedaj polinom $q(x) = p(x) - \\frac{1}{2} x(x+1)$ same celoštevilske koeficiente.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo $P(x) = \\alpha x^{2} + \\beta x + \\gamma$. Tedaj morajo biti števila $P(0) = \\gamma$, $P(1) - P(0) = (\\alpha + \\beta + \\gamma) - \\gamma = \\alpha + \\beta$ in $P(2) - 2 P(1) + P(0) = (4\\alpha + 2\\beta + \\gamma) - 2(\\alpha + \\beta + \\gamma) + \\gamma = 2\\alpha$ cela.\n\nOznačimo $2\\alpha = a \\in \\mathbb{Z}$, torej $\\alpha = \\frac{a}{2}$. Če je $a$ sod, je število $\\alpha$ celo, zato mora biti tudi število $\\beta$ celo, to pa je v protislovju s predpostavko naloge, saj so vsi koeficienti polinoma $P$ celoštevilski. Zato mora biti $a$ lih, torej oblike $a = 2a' + 1$, $a' \\in \\mathbb{Z}$. Tako je $\\alpha = a' + \\frac{1}{2}$. Od tod sledi, da mora biti tudi $\\beta$ oblike $\\beta = b + \\frac{1}{2}$, $b \\in \\mathbb{Z}$. Zato je $P(x) = a' x^{2} + \\frac{1}{2} x^{2} + b x + \\frac{1}{2} x + \\gamma$, torej ima polinom $Q(x) = a' x^{2} + b x + \\gamma$ res same celoštevilske koeficiente.\n\n\n2. način.\n\nOznačimo $P(x) = \\alpha x^{2} + \\beta x + \\gamma$. Tedaj morajo biti števila $P(0) = \\gamma$, $P(1) - P(0) = (\\alpha + \\beta + \\gamma) - \\gamma = \\alpha + \\beta$ in $P(-1) - P(0) = (\\alpha - \\beta + \\gamma) - \\gamma = \\alpha - \\beta$ cela. Torej vemo, da $\\alpha + \\beta, \\alpha - \\beta \\in \\mathbb{Z}$. Če zadnji dve števili seštejemo in odštejemo, ugotovimo, da morata biti celi tudi števili $2\\alpha$ in $2\\beta$. Če je $2\\alpha$ sodo število, je $\\alpha$ celo število. Ker $\\alpha + \\beta \\in \\mathbb{Z}$, je tudi $\\beta$ celo število - pogoj naloge zato ni izpolnjen. Tako mora biti $2\\alpha$ liho število. S podobnim sklepom ugotovimo, da je tudi $2\\beta$ liho število.\n\nTako lahko zapišemo $2\\alpha = 2a + 1$, $a \\in \\mathbb{Z}$ in $2\\beta = 2b + 1$, $b \\in \\mathbb{Z}$. To vstavimo v polinom $Q$ in dobimo $Q(x) = a x^{2} + b x + \\gamma$. Vsi njegovi koeficienti so res cela števila.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41532,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let a point $E$ lie on the side $CD$ of a square $ABCD$. Let a point $F$ lie on the line $AB$, but not on the line segment $AB$, and let $|BF| = |DE|$. Prove that the lines $AC$ and $EF$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $T$ be the intersection point of the lines $AC$ and $EF$. Because $|BF| = |DE|$, $|BC| = |DA|$ and $\\angle CBF = \\angle ADE = \\frac{\\pi}{2}$, the triangles $ADE$ and $CBF$ are congruent. We thus have $|AE| = |CF|$, and the quadrilateral $AFCE$ is an isosceles trapezoid. Hence $\\angle EFA = \\angle CAF$, which is also equal to $\\frac{\\pi}{4}$ because $AC$ is a diagonal of the square $ABCD$. We get $\\angle ATF = \\pi - \\angle EFA - \\angle CAF = \\frac{\\pi}{2}$, from which we conclude that the lines $AC$ and $EF$ are perpendicular.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41533,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDE$ be a cyclic pentagon in which $|CD| = |DE|$. Let the diagonals $AD$ and $BE$ intersect at point $K$, and let the diagonals $AC$ and $BD$ intersect at point $L$. Prove that the lines $KL$ and $EC$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "According to the inscribed angle theorem, we have $\\angle CED = \\angle CAD$ and $\\angle DCE = \\angle DBE$. Because $ECD$ is an isosceles triangle with the top angle at $D$, we have $\\angle CED = \\angle DCE$. Due to the collinearity of the points $A, L, C$ and the collinearity of the points $A, K, D$ we also have $\\angle LAK = \\angle LBK$, hence points $A, B, L, K$ are concyclic. From this we get $\\angle KLA = \\angle KBA = \\angle EBA = \\angle ECA$. The lines $KL$ and $EC$ are thus parallel.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41534,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of natural numbers $m$ and $n$ such that $2m^2 + n^2$ divides $3mn + 3m$.",
    "options": [],
    "answer": "(1, 1), (4, 2), (4, 10)",
    "solution": "Suppose $3mn + 3m = k(2m^2 + n^2)$ where $k$ is a natural number. The inequality between the arithmetic and geometric means gives\n$$\n3mn + 3m = 3m(n + 1) = k(2m^2 + n^2) \\geq 2\\sqrt{2}kmn,\n$$\nhence\n$$\nk \\leq \\frac{3(n + 1)}{2\\sqrt{2}n} = \\frac{3}{2\\sqrt{2}}\\left(1 + \\frac{1}{n}\\right) \\leq \\frac{3}{\\sqrt{2}}.\n$$\nWe conclude $k \\leq 2$.\n\nThe equation $k(2m^2 + n^2) = 3mn + 3m$ may be considered a quadratic equation for $n$:\n$$\nkn^2 - 3mn + 2km^2 - 3m = 0. \\quad (6)\n$$\nThe discriminant of this equation is equal to $D = 9m^2 - 8k^2m^2 + 12km$.\n\nIf $k = 2$, then $D = 24m - 23m^2$. Since the discriminant must be non-negative, the only solution is $m = 1$ and from the quadratic equation (6) we calculate $n = 1$.\n\nNow let $k = 1$. The discriminant of the quadratic equation (6) must be equal to a square of a non-negative integer $t$:\n$$\nt^2 = D = m^2 + 12m = (m + 6)^2 - 36.\n$$\nWe reorganize the above equation to get $36 = (m + 6 + t)(m + 6 - t)$. Because $t \\geq 0$ and $m > 0$, we get $m + 6 + t > 0$ and hence $m + 6 - t > 0$. The numbers $m + 6 + t$ and $m + 6 - t$ obviously have the same parity. We conclude $m + 6 + t = m + 6 - t = 6$ or $m + 6 + t = 18$ and $m + 6 - t = 2$. In the first case, we get $m = 0$, which is not a natural number. In the second case, we get $m = 4$ and $t = 8$, and the equation (6) has two solutions: $n = 2$ and $n = 10$.\n\nThe solutions of the problem are the following pairs of natural numbers: $(1, 1)$, $(4, 2)$ and $(4, 10)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41535,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $BE$ and $CF$ be heights of an acute triangle $ABC$. Two circles through points $A$ and $F$ touch the line $BC$ in points $P$ and $Q$, respectively, where $B$ lies between the points $C$ and $Q$. Prove that the lines $PE$ and $QF$ intersect on the circumscribed circle of the triangle $AEF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let all the angles in the solution be directed. Given an acute triangle $ABC$, denote $G$ the foot (lying on the side $BC$) of the altitude from $A$. The powers of the point $B$ with respect to the circles through points $A$ and $F$ give us $|BP|^2 = |BA| \\cdot |BF| = |BQ|^2$, hence $|BP| = |BQ|$.\n\nAccording to the Tales' theorem, the points $A$, $E$, $G$ and $B$ are concyclic, the same holds for the points $A$, $F$, $G$ and $C$. From the theorem about the power of a point we derive\n$$\n\\begin{aligned}\n|CA| \\cdot |CE| &= |CG| \\cdot |CB| = |BC|^2 - |BC| \\cdot |BG| = \\\\\n&= |BC|^2 - |BA| \\cdot |BF| = |BC|^2 - |BP|^2 = \\\\\n&= (|BC| + |BP|)(|BC| - |BP|) = (|BC| + |BQ|)(|BC| - |BP|).\n\\end{aligned}\n$$\nThe expression in the first brackets is positive, hence the expression in the second brackets is positive as well. We get $|BC| > |BP|$, which means that $P$ lies between $B$ and $C$, hence $|CA| \\cdot |CE| = |CQ| \\cdot |CP|$. The points $A$, $E$, $P$ and $Q$ are thus concyclic and $\\angle CPE = \\angle QAE$.\n\nAccording to the theorem about the angle formed by a chord and a tangent, we have $\\angle BQF = \\angle QAF$, hence $\\angle FAE = \\angle QAE - \\angle QAF = \\angle CPE - \\angle PQF$. Since the angle $\\angle FAE$ is positively directed, we have $\\angle CPE > \\angle CQF$, which means that the lines $QF$ and $PE$ intersect. Denote $X$ their point of intersection. Then\n$$\n\\begin{aligned}\n\\angle EXF &= \\angle PXQ = -\\angle QPX - \\angle XQP = -\\angle CPE - \\angle FQP = \\\\\n&= \\angle PQF - \\angle CPE = -\\angle FAE = \\angle EAF,\n\\end{aligned}\n$$\nfrom which we conclude that the points $A$, $E$, $F$ and $X$ are concyclic. The lines $QF$ and $PE$ thus intersect on the circumscribed circle of the triangle $AEF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41536,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $K$ be the circumscribed circle of a triangle $ABC$. Let $D$ be the midpoint of the arc $AB$ that does not contain the point $C$, and let $E$ the midpoint of the arc $AC$ that does not contain the point $B$. Let $F$ and $G$ denote the points of tangency of the inscribed circle of the triangle $ABC$ with the sides $AB$ and $AC$, respectively. Let $X$ be the intersection point of the lines $EG$ and $DF$. Suppose that $DEX$ is an isosceles triangle with the top angle at $X$. Prove that the triangle $ABC$ is also isosceles with the top angle at $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $I$ be the center of the inscribed circle of the triangle $ABC$, and let $K$ and $L$ be the intersection points of the line $DE$ with the lines $AB$ and $AC$, respectively. Because the bisector of an angle of a triangle goes through the center of the inscribed circle and the midpoint of the opposite arc, the points $B, I$ and $E$ lie on a common line (the bisector of the angle at $B$). Hence $\\angle IBK = \\angle CBI = \\angle CDE$, and the points $B, I, K$ and $D$ are concyclic. It may be shown similarly that the points $C, I$ and $D$ lie on a common line (the bisector of the angle at $C$) and that the points $C, E, L$ and $I$ are concyclic.\n\n![](attached_image_1.png)\n\nFrom this we conclude $\\angle LKA = \\angle DKB = \\angle DIB = \\angle CIE = \\angle CLE = \\angle ALK$. The triangle $KLA$ is thus isosceles with the top angle at $A$. Because $AF$ and $AG$ are tangent segments, the triangle $FGA$ is also isosceles with the top angle at $A$. The\n\nlines $DE$ and $FG$ are thus parallel. According to the assumptions of the problem, the triangle $EDX$ is isosceles with the top angle at $X$, hence the triangle $GFX$ is also isosceles with the top angle at $X$. The quadrilateral $AFXG$ is thus a deltoid, and the line $AX$ is perpendicular to $GF$ and, consequently, it is also perpendicular to $DE$. From this we conclude that $ADXE$ is a deltoid, hence $ADE$ is an isosceles triangle with the top angle at $A$. The angles $\\angle ACD$ and $\\angle EBA$ are equal because they are inscribed angles over chords of equal length. The triangle $ABC$ is thus an isosceles triangle with the top angle at $A$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41537,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ and $E$ be such points on the sides $BC$ and $AC$ of a triangle $ABC$, respectively, that the points $A$, $B$, $D$ and $E$ lie on the same circle. Let $L$ denote the center of the inscribed circle of the triangle $BCE$, and let $G$ denote the point of tangency of this inscribed circle with the side $EC$. Let $K$ denote the center of the inscribed circle of the triangle $DCA$, and let $F$ denote the point of tangency of this inscribed circle with the side $DC$. Let $N$ be the intersection point of the lines $EL$ and $DK$, and let $M$ be the intersection point of the lines $KF$ and $LG$. Prove that the points $A$, $B$, $D$ and $N$ lie on the same circle and that $KMLN$ is a deltoid.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K$ be a circle containing points $A$, $B$, $D$ and $E$. Because $EL$ is the bisector of the angle $\\angle BEC$, it bisects the arc $\\overarc{AB}$ that contains $E$. Similarly, the line $DK$ bisects the arc $\\overarc{AB}$ that contains $D$. Because $E$ and $D$ lie on the same side of the line $\\overarc{AB}$, the lines $EL$ and $DK$ bisect the same arc $\\overarc{AB}$ and their point of intersection lies on the circle $K$ (it is the point $N$). Hence, the points $A$, $B$, $D$, $N$ are concyclic.\n\nBecause the points $A$, $B$, $D$, $E$ are concyclic, the angles $\\angle CDA$ and $\\angle BEC$ are equal. If we thus denote $\\alpha = \\angle CDK$, we have $\\angle CDK = \\angle KDA = \\angle BEL = \\angle LEC = \\alpha$. Since the points $K$, $L$ and $C$ are colinear, the following holds:\n$$\n\\begin{align*} \n\\angle NLK &= 180^\\circ - CLE = \\angle ECL + \\alpha = \\\\ \n&= \\angle KCD + \\alpha = 180^\\circ - \\angle DKC = \\\\ \n&= \\angle LKN. \n\\end{align*}\n$$\nWe can show similarly that $\\angle MKL = \\angle KLM$. The quadrilateral $KMLN$ is thus a deltoid.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41538,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$ (x + f(x)^2)f(y) = f(yf(x)) + xyf(x) $$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = 0 for all x and f(x) = x for all x",
    "solution": "If we substitute $x = 1$ into the functional equation, we get\n$$\nf(y) + f(1)^2 f(y) = f(yf(1)) + yf(1). \\quad (1)\n$$\nWe successively substitute $1$, $f(1)$ and $f(1)^2$ instead of $y$ into this equation to get\n$$\nf(f(1)) = f(1)^3, \\tag{2}\n$$\n$$\nf(1)^3 + f(1)^5 = f(f(1)^2) + f(1)^2, \\quad (3)\n$$\n$$\nf(1)^7 + 2f(1)^5 - f(1)^4 - f(1)^2 = f(f(1)^3). \\quad (4)\n$$\nIn the last two equations, we have already considered (2), and in the last equation, we have also considered (3). If we now substitute $y = 1$ and $x = f(1)$ into the functional equation and use (2), we get\n$$\nf(1)^2 + f(1)^7 = f(f(1)^3) + f(1)^4. \\quad (5)\n$$\nFrom (4) and (5) we derive $f(1)^5 = f(1)^2$. Hence $f(1) = 0$ or $f(1) = 1$. If $f(1) = 0$, then from (1) we get $f(y) = f(0)$ for any $y$, and we conclude that $f$ is a constant function. The only constant function that satisfies the equation is the zero function.\nIf, on the other hand, $f(1) = 1$, then from (1) we get $f(y) = y$ for any $y$, and we verify that this function satisfies the functional equation. The solutions are $f(x) = 0$ and $f(x) = x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41539,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZapiši vsa naravna števila $x$, za katera je vrednost izraza $6(x-1)-2(3 x+2(x-1))$ večja od vrednosti izraza $5(x+1)-3 x-57$.",
    "options": [],
    "answer": "1, 2, 3, 4, 5, 6, 7, 8",
    "solution": "Solution:\n\nZapišemo neenačbo $6(x-1)-2(3 x+2(x-1))>5(x+1)-3 x-57$.\n\nOdpravimo oklepaje na levi in desni strani neenačbe. Dobimo $6 x-6-10 x+4>2 x-52$.\n\nIzračunamo $-6 x>-50$ oziroma $x<\\frac{25}{3}$.\n\nZa naravna števila $1,2,3,4,5,6,7$ in $8$ je vrednost prvega izraza večja od vrednosti drugega izraza.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41540,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV trikotniku $ABC$ je kot $\\alpha$ velik $30^{\\circ}$, stranica $a$ je dolga $4~\\mathrm{cm}$, stranica $c$ je dolga dvakrat toliko kot težiščnica na stranico $c$. Natančno izračunaj dolžine stranic trikotnika $ABC$. Nariši skico.",
    "options": [],
    "answer": "a = 4 cm, b = 4√3 cm, c = 8 cm",
    "solution": "Solution:\n\nTočka $D$ naj bo razpolovišče stranice $c$. Trikotnik $CAD$ je enakokrak, zato velja $\\angle ACD = 30^{\\circ}$ in $\\angle CDA = 120^{\\circ}$. Torej je $\\angle BDC = 60^{\\circ}$. Ker je tudi trikotnik $BCD$ enakokrak, velja $\\angle CBD = \\angle DCB = 60^{\\circ}$. To pomeni, da je trikotnik $ABC$ pravokoten ($\\gamma = 90^{\\circ}$).\n\nS pomočjo kotnih funkcij izračunamo dolžini stranic $b$ in $c$, npr. $\\tan 30^{\\circ} = \\frac{a}{b}$ in $\\sin 30^{\\circ} = \\frac{a}{c}$.\n\nIzračunamo\n$$\nb = \\frac{a}{\\tan 30^{\\circ}} = 4 \\sqrt{3}~\\mathrm{cm}\n$$\nin\n$$\nc = \\frac{a}{\\sin 30^{\\circ}} = 8~\\mathrm{cm}\n$$\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41541,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj leži točka $E$ na stranici $C D$ kvadrata $A B C D$. Točka $F$ leži na premici $A B$, a ne na daljici $A B$, in zadošča $|B F|=|D E|$. Dokaži, da sta premici $A C$ in $E F$ pravokotni.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo s $T$ presečišče premic $A C$ in $E F$. Ker je $|B F|=|D E|$, $|B C|=|D A|$ in $\\angle C B F=\\angle A D E=\\frac{\\pi}{2}$, sta trikotnika $A D E$ in $C B F$ skladna. Zato je $|A E|=|C F|$ in štirikotnik $A F C E$ je enakokrak trapez. Sledi $\\angle E F A=\\angle C A F$, ta kot pa je enak $\\frac{\\pi}{4}$, saj je $A C$ diagonala kvadrata $A B C D$. Dobimo $\\angle A T F=\\pi-\\angle E F A-\\angle C A F=\\frac{\\pi}{2}$, zato sta premici $A C$ in $E F$ pravokotni.\n\n![](attached_image_1.png)\n\n2. način. Zaradi $|B F|=|E D|$ in vzporednosti premic $B F$ in $D E$, je štirikotnik $B F E D$ paralelogram. Premica $E F$ je vzporedna premici $B D$. Ker sta $A C$ in $B D$ diagonali kvadrata $A B C D$, se sekata pravokotno. Torej se tudi premici $A C$ in $E F$ sekata pravokotno.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41542,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $ABCDE$ tetiven petkotnik, v katerem je $|CD| = |DE|$. Diagonali $AD$ in $BE$ se sekata v točki $K$, diagonali $AC$ in $BD$ pa v točki $L$. Dokaži, da sta premici $KL$ in $EC$ vzporedni.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPo izreku o obodnih kotih v krogu je $\\angle CED = \\angle CAD$ in $\\angle DCE = \\angle DBE$. Ker je $ECD$ enakokrak trikotnik z vrhom pri $D$, je $\\angle CED = \\angle DCE$.\n\nZaradi kolinearnosti točk $A$, $L$, $C$ ter zaradi kolinearnosti točk $A$, $K$, $D$ je tudi $\\angle LAK = \\angle LBK$ in so zato točke $A$, $B$, $L$, $K$ konciklične.\n\nSledi $\\angle KLA = \\angle KBA = \\angle EBA = \\angle ECA$. Torej sta premici $KL$ in $EC$ vzporedni.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41543,
    "subject": "Mathematics (Multi-modal)",
    "question": "A mole named Črt has 5 rooms in his lair. The rooms are numbered with numbers from 1 to 5. Črt has drilled tunnels between some of the rooms so that he can crawl from every room to any other room using some of the tunnels. No two tunnels intersect. Every tunnel starts in one room and ends in another room (different from the first one), and it does not pass through any other rooms. Rooms that are directly connected by a tunnel we call *neighbouring*. List all the pairs of neighbouring rooms if you know the following.\n\n* By crawling through exactly three (not necessarily different) tunnels Črt cannot reach rooms 1 and 5 from room 5.\n* By crawling through exactly two tunnels Črt can reach rooms 2 and 3 from room 5.\n* By crawling through exactly three (not necessarily different) tunnels Črt can reach room 1 from room 3.",
    "options": [],
    "answer": "Neighbouring pairs: (1,4), (2,3), (2,4), (3,4), (4,5).",
    "solution": "We first notice that rooms $5$ and $1$ cannot be neighbouring. If they were, Črt could reach room $1$ from room $5$ by crawling through the connecting tunnel exactly three times—first from room $5$ to room $1$, then back to room $5$, and then again to room $1$. This would be in contradiction with the known facts about the rooms in the mole's lair. Similarly, room $5$ cannot be directly connected with rooms $2$ and $3$. If room $5$ was neighbouring to room $2$, for instance, Črt could crawl from room $5$ back to room $5$ through exactly three tunnels—first from room $5$ to room $2$ through exactly two tunnels (a fact about the lair) and then through the tunnel between rooms $2$ and $5$ back to room $5$.\n\nBecause Črt can crawl through exactly two tunnels from room $5$ to room $2$ and because room $5$ can be directly connected only with room $4$, the following pairs of rooms must be directly connected with tunnels: rooms $5$ and $4$, rooms $4$ and $2$, rooms $4$ and $3$.\n\nBecause Črt cannot reach room $1$ from room $5$ by crawling through exactly three tunnels (a fact about the lair), room $1$ cannot be neighbouring to rooms $2$ and $3$. If this was the case, Črt could crawl from room $5$ to room $1$ through exactly three tunnels passing rooms $4$ and $2$ or rooms $4$ and $3$.\n\nBecause Črt can crawl through exactly three tunnels from room $3$ to room $1$ (a fact about the lair), two more pairs of rooms must be directly connected with tunnels: rooms $1$ and $4$, rooms $2$ and $3$.\n\nAll pairs of neighbouring rooms are thus $(1, 4)$, $(2, 3)$, $(2, 4)$, $(3, 4)$ and $(4, 5)$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 41544,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV trgovino so pripeljali $475~\\mathrm{kg}$ sadja: jabolka, mandarine in banane. Koliko kg tehta posamezna vrsta sadja, če polovica jabolk tehta dvakrat toliko kot tehta $30\\%$ mandarin, banan pa je za $35~\\mathrm{kg}$ manj kot jabolk?",
    "options": [],
    "answer": "apples 180 kg, mandarins 150 kg, bananas 145 kg",
    "solution": "Solution:\n\nNaj bodo neznane količine označene: npr. $j$ pomeni količino jabolk v $\\mathrm{kg}$, $m$ pomeni količino mandarin v $\\mathrm{kg}$ in $b$ pomeni količino banan v $\\mathrm{kg}$. Zapišemo zvezo med količinami:\n\n$$\nj + m + b = 475\n$$\n\nin enačbi\n\n$$\n\\frac{1}{2}j = 2 \\cdot \\frac{30}{100}m\n$$\n\nin\n\n$$\nb = j - 35.\n$$\n\nRešimo sistem treh enačb s tremi neznankami.\n\nRešitev je $j = 180~\\mathrm{kg}$, $m = 150~\\mathrm{kg}$ in $b = 145~\\mathrm{kg}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41545,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKrt Črt ima v svojem brlogu 5 sob, oštevilčenih s številkami od $1$ do $5$. Med nekaterimi izmed njih je Črt izvrtal rove, tako da se iz vsake sobe lahko po nekaj rovih splazi v vsako drugo sobo. Nobena dva rova se ne sekata. Vsak rov se začne v neki sobi, konča pa v neki drugi (od začetne različni) sobi in vmes ne gre skozi nobeno sobo. Sobi, ki ju rov neposredno povezuje, imenujemo sosednji. Naštej pare sosednjih sob, če velja:\n\n- če se Črt splazi po natanko treh (ne nujno različnih) rovih, iz sobe $5$ ne more priti v sobi $1$ in $5$;\n- če se Črt splazi po natanko dveh rovih, lahko iz sobe $5$ pride v sobi $2$ in $3$;\n- če se Črt splazi po natanko treh (ne nujno različnih) rovih, lahko iz sobe $3$ pride v sobo $1$.",
    "options": [],
    "answer": "(1,4), (2,3), (2,4), (3,4), (4,5)",
    "solution": "Solution:\n\nGotovo sobi $5$ in $1$ nista sosednji, saj bi se Črt sicer lahko trikrat sprehodil po tem rovu iz sobe $5$ v sobo $1$, od tam nazaj v sobo $5$ in nato še enkrat v sobo $1$. S tem bi iz sobe $5$ prišel v sobo $1$ po natanko treh rovih, kar je protislovno z navodili naloge.\n\nPrav tako soba $5$ ni neposredno povezana s sobama $2$ in $3$. Če bi bila neposredno povezana na primer s sobo $2$, bi se Črt iz sobe $5$ lahko splazil po natanko treh rovih nazaj v sobo $5$, tako da bi se po natanko dveh rovih splazil iz sobe $5$ v sobo $2$ in za tem po neposrednem rovu iz $2$ v $5$.\n\nKer se Črt lahko po natanko dveh rovih splazi iz sobe $5$ v sobo $2$ in ker je soba $5$ lahko sosednja le sobi $4$, morajo biti neposredno povezani pari sob $5$ in $4$, $4$ in $2$ ter $4$ in $3$.\n\nKer se iz sobe $5$ Črt ne more splaziti v sobo $1$ po natanko treh rovih, soba $1$ ne more biti sosednja sobama $2$ in $3$. V nasprotnem primeru bi se Črt lahko splazil iz sobe $5$ v sobo $1$ preko sob $4$ in $2$ ali $4$ in $3$.\n\nKer se iz sobe $3$ Črt lahko splazi v sobo $1$ po natanko treh hodnikih, morata biti sosednja tudi para sob $1$ in $4$ ter $2$ in $3$.\n\nVsi sosednji pari sob so torej $(1,4)$, $(2,3)$, $(2,4)$, $(3,4)$ in $(4,5)$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41546,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNatančno izračunaj dolžini stranic pravokotnika, katerega obseg je $4~\\mathrm{cm}$, kot med diagonalama pa $60^\\circ$.",
    "options": [],
    "answer": "Side lengths: 3 − √3 cm and √3 − 1 cm",
    "solution": "Solution:\n\nNarišemo skico in zapišemo zvezi med dolžinami stranic: $\\tan 30^\\circ = \\frac{b}{a}$ in $2a + 2b = 4$.\n\nIzrazimo npr. $b = \\frac{\\sqrt{3}}{3} a$ in vstavimo v $2a + 2b = 4$. Dobimo $2a + 2 \\frac{\\sqrt{3}}{3} a = 4$.\n\nIzračunamo\n\n$$\na = 3 - \\sqrt{3}~\\mathrm{cm}\n$$\n\nin\n\n$$\nb = \\sqrt{3} - 1~\\mathrm{cm}\n$$\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41547,
    "subject": "Mathematics (Multi-modal)",
    "question": "We have $10$ intervals of numbers $I_1 = [a_1, b_1], I_2 = [a_2, b_2], \\dots, I_{10} = [a_{10}, b_{10}]$ such that $I_1 \\cup I_2 \\cup \\dots \\cup I_{10} = [0, 1]$. Prove that there is a $k \\le 10$ such that we can choose $k$ intervals $I_{n_1}, I_{n_2}, \\dots, I_{n_k}$ among $I_1, I_2, \\dots, I_{10}$ for which no two different intervals $I_{n_i}$ and $I_{n_j}$ intersect and $(b_{n_1} - a_{n_1}) + (b_{n_2} - a_{n_2}) + \\dots + (b_{n_k} - a_{n_k}) \\ge \\frac{1}{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We shall construct two sets of intervals $A$ and $B$ such that the intervals in $A$ will be pairwise disjoint, the intervals in $B$ will be pairwise disjoint, and the union of all the intervals in $A$ and $B$ will be equal to $[0, 1]$. This construction will prove the claim of the problem because then either the sum of the lengths of all the intervals in $A$ or the sum of the lengths of all the intervals in $B$ will be greater than $\\frac{1}{2}$. If not, the sum of the lengths of all the intervals in both sets would be smaller than $1$, which would contradict the fact that the union of all the intervals in both sets is equal to $[0, 1]$. The elements of one of the sets $A$ and $B$ will solve the problem.\n\nLet $A$ and $B$ be empty sets. Among the intervals $I_1 = [a_1, b_1], I_2 = [a_2, b_2], \\dots, I_{10} = [a_{10}, b_{10}]$ choose the one that contains $0$ and has the greatest right endpoint, $b_{A_1}$. Add this interval $I_{A_1}$ in the set $A$. Among the other intervals (different than $I_{A_1}$) choose the one that has a non-empty intersection with $I_{A_1}$ and the greatest right endpoint, $b_{B_1}$. Add this interval $I_{B_1}$ in the set $B$. Among the rest (intervals different than $I_{A_1}$ and $I_{B_1}$) choose the one that has a non-empty intersection with $I_{B_1}$ and the greatest right endpoint, $b_{A_2}$. Add this interval $I_{A_2}$ in the set $A$. Repeat the process as long as the right endpoints of the intervals are all different than $1$. Suppose, at the end of the process, the sets of intervals $A = \\{I_{A_1}, I_{A_2}, \\dots, I_{A_n}\\}$ and $B = \\{I_{B_1}, I_{B_2}, \\dots, I_{B_m}\\}$, where $n - 1 \\le m \\le n$ and $n + m \\le 10$, are equal.\n\nLet us prove that such sets have the desired properties. From the above construction we see that the right endpoints of the successively constructed intervals increase, i.e., for $i \\le n - 1$ we have $b_{A_i} < b_{B_i}$ and $b_{B_i} < b_{A_{i+1}}$. Clearly, the union of all the intervals in both sets is equal to $[0, 1]$. Let us also prove that the elements in any of the sets are pairwise disjoint. Let $I_{A_j}$ and $I_{A_{j+1}}$ be two successively constructed intervals in $A$, and suppose that their intersection is empty. According to the definition, $I_{B_j}$ has the greatest right endpoint among all the intervals (different than $I_{B_i}$ and $I_{A_i}$ for $i < j$) that intersect $I_{A_j}$. The right endpoint of the interval $I_{B_j}$ thus needs to be greater than the right endpoint of the interval $I_{A_{j+1}}$. It needs to hold $b_{B_i} > b_{A_{i+1}}$, which is a contradiction. The elements of $A$ must be pairwise disjoint. We may show similarly that the elements of $B$ are pairwise disjoint. This proves the claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41548,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJanja se je odločila za ekološko vzrejo kokoši in gosk. Ograditi namerava parcelo v obliki pravokotnika s ploščino $1632~\\mathrm{m}^2$. Goske in kokoši bo imela ločene, tako da bo parcelo po širini pregradila na pol (glej sliko). Za ograditev bo skupno potrebovala $198~\\mathrm{m}$ ograde. Kolikšna je širina in kolikšna dolžina parcele?\n\n![](attached_image_1.png)\n\n(8 točk)",
    "options": [],
    "answer": "Length 51 and width 32, or length 48 and width 34",
    "solution": "Solution:\n\nDolžino parcele označimo z $x$ in širino z $y$. Ploščina parcele (ploščina pravokotnika) je $x \\cdot y = 1632$.\n\nOgrado sestavljajo vse 4 stranice in prečna ograda: $2x + 3y = 198$.\n\nDobili smo sistem enačb, najlažje ga rešimo z zamenjalnim načinom. Iz prve enačbe izrazimo npr. $y = \\frac{1632}{x}$ in vstavimo v drugo enačbo:\n\n$$\n2x + 3 \\left( \\frac{1632}{x} \\right) = 198\n$$\n\nMnožimo z $x$:\n\n$$\n2x^2 + 3 \\cdot 1632 = 198x\n$$\n$$\n2x^2 + 4896 = 198x\n$$\n$$\n2x^2 - 198x + 4896 = 0\n$$\n$$\nx^2 - 99x + 2448 = 0\n$$\n\nDobimo dve rešitvi:\n\n$$\nx_1 = 51 \\quad (y_1 = 32)\n$$\n$$\nx_2 = 48 \\quad (y_2 = 34)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41549,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je polinom $p$ s predpisom $p(x)=x^{4}+3 x^{3}+a x^{2}+b x+7$. Določi števili $a$ in $b$ tako, da bo premica $z$ enačbo $y=2 x+7$ sekala graf polinoma $p$ v točkah $z$ abscisama 1 in -4.",
    "options": [],
    "answer": "a = -4, b = 2",
    "solution": "Solution:\n\nZapišemo enačbo $2 x+7 = x^{4} + 3 x^{3} + a x^{2} + b x + 7$. Enačbo uredimo in dobimo $x^{4} + 3 x^{3} + a x^{2} + (b-2) x = 0$.\n\nV enačbo vstavimo $x=1$ in dobimo $1 + 3 + a + b - 2 = 0$ oziroma $a + b + 2 = 0$.\n\nV enačbo vstavimo še $x = -4$ in dobimo $256 - 192 + 16 a - 4 b + 8 = 0$ oziroma $4 a - b + 18 = 0$.\n\nRešimo sistem enačb:\n\n$a + b + 2 = 0$\n\n$4 a - b + 18 = 0$\n\nSeštejemo obe enačbi:\n\n$a + b + 2 = 0 \\implies b = -a - 2$\n\nVstavimo v drugo enačbo:\n\n$4a - (-a - 2) + 18 = 0$\n\n$4a + a + 2 + 18 = 0$\n\n$5a + 20 = 0$\n\n$a = -4$\n\n$b = -(-4) - 2 = 4 - 2 = 2$\n\nTorej sta $a = -4$ in $b = 2$.\n\n\nDrugi način:\n\nIzračunamo koordinate presečišč premice in grafa polinoma $(1,9)$ in $(-4,-1)$. Koordinate obeh točk vstavimo v predpis za polinom. Dobimo enačbi:\n\nZa $x=1$: $p(1) = 1 + 3 + a + b + 7 = 11 + a + b = 9 \\implies a + b = -2$\n\nZa $x=-4$: $p(-4) = 256 - 192 + 16a - 4b + 7 = 71 + 16a - 4b = -1 \\implies 16a - 4b = -72 \\implies 4a - b = -18$\n\nRešimo sistem:\n\n$a + b = -2$\n\n$4a - b = -18$\n\nSeštejemo obe enačbi:\n\n$a + b = -2 \\implies b = -2 - a$\n\n$4a - (-2 - a) = -18$\n\n$4a + a + 2 = -18$\n\n$5a = -20$\n\n$a = -4$\n\n$b = -2 - (-4) = 2$\n\nTorej sta $a = -4$ in $b = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41550,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKoliko odstotkov od\n$$\n\\frac{\\left(\\frac{1}{2}\\right)^{2}+\\left(\\frac{1}{2}\\right)^{3}}{\\frac{1}{4} \\cdot\\left(1-\\frac{1}{3}\\right)}-\\frac{1}{1+\\frac{1}{2}}\n$$\nje\n$$\n\\frac{1}{12} \\cdot \\frac{\\left(0,5-\\frac{1}{4}\\right)^{2} \\cdot 8}{0,008 \\cdot\\left(2+\\frac{1}{2}\\right)^{4}}\n$$\n? Rezultat zaokroži na stotine. Nalogo reši brez uporabe žepnega računala.",
    "options": [],
    "answer": "8.42%",
    "solution": "Solution:\n\nPoenostavimo prvi izraz\n$$\n\\frac{\\frac{1}{4}+\\frac{1}{8}}{\\frac{1}{4} \\cdot \\frac{2}{3}}-\\frac{1}{\\frac{3}{2}} = \\frac{\\frac{3}{8}}{\\frac{2}{12}}-\\frac{2}{3} = \\frac{9}{4}-\\frac{2}{3} = \\frac{19}{12}.\n$$\n\nPoenostavimo tudi drugi izraz\n$$\n\\frac{1}{12} \\cdot \\frac{\\left(\\frac{1}{2}-\\frac{1}{4}\\right)^{2} \\cdot 8}{\\frac{1}{125} \\cdot\\left(\\frac{5}{2}\\right)^{4}} = \\frac{1}{12} \\cdot \\frac{\\frac{1}{4}^{2} \\cdot 8}{\\frac{1}{125} \\cdot \\frac{625}{16}} = \\frac{1}{12} \\cdot \\frac{8}{5} = \\frac{2}{15}.\n$$\n\nZapišemo enačbo\n$$\n\\frac{x}{100} \\cdot \\frac{19}{12} = \\frac{2}{15}.\n$$\nIzračunamo\n$$\nx = \\frac{2 \\cdot 12}{19 \\cdot 15} \\cdot 100 = 8,42.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41551,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDana je premica z enačbo $2x - 3y + 15 = 0$ in točka $T(3, y)$ na njej. Izračunaj $x$, tako da bo razdalja med točkama $A(x, 4)$ in $T$ enaka $3 \\sqrt{5}$.",
    "options": [],
    "answer": "x = 9 or x = -3",
    "solution": "Solution:\nIzračunamo neznano koordinato točke $T$, tako da vstavimo $x = 3$ v enačbo premice in rešimo enačbo $2 \\cdot 3 - 3y + 15 = 0$. Izračunamo $y = 7$.\n\nKoordinato $x$ točke $A$ izračunamo z uporabo formule za razdaljo med točkama oziroma Pitagorovega izreka:\n$$\nd(A, T) = \\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\n$$\nVstavimo podatke:\n$$(3 \\sqrt{5})^2 = (4 - 7)^2 + (x - 3)^2$$\nin poenostavimo:\n$$\n45 = (-3)^2 + (x - 3)^2\n$$\noziroma\n$$(x - 3)^2 = 36$$\nRešimo enačbo in dobimo rešitvi $x_1 = 9$ in $x_2 = -3$.\n\nI. način:\nIzračun $(3 \\sqrt{5})^2 = 45$\n\nPoenostavitev enačbe $(x - 3)^2 = 36$\n\nZapis $x - 3 = \\pm \\sqrt{36}$\n\nIzračunani rešitvi $x_1 = 9$ in $x_2 = -3$\n\nII. način:\nKvadriranje $(x - 3)^2 = x^2 - 6x + 9$ in $(3 \\sqrt{5})^2 = 45$\n\nUreditev enačbe $x^2 - 6x - 27 = 0$\n\nUporaba Vietovega pravila $(x - 9)(x + 3) = 0$\n\nIzračunani rešitvi $x_1 = 9$ in $x_2 = -3$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41552,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many polynomials of degree $5$ with the coefficients being $1$ or $-1$ have a root at $1$?\n(A) $5$\n(B) $10$\n(C) $15$\n(D) $20$\n(E) $24$",
    "options": [],
    "answer": "D",
    "solution": "The value of the polynomial at $1$ is equal to the sum of all the coefficients of the polynomial. Since this value must be $0$, exactly three coefficients must be $1$ and three must be $-1$. The number of such polynomials is thus $\\binom{6}{3} = 20$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41553,
    "subject": "Mathematics (Multi-modal)",
    "question": "We have a cylinder with a radius of $1$ cm and a height of $4$ cm. While the point $P$ lies on the bottom circle of the cylinder, the point $Q$ lies on the top circle right above $P$. A string connecting the point $P$ and $Q$ makes a complete turn round the cylinder (see the picture). How long is the shortest such string in cm?\n![](attached_image_1.png)\n\n(A) $2\\pi$\n(B) $4\\pi$\n(C) $\\pi\\sqrt{2}$\n(D) $2\\sqrt{\\pi^2 + 4}$",
    "options": [],
    "answer": "D",
    "solution": "If we cut the cylinder along the vertical line segment $PQ$ and unfold it, we get a rectangle with sides of lengths $4$ cm and $2\\pi$ cm. Now the problem asks for the shortest path between two opposite vertices of the rectangle. The length of such a path is the same as the length of the diagonal, hence $\\sqrt{(2\\pi)^2 + 4^2} = 2\\sqrt{\\pi^2 + 4}$ cm.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 41554,
    "subject": "Mathematics (Multi-modal)",
    "question": "Peter breeds horses and cows on his farm. One day, the number of horses was the same as the number of cows and greater than $0$. Then Peter bought some more cows so that the number of cows increased by $50\\%$. Now only $30\\%$ of the animals are horses. How many horses does Peter have on his farm?\n(A) $8$ (B) $9$ (C) $10$\n(D) Some other number. (E) The problem has no solution.",
    "options": [],
    "answer": "E",
    "solution": "Say at the beginning there were $x$ cows and $x$ horses on the farm. When the number of cows increased by $50\\%$, there were $x + \\frac{1}{2}x = \\frac{3}{2}x$ cows. Hence the ratio of horses was equal to $x/(x + \\frac{3}{2}x) = \\frac{2}{5}$, which is $40\\%$ and not $30\\%$. The correct answer is thus (E).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41555,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the first circle with diameter $4$ we inscribe a square. In the square we inscribe a new circle. In the last circle we again inscribe a square and repeat the process. What is the diameter of the fourth circle?\n(A) $\\frac{1}{\\sqrt{2}}$ (B) $\\frac{1}{2}$ (C) $1$ (D) $\\sqrt{2}$ (E) $2\\sqrt{2}$",
    "options": [],
    "answer": "D",
    "solution": "The ratio of the diameters of two consecutive circles is equal to the ratio of the diagonal and the side of the square, hence $\\sqrt{2}$. The ratio of the diameters of the first and the fourth circle is thus equal to $(\\sqrt{2})^3 = 2\\sqrt{2}$. The diameter of the fourth circle is $\\frac{4}{2\\sqrt{2}} = \\sqrt{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41556,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the sum of all the real numbers that solve the equation $|x - 2011| + |x - 2012| = 3$?\n(A) 2011 (B) 2012 (C) 2013 (D) 4021 (E) 4023",
    "options": [],
    "answer": "E",
    "solution": "There are only two such real numbers: $2010$ and $2013$. Their sum is equal to $4023$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41557,
    "subject": "Mathematics (Multi-modal)",
    "question": "Among the functions below, which takes the value $0$ exactly twice?\n(A) $f(x) = \\sin x - 1$\n(B) $f(x) = |x^2 - 1| - 2$\n(C) $f(x) = e^x - 1$\n(D) $f(x) = |2x - 1|$\n(E) $f(x) = x - 1$",
    "options": [],
    "answer": "B",
    "solution": "The graph of the function given in (A) intersects the $x$ axis infinitely many times, the graph of the function in (B) intersects the $x$ axis in $(\\sqrt{3}, 0)$ and $(-\\sqrt{3}, 0)$, the graph of the function in (C) intersects the $x$ axis only in $(0, 0)$, the graph of the function in (D) intersects the $x$ axis only in $(\\frac{1}{2}, 0)$, and the graph of the function given in (E) intersects the $x$ axis only in $(1, 0)$. The correct answer is thus (B).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41558,
    "subject": "Mathematics (Multi-modal)",
    "question": "At most how many interior angles of an $n$-gon can be greater than $180^\\circ$?\n(A) $n-1$ (B) $n-2$ (C) $n-3$ (D) $n-4$ (E) $n-5$",
    "options": [],
    "answer": "C",
    "solution": "The answer is $n - 3$. It can be seen from the picture that only $n - 3$ interior angles of an $n$-gon can be greater than $180^\\circ$. The sum of all the interior angles of an $n$-gon is equal to $(n - 2) \\cdot 180^\\circ$. If at least $n - 2$ interior angles were greater than $180^\\circ$, the sum of all the interior angles would be greater than $(n - 2) \\cdot 180^\\circ$, which would be a contradiction.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41559,
    "subject": "Mathematics (Multi-modal)",
    "question": "Jakob is reading a book with $630$ pages. On the first day he read one third of the book. The sum of the numbers labelling pages that Jakob read on the second day is $4400$. How many pages does Jakob still need to read to come to the end of the book? (The first page of the book is numbered $1$.)\n\n(A) $210$\n(B) $211$\n(C) $230$\n(D) $390$\n(E) $400$",
    "options": [],
    "answer": "E",
    "solution": "On the first day, Jakob read $210$ pages of the book. Suppose that on the second day he read $n$ pages. Then the sum of the numbers on these pages was equal to $211 + 212 + 213 + \\ldots + (210 + n) = 210n + \\frac{n(n+1)}{2}$, hence $210n + \\frac{n(n+1)}{2} = 4410$. From this we get $210n \\leq 4410$ or $n \\leq 20$. Taking $n = 20$ gives us $210n + \\frac{n(n+1)}{2} = 4410$, from which we conclude that on the second day he read $20$ pages, and he still needs to read $630 - 210 - 20 = 400$ pages to come to the end of the book.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41560,
    "subject": "Mathematics (Multi-modal)",
    "question": "Which two points lie on the graph of a linear function $y = bx + 1$ where $b$ is some non-zero real number?\n(A) $(0,1)$ and $(\\frac{1}{b},0)$\n(B) $(0,b)$ and $(-\\frac{1}{b},0)$\n(C) $(0,1)$ and $(b,0)$\n(D) $(0,1)$ and $(-\\frac{1}{b},0)$\n(E) $(0,-\\frac{1}{b})$ and $(1,0)$",
    "options": [],
    "answer": "D",
    "solution": "Only two among the given points always fulfill the equation. These are $(0, 1)$ and $(-\\frac{1}{b}, 0)$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 41561,
    "subject": "Mathematics (Multi-modal)",
    "question": "We multiplied a number whose cube is equal to $2012^{12}$ by the square of $2012^{11}$. Which number did we get?\n(A) $2012^{58}$ (B) $2012^{26}$ (C) $2012^{88}$ (D) $2012^{15}$ (E) $2012^{12}$",
    "options": [],
    "answer": "B",
    "solution": "The number $2012^4$ was multiplied by $(2012^{11})^2$, hence we got $2012^4 \\cdot 2012^{22} = 2012^{26}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41562,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right-angled triangle with the right angle at $C$. Let the points $K$, $L$ and $M$ lie on the sides $CA$, $AB$ and $BC$, respectively, so that the angle $\\angle MLK$ is a right angle and $|KC| = |KL|$. Prove that the bisectors of the angles $\\angle AKL$ and $\\angle LMB$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K'$ and $L'$ be the intersection points of the side $AB$ with the bisectors of the angles $\\angle AKL$ and $\\angle LMB$, respectively. The triangles $KMC$ and $KML$ coincide in two sides and the angle that lies opposite the longer among the two sides, hence the triangles are congruent.\n![](attached_image_1.png)\nThe line $KM$ is thus the bisector of the angles $\\angle LKC$ and $\\angle CML$. From this we conclude that the angles $\\angle K'KM$ and $\\angle KML'$ are right angles. Now the claim is straightforward.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41563,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. Let $D$ denote the intersection point of the bisector of the angle $\\angle BAC$ and the side $BC$, and let $E$ denote the intersection point of the bisector of the angle $\\angle CBA$ and the side $AC$. Suppose $|CD| = |CE|$. Prove that the triangle $ABC$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $F$ the intersection point of the lines $AD$ and $BE$. Then the line $CF$ is the bisector of the angle $\\angle ACB$. The triangles $CFD$ and $CFE$ coincide in two sides and the angle between them, hence they are congruent. The angles $\\angle CDF$ and $\\angle FEC$ are thus equal. The triangles $ADC$ and $BEC$ coincide in one side and the adjacent angles, so they are congruent. Hence the angles $\\angle DAC$ and $\\angle CBE$ are equal. From this we may obviously conclude that the triangle $ABC$ is isosceles with the top angle at $C$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41564,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square and $M$ the midpoint of the side $BC$. Let $P$ be the orthogonal projection of point $C$ onto the line segment $DM$. Prove that the triangle $DAP$ is an isosceles triangle with the top angle at $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $N$ be the midpoint of the side $CD$, and let $R$ be the intersection point of the lines $DP$ and $AN$. Then $\\angle DRA = 180^\\circ - \\angle ADR - \\angle NAD = 180^\\circ - \\angle ADR - \\angle MDC = 180^\\circ - 90^\\circ = 90^\\circ$. The line $RN$ is thus parallel to the line $PC$, and from $|DN| = |NC|$ we get $|DR| = |RP|$. Since the triangles $DRA$ and $PRA$ coincide in two sides and the angle between them, they are congruent. We have $|DA| = |AP|$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41565,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral with longest side $AB$. Let the bisectors of the angles $\\angle DCB$ and $\\angle ADC$ additionally intersect the circumscribed circle of the quadrilateral $ABCD$ in points $E$ and $F$. Call $G$ the intersection point of the lines $CE$ and $DF$, and $H$ the intersection point of the lines $AE$ and $BF$.\nProve that the lines $EF$ and $GH$ intersect at right angles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because the points $C$, $D$, $E$ and $F$ are concyclic, the angles $\\angle FDC$ and $\\angle FEC$ are equal. Because the points $A$, $E$, $F$ and $D$ are concyclic, the angles $\\angle ADF$ and $\\angle HEF$ are equal. Hence the angles $\\angle HEF$ and $\\angle FEG$ are equal. We can show similarly that the angles $\\angle GFE$ and $\\angle EFH$ are equal. The triangles $EFG$ and $EFH$ coincide in one side and the adjacent angles, so they are congruent. We have $|EG| = |EH|$, hence the triangle $HGE$ is isosceles with the top angle at $E$. From this we conclude that the lines $EF$ and $GH$ intersect at right angles.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41566,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $n$ and prime numbers $p$ such that $\\sqrt[3]{n} + \\frac{p}{\\sqrt[3]{n}}$ is the square of a natural number.",
    "options": [],
    "answer": "(n, p) = (1, 3) and (27, 3)",
    "solution": "Denote $\\sqrt[3]{n} + \\frac{p}{\\sqrt[3]{n}} = k^2$ where $k$ is a natural number. We raise the equation to the 3rd power and get $n + 3p\\sqrt[3]{n} + 3\\frac{p^2}{\\sqrt[3]{n}} + \\frac{p^3}{n} = k^6$, which is $n + 3pk^2 + \\frac{p^3}{n} = k^6$. From this we see that $n$ must divide $p^3$. Since $p$ is prime, we conclude $n = 1$, $n = p$, $n = p^2$ or $n = p^3$. If $n = p$ or $n = p^2$, we get the equation $p + 3pk^2 + p^2 = k^6$. Hence $k$ must be divisible by $p$, so the right side of the equation is divisible by $p^2$, but the left is not. We still have to check $n = 1$ and $n = p^3$. If we substitute them into the equation, we get $1 + p = k^2$, hence $p = (k-1)(k+1)$. From this we conclude that $k=2$ and $p=3$. We get two solutions: $n = 1$ and $p = 3$, $n = 27$ and $p = 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41567,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $n$ and prime numbers $p$ such that $\\sqrt[n]{n+\\frac{2}{p}}$ is a natural number.",
    "options": [],
    "answer": "n = 1, p = 2",
    "solution": "Denote $\\sqrt[n]{n + \\frac{2}{p}} = k$ where $k$ is a natural number, hence $n + \\frac{2}{p} = k^n$. Thus $p$ is a divisor of $2$. Since $p$ is prime, we have $p = 2$ or $p = 1$ (but $1$ is not prime). So $p = 2$.\n\nNow, $n + \\frac{2}{2} = k^n \\implies n + 1 = k^n$.\n\nTry $n = 1$: $1 + 1 = 2 = k^1 \\implies k = 2$.\nTry $n = 2$: $2 + 1 = 3 = k^2 \\implies k = \\sqrt{3}$ (not integer).\nTry $n = 3$: $3 + 1 = 4 = k^3 \\implies k = \\sqrt[3]{4}$ (not integer).\n\nSo only $n = 1$, $p = 2$, $k = 2$ is a solution.\n\nAlternatively, if the original problem was $\\sqrt[n]{n + \\frac{2}{n}}$ (as in the solution context), then $n$ must be a divisor of $2$, so $n = 1$ or $n = 2$.\n\nFor $n = 1$: $1 + \\frac{2}{1} = 3 = k^1 \\implies k = 3$.\nFor $n = 2$: $2 + \\frac{2}{2} = 3 = k^2 \\implies k = \\sqrt{3}$ (not integer).\n\nThus, the only solution is $n = 1$, $p = 2$, $k = 2$.\n\nTherefore, the solutions are $n = 1$, $p = 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41568,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $n$ and prime numbers $p$ such that $\\sqrt{n} + \\frac{p}{\\sqrt{n}}$ is the square of a natural number.",
    "options": [],
    "answer": "(n, p) = (1, 3) and (9, 3)",
    "solution": "Denote $\\sqrt{n} + \\frac{p}{\\sqrt{n}} = k^2$ where $k$ is a natural number.\n\nSquaring both sides of the equation gives us\n$$\nn + 2p + \\frac{p^2}{n} = k^4.\n$$\nHence $n$ must divide $p^2$. Since $p$ is prime, we conclude $n = 1$, $n = p$ or $n = p^2$.\n\nIf $n = p$, we get the equation $p + 2p + p = k^4$ or $4p = k^4$. Hence $k$ must be even and $p$ must be divisible by $4$, which is impossible.\n\nIf $n = 1$ or $n = p^2$, we get the equation $1 + 2p + p^2 = k^4$ or $1 + p = k^2$, hence $p = (k-1)(k+1)$.\n\nWe conclude that $k=2$, hence $p=3$. This gives us two solutions: $n=1$ and $p=3$, $n=9$ and $p=3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41569,
    "subject": "Mathematics (Multi-modal)",
    "question": "Jure put $2012$ black marbles into a sequence. Then he replaced every third marble in the sequence by a red marble. After that he replaced every fifth marble in the sequence by a yellow marble. At the end, he replaced every seventh black marble in the sequence by a blue marble. How many black marbles were there in the sequence at the end?",
    "options": [],
    "answer": "921",
    "solution": "Number the marbles with numbers from $1$ to $2012$ in the order they stand in the sequence. Let us first calculate how many black marbles were there in the sequence after the second step. Jure replaced exactly those black marbles (by red or yellow ones) whose number was divisible by $3$ or $5$. Because $2012 = 670 \\cdot 3 + 2 = 402 \\cdot 5 + 2 = 134 \\cdot 15 + 2$, exactly $670 + 402 - 134 = 938$ black marbles were replaced. Right after the second step there were thus $1074$ black marbles in the sequence. Because $1074 = 153 \\cdot 7 + 3$, Jure replaced $153$ black marbles (by blue ones) in the third step. At the end, there were $1074 - 153 = 921$ black marbles in the sequence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41570,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $x$ that solve the equation\n$$\n\\cos(\\pi \\sin^2 x) + \\sin(\\pi \\cos^2 x) = 1.\n$$",
    "options": [],
    "answer": "x = n·π or x = π/4 + n·(π/2), for any integer n",
    "solution": "Denote $y = \\pi \\sin^2 x$. Because $\\sin(\\pi \\cos^2 x) = \\sin(\\pi(1 - \\sin^2 x)) = \\sin(\\pi - \\pi \\sin^2 x) = \\sin(\\pi \\sin^2 x)$, we get the equation $\\cos y + \\sin y = 1$.\n\nBecause $\\cos y + \\sin y = \\sin y + \\sin\\left(\\frac{\\pi}{2} - y\\right) = 2\\sin\\frac{\\pi}{4}\\cos\\frac{2y-\\frac{\\pi}{2}}{2} = 1$, we get $\\cos\\left(y - \\frac{\\pi}{4}\\right) = \\frac{\\sqrt{2}}{2}$, and hence $y = \\frac{\\pi}{4} \\pm \\frac{\\pi}{4} + 2k\\pi = \\pi \\sin^2 x \\in [0, \\pi]$, it must be $k=0$. Thus $y=0$ or $y=\\frac{\\pi}{2}$.\n\nIf $y=0$, we get $x = n\\pi$ where $n$ is an integer.\n\nFor $y = \\frac{\\pi}{2}$ it must hold $\\sin^2 x = \\frac{1}{2}$, from which we derive $\\sin x = \\pm\\frac{\\sqrt{2}}{2}$ or $x = \\frac{\\pi}{4} + n\\frac{\\pi}{2}$ where $n$ is an integer.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41571,
    "subject": "Mathematics (Multi-modal)",
    "question": "Lara and Sara will draw $n$ straight lines on a rectangular sheet of paper, one after the other, in turn, and one line each time. Every line will be parallel to one of the edges of the sheet and will go from edge to edge. No line will coincide with an edge or an already drawn line. At the end, the sheet of paper will be divided into several rectangles. If the number of these rectangles will be odd, the winner will be Laura, and if it will be even, the winner will be Sara. Determine the winner depending on $n$ and the person who starts drawing a line.",
    "options": [],
    "answer": "If the number of lines is odd, Sara wins regardless of who starts. If the number of lines is even, the player who does not start (the second player) wins.",
    "solution": "If $n$ is odd, the winner is Sara, no matter who starts. If $n$ is even, the winner is the one who does not start. Suppose at the end there are $p$ vertical and $r$ horizontal lines on the sheet of paper, where $p + r = n$. Then the sheet of paper is divided into $(p+1)(r+1)$ rectangles. If $n$ is odd, one of the numbers $p$ and $r$ is odd, hence $(p+1)(r+1)$ is even. In this case, Sara is the winner, no matter who starts and how they play the game.\n\nNow suppose $n$ is even. Suppose after $n-1$ steps there are $s$ vertical and $t$ horizontal lines on the sheet of paper, where $s + t = n - 1$. Since $n-1$ is odd, numbers $s$ and $t$ must have different parities. After the last line is drawn, the sheet of paper is divided into $(s+1)t$ or $s(t+1)$ rectangles. Because $s$ and $t$ have different parities, the numbers $(s+1)t = st+t$ and $s(t+1) = st+s$ have different parities as well. The one who makes the last move may thus decide whether the final number of rectangles is odd or even. This player can win, no matter how the game has been played before. Since $n$ is even, the winner is the one who makes the last move, this is the one who does not start the game.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 41572,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the greatest common divisor of the numbers $11n + 4$ and $7n + 2$, where $n$ is a positive integer?",
    "options": [],
    "answer": "6",
    "solution": "The greatest common divisor of $11n+4$ and $7n+2$ also divides $7(11n+4) - 11(7n+2) = 6$, so it can be at most $6$. If $n=4$, then $11n+4 = 48$ and $7n+2 = 30$, and the greatest common divisor of these two numbers is $6$. Hence, the answer is $6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41573,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle. The bisector of the side $AB$ intersects the lines $BC$ and $CA$ at the points $X$ and $Y$, respectively, and the bisector of the side $AC$ intersects the lines $BC$ and $AB$ at the points $Z$ and $W$, respectively. Prove that the points $X, Y, Z$ and $W$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The solution uses directed angles. Let $E$ be the midpoint of the side $AB$ and let $F$ be the midpoint of the side $AC$. Since $EF$ is a midline of the triangle $ABC$, $EF$ is parallel to $BC$. Because the lines $XY$ and $ZW$ are bisectors of the sides $AB$ and $AC$, respectively, the points $E, X$ and $Y$ as well as the points $F, Z$ and $W$ are collinear. The points $E, F, Y$ and $W$ are concyclic because\n$$\n\\angle WEY = \\angle BEY = \\frac{\\pi}{2} = \\angle WFC = \\angle WFY.\n$$\n\nFrom concyclicity of the points $E, F, Y$ and $W$, collinearity of the points $F, Z$ and $W$, and collinearity of the points $E, X$ and $Y$ we now derive\n$$\n\\angle XYW = \\angle EYW = \\angle EFW = \\angle EFZ.\n$$\nFurther, from parallelism of the lines $EF$ and $BC$, collinearity of the points $F, W$ and $Z$, and collinearity of the points $B, C, X$ and $Y$ we get\n$$\n\\angle EFZ = \\angle BZF = \\angle XZF = \\angle XZW.\n$$\nThe above two equalities say that $\\angle XYW = \\angle XZW$, hence the points $X, Y, Z$ and $W$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41574,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there do not exist natural numbers $a$ and $b$ that satisfy the equation\n$$\na! + 48 = 48(a + 1)^b.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose such two numbers $a$ and $b$ exist.\nBecause $48 \\mid 48(a+1)^b$, it also holds that $48 \\mid a! + 48$ or $48 \\mid a!$. Because $48 = 2^4 \\cdot 3$, we have $48 \\mid a!$ only for $a \\ge 6$. The cases $a=6$ and $a=7$ are considered separately.\n\nWhen $a=6$, we search for a solution of the equation $6! + 48 = 48 \\cdot 7^b$. Hence $15+1=7^b$, which has no solutions in natural numbers since the left side is even and the right side is odd.\n\nWhen $a=7$, we get the equation $7! + 48 = 48 \\cdot 8^b$ or $105+1 = 8^b$. This equation also has no solutions because $106$ is not a power of $8$.\n\nHence it must hold true that $a \\ge 8$. We now distinguish 2 cases:\n\n* $a+1$ is not a prime number. Because $a \\ge 8$, the equation can be divided by $48$ and written as\n$$\n3 \\cdot 5 \\cdot 7 \\cdot 8 \\dots a+1 = (a+1)^b. \\quad (1)\n$$\nThe number $3 \\cdot 5 \\cdot 7 \\cdot 8 \\dots a$ has the property that it is divisible by all prime numbers smaller than or equal to $a$. Now suppose $p$ is a prime number that divides $a+1$. Because $p+1$ is a composite number, we have $p \\le a$. Hence $3 \\cdot 5 \\cdot 7 \\cdot 8 \\dots a \\equiv 0 \\pmod p$. Because we also have $(a+1)^b \\equiv 0 \\pmod p$, from (1) we derive $1 \\equiv 0 \\pmod p$. This is a contradiction.\n\n* $a+1$ is a prime number. According to Wilson's theorem we now have $a! \\equiv -1 \\pmod{(a+1)}$. It also holds that $48(a+1)^b \\equiv 0 \\pmod{(a+1)}$, hence\n$$\n0 \\equiv a! + 48 \\equiv -1 + 48 \\equiv 47 \\pmod{(a+1)}.\n$$\nFrom this we get $a+1 \\mid 47$ and, further, because $47$ is prime, $a+1=47$. We obtain the equation $46!+48 = 48 \\cdot 47^b$ or $15 \\cdot (7 \\cdot 8 \\dots 46) + 1 = 47^b$. The next estimate defines a range of possible values of $b$:\n$$\n15 \\cdot (7 \\cdot 8 \\dots 46) + 1 = 24 \\cdot 35 \\cdot \\underbrace{(9 \\cdot 10 \\dots 46)}_{38 \\text{ stevil}} + 1 < 47^{40}\n$$\nHence $b < 40$.\n\nSince it holds modulo $41$ that $15 \\cdot (7 \\cdot 8 \\dots 41 \\dots 46) + 1 \\equiv 1 \\pmod{41}$, it must also hold that $47^b \\equiv 6^b \\equiv 1 \\pmod{41}$. We now only need to check if such a number exists.\n\nThis can be done using Fermat's little theorem. The theorem gives $6^{40} \\equiv 1 \\pmod{41}$. Let now $c$ be the smallest exponent such that $6^c \\equiv 1 \\pmod{41}$. It is known that $c \\mid 40$, so we just need to observe the divisors of $40$. Because $6^8 \\equiv 10 \\pmod{41}$ and $6^{20} \\equiv -1 \\pmod{41}$, neither of the divisors smaller than $40$ is good. The smallest solution is thus $c=40$, which means that equation $6^b \\equiv 1 \\pmod{41}$ has no solutions for $b < 40$.\n\nHence, the initial equation has no solutions in natural numbers.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41575,
    "subject": "Mathematics (Multi-modal)",
    "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": "p=3, q=2",
    "solution": "Write $p^4 - q^6 = r^n$ for some prime $r$ and some positive integer $n$. The expression $p^4 - q^6$ can be factored as $p^4 - q^6 = (p^2 - q^3)(p^2 + q^3)$. Since $p^4 - q^6 > 0$, the primes $p$ and $q$ cannot be equal and are therefore relatively prime. Let $d$ be the greatest common divisor of the numbers $p^2 - q^3$ and $p^2 + q^3$. Then $d$ divides $2p^2$ and $2q^3$. Since $p$ and $q$ are relatively prime, it follows that $d$ can be at most $2$.\n\nIf $d = 2$, then $r = 2$ as well as $p^2 - q^3 = 2$ and $p^2 + q^3 = 2^{n-1}$. Adding the equations and dividing by $2$ we get $p^2 = 1 + 2^{n-2}$, or $2^{n-2} = p^2 - 1 = (p-1)(p+1)$. The numbers $p-1$ and $p+1$ have to be powers of $2$, but this is only possible if $p=3$. Now, we get a contradiction with $q^3 = 7$.\n\nSo, $d=1$. Since $p^2 - q^3$ and $p^2 + q^3$ are relatively prime and their product is a power of a prime, we have $p^2 - q^3 = 1$, or $p^2 = q^3 + 1 = (q+1)(q^2 - q + 1)$. Since $q$ is prime, we have $q+1 > 2$ and $q^2 - q + 1 > 2$, so neither of the factors $q+1$ and $q^2 - q + 1$ is equal to $1$. If their product is to be the square of a prime, the two factors have to be equal. The equality $q+1 = q^2 - q + 1$ implies $q = 2$ and from $p^2 = 1+q^3$ we then obtain $p=3$. Finally, we check that $3^4 - 2^6 = 17$ is indeed a power of a prime. The only solution is $p=3, q=2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41576,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$, $q$ and $r$ such that $p + q^2 = r^4$.",
    "options": [],
    "answer": "p = 7, q = 3, r = 2",
    "solution": "Rewrite the equations as $p = r^4 - q^2 = (r^2 - q)(r^2 + q)$. Since $p$ is prime, we have $r^2 - q = 1$ and $r^2 + q = p$. From the first equality we obtain $q = r^2 - 1 = (r-1)(r+1)$. Since $q$ is prime we get $r-1=1$. So, $r=2$ and $q=3$. Finally, from $r^2 + q = p$ we find $p=7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41577,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 5$. Prove that a square can be divided into $n$ quadrilaterals having the side ratio of $2 : 3$. The quadrilaterals need not be of the same size.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We shall call a quadrilateral suitable if it has side ratio of $2 : 3$. Suppose that we are able to divide a square into $k$ suitable quadrilaterals for some $k \\in \\mathbb{N}$. Then we are also able to divide it into $k+3$ suitable quadrilaterals by additionally dividing one of the $k$ suitable quadrilaterals into four suitable quadrilaterals.\n\n![](attached_image_1.png)\n\nTo solve the problem it suffices to prove that a square can be divided into $6$, $7$ and $8$ suitable quadrilaterals. A square can be divided into $6$ suitable parts as follows.\n\n![](attached_image_2.png)\n\nWe notice that a square can be divided into $3$ suitable quadrilaterals.\n\n![](attached_image_3.png)\n\nUsing this division we can also divide a square into $7$ or $8$ suitable parts. This can be done as shown in the two figures below.\n\n![](attached_image_4.png)\n\n![](attached_image_5.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41578,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\setminus \\{-1\\} \\to \\mathbb{R}$ such that\n$$\nf(x) + f(y) = (x + y + 2)f(x)f(y)\n$$\nfor all $x, y \\in \\mathbb{R} \\setminus \\{-1\\}$.",
    "options": [],
    "answer": "f(x) = 0 for all x, or f(x) = 1/(x+1) for all x ≠ −1",
    "solution": "Plug in $x = y = 0$ to find $f(0) = f(0)^2$. So, either $f(0) = 0$ or $f(0) = 1$.\n\nIf $f(0) = 0$, then plug in $y = 0$ to see that $f(x) = 0$ for all $x$. Obviously, this function satisfies the conditions of the problem.\n\nIf, on the other hand, $f(0) = 1$, plug in $y = 0$ to see that $f(x) = \\frac{1}{x+1}$. It is easy to check that this function also satisfies the conditions.\n\nHence the two solutions are $f(x) = 0$ and $f(x) = \\frac{1}{x+1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41579,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse funkcije $f: \\mathbb{R} \\setminus \\{-1\\} \\rightarrow \\mathbb{R}$, za katere velja\n$$\nf(x) + f(y) = (x + y + 2) f(x) f(y)\n$$\nza vse $x, y \\in \\mathbb{R} \\setminus \\{-1\\}$.",
    "options": [],
    "answer": "f(x) = 0 for all allowed x; f(x) = 1/(x+1) for all allowed x",
    "solution": "Solution:\n\nV funkcijsko enačbo vstavimo $x = y = 0$, da dobimo $f(0) = f(0)^2$. Torej je $f(0) = 0$ ali $f(0) = 1$.\n\nČe je $f(0) = 0$, v funkcijsko enačbo vstavimo $y = 0$ in dobimo $f(x) = 0$ za vsak $x$. Očitno ta funkcija zadošča pogojem naloge.\n\nČe pa je $f(0) = 1$, v funkcijsko enačbo spet vstavimo $y = 0$ in izpeljemo $f(x) = \\frac{1}{x+1}$. Prepričamo se lahko, da tudi ta funkcija zadošča pogojem naloge.\n\nRešitvi naloge sta torej funkciji $f(x) = 0$ in $f(x) = \\frac{1}{x+1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41580,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNajveč koliko je lahko največji skupni delitelj števil $11 n+4$ in $7 n+2$, če je $n$ naravno število?",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nNajvečji skupni delitelj števil $11 n+4$ in $7 n+2$ deli tudi $7(11 n+4)-11(7 n+2)=6$, torej je lahko največ $6$. Če je $n=4$, je $11 n+4=48$ in $7 n+2=30$, največji skupni delitelj teh dveh števil pa je natanko $6$. Odgovor je torej $6$.\n\n\n2. način. Največji skupni delitelj števil $11 n+4$ in $7 n+2$ poiščimo z Evklidovim algoritmom.\n$$\n\\begin{aligned}\n11 n+4 & = 1 \\cdot (7 n+2) + (4 n+2) \\\\\n7 n+2 & = 1 \\cdot (4 n+2) + 3 n \\\\\n4 n+2 & = 1 \\cdot 3 n + (n+2) \\\\\n3 n & = 3 \\cdot (n+2) - 6\n\\end{aligned}\n$$\nNajvečji skupni delitelj zgornjih dveh števil deli $6$. Če je $n=4$, je $11 n+4=48$ in $7 n+2=30$, največji skupni delitelj teh dveh števil pa je natanko $6$. Odgovor je torej $6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41581,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa praštevila $p, q$ in $r$, za katera velja $p+q^{2}=r^{4}$.",
    "options": [],
    "answer": "p=7, q=3, r=2",
    "solution": "Solution:\n\nEnačbo preoblikujemo v $p = r^{4} - q^{2} = (r^{2} - q)(r^{2} + q)$. Ker je $p$ praštevilo, mora biti $r^{2} - q = 1$ in $r^{2} + q = p$. Prvo enačbo preoblikujemo v $q = r^{2} - 1 = (r - 1)(r + 1)$. Ker je $q$ praštevilo, mora biti $r - 1 = 1$. Torej je $r = 2$ in $q = 3$. Iz enačbe $r^{2} + q = p$ dobimo še $p = 7$.\n\n\n2. način. Ugotovimo, da mora biti eno od praštevil $2$. Izločimo možnosti, da je $p = 2$ ali $q = 2$ (s kongruencami ali preoblikovanjem enačbe podobno kot pri prvi rešitvi). Dobimo, da je $r = 2$ in $p + q^{2} = 16$. Preverimo možnosti in dobimo $q = 3$ in $p = 7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41582,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNajvečja števka števila $13$ je $3$, največja števka števila $77$ je $7$. Za posamezne neničelne števke ugotovite, kolikokrat nastopajo kot največje števke v naravnih številih od $1$ do $99$, in sestavite ustrezno frekvenčno tabelo. To frekvenčno porazdelitev prikažite s histogramom in s frekvenčnim poligonom. Dokažite, da je vsota ploščin pravokotnikov v histogramu enaka ploščini lika, ki ga omejujeta lomljena krivulja frekvenčnega poligona in abscisna os.",
    "options": [],
    "answer": "For digit d from 1 to 9, the frequency is 2d + 1. Thus the table is: 1→3, 2→5, 3→7, 4→9, 5→11, 6→13, 7→15, 8→17, 9→19. The histogram has unit-width classes with these heights; the frequency polygon connects the midpoints of the tops of adjacent bars and is closed to the axis at the ends.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41583,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNaj bo $a < 0$. Poenostavi izraz\n$$\n\\frac{\\sqrt{27 a^{2}} - \\sqrt{12 a^{2}} + a \\sqrt{108} + 5 \\sqrt{3}}{5 \\sqrt{3} a - 5 \\sqrt{3}}\n$$\nIzračunaj vrednost izraza za $a = -2^{-1}$.",
    "options": [],
    "answer": "(a+1)/(a-1); at a = -2^{-1}: -1/3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41584,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n1. Na sliki sta graf logaritemske funkcije $f$ s predpisom $f(x)=2 \\log (x-2)+1$ in premica $p$.\n![](attached_image_1.png)\n\na) Zapiši enačbo premice $p$.\n\nb) Izračunaj ničlo logaritemske funkcije. Rezultat naj bo natančen.",
    "options": [],
    "answer": "a) Requires the diagram to read off specific points. b) x = 2 + 10^{-1/2}",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41585,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nReši neenačbo $\\sqrt{2x + y - 4} + \\sqrt{3x - 2y - 13} \\leq 0$.",
    "options": [],
    "answer": "x = 3, y = -2",
    "solution": "Solution:\n\nNaj bo $A = \\sqrt{2x + y - 4}$ in $B = \\sqrt{3x - 2y - 13}$. Ker sta $A$ in $B$ kvadratna korena, morata biti njuna podkoreninska izraza nenegativna:\n\n$$\n2x + y - 4 \\geq 0\n$$\n$$\n3x - 2y - 13 \\geq 0\n$$\n\nPoleg tega je vsota dveh nič ali večjih števil manjša ali enaka $0$ le, če sta oba enaka $0$:\n\n$$\n\\sqrt{2x + y - 4} = 0 \\implies 2x + y - 4 = 0\n$$\n$$\n\\sqrt{3x - 2y - 13} = 0 \\implies 3x - 2y - 13 = 0\n$$\n\nRešimo sistem enačb:\n\n1. $2x + y - 4 = 0$\n2. $3x - 2y - 13 = 0$\n\nIz prve enačbe:\n$$\ny = 4 - 2x\n$$\nVstavimo v drugo enačbo:\n$$\n3x - 2(4 - 2x) - 13 = 0 \\\\\n3x - 8 + 4x - 13 = 0 \\\\\n7x - 21 = 0 \\\\\nx = 3\n$$\n\nSedaj $y = 4 - 2 \\cdot 3 = 4 - 6 = -2$.\n\nTorej je rešitev $x = 3$, $y = -2$.\n\nPreverimo še pogoje pod korenom:\n\n$2x + y - 4 = 2 \\cdot 3 + (-2) - 4 = 6 - 2 - 4 = 0$\n\n$3x - 2y - 13 = 3 \\cdot 3 - 2 \\cdot (-2) - 13 = 9 + 4 - 13 = 0$\n\nOba pogoja sta izpolnjena.\n\nTorej je rešitev neenačbe natanko $x = 3$, $y = -2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41586,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa praštevila $p$ in $q$, za katera je $p^{4}-q^{6}$ potenca praštevila. (Števili 7 in 8 sta potenci praštevila, 6 pa ne.)",
    "options": [],
    "answer": "p=3, q=2",
    "solution": "Solution:\n\nPišimo $p^{4}-q^{6}=r^{n}$ za neko praštevilo $r$ in naravno število $n$. Izraz $p^{4}-q^{6}$ najprej razstavimo kot $p^{4}-q^{6}=(p^{2}-q^{3})(p^{2}+q^{3})$. Ker je $p^{4}-q^{6}>0$, praštevili $p$ in $q$ ne moreta biti enaki, torej sta si tuji. Naj bo $d$ največji skupni delitelj števil $p^{2}-q^{3}$ in $p^{2}+q^{3}$. Potem $d$ deli tudi $2p^{2}$ in $2q^{3}$, zato je zaradi tujosti $p$ in $q$ lahko $d$ največ $2$.\n\nČe je $d=2$, je tudi $r=2$ in je $p^{2}-q^{3}=2$ in $p^{2}+q^{3}=2^{n-1}$. Enačbi seštejemo, delimo z $2$ in dobimo $p^{2}=1+2^{n-2}$ oziroma $2^{n-2}=p^{2}-1=(p-1)(p+1)$. Števili $p-1$ in $p+1$ morata biti obe potenci števila $2$, kar pa je možno le v primeru $p=3$. Toda zdaj dobimo protislovje $q^{3}=7$.\n\nTorej je $d=1$. Ker sta števili $p^{2}-q^{3}$ in $p^{2}+q^{3}$ tuji, njun produkt pa je potenca praštevila, mora biti $p^{2}-q^{3}=1$ oziroma $p^{2}=q^{3}+1=(q+1)(q^{2}-q+1)$. Ker je $q$ praštevilo, je $q+1>2$ in $q^{2}-q+1>2$, torej noben od faktorjev $q+1$ in $q^{2}-q+1$ ni enak $1$. Če želimo, da bo njun produkt kvadrat praštevila, morata biti torej faktorja enaka. Iz enakosti $q+1=q^{2}-q+1$ sledi $q=2$ in iz $p^{2}=1+q^{3}$ dobimo še $p=3$. Nazadnje še preverimo, da je $3^{4}-2^{6}=17$ res potenca praštevila. Edina rešitev je torej $p=3, q=2$.\n\n\n2. način. Spet pišimo $p^{4}-q^{6}=r^{n}$. Z razmislekom o sodosti in lihosti ugotovimo, da vsi trije izmed $p, q, r$ hkrati ne morejo biti lihi. Ker je $r>0$, velja še $p>q$. Zaradi pozitivnosti vseh členov pridemo tudi do sklepa, da so si $p, q, r$ paroma različni. Ker je vsaj en sod, lahko obravnavamo 3 možnosti. Če je $p=2$, zaradi $p>q$ sledi $q=1$. V tem primeru torej ni rešitev. Če je $q=2$, lahko izraz razstavimo: $(p^{2}-8)(p^{2}+8)=r^{n}$, oziroma $p^{2}-8=r^{a}$ in $p^{2}+8=r^{b}$, kjer sta $a, b \\in \\mathbb{N}_{0}$, in $b>a$. Spet ločimo dve možnosti: če $a=0$, dobimo rešitev $p=3, q=2, r=17$. Če pa je $a>0$, lahko enačbi odštejemo in izpostavimo $r$: $16=r(r^{b-1}-r^{a-1})$. Dobimo torej, da $r \\mid 16$, od koder sledi $r=2=q$, kar spet ne da rešitev. Ostane še obravnava primera, ko je le $r=2$. Spet zapišemo $p^{2}-q^{3}=2^{a}$ in $p^{2}+q^{3}=2^{b}$, kjer sta $a, b \\in \\mathbb{N}_{0}$ in $a<b$. Tokrat ločimo tri primere: če je $a=0$, dobimo $(p-1)(p+1)=q^{3}$, od koder (zaradi $p \\neq 2$) sledi $p-1=q$ in $p+1=q^{2}$ oziroma $q=2$. Torej spet ni rešitev. Če je $a=1$ pa lahko enačbi seštejemo in izpostavimo $2$. Po razcepu dobimo: $(p-1)(p+1)=2^{b-1}$, kjer $b>1$. Sledi, da sta tako $p-1$ kot $p+1$ potenci dvojke, kar je možno le, če $p=3$, od koder pa dobimo, da je $q^{3}=7$. Spet ni rešitev. Če pa je $a>1$, spet seštejemo enačbi in izpostavimo štirico: $2p^{2}=4(2^{b-2}-2^{a-2})$. Od tod pa sledi $p=2$, kar spet ne da rešitev.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41587,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be the midpoint of the side $AB$, let $E$ be the point of intersection of the side $BC$ and the bisector of the angle $\\angle BAC$, and let $F$ be the orthogonal projection of the point $E$ onto the side $AB$ of the triangle $ABC$. Suppose $\\angle CDA = \\angle ACB$ and $|CE| = |BF|$. Find the sizes of the angles of the triangle $ABC$.",
    "options": [],
    "answer": "Angle at C is 90 degrees, angles at A and B are 45 degrees each.",
    "solution": "The equality $\\angle CDA = \\angle ACB$ implies that the triangles $ADC$ and $ACB$ are similar, so\n$$\n\\frac{|AC|}{|AB|} = \\frac{|AD|}{|AC|} = \\frac{|AB|}{2|AC|}\n$$\nwhich implies $|AB| = |AC|\\sqrt{2}$. Since $AE$ is the bisector of the angle $\\angle BAC$, we get\n![](attached_image_1.png)\n$$\n\\frac{|BE|}{|CE|} = \\frac{|AB|}{|AC|} = \\sqrt{2}\n$$\nand $|CE| = |BF|$ implies $|BE| = \\sqrt{2}|BF|$. Since $EFB$ is a right triangle with the right angle at $F$, the Pythagoras' theorem implies that $|EF| = |BE|\\sqrt{2}$, so $EFB$ is also an isosceles triangle and $\\angle CBA = \\angle EBF = \\frac{\\pi}{4}$. We know already that the triangles $ABC$ and $ACD$ are similar, so $\\angle ACD = \\angle CBA = \\frac{\\pi}{4}$. Let $D'$ be the orthogonal projection of the point $A$ onto the line $CD$. Then $AD'C$ is an isosceles right triangle, so $|AD'| = \\frac{|AC|}{\\sqrt{2}}$. On the other hand we have already determined that $|AD| = \\frac{1}{2}|AB| = \\frac{|AC|}{\\sqrt{2}}$. Since the points $D$ and $D'$ both lie on the line $CD$, where $D'$ is the orthogonal projection of the point $A$ onto $CD$, we have $D' = D$. So, $\\angle CDA = \\frac{\\pi}{2}$. We conclude that $\\angle ACB = \\frac{\\pi}{2}$ and $\\angle BAC = \\frac{\\pi}{4}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41588,
    "subject": "Mathematics (Multi-modal)",
    "question": "Natural numbers $m$ and $n$ satisfy $n > m \\ge 1$. Let $S$ be the set of all pairs of natural numbers $(x, y)$ where $1 \\le x, y \\le n$. Determine the least natural number $k$ such that for each subset $P \\subseteq S$ with cardinality $k$ there exist $m+1$ pairs $(x_1, y_1), (x_2, y_2), \\dots, (x_{m+1}, y_{m+1}) \\in P$ where the numbers $x_1, x_2, \\dots, x_{m+1}$ are pairwise distinct and the numbers $y_1, y_2, \\dots, y_{m+1}$ are pairwise distinct.",
    "options": [],
    "answer": "nm + 1",
    "solution": "We shall prove that the solution of the problem is $k = nm + 1$.\n\nSuppose that the problem were solvable for a $k \\le nm$. Let $A$ be a subset of $S$ that contains all pairs of natural numbers $(x, y)$ such that $1 \\le x \\le m$ and $1 \\le y \\le n$. Then $|A| = mn$. Consider $m+1$ arbitrary elements of this set. Since the first coordinate of any of them is between $1$ and $m$, according to Dirichlet's principle there would exist two pairs with the same first coordinate. This would be a contradiction with the conditions of the problem. The same would hold for any subset of $A$ with $k$ elements, hence $k$ cannot be less than or equal to $nm$.\n\nNow consider any set $P \\subseteq S$ with $nm + 1$ elements. Divide $S$ into $n$ sets $B_i = \\{(x, y) ; x + y \\equiv i \\pmod{n}, 1 \\le x, y \\le n\\}$ for $i = 1, 2, \\dots, n$. The sets are pairwise disjoint, each has exactly $n$ elements, and their union is the entire set $S$. Because $|P| = nm + 1$, according to Dirichlet's principle there must exist a set $B_{i_0}$, which contains at least $m + 1$ elements from $P$. Suppose $(x_1, y_1), (x_2, y_2) \\in B_{i_0}$ are two different elements. If $x_1 = x_2$, then from $x_1 + y_1 \\equiv i_0 \\equiv x_2 + y_2 \\pmod{n}$ we derive $y_1 \\equiv y_2 \\pmod{n}$. Since $1 \\le y_1, y_2 \\le n$, we have $y_1 = y_2$. This is in contradiction with the assumption that the elements are different. We conclude that the elements of $B_{i_0}$ have pairwise different first coordinates and pairwise different second coordinates. Hence, if we chose $m + 1$ elements from $P$ such that they all are also elements of $B_{i_0}$, they do fulfil the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41589,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ and $m$ be natural numbers. We want to color each cell of a $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": "It is possible if and only if at least one of the two dimensions is even.",
    "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, as shown in the figure below. 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.\n\n![](attached_image_1.png)\n\nNow observe the case when both numbers are odd. Suppose the table is colored in the desired way. We search for a contradiction. Because the number of all cells in the table is odd, there must be an odd number of cells of one of the two colors, say black. Number the black cells with numbers $1$ to $2z-1$ for some natural number $z$. Let the number of all black neighbours of the cell numbered $i$ be $k_i \\in \\mathbb{N}$. The number of all unordered pairs of neighbouring black-colored cells must then be equal to\n$$\n\\frac{k_1 + k_2 + \\dots + k_{2z-1}}{2}\n$$\nThis holds due to the following reasoning: if we count the number of all black neighbours of all the cells in the table, we count every pair of neighbouring black cells twice. We have assumed that the table is colored in the desired way, hence all the numbers $k_i$ are even. The numerator of the above fraction thus contains a sum of an odd number of odd numbers, which again is an odd number. From this we conclude that the number of all pairs of neighbouring black cells is not an integer, which is an obvious contradiction. Hence, if numbers $m$ and $n$ are odd, the table cannot be colored in the desired way.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 41590,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive real numbers $x$ and $y$ satisfy\n$$\n2013^{\\log_3 x} = y^{\\log_5 2013} \\quad \\text{and} \\quad \\log_{\\frac{1}{2}} x + \\log_{\\frac{1}{2}} y > 0.\n$$\nWhich of the numbers $x$ and $y$ is greater?",
    "options": [],
    "answer": "x > y",
    "solution": "Taking logarithms on both sides of the equation and taking into account that $\\log a^b = b \\log a$ and $\\log_a b = \\frac{\\log b}{\\log a}$, we get\n$$\n\\frac{\\log x \\log 2013}{\\log 3} = \\frac{\\log y \\log 2013}{\\log 5},\n$$\nor\n$$\n\\log y = \\frac{\\log 5}{\\log 3} \\log x = \\log_3 5 \\log x.\n$$\nFrom here we can conclude that $\\log y$ and $\\log x$ have the same sign, so $x$ and $y$ are either both less than $1$, both equal to $1$ or both greater than $1$.\n\nOn the other hand the inequality can be rewritten as $\\log_{\\frac{1}{2}}(xy) > 0$ and since $\\frac{1}{2} < 1$ we have $xy < 1$.\n\nBoth together imply that $x$ and $y$ are less than $1$, so $\\log x$ and $\\log y$ are negative. But $\\log_3 5 > 1$ implies that $\\log y = \\log_3 5 \\log x < \\log x$, or $y < x$.\n\nSo, $x$ is greater.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41591,
    "subject": "Mathematics (Multi-modal)",
    "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 $\\left[\\frac{x+23}{8}\\right] = [\\sqrt{x}] = y$.",
    "options": [],
    "answer": "4",
    "solution": "Let $y$ be such a number. Then $\\sqrt{x} \\ge |\\sqrt{x}| = y$. Since $\\sqrt{x} \\ge 0$ we have $y = [\\sqrt{x}] \\ge 0$, so we may square the inequality to get $x \\ge y^2$. Also, $\\frac{x+23}{8} < [\\frac{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": 41592,
    "subject": "Mathematics (Multi-modal)",
    "question": "At most how many prime numbers can be contained in a non-constant geometric sequence of positive real numbers?",
    "options": [],
    "answer": "2",
    "solution": "The answer is $2$. As an example of such a sequence we can take $a_n = 2\\left(\\frac{3}{2}\\right)^{n-1}$, which contains two primes, $2$ and $3$.\n\nAssume that a geometric sequence $a_n = a q^{n-1}$, where $q \\neq 1$, contains three primes. Assume also that these primes are $a_k$, $a_m$ and $a_n$, where $k < m < n$. Since the sequence is not constant, the three primes are distinct. Since $\\frac{a_n}{a_m} = q^{n-m}$ and $\\frac{a_m}{a_k} = q^{m-k}$, we have $\\left(\\frac{a_n}{a_m}\\right)^{m-k} = \\left(\\frac{a_m}{a_k}\\right)^{n-m}$, or\n\n$$\na_n^{m-k} a_k^{n-m} = a_m^{n-k}.\n$$\n\nBut $m-k, n-m, n-k > 0$ and $a_k, a_m, a_n$ are distinct, so this is not possible.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41593,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$, $C$ and $D$ be points on a circle $\\Gamma$ such that the lines $AB$ and $CD$ intersect at the point $T$, where $A$ lies between $B$ and $T$ and $D$ lies between $C$ and $T$. Let the line through $D$ that is parallel to the line $AB$ intersect the circle $\\Gamma$ again at the point $E$, and let the line $ET$ intersect $\\Gamma$ again at the point $F$. Let $G$ denote the point of intersection of the lines $CF$ and $AB$. Let $X$ be the midpoint of the line segment $AB$ and let $Y$ be the point that we obtain by reflecting the point $T$ through the point $G$. Prove that the points $X$, $Y$, $C$ and $D$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The points $C$, $D$, $E$ and $F$ are concyclic and the lines $AB$ and $DE$ are parallel, so we have $\\angle GTF = \\angle DEF = \\angle DCF$. Hence, triangles $\\Delta TGF$ and $\\Delta CGT$ are similar and they have an equal side ratio. We notice that\n$$\n\\frac{|GF|}{|GT|} = \\frac{|GT|}{|GC|} \\text{ or } |GT|^2 = |GF||GC|.\n$$\nAccording to the power-of-a-point theorem for the point $G$ and the circle $\\Gamma$ we also have $|GF||GC| = |GA||GB|$. If we take into account that $|GA| = |TA| - |GT|$ and $|GB| = |TB| - |GT|$, we get\n$$\n|GT|^2 = (|TA| - |GT|)(|TB| - |GT|) = |TA||TB| - |GT|(|TA| + |TB|) + |GT|^2,\n$$\nwhich means that $|TA||TB| = |GT|(|TA| + |TB|)$.\nBecause $X$ is the midpoint of the line segment $AB$, we have $|TX| = \\frac{|TA|+|TB|}{2}$, and from the definition of the point $Y$ we get $|TY| = 2|GT|$. Summing it all up, we derive\n$$\n|TX||TY| = |GT|(|TA| + |TB|) = |TA||TB|.\n$$\nThe power-of-a-point theorem for the point $T$ and the circle $\\Gamma$ now says that $|TA||TB|$ is equal to $|TD||TC|$. We now derive $|TX||TY| = |TD||TC|$. The points $X$, $Y$, $C$ and $D$ are thus concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41594,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNaj bo $n$ naravno število. Pokaži, da je vrednost izraza\n$$\n\\frac{\\left(9^{n-1}+3^{2 n-1}\\right)^{3}}{\\left(3^{3 n-1}+27^{n}\\right)^{2}}\n$$\nneodvisna od $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41595,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMeta je včeraj kupila 5 litrov mleka in 4 hlebce kruha ter plačala 9,10 evra. Danes je mleko cenejše za $10\\%$, kruh pa za $15\\%$. Za enako količino mleka in kruha bi danes Meta plačala 7,94 evra. Koliko sta stala liter mleka in hlebec kruha včeraj?",
    "options": [],
    "answer": "Milk: 0.82 € per liter; Bread: 1.25 € per loaf",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41596,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n2. Dan je pokončni valj, za katerega velja, da ploščina osnovne ploskve, ploščina plašča in površina tvorijo aritmetično zaporedje.\n\na) Naj bo $v$ višina danega valja. Izrazi polmer $r$ tega valja $z$ $v$.\n\nb) Denimo, da je dani valj visok $3~\\mathrm{m}$. V kolikem času bi se $v$ tem primeru valj napolnil do polovice $z$ vodo, če bi vsako sekundo priteklo vanj $0,5$ litra vode? Čas zapiši v urah, minutah in sekundah.",
    "options": [],
    "answer": "a) r = (2/3)·v. b) Time to half-fill = 12000·π seconds ≈ 10 hours 28 minutes 19 seconds.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41597,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa realno število $a$ naj $[a]$ označuje največje celo število, ki ni večje od $a$. Poišči vsa cela števila $y$, za katera obstaja realno število $x$, da velja $\\left[\\frac{x+23}{8}\\right]=[\\sqrt{x}]=y$.",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nNaj bo $y$ tako število. Potem velja $\\sqrt{x} \\geq [\\sqrt{x}] = y$. Ker je $\\sqrt{x} \\geq 0$, je tudi $y = [\\sqrt{x}] \\geq 0$, torej lahko neenakost kvadriramo, da dobimo $x \\geq y^{2}$. Poleg tega je $\\frac{x+23}{8} < \\left[\\frac{x+23}{8}\\right] + 1 = y + 1$ oziroma $x < 8y - 15$. Od tod sledi $y^{2} < 8y - 15$ oziroma $(y-3)(y-5) < 0$, kar pomeni, da je $3 < y < 5$. Ker pa mora biti $y$ celo število, je $y = 4$. V tem primeru res obstaja $x$, ki ustreza pogoju, na primer $x = 16$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41598,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa pozitivni realni števili $x$ in $y$ velja\n$$\n2013^{\\log_{3} x} = y^{\\log_{5} 2013} \\quad \\text{in} \\quad \\log_{\\frac{1}{2}} x + \\log_{\\frac{1}{2}} y > 0\n$$\nKatero od števil $x$ in $y$ je večje?",
    "options": [],
    "answer": "x",
    "solution": "Solution:\n\nČe enakost logaritmiramo, dobimo\n$$\n\\frac{\\log x \\log 2013}{\\log 3} = \\frac{\\log y \\log 2013}{\\log 5}\n$$\noziroma\n$$\n\\log y = \\frac{\\log 5}{\\log 3} \\log x = \\log_{3} 5 \\log x.\n$$\nOd tod med drugim sledi, da sta $\\log y$ in $\\log x$ istega predznaka, torej sta $x$ in $y$ bodisi oba manjša od $1$, bodisi oba enaka $1$, bodisi oba večja od $1$.\n\nPo drugi strani lahko neenakost preuredimo v\n$$\n\\log_{\\frac{1}{2}}(x y) > 0\n$$\nin ker je $\\frac{1}{2} < 1$, sledi $x y < 1$.\nIz obojega sledi, da sta $x$ in $y$ manjša od $1$, torej sta $\\log x$ in $\\log y$ negativna.\n\nKer pa je $\\log_{3} 5 > 1$, sledi $\\log y = \\log_{3} 5 \\log x < \\log x$, oziroma $y < x$.\nTorej je večje število $x$.\n\n\n2. način.\n\nKer sta $x$ in $y$ pozitivni števili, ju lahko zapišemo v obliki $x = 3^{\\alpha}$ in $y = 5^{\\beta}$, kjer sta $\\alpha$ in $\\beta$ realni števili.\nČe slednje vstavimo v enakost, dobimo\n$$\n2013^{\\alpha} = \\left(5^{\\log_{5} 2013}\\right)^{\\beta} = 2013^{\\beta},\n$$\ntorej je $\\alpha = \\beta$.\nČe $x$ in $y$ vstavimo še v neenakost, dobimo\n$$\n\\alpha \\left(\\log_{\\frac{1}{2}} 3 + \\log_{\\frac{1}{2}} 5\\right) > 0.\n$$\nKer je $\\log_{\\frac{1}{2}} 3 + \\log_{\\frac{1}{2}} 5 < 0$, sledi $\\alpha < 0$.\nTorej je $x = 3^{\\alpha} > 5^{\\alpha} = y$.\nVečje število je torej $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41599,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $D$ razpolovišče stranice $AB$, $E$ presečišče stranice $BC$ in simetrale kota $\\angle BAC$, $F$ pa pravokotna projekcija točke $E$ na stranico $AB$ trikotnika $ABC$. Denimo, da je $\\angle CDA = \\angle ACB$ in $|CE| = |BF|$. Določi velikost kotov trikotnika $ABC$.",
    "options": [],
    "answer": "∠ACB = π/2, ∠BAC = π/4, ∠CBA = π/4",
    "solution": "Solution:\n\nKer je $\\angle CDA = \\angle ACB$, sta si trikotnika $ADC$ in $ACB$ podobna, zato je\n$$\n\\frac{|AC|}{|AB|} = \\frac{|AD|}{|AC|} = \\frac{|AB|}{2|AC|}\n$$\nod koder sledi $|AB| = |AC| \\sqrt{2}$. Ker je $AE$ simetrala kota $\\angle BAC$, je\n$$\n\\frac{|BE|}{|CE|} = \\frac{|AB|}{|AC|} = \\sqrt{2}\n$$\nin iz $|CE| = |BF|$ sledi $|BE| = \\sqrt{2}|BF|$. Ker je $EFB$ pravokotni trikotnik s pravim kotom pri oglišču $F$, po Pitagorovem izreku sledi, da je tudi $|EF| = |BE| \\sqrt{2}$, kar pomeni, da je trikotnik $EFB$ tudi enakokrak in je $\\angle CBA = \\angle EBF = \\frac{\\pi}{4}$. Vemo že, da sta si trikotnika $ABC$ in $ACD$ podobna, zato je $\\angle ACD = \\angle CBA = \\frac{\\pi}{4}$. Naj bo $D'$ pravokotna projekcija točke $A$ na premico $CD$. Potem je $AD'C$ enakokrak pravokotni trikotnik, zato je $|AD'| = \\frac{|AC|}{\\sqrt{2}}$. Po drugi strani pa smo že izračunali, da je $|AD| = \\frac{1}{2}|AB| = \\frac{|AC|}{\\sqrt{2}}$. Ker točki $D$ in $D'$ obe ležita na premici $CD$, pri čemer je $D'$ pravokotna projekcija točke $A$ na $CD$, sledi, da je $D' = D$. Torej je $\\angle CDA = \\frac{\\pi}{2}$. Sledi $\\angle ACB = \\frac{\\pi}{2}$ in $\\angle BAC = \\frac{\\pi}{4}$.\n\n\n2. način. Kot v prvi rešitvi pokažemo, da je $|AB| = \\sqrt{2}|AC|$ in $\\angle CBA = \\frac{\\pi}{4}$. Od tod po kosinusnem izreku sledi\n$$\n|AC|^2 = |AB|^2 + |BC|^2 - 2|AB||BC| \\frac{\\sqrt{2}}{2} = 2|AC|^2 + |BC|^2 - 2|AC||BC|\n$$\nkar lahko preoblikujemo v $(|AC| - |BC|)^2 = 0$. Tore je $|AC| = |BC|$ in zato $\\angle BAC = \\angle CBA = \\frac{\\pi}{4}$ in $\\angle ACB = \\frac{\\pi}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41600,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nReši enačbo $25^{\\sqrt{x}} - 124 \\cdot 5^{\\sqrt{x}} = 125$.",
    "options": [],
    "answer": "9",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41601,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNajveč koliko praštevil lahko vsebuje nekonstantno geometrijsko zaporedje pozitivnih realnih števil?",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nOdgovor je $2$. Primer takega zaporedja je na primer zaporedje $a_{n} = 2 \\left(\\frac{3}{2}\\right)^{n-1}$, ki vsebuje praštevili $2$ in $3$.\n\nDenimo, da geometrijsko zaporedje $a_{n} = a q^{n-1}$, kjer je $q \\neq 1$, vsebuje tri praštevila. Denimo, da so to $a_{k}$, $a_{m}$ in $a_{n}$, kjer je $k < m < n$. Ker je zaporedje nekonstantno, so ta tri praštevila različna.\n\nKer je $\\frac{a_{n}}{a_{m}} = q^{n-m}$ in $\\frac{a_{m}}{a_{k}} = q^{m-k}$, velja $\\left(\\frac{a_{n}}{a_{m}}\\right)^{m-k} = \\left(\\frac{a_{m}}{a_{k}}\\right)^{n-m}$ oziroma $a_{n}^{m-k} a_{k}^{n-m} = a_{m}^{n-k}$.\n\nKer pa so $m-k, n-m, n-k > 0$ in so $a_{k}$, $a_{m}$, $a_{n}$ različna praštevila, to ni mogoče.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41602,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $D$ a point on the side $AB$ and $E$ a point on the side $AC$ such that $|AE| = |ED| = |DB|$ and $|AD| = |DC| = |CB|$. Determine the sizes of the angles of the triangle $ABC$.",
    "options": [],
    "answer": "∠A = 36°, ∠B = 72°, ∠C = 72°",
    "solution": "Write $\\angle EBD = \\alpha$. Then $\\angle DEB = \\alpha$, so $\\angle EDA = 2\\alpha$ and $\\angle DAE = 2\\alpha$. This implies $\\angle DEC = 4\\alpha$, or $\\angle BEC = 3\\alpha$. At the same time we have $\\angle ACD = \\angle DAC = 2\\alpha$, so $\\angle BDC = 4\\alpha$ and $\\angle CBD = 4\\alpha$, or $\\angle CBE = 3\\alpha$. It follows that the triangle $EBC$ is isosceles with the apex at $C$, so $|CE| = |CD|$ and therefore $\\angle CDE = \\angle DEC = 4\\alpha$. So, $180^\\circ = \\angle BDC + \\angle CDE + \\angle EDA = 4\\alpha + 4\\alpha + 2\\alpha = 10\\alpha$,\nor $\\alpha = 18^\\circ$. From here we conclude that $\\angle BAC = 2\\alpha = 36^\\circ$, $\\angle CBA = 4\\alpha = 72^\\circ$ and $\\angle ACB = 180^\\circ - \\angle BAC - \\angle CBA = 72^\\circ$.\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41603,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be the midpoint of the side $BC$, $E$ the midpoint of the side $CA$ and $T$ the center of mass of a triangle $ABC$. Let the lines $AT$, $BT$ and $CT$ also intersect the circumcircle of the triangle $ABC$ at the points $P$, $Q$ and $R$, respectively. Suppose $\\angle ACB = \\angle RQP$. Prove that $DCET$ is a cyclic quadrilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The quadrilateral $RBCQ$ is cyclic, so $\\angle RQB = \\angle RCB$, and the assumptions of the problem imply $\\angle ECT = \\angle ACR = \\angle BQP$. The quadrilateral $QABP$ is also cyclic, so $\\angle BQP = \\angle BAP$. Since $E$ and $D$ are the midpoints of the two sides, the lines $AB$ and $DE$ are parallel. This implies $\\angle BAP = \\angle EDA = \\angle EDT$. We can conclude that $\\angle ECT = \\angle EDT$, so $DCET$ is a cyclic quadrilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41604,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a parallelogram. Let $E$ and $F$ denote such points on the sides $AB$ and $BC$ that satisfy $|AE| = |CF|$. Let $G$ be obtained by reflection of the point $E$ through the point $A$. Denote by $M$ the point of intersection of the lines $AF$ and $EC$, and denote by $N$ the point of intersection of the lines $AF$ and $GC$. Prove that $\\angle NDM = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove that the lines $DM$ and $DN$ are the interior and the exterior bisector of the angle $\\angle ADC$. Let $X$ be the point of intersection of the lines $AF$ and $CD$.\n\nMenelaus' theorem for the triangle $EBC$ and the collinear points $A, M$ and $F$ says\n$$\n\\frac{|EA| \\ |BF| \\ |CM|}{|AB| \\ |FC| \\ |ME|} = 1. \\qquad (2)\n$$\nBecause $AB \\parallel CD$, the triangles $AEM$ and $XCM$ have identical angles, hence they are similar and $\\frac{|CM|}{|ME|} = \\frac{|MX|}{|AM|}$. From $|AE| = |FC|$ and (2) we now derive\n$$\n\\frac{|BF|}{|AB|} = \\frac{|AM|}{|MX|}. \\qquad (3)\n$$\nFrom $AB \\parallel CD$ we get $\\angle BAX = \\angle DXA$, and from $AD \\parallel BC$ we get $\\angle AFB = \\angle XAD$. The triangles $ABF$ and $XDA$ are thus similar and $\\frac{|BF|}{|AB|} = \\frac{|AD|}{|DX|}$. From (3) we derive $\\frac{|AM|}{|MX|} = \\frac{|AD|}{|DX|}$, which means that $DM$ is the interior bisector of $\\angle ADX$.\n\nAn analogous proof follows for the exterior bisector. Menelaus' theorem for the triangle $GBC$ and the collinear points $N, A$ and $F$ says\n$$\n\\frac{|GA| \\ |BF| \\ |CN|}{|AB| \\ |FC| \\ |NG|} = 1. \\qquad (4)\n$$\nBecause $AB \\parallel CD$, the triangles $AGN$ and $XCN$ have identical angles, hence they are similar and $\\frac{|CN|}{|NG|} = \\frac{|NX|}{|AN|}$. From $|AG| = |FC|$ and (4) we now derive\n$$\n\\frac{|BF|}{|AB|} = \\frac{|AN|}{|NX|}. \\qquad (5)\n$$\nFrom the ratio $\\frac{|BF|}{|AB|} = \\frac{|AD|}{|DX|}$ we get $\\frac{|AN|}{|NX|} = \\frac{|AD|}{|DX|}$. Hence, $DN$ is the exterior bisector of the angle $\\angle ADX$.\n\nSince the exterior and interior bisector are perpendicular, we have $\\angle NDM = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41605,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $n$ for which there exist $a_1, a_2, \\dots, a_n \\in \\mathbb{N}$ such that\n$$\na_1 + 2a_2 + \\dots + n a_n = 6n\n$$\nand\n$$\n\\frac{1}{a_1} + \\frac{2}{a_2} + \\dots + \\frac{n}{a_n} = 2 + \\frac{1}{n}.\n$$",
    "options": [],
    "answer": "3",
    "solution": "Each of the sums $a_1 + 2a_2 + \\dots + n a_n$ and $\\frac{1}{a_1} + \\frac{2}{a_2} + \\dots + \\frac{n}{a_n}$ contains $1 + 2 + \\dots + n = \\frac{n(n+1)}{2}$ numbers. From the given data, the arithmetic mean $A$ and the harmonic mean $H$ of numbers $a_1, a_2, a_3, a_3, a_3, \\dots, a_n, a_n, \\dots, a_n$ can be calculated. We get\n$$\nA = \\frac{a_1 + 2a_2 + \\dots + n a_n}{\\frac{n(n+1)}{2}} = \\frac{6n}{\\frac{n(n+1)}{2}} = \\frac{12}{n+1}\n$$\nand\n$$\nH = \\frac{\\frac{n(n+1)}{2}}{\\frac{1}{a_1} + \\frac{2}{a_2} + \\dots + \\frac{n}{a_n}} = \\frac{\\frac{n(n+1)}{2}}{2 + \\frac{1}{n}} = \\frac{n^2(n+1)}{2(2n+1)}\n$$\nSince all the numbers are positive, the inequality of arithmetic and geometric means states\n$$\n\\frac{n^2(n+1)}{2(2n+1)} = H \\le A = \\frac{12}{n+1},\n$$\nwhich is equivalent to\n$$\nn^2(n+1)^2 \\le 24(2n+1).\n$$\nBut for $n \\ge 4$ we have $\\frac{n^3}{2} \\ge \\frac{64}{2} = 32 > 24$ and hence\n$$\nn^2(n+1)^2 = \\frac{n^2(n+1)(2n+2)}{2} \\ge (2n+1)\\frac{n^3}{2} > 24(2n+1),\n$$\nfrom which we conclude that $n$ may be at most $3$.\n\nFor $n=1$ we get the conditions $a_1 = 6$ and $\\frac{1}{a_1} = 3$, which is impossible. For $n=2$ we get the conditions $a_1 + 2a_2 = 12$ and $\\frac{1}{a_1} + \\frac{2}{a_2} = \\frac{5}{2}$. The second condition can now only be satisfied if one of the numbers is equal to $1$. If this is not the case, when $a_1 \\ge 2$ and $a_2 \\ge 2$, the left side of the equality is too small. The first condition says that only $a_2$ can be odd, hence $a_2 = 1$ and $a_1 = 10$. This is inconsistent with the second condition. We conclude that for $n=2$ again such numbers $a_1$ and $a_2$ do not exist.\n\nIn contrast, for $n=3$ such three numbers can be found. Consider $a_1 = 6$, $a_2 = 3$ and $a_3 = 2$. It holds true\n$$\na_1 + 2a_2 + 3a_3 = 6 + 2 \\cdot 3 + 3 \\cdot 2 = 18 = 6 \\cdot 3\n$$\nand\n$$\n\\frac{1}{a_1} + \\frac{2}{a_2} + \\frac{3}{a_3} = \\frac{1}{6} + \\frac{2}{3} + \\frac{3}{2} = \\frac{1+4+9}{6} = \\frac{14}{6} = 2 + \\frac{1}{3}.\n$$\nThe only natural number that solves the problem is thus $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41606,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $E$ be a point on the side $CD$ of a rectangle $ABCD$ such that $\\angle AEB$ is a right angle and $3|EA| = 2|EC|$. Determine the ratio of the lengths of the sides of the rectangle $ABCD$.",
    "options": [],
    "answer": "4*sqrt(3)/3",
    "solution": "Denote $|AB| = |CD| = a$, $|BC| = |DA| = b$ and $|EC| = c$. Then $|EA| = \\frac{2}{3}c$ and $|ED| = a - c$. By Pythagoras' theorem for the triangle $AED$ we have $b^2 + (a-c)^2 = \\frac{4}{9}c^2$, or $b^2 = -a^2 + 2ac - \\frac{5}{9}c^2$.\n\nBy Pythagoras' theorem for the triangles $BCE$ and $ABE$ we have $c^2 + b^2 = |EB|^2 = a^2 - \\frac{4}{9}c^2$ and $b^2 = a^2 - \\frac{13}{9}c^2$.\n\n![](attached_image_1.png)\n\nComparing these two equations we get $-a^2+2ac-\\frac{5}{9}c^2 = a^2-\\frac{13}{9}c^2$, or $a^2-ac-\\frac{4}{9}c^2 = 0$. The latter can be factored as $(a - \\frac{4}{3}c)(a + \\frac{1}{3}c) = 0$. Since $a$ and $c$ are positive, we have $a = \\frac{4}{3}c$. This implies $b^2 = \\frac{1}{3}c^2$, or $b = \\frac{1}{\\sqrt{3}}c$. So, the ratio of the lengths of the two sides is $\\frac{a}{b} = \\frac{4\\sqrt{3}}{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41607,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $K_1$ be a circle with the center $S_1$ and the radius $r$. Let $K_2$ be a circle with the center $S_2$, lying on the circle $K_1$, and the radius $\\frac{2}{3}r$. Let $A$ be the point of intersection of the line $S_1S_2$ and the circle $K_2$ that lies in the exterior of the circle $K_1$. Let $C$ denote one of the intersection points of the circles $K_1$ and $K_2$. The line $AC$ also intersects the circle $K_1$ at the point $D$. Let $H$ be the orthogonal projection of the point $D$ onto the line $S_1S_2$. Prove that the point $H$ lies on the circle $K_2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The quadrilateral $ES_2CD$ is cyclic, so $\\angle S_2ED = \\angle S_2CA = \\angle CAS_2$ and $EAD$ is an isosceles triangle with the apex at $D$. This implies $|AH| = |EH|$, or $|AH| = \\frac{1}{2}|EA| = \\frac{1}{2}(2r + \\frac{2}{3}r) = \\frac{4}{3}r$. Since $\\frac{4}{3}r$ is precisely the radius of the circle $K_2$, we conclude that the point $H$ lies on $K_2$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41608,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV splošni obliki zapiši predpise za vse polinome četrte stopnje, ki imajo ničle $-2, 0, 1$ (ena izmed njih je dvojna) in njihovi grafi potekajo skozi točko $M(-1, 5)$.",
    "options": [],
    "answer": "The three polynomials are:\n1) -(5/2)x^4 -(5/2)x^3 + 5x^2\n2) (5/2)x^4 + (15/2)x^3 - 10x\n3) -(5/4)x^4 + (15/4)x^2 - (5/2)x",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41609,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIzberimo neki osni presek enakostraničnega stožca s polmerom $2\\ \\mathrm{dm}$ in nanj postavimo pravokotni koordinatni sistem tako, da je koordinatno izhodišče v središču osnovne ploskve stožca, vrh pa leži na pozitivnem delu ordinatne osi (enoti na abscisni in ordinatni osi sta dolgi $1\\ \\mathrm{dm}$). Izračunaj koordinati vrha stožca. V odsekovni obliki zapiši enačbi nosilk tistih stranic stožca, ki ležita na izbranem osnem preseku. Izračunaj površino in prostornino stožca. Rezultati naj bodo natančni.",
    "options": [],
    "answer": "Apex coordinates: (0, 2√3).\nSide lines (intercept form): x/2 + y/(2√3) = 1 and x/(-2) + y/(2√3) = 1.\nSurface area (total): 12π dm^2.\nVolume: (8√3/3)π dm^3.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41610,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoenostavi izraz\n$$\n\\frac{a-2}{a \\cdot (a-2)+4} + \\frac{11-(1+a)}{8+a^{3}} - \\frac{1}{a+2}.\n$$\nZa katere realne vrednosti $a$ izraz nima pomena?",
    "options": [],
    "answer": "1/(a^2 - 2a + 4), with the expression undefined at a = -2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41611,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $E$ taka točka na stranici $CD$ pravokotnika $ABCD$, da je kot $\\angle AEB$ pravi in velja $3|EA|=2|EC|$. Določi razmerje med dolžinama stranic pravokotnika $ABCD$.",
    "options": [],
    "answer": "4*sqrt(3)/3",
    "solution": "Solution:\n\nOznačimo $|AB|=|CD|=a$, $|BC|=|DA|=b$ in $|EC|=c$. Tedaj je $|EA|=\\frac{2}{3}c$ in $|ED|=a-c$. Po Pitagorovem izreku za trikotnik $AED$ velja\n$$\nb^2 + (a-c)^2 = \\frac{4}{9}c^2\n$$\n oziroma\n$$\nb^2 = -a^2 + 2ac - \\frac{5}{9}c^2.\n$$\nPo Pitagorovem izreku za trikotnika $BCE$ in $ABE$ velja\n$$\nc^2 + b^2 = |EB|^2 = a^2 - \\frac{4}{9}c^2\n$$\n oziroma\n$$\nb^2 = a^2 - \\frac{13}{9}c^2.\n$$\nTi dve enačbi primerjamo in dobimo\n$$\n-a^2 + 2ac - \\frac{5}{9}c^2 = a^2 - \\frac{13}{9}c^2\n$$\n oziroma\n$$\na^2 - ac - \\frac{4}{9}c^2 = 0.\n$$\nSlednje lahko razstavimo kot\n$$\n\\left(a - \\frac{4}{3}c\\right)\\left(a + \\frac{1}{3}c\\right) = 0.\n$$\nKer sta $a$ in $c$ pozitivna, mora torej veljati $a = \\frac{4}{3}c$. Od tod dobimo še\n$$\nb^2 = \\frac{1}{3}c^2\n$$\n oziroma\n$$\nb = \\frac{1}{\\sqrt{3}}c.\n$$\nTorej je razmerje med dolžinama stranic pravokotnika enako\n$$\n\\frac{a}{b} = \\frac{4\\sqrt{3}}{3}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41612,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKvadratu $ABCD$ s stranico $a$ je včrtan enakokrak trikotnik $ABE$ tako, da je vrh $E$ razpolovišče stranice $CD$. Točka $F$ je nožišče višine na krak $AE$ trikotnika $ABE$ iz oglišča $B$. Dokaži, da je $|EF| : |FB| : |BE| = 3 : 4 : 5$. Nariši ustrezno skico.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41613,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDenimo, da obstajata taki točki $D$ na stranici $AB$ in $E$ na stranici $AC$ trikotnika $ABC$, da je $|AE| = |ED| = |DB|$ in $|AD| = |DC| = |CB|$. Določi velikosti kotov trikotnika $ABC$.",
    "options": [],
    "answer": "Angle at A is 36 degrees, angle at B is 72 degrees, angle at C is 72 degrees.",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOznačimo $\\angle EBD = \\alpha$. Potem je $\\angle DEB = \\alpha$, torej je $\\angle EDA = 2\\alpha$ in $\\angle DAE = 2\\alpha$. Sledi $\\angle DEC = 4\\alpha$ oziroma $\\angle BEC = 3\\alpha$. Hkrati je $\\angle ACD = \\angle DAC = 2\\alpha$, torej je $\\angle BDC = 4\\alpha$ in $\\angle CBD = 4\\alpha$ oziroma $\\angle CBE = 3\\alpha$. Sledi, da je trikotnik $EBC$ enakokrak z vrhom pri $C$, torej je $|CE| = |CD|$ in zato $\\angle CDE = \\angle DEC = 4\\alpha$. Torej je\n$$\n180^\\circ = \\angle BDC + \\angle CDE + \\angle EDA = 4\\alpha + 4\\alpha + 2\\alpha = 10\\alpha\n$$\noziroma $\\alpha = 18^\\circ$. Od tod izračunamo $\\angle BAC = 2\\alpha = 36^\\circ$, $\\angle CBA = 4\\alpha = 72^\\circ$ ter $\\angle ACB = 180^\\circ - \\angle BAC - \\angle CBA = 72^\\circ$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41614,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $\\mathcal{K}_1$ krožnica s središčem $S_1$ in polmerom $r$. Naj bo $\\mathcal{K}_2$ krožnica s središčem $S_2$ na krožnici $\\mathcal{K}_1$ in polmerom $\\frac{2}{3} r$. Presečišče premice $S_1 S_2$ s krožnico $\\mathcal{K}_2$, ki leži zunaj kroga, omejenega s krožnico $\\mathcal{K}_1$, označimo z $A$. Eno izmed presečišč krožnic $\\mathcal{K}_1$ in $\\mathcal{K}_2$ označimo s $C$. Premica $AC$ naj seka krožnico $\\mathcal{K}_1$ še v točki $D$. Naj bo $H$ pravokotna projekcija točke $D$ na premico $S_1 S_2$. Dokaži, da točka $H$ leži na krožnici $\\mathcal{K}_2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nKer je štirikotnik $E S_2 C D$ tetiven, je $\\angle S_2 E D=\\angle S_2 C A=\\angle C A S_2$. Torej je trikotnik $E A D$ enakokrak z vrhom $D$. Od tod sledi $|A H|=|E H|$ oziroma $|A H|=\\frac{1}{2}|E A|=\\frac{1}{2}\\left(2 r+\\frac{2}{3} r\\right)=\\frac{4}{3} r$. Ker je $\\frac{4}{3} r$ natanko premer krožnice $\\mathcal{K}_2$, točka $H$ leži na krožnici $\\mathcal{K}_2$.\n\n\n2. način.\n\nPostavimo problem v pravokotni koordinatni sistem z izhodiščem v $S_1$. Brez škode za splošnost lahko izberemo koordinatni sistem tako, da je $r=1$ in da tudi $S_2$ leži na x-osi. Tedaj velja $S_1(0,0), S_2(1,0), A\\left(\\frac{5}{3}, 0\\right), \\mathcal{K}_1: x^{2}+y^{2}=1, \\mathcal{K}_2:(x-1)^{2}+y^{2}=\\left(\\frac{2}{3}\\right)^{2}$. Presečišče $\\mathcal{K}_1$ in $\\mathcal{K}_2$: $x^{2}-2 x+1+y^{2}=\\frac{4}{9}$, upoštevamo $x^{2}+y^{2}=1$, torej $x=\\frac{7}{9}$ in $y= \\pm \\frac{4 \\sqrt{2}}{9}$. Brez škode za splošnost lahko izberemo pozitivni $y$, torej dobimo $C\\left(\\frac{7}{9}, \\frac{4 \\sqrt{2}}{9}\\right)$. Premica $p$ skozi $A C$: $y=-\\frac{\\sqrt{2}}{2} x+\\frac{5 \\sqrt{2}}{6}$. Izračunamo presečišče $p$ in $\\mathcal{K}_1$: $x^{2}+\\left(-\\frac{\\sqrt{2}}{2} x+\\frac{5 \\sqrt{2}}{6}\\right)^{2}=1$, torej $\\frac{3}{2} x^{2}-\\frac{5}{3} x+\\frac{7}{18}=0$, in dobimo $x_1=\\frac{7}{9}$ in $x_2=\\frac{1}{3}$. Prvi je x koordinata točke $C$, drugi pa x koordinata točke $D$. Koordinate točke $H$, ki je pravokotna projekcija točke $D$ na x-os, so torej $H\\left(\\frac{1}{3}, 0\\right)$. Ta točka ustreza enačbi za $\\mathcal{K}_2$, torej leži na $\\mathcal{K}_2$.\n\n\n3. način.\n\nNaj bo $X$ od $A$ različno presečišče $\\mathcal{K}_2$ z $S_1 S_2$. $\\angle C S_2 B=2 \\angle C A X$ (obodni in središčni kot). $\\angle A C S_2=\\angle A C S_2=\\angle D B S_2$ torej $\\angle S_2 S_1 D=2 \\angle D B S_2=\\angle C S_2 B$. $\\frac{|S_1 D|}{|S_1 X|}=\\frac{r}{r-\\frac{2}{3} r}=3$ in $\\frac{|B S_2|}{S_2 C}=\\frac{2 r}{\\frac{2}{3} r}=3$, torej sta trikotnika $B S_2 C$ in $D S_1 X$ podobna. Ker pa je $\\angle B C S_2=\\frac{\\pi}{2}$, je $\\angle D X S_1=\\frac{\\pi}{2}$ in torej $H=X$.\n\n\n4. način.\n\nBrez škode za splošnost si lahko izberemo enote tako, da je $r=1$. Naj bo $F$ drugo presečišče $\\mathcal{K}_1$ s $S_1 S_2$ in naj bo $P$ pravokotna projekcija točke $C$ na $S_1 S_2$. Tedaj velja $|S_1 P|+|P S_2|=1$. Uporabimo Pitagorov izrek v trikotniku $C P S_2$: $|C P|^{2}=\\left(\\frac{2}{3}\\right)^{2}-|P S_2|^{2}$ in v trikotniku $S_1 C P$: $|C P|^{2}=1-|S_1 P|^{2}=1-(1-|S_2 P|)^{2}$. Iz teh dveh enačb sedaj dobimo $|P S_2|=\\frac{2}{9}$ in $|C P|=\\sqrt{\\frac{32}{81}}$. Pitagorov izrek v trikotniku $A C P$: $|A C|^{2}=|C P|^{2}+|P A|^{2}=\\frac{32}{27}$, torej $|A C|=\\sqrt{\\frac{32}{27}}$. Potenca točke $A$ na $\\mathcal{K}_1$: $|A C| \\cdot|A D|=|A S_2| \\cdot|A F|=\\frac{16}{9}$, torej $|A D|=\\frac{16}{9 \\sqrt{\\frac{32}{27}}}$. Trikotnika $P A C$ in $D H A$ sta podobna, torej velja $\\frac{|A C|}{|A D|}=\\frac{|A P|}{|A H|}$, torej $|A H|=\\frac{|A D| \\cdot|A P|}{|A C|}=\\frac{4}{3}$, kar je ravno premer $\\mathcal{K}_2$, torej $H$ leži na $\\mathcal{K}_2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41615,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $D$ razpolovišče stranice $BC$, $E$ razpolovišče stranice $CA$ in $T$ težišče trikotnika $ABC$. Premice $AT$, $BT$ in $CT$ naj sekajo trikotniku $ABC$ očrtano krožnico še v točkah $P$, $Q$ in $R$. Denimo, da je $\\angle ACB = \\angle RQP$. Dokaži, da je štirikotnik $DCE T$ tetiven.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nZaradi tetivnosti štirikotnika $RBCQ$ je $\\angle RQB = \\angle RCB$, torej iz pogoja naloge sledi $\\angle ECT = \\angle ACR = \\angle BQP$. Ker je štirikotnik $QABP$ tetiven, pa je $\\angle BQP = \\angle BAP$. Ker sta $E$ in $D$ razpolovišči stranic, sta premici $AB$ in $DE$ vzporedni, torej je $\\angle BAP = \\angle EDA = \\angle EDT$. Iz vsega sledi $\\angle ECT = \\angle EDT$, torej je štirikotnik $DCET$ tetiven.\n\n\n2. način. Označimo dolžine stranic trikotnika in njegove kote kot običajno in naj bo $F$ razpolovišče stranice $AB$. Zaradi tetivnosti šestkotnika $ARBPCQ$ in predpostavke naloge je $\\angle RAT = \\angle RAP = \\angle RQP = \\gamma$ in $\\angle TRA = \\angle CRA = \\beta$. To pomeni, da sta si trikotnika $ABC$ in $TRA$ podobna in velja $\\angle ATR = \\alpha$. Po drugi strani pa je zaradi tetivnosti tudi $\\angle BRT = \\angle BRC = \\alpha$. Torej sta si trikotnika $AFT$ in $BFR$ podobna in zaradi $|AF| = |BF|$ sta celo skladna. Zato je $ARBT$ paralelogram, od koder sledi $\\angle DTE = \\pi - \\angle BTP = \\pi - \\angle TBR = \\pi - \\angle RAT = \\pi - \\gamma$, torej je štirikotnik $TDCE$ res tetiven.\n\n\n3. način. Kot v drugi rešitvi ugotovimo, da je $\\angle ATR = \\alpha$. Od tod pa sledi podobnost trikotnikov $AFT$ in $CFA$. Zato je\n$$\n\\frac{|AF|}{|FT|} = \\frac{|CF|}{|AF|}\n$$\nKer težišče deli težiščnico v razmerju $2:1$, je $|FT| = \\frac{1}{3}|CF|$, in ker je $|AF| = \\frac{c}{2}$, sledi $|CF| = \\frac{\\sqrt{3}}{2}c$. Prav tako zaradi podobnosti trikotnikov $AFT$ in $CFA$ velja\n$$\n\\frac{|AT|}{|AC|} = \\frac{|AF|}{|CF|}\n$$\nUpoštevamo, da je $|AF| = \\frac{c}{2}$ in $|CF| = \\frac{\\sqrt{3}}{2}c$, in dobimo $|AT| = \\frac{b}{\\sqrt{3}}$ in $|AD| = \\frac{3}{2}|AT| = \\frac{\\sqrt{3}}{2}b$. Sledi\n$$\n\\frac{|AT|}{|AE|} = \\frac{2}{\\sqrt{3}} = \\frac{|AC|}{|AD|}\n$$\ntorej sta si trikotnika $ATE$ in $ACD$ podobna. Torej je $\\angle ECD = \\pi - \\angle DTE$, kar pomeni, da je štirikotnik $EC DT$ tetiven.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41616,
    "subject": "Mathematics (Multi-modal)",
    "question": "A table tennis tournament is conducted in the following way. In each round, if the number of contestants is odd, one contestant is drawn to proceed automatically to the next round. After that, pairs are drawn from the other contestants. The contestants from each pair compete against each other, and the winner proceeds to the next round. Let $f(n)$ denote the number of rounds in the tournament with $n$ contestants. (For example, $f(5) = 3$.) Determine $f(2013)$ and find the least positive integer $n$ such that $f(n) = f(2013)$.",
    "options": [],
    "answer": "f(2013) = 11; least n with f(n) = 11 is 1025",
    "solution": "Starting with $2013$ contestants $1 + \\frac{2012}{2} = 1007$ of them go to the second round and $1 + \\frac{1006}{2} = 504$ make it to the third round. For the fourth and the fifth round we get $\\frac{504}{2} = 252$ and $\\frac{252}{2} = 126$ contestants, respectively. For the sixth and the seventh round we have $\\frac{126}{2} = 63$ and $1 + \\frac{62}{2} = 32$ contestants. For the eighth and the ninth round $\\frac{32}{2} = 16$ and $\\frac{16}{2} = 8$ contestants remain and for the tenth round there are $\\frac{8}{2} = 4$. Finally, in the eleventh round only $2$ contestants remain and compete for the first place. So, $f(2013) = 11$.\n\nIf the number of contestants is less than or equal to $1024 = 2^{10}$ then the tournament will consist of at most $10$ rounds. Indeed, after the $j$th round there will be at most $\\frac{2^{10}}{2^j} = 2^{10-j}$ contestants left. So, after $10$ rounds we can have at most $2^{10-10} = 1$ and the tournament ends. Let us show that $f(1025) = 11$. Since $1025 = 1 + 2^{10}$ there are $1+2^9$ contestants in the second round, $1+2^8$ in the third, $1+2^7$ in the fourth and so on, all the way to the tenth round, where there are $1+2^1 = 3$ contestants. The eleventh round will see two of these three contestants competing for first place. So, the smallest natural number such that $f(n) = f(2013)$ is $1025$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41617,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be coprime natural numbers of different parity. Prove that the numbers $2^{2m} + 2^{m+1} + 1$ and $2^{2n} + 2^{n+1} + 1$ are coprime as well.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $a = 2^{2n} + 2^{n+1} + 1$ and $b = 2^{2m} + 2^{m+1} + 1$. We notice that $a = (2^n + 1)^2$ and $b = (2^m + 1)^2$. Because we have $(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)$. We 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 numbers $m$ and $n$ are coprime. The greatest common divisor $D(a,b)$ can thus only be 1, 3 or 9.\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 is to say, number $a$ is not divisible by 3, it is coprime to 3. The greatest common divisor $D(a,b)$ of the numbers $a$ and $b$ must thus be 1, which means that $a$ and $b$ are coprime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41618,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each hexagon of an infinite honeycomb contains one natural number. This number equals the average number of five of the six numbers written in the six neighbouring hexagons. Prove that all the numbers in the honeycomb are the same.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since all the numbers in the honeycomb are positive integers their minimum exists. Let us denote it by $n$. It suffices to show the following: if one of the hexagons contains the number $n$, then all of the neighbouring hexagons also contain $n$. Since there exists at least one hexagon containing $n$, this will imply that all hexagons contain $n$.\n\nAssume that a hexagon contains $n$. Since $n$ is the smallest number in the honeycomb, the neighbouring hexagons have to contain numbers that are greater than or equal to $n$. But the average of five of them is equal to $n$. Assume that the number in the sixth hexagon is equal to $k$ and $k \\neq n$. Denote the remaining three numbers on the hexagons neighbouring the hexagon containing $k$ by $n_1, n_2$ and $n_3$ (see figure).\n\n![](attached_image_2.png)\n\nIn the figure, consider the hexagon containing $n$ and neighbouring the hexagon containing $n_1$. Since it contains $n$ we can conclude as before that five of the neighbouring hexagons contain $n$. But $k > n$, so $n_1 = n$. Now, consider the hexagon containing $n_1 = n$. As before we can conclude that $n_2 = n$. Finally, by considering the hexagon containing $n_2 = n$, we can conclude that $n_3 = n$. All of the numbers contained in the hexagons neighbouring the hexagon containing $k$ are equal to $n$, so $k$ should be equal to $n$ as well. This contradicts the assumption that $k \\neq n$, which implies that $k = n$ and that concludes the proof.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 41619,
    "subject": "Mathematics (Multi-modal)",
    "question": "We want to cover a table of size $4 \\times 4$ with dominoes of the following shape\n![](attached_image_1.png)\n(can also be reflected or rotated),\nwhere the dominoes may overlap or extend over the edges of the table.\nAt least how many dominoes do we need?",
    "options": [],
    "answer": "5",
    "solution": "We need at least 5 dominoes. If it is possible to cover the table with four dominoes, then they cannot overlap and cannot extend over the edges of the table since four dominoes cover exactly 16 squares. In this case there are only two possible ways of covering the top left corner as shown in the figures. But in both cases the field marked with * cannot be covered.\n![](attached_image_2.png)\n![](attached_image_3.png)\n![](attached_image_4.png)\nIt is easy to see that 5 dominoes suffice as demonstrated by the figure above.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 41620,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTurnir v namiznem tenisu poteka po naslednjem pravilu. V vsakem krogu v primeru lihega števila tekmovalcev najprej izžrebajo enega, ki se avtomatično uvrsti v naslednji krog. Ostale tekmovalce $z$ žrebom razporedijo v pare. Tekmovalca vsakega para se pomerita med seboj, zmagovalec iz vsakega para pa se uvrsti v naslednji krog. Naj $f(n)$ označuje število krogov na turnirju z $n$ tekmovalci. (Tako je npr. $f(5)=3$.) Določi $f(2013)$ in poišči najmanjše naravno število $n$, za katerega je $f(n)=f(2013)$.",
    "options": [],
    "answer": "f(2013) = 11; smallest n with f(n) = 11 is 1025",
    "solution": "Solution:\n\nČe je na turnirju 2013 tekmovalcev, potem gre v drugi krog $1 + \\frac{2012}{2} = 1007$ tekmovalcev. V tretji krog se uvrsti $1 + \\frac{1006}{2} = 504$ tekmovalcev, v četrti krog pa $\\frac{504}{2} = 252$ tekmovalcev. V peti krog nato napreduje $\\frac{252}{2} = 126$ tekmovalcev, v šestega $\\frac{126}{2} = 63$ tekmovalcev, v sedmega pa $1 + \\frac{62}{2} = 32$ tekmovalcev. V osmi krog gre $\\frac{32}{2} = 16$ tekmovalcev, v devetega $\\frac{16}{2} = 8$ tekmovalcev, v desetega pa $\\frac{8}{2} = 4$ tekmovalci. V enajsti krog se uvrstita 2 tekmovalca, ki se v tem krogu pomerita za zmago. Torej je $f(2013) = 11$.\n\nČe je na turnirju manj ali enako $1024 = 2^{10}$ tekmovalcev, potem bo na turnirju največ 10 krogov. Po $j$-tem krogu bo namreč ostalo največ $\\frac{2^{10}}{2^{j}} = 2^{10-j}$ tekmovalcev. Torej jih po 10-ih krogih ostane največ $2^{10-10} = 1$ in turnir je že zaključen. Poračunajmo, da velja $f(1025) = 11$. Ker je $1025 = 1 + 2^{10}$, se v drugi krog uvrsti $1 + 2^{9}$ tekmovalcev, v tretji krog $1 + 2^{8}$ tekmovalcev, v četrtega $1 + 2^{7}$ tekmovalcev in tako dalje, vse do desetega kroga, v katerega se uvrstijo $1 + 2^{1} = 3$ tekmovalci. V enajsti krog se uvrstita dva tekmovalca, ki se pomerita za zmago. Najmanjše naravno število $n$, pri katerem je $f(n) = f(2013)$, je torej $1025$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41621,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV vsakem šestkotniku neskončnega satovja je zapisano naravno število, ki je enako povprečju števil, zapisanih v petih izmed sosednjih šest šestkotnikov. Dokaži, da so vsa števila v satovju enaka.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nKer so vsa števila v satovju naravna, obstaja najmanjše izmed njih. Označimo ga z $n$. Dovolj je pokazati, naslednje: če je v nekem šestkotniku zapisano število $n$, potem so v sosednjih šestkotnikih tudi zapisana števila $n$. Ker obstaja šestkotnik s številom $n$, bo potem namreč sledilo, da je v vseh šestkotnikih zapisano število $n$.\n\nNaj bo v nekem šestkotniku zapisano število $n$. Ker je $n$ najmanjše število v satovju, so na sosednjih šestkotnikih zapisana števila večja ali enaka $n$. Ker pa je povprečje petih izmed njih enako $n$, mora biti teh pet števil enakih $n$. Naj bo število na šestem sosednjem šestkotniku enako $k$. Denimo, da je $k \\neq n$. Označimo preostala tri števila na šestkotnikih sosednjih šestkotniku s številom $k$ z $n_1$, $n_2$ in $n_3$ (glej sliko).\n\n![](attached_image_1.png)\n\nNa sliki si poglejmo šestkotnik s številom $n$, ki je soseden šestkotniku s številom $n_1$. Ker je na njem zapisano število $n$, podobno kot prej sklepamo, da je pet izmed števil na sosednjih šestkotnikih enakih $n$. Ker pa je $k>n$, sledi, da je $n_1=n$. Če sedaj pogledamo šestkotnik s številom $n_1=n$, podobno sklepamo, da je tudi $n_2=n$. Če nazadnje pogledamo še šestkotnik s številom $n_2=n$, spet lahko sklepamo, da je tudi $n_3=n$. Torej so vsa števila na šestkotnikih sosednjih šestkotniku s številom $k$ enaka $n$, zato bi moralo biti tudi število $k$ enako $n$, kar pa je protislovje. Torej je $k=n$ in s tem je dokaz končan.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41622,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nReši enačbo\n$$\n3 + \\frac{11}{x-1} + \\frac{10}{1-2x} - \\frac{2}{4x^2 - 4x + 1} = 0\n$$\nRešitve zapiši v obliki okrajšanih ulomkov z racionaliziranimi imenovalci.",
    "options": [],
    "answer": "x = 0, x = sqrt(3)/6, x = -sqrt(3)/6",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41623,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $m$ realno število in $f(x) = x^{2} - m x + m - 1$ ter $g(x) = x^{2} - 2 x - m$. Izračunaj vrednost parametra $m$ tako, da bosta najmanjši vrednosti funkcij $f$ in $g$ enaki. Pri največjem izmed tako izračunanih vrednostih parametra $m$ reši neenačbo $2 f(x) \\geq g(x-1)$.",
    "options": [],
    "answer": "m = 0 or m = 8; for m = 8, the inequality 2 f(x) ≥ g(x − 1) reduces to x^2 − 12x + 19 ≥ 0, so the solution set is x ≤ 6 − sqrt(17) or x ≥ 6 + sqrt(17).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41624,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Poenostavi izraz\n$$\n\\left(\\frac{\\cos \\alpha}{\\cos \\beta}\\right)^2-(\\cos \\alpha \\cdot \\tan \\beta)^2+\\sin ^2 \\alpha\n$$\nče je $\\cos \\beta \\neq 0$.\n\nb) Izračunaj vrednost $\\alpha+\\beta$, če je $\\sin \\alpha=\\frac{\\sqrt{5}}{5}$ in $\\sin \\beta=\\frac{\\sqrt{10}}{10}$ ter je $0^{\\circ}<\\alpha, \\beta<45^{\\circ}$. Nalogo reši brez uporabe žepnega računala.",
    "options": [],
    "answer": "a) 1; b) 45°",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41625,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIzračunaj koordinate točke, ki leži na premici skozi točki $A(-5,-2)$ in $B(-3,-1)$ ter je enako oddaljena od točk $C(-1,0)$ in $D(4,-1)$.",
    "options": [],
    "answer": "(17/9, 13/9)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41626,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTabelo velikosti $4 \\times 4$ želimo pokriti z dominami oblike\n\n![](attached_image_1.png)\n\n(lahko tudi zrcaljena ali zasukana),\npri čemer se domine lahko prekrivajo ali segajo čez rob tabele. Najmanj koliko domin potrebujemo?",
    "options": [],
    "answer": "5",
    "solution": "Solution:\n\nPotrebujemo vsaj 5 domin. Če bi tabelo lahko pokrili s štirimi dominami, se te med sabo ne bi smele prekrivati in ne bi smele segati čez rob tabele, saj imajo štiri domine skupaj natanko 16 polj. Če to upoštevamo, potem lahko polje v levem zgornjem kotu pokrijemo le na dva načina, kot prikazuje skica, vendar potem polja označenega $*$ ne moremo nikakor pokriti.\n\n![](attached_image_2.png)\n\nDa 5 domin zadostuje, prikazuje naslednja skica.\n\n![](attached_image_3.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41627,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ and $w$ be nonnegative real numbers such that $x + y + z + w = 1$. Find the greatest and least value of the expression\n$$\n(x + 3y)(y + 2z) + (y + 3z)(z + 2w) + (z + 3w)(w + 2x) + (w + 3x)(x + 2y).\n$$",
    "options": [],
    "answer": "least value 5/2, greatest value 13/4",
    "solution": "After eliminating parentheses and combining like terms in the expression we get\n$$\n3(x^2 + y^2 + z^2 + w^2) + 7(xy + yz + zw + wx) + 4(xz + yw),\n$$\nwhich can be further transformed to\n$$\n3(x + y + z + w)^2 + (xy + yz + zw + wx) - 2(xz + yw).\n$$\nBecause $x + y + z + w = 1$, the expression given in the problem is now equal to\n$$\n3 + (xy + yz + zw + wx) - 2(xz + yw).\n$$\nThe inequality of arithmetic and geometric means gives $xy + yz + zw + wx = (x + z)(y + w) \\leq \\left(\\frac{x+z}{2} + \\frac{y+w}{2}\\right)^2 = \\frac{1}{4}$.\n\nWe can now estimate\n$$\n3 + (xy + yz + zw + wx) - 2(xz + yw) \\leq 3 + (xy + yz + zw + wx) \\leq 3 + \\frac{1}{4} = \\frac{13}{4}.\n$$\nAn equality holds, for instance, if $(x, y, z, w) = (\\frac{1}{2}, \\frac{1}{2}, 0, 0)$. The greatest value of the expression given in the problem is thus $\\frac{13}{4}$.\n\nTwo arbitrary non-negative real numbers satisfy inequality $(x-z)^2 \\geq 0$, which can be easily transformed to $\\frac{(x+z)^2}{2} \\geq 2xz$. Hence\n$$\n2(xz + yw) \\leq \\frac{(x+z)^2}{2} + \\frac{(y+w)^2}{2} \\leq \\frac{(x+z)^2}{2} + \\frac{2(x+z)(y+w)}{2} + \\frac{(y+w)^2}{2} = \\frac{((x+z) + (y+w))^2}{2} = \\frac{1}{2}.\n$$\nWe can now estimate\n$$\n3 + (xy + yz + zw + wx) - 2(xz + yw) \\geq 3 - 2(xz + yw) \\geq 3 - \\frac{1}{2} = \\frac{5}{2}.\n$$\nAn equality holds, for instance, if $(x, y, z, w) = (0, \\frac{1}{2}, 0, \\frac{1}{2})$. The least value of the expression given in the problem is thus $\\frac{5}{2}$. We have found that\n$$\n\\frac{5}{2} \\leq 3 + (xy + yz + zw + wx) - 2(xz + yw) \\leq \\frac{13}{4}\n$$\nwhere the left equality holds, for instance, if $(x, y, z, w) = (0, \\frac{1}{2}, 0, \\frac{1}{2})$, and the right equality holds, for instance, if $(x, y, z, w) = (\\frac{1}{2}, \\frac{1}{2}, 0, 0)$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41628,
    "subject": "Mathematics (Multi-modal)",
    "question": "If at the cinema three box offices are open, the visitors have to wait 15 min to buy a ticket. By how many minutes is the waiting time reduced if two more box offices are opened?\n(A) 3\n(B) 5\n(C) 6\n(D) 7\n(E) 10",
    "options": [],
    "answer": "C",
    "solution": "With 3 box offices open the waiting time is 15 minutes. If 1 box office is open the waiting time is 3 times longer, i.e. 45 minutes. If 5 box offices are open the waiting time is $\\frac{45}{5} = 9$ minutes. So, if two additional box offices are opened the waiting time is reduced by $15 - 9 = 6$ minutes. The correct answer is C.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41629,
    "subject": "Mathematics (Multi-modal)",
    "question": "Function $f$ satisfies $3f(x) + f(-x) = 4 \\sin x \\cos x$ for any real number $x$. Find the correct functional form of $f$.\n(A) $\\sin x$\n(B) $\\cos x$\n(C) $\\cos x \\sin x$\n(D) $\\sin 2x$\n(E) $\\cos 2x$",
    "options": [],
    "answer": "D",
    "solution": "Rearrange the equality to obtain $f(x) = \\frac{4}{3} \\sin x \\cos x - \\frac{1}{3} f(-x)$. Inserting $-x$ instead of $x$ and noting that the sine function is odd and the cosine function is even, we get $f(-x) = \\frac{4}{3} \\sin(-x) \\cos(-x) - \\frac{1}{3} f(x) = -\\frac{4}{3} \\sin x \\cos x - \\frac{1}{3} f(x)$. Plugging this into the first equality yields $f(x) = \\frac{4}{3} \\sin x \\cos x - \\frac{1}{3}(-\\frac{4}{3} \\sin x \\cos x - \\frac{1}{3} f(x)) = (\\frac{4}{3} + \\frac{4}{9}) \\sin x \\cos x + \\frac{1}{9} f(x)$. This can be rearranged into $\\frac{8}{9} f(x) = \\frac{16}{9} \\sin x \\cos x$. From here we conclude that $f(x) = 2 \\sin x \\cos x = \\sin 2x$. The correct answer is D.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41630,
    "subject": "Mathematics (Multi-modal)",
    "question": "When a factory modernized its equipment, the productivity grew by $25\\%$. Before, the management decided to fire $20\\%$ of the employees. By how many $\\%$ has the number of final products in this factory changed after both actions?\n(A) It has decreased by $5\\%$.\n(B) It has decreased by $2.5\\%$.\n(C) It has decreased by $2\\%$.\n(D) It has not changed.\n(E) It has grown by $5\\%$.",
    "options": [],
    "answer": "D",
    "solution": "Denote the number of products produced by the factory before the changes by $x$. Then they produced $\\frac{125}{100} \\cdot x$ products after the modernization of the equipment. After firing the employees the number has dropped to $\\frac{80}{100} \\cdot \\frac{125}{100} \\cdot x = x$ products. Hence, the number of products has not changed. The correct answer is **D**.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41631,
    "subject": "Mathematics (Multi-modal)",
    "question": "We have $10$ balls in a bowl, some of them are blue, some of them are yellow and the others are green. They can be put in a line in $360$ different ways. At most how many blue balls are there in the bowl?\n(A) $4$\n(B) $5$\n(C) $6$\n(D) $7$\n(E) $8$",
    "options": [],
    "answer": "D",
    "solution": "Denote the numbers of blue, yellow and green balls by $b$, $y$ and $g$, where $b + y + g = 10$. The balls can be put in a line in $\\frac{10!}{b!y!g!}$ different ways. So, $\\frac{10!}{b!y!g!} = 360$. This equality can be rewritten as $10 \\cdot 9 \\cdots (b+1) = 360 \\cdot y! \\cdot g!$. This implies $10 \\cdot 9 \\cdots (b+1) \\ge 360 = 10 \\cdot 9 \\cdot 4$, so $b+1 \\le 8$, or $b \\le 7$. If, for example, we have $b=7$, $y=2$ and $g=1$, then $\\frac{10!}{7!2!1!} = \\frac{10 \\cdot 9 \\cdot 8}{2} = 360$. There can be at most $7$ blue balls in the bowl. The correct answer is $D$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41632,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the value of the product $x \\cdot y$ if $3^x = a$ and $a^y = 81$?\n(A) 4\n(B) 3\n(C) 12\n(D) 0\n(E) 1",
    "options": [],
    "answer": "A",
    "solution": "Since $81 = a^y = (3^x)^y = 3^{x y}$, and $81 = 3^4$, we have $x y = 4$. The correct answer is **A**.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41633,
    "subject": "Mathematics (Multi-modal)",
    "question": "Function $f$ satisfies $f(x) = x^2 + 1$. What is the value of $\\frac{f(f(x)+x)}{f(x)}$?\n\n(A) $x^2 + x + 1$\n\n(B) $x^2 + 2x + 2$\n\n(C) $x^2 + 1$\n\n(D) $x^2 + 2x + 1$\n\n(E) $x^2 + x$",
    "options": [],
    "answer": "B",
    "solution": "From the definition of $f$ we get\n$$\n\\frac{f(f(x)+x)}{f(x)} = \\frac{(x^2+1+x)^2+1}{x^2+1} = \\frac{x^4+2x^3+3x^2+2x+2}{x^2+1}.\n$$\nDivision of polynomials gives us the result, $x^2+2x+2$. Hence, the correct answer is $B$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41634,
    "subject": "Mathematics (Multi-modal)",
    "question": "Into a square with the side of length $2$ we draw two semicircles whose diameters are the sides of the square as shown in the figure. What is the area of the unshaded part of the square?\n![](attached_image_1.png)\n(A) $\\frac{\\pi}{2}$\n(B) $2$\n(C) $\\frac{3}{2} + \\frac{\\pi}{4}$\n(D) $\\frac{3\\pi}{4} - \\frac{1}{2}$\n(E) $\\frac{3\\pi}{4}$",
    "options": [],
    "answer": "B",
    "solution": "Adding both diagonals onto the figure we notice that the parts $A$, $B$, $C$ and $D$ have equal area. The area of the unshaded part is therefore equal to one half of the area of the square, i.e. $\\frac{4}{2} = 2$. The correct answer is $B$.\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41635,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x = 2^{2013}$. Then the value of the expression\n$$\nx - \\sqrt{x^2 + 1} + \\frac{1}{\\sqrt{x^2 + 1} + x}\n$$\nis equal to\n(A) -1\n(B) 0\n(C) 1\n(D) $2^{2013}$\n(E) 2",
    "options": [],
    "answer": "B",
    "solution": "After finding the common denominator and rearranging the expression we get\n$$\n\\frac{(x + \\sqrt{x^2 + 1})(x - \\sqrt{x^2 + 1}) + 1}{\\sqrt{x^2 + 1} + x} = \\frac{(x^2 - (x^2 + 1)) + 1}{\\sqrt{x^2 + 1} + x} = 0.\n$$\nThe correct answer is B.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41636,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$, the bisectors of the angles $\\angle BAC$ and $\\angle CBA$ intersect at the point $T$. Let $\\gamma$ denote the size of the angle $\\angle ACB$. What is the size of the angle $\\angle ATB$?\n![](attached_image_1.png)\n\n(A) $2\\gamma$\n(B) $180^\\circ - \\gamma$\n(C) $360^\\circ - 4\\gamma$\n(D) $60^\\circ + \\gamma$\n(E) $90^\\circ + \\frac{\\gamma}{2}$",
    "options": [],
    "answer": "E",
    "solution": "We have $\\angle ATB = 180^\\circ - \\angle BAT - \\angle TBA = 180^\\circ - \\frac{\\alpha}{2} - \\frac{\\beta}{2} = 180^\\circ - \\frac{1}{2}(\\alpha + \\beta)$. Since $\\alpha + \\beta + \\gamma = 180^\\circ$, we get $\\alpha + \\beta = 180^\\circ - \\gamma$. So, $\\angle ATB = 180^\\circ - \\frac{1}{2}(180^\\circ - \\gamma) = 90^\\circ + \\frac{\\gamma}{2}$. The correct answer is E.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41637,
    "subject": "Mathematics (Multi-modal)",
    "question": "A herd of deer consists of harts and hinds. Hinds represent $55\\%$ of the herd, and their weight is $45\\%$ of the total weight of the herd. How many times is the average weight of a hart greater than the average weight of a hind?\n(A) $\\frac{81}{40}$\n(B) $\\frac{3}{2}$\n(C) $\\frac{121}{81}$\n(D) $\\frac{11}{9}$\n(E) $\\frac{6}{5}$",
    "options": [],
    "answer": "C",
    "solution": "Assume that the herd consists of $x$ animals with the combined weight of $y$. Then there are $\\frac{55}{100}x$ hinds with the combined weight of $\\frac{45}{100}y$ and $\\frac{45}{100}x$ harts with the combined weight of $\\frac{55}{100}y$. The average weight of a hind is then $\\frac{45}{100}y : \\frac{55}{100}x = \\frac{9y}{11x}$, and the average weight of a hart is $\\frac{55}{100}y : \\frac{45}{100}x = \\frac{11y}{9x}$. Hence, the average weight of a hart is $\\frac{11y}{9x} : \\frac{9y}{11x} = \\frac{121}{81}$ times larger than the average weight of a hind. The correct answer is C.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41638,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the unit square $ABCD$, the two rays from the vertex $A$ divide the right angle in three: the middle angle measures $30^\\circ$ and the other two angles are of the same size. One of the rays intersects the side $BC$ at the point $T$. Find the length of the line segment $BT$.\n\n(A) $\\frac{1}{2}$\n\n(B) $\\frac{\\sqrt{3}}{2}$\n\n(C) $\\frac{1}{3}$\n\n(D) $\\frac{\\sqrt{3}}{3}$\n\n(E) $\\frac{\\sqrt{2}}{2}$",
    "options": [],
    "answer": "D",
    "solution": "From the given data we determine $\\angle BAT = 30^\\circ$. Since $\\tan(\\angle BAT) = \\frac{|TB|}{|AB|}$, we have $|TB| = |AB| \\tan 30^\\circ = 1 \\cdot \\frac{\\sqrt{3}}{3} = \\frac{\\sqrt{3}}{3}$. The correct answer is $D$.\n\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 41639,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $n$ of the form $n = \\overline{23ab16c}$, such that all their digits are different and they are divisible by $9$ and $11$. Here, $a$, $b$ and $c$ are digits.",
    "options": [],
    "answer": "2379168 and 2389167",
    "solution": "The number $n$ is divisible by $99$. We may write\n$$\n\\begin{aligned}\nn &= \\overline{23ab1} \\cdot 100 + 60 + c = \\overline{23ab1} \\cdot 99 + \\overline{23ab1} + 60 + c = \\\\\n&= \\overline{23ab1} \\cdot 99 + \\overline{23a} \\cdot 100 + 10b + 1 + 60 + c = \\\\\n&= (\\overline{23ab1} \\cdot 99 + \\overline{23a}) \\cdot 99 + \\overline{23a} + 10b + c + 61 = \\\\\n&= (\\overline{23ab1} \\cdot 99 + \\overline{23a}) \\cdot 99 + 230 + a + 10b + c + 61 = \\\\\n&= (\\overline{23ab1} \\cdot 99 + \\overline{23a} + 2) \\cdot 99 + a + 10b + c + 93,\n\\end{aligned}\n$$\nso $99$ divides $a + 10b + c + 93$. Since $a$, $b$ and $c$ are different digits, which are also not equal to $1$, $2$, $3$ or $6$, we get $a + 10b + c \\ge 4 + 0 + 5 = 9$ and $a + 10b + c \\le 7 + 90 + 8 = 105$. So, $102 \\le a + 10b + c + 93 \\le 198$. On the other hand, $a + 10b + c + 93$ is divisible by $99$, so $a + 10b + c + 93 = 198$, which implies $b = 9$ and $\\{a, c\\} = \\{7, 8\\}$. There are two solutions, $n = 2379168$ and $n = 2389167$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41640,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all quadruples of non-zero digits $a$, $b$, $c$ and $d$ such that $\\overline{ab20} - \\overline{13cd} = \\overline{cdab}$.",
    "options": [],
    "answer": "(a, b, c, d) = (6, 7, 5, 3)",
    "solution": "Rewrite the equation as $cdab + 13cd = ab20$. Since $b$ and $d$ are non-zero digits the equation for the ones is $b + d = 10$. Subtracting this on both sides of the equation we get $cda0 + 13c0 = ab10$. Dividing the equation by $10$ we obtain $cda + 13c = ab1$. Since $a$ and $c$ are non-zero digits, their sum cannot be equal to $1$, so the equation for the ones yields $a + c = 11$. Once more we subtract this expression from both sides of the equation and divide by $10$. We get $cd + 13 = a(b - 1)$ since $b - 1 \\ge 0$. We consider two cases. If $d \\le 6$, then $d + 3 = b - 1$ and $c + 1 = a$. We can conclude that $a = 6$, $b = 7$, $c = 5$ and $d = 3$. If on the other hand $d \\ge 7$, then\n$d + 3 = 10 + (b - 1)$ and $c + 1 = a - 1$. But in this case we get a contradiction $c = \\frac{9}{2}$. Hence, the only solution is $(a, b, c, d) = (6, 7, 5, 3)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41641,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be positive real numbers, such that their product is $1$ and the sum of their squares is $4$. Find the exact value of the expression $a^{-3} + b^{-3}$.",
    "options": [],
    "answer": "3*sqrt(6)",
    "solution": "From $a^2 + b^2 = 4$ and $ab = 1$ we get $(a+b)^2 = a^2 + b^2 + 2ab = 4 + 2 = 6$, or $a+b = \\sqrt{6}$ since $a$ and $b$ are positive. This implies\n$$\n\\frac{1}{a^3} + \\frac{1}{b^3} = \\frac{a^3 + b^3}{a^3 b^3} = \\frac{(a+b)^3 - 3ab(a+b)}{a^3 b^3} = \\frac{6\\sqrt{6} - 3 \\cdot 1 \\cdot \\sqrt{6}}{1} = 3\\sqrt{6}.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41642,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integer solutions of the equation $m^4 + 2n^2 = 9mn$.",
    "options": [],
    "answer": "(-3, -9), (-2, -8), (-2, -1), (0, 0), (2, 1), (2, 8), (3, 9)",
    "solution": "If the pair $(m, n)$ is a solution of the equation, then so is the pair $(-m, -n)$. So, we may assume that $m$ is non-negative.\n\nRearrange the equation into $2m^2 - 9mn + m^4 = 0$ and treat it as a quadratic equation in $n$. Its discriminant is $81m^2 - 8m^4$. If the equation is to have integer solutions, the discriminant has to be a perfect square. So, $81 - 8m^2$ is a perfect square. In particular, $81 - 8m^2 \\ge 0$, so $m \\le 3$.\n\nIt is easy to verify that $81 - 8m^2$ is a perfect square for $m = 0, m = 2$ and $m = 3$. When $m = 0$ the solution is $n = 0$, when $m = 2$ the two solutions are $n = 1$ and $n = 8$, for $m = 3$ the only integer solution is $n = 9$.\n\nAll the integer solutions of the equation are $(-3, -9), (-2, -8), (-2, -1), (0, 0), (2, 1), (2, 8)$ and $(3, 9)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41643,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that every tangential quadrilateral whose diagonals intersect at right angles is a deltoid.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us use the notation from the figure. By Pythagoras' theorem we have $a^2 = x^2 + y^2$, $b^2 = y^2 + z^2$, $c^2 = z^2 + w^2$ and $d^2 = w^2 + x^2$. This implies $a^2 + c^2 = b^2 + d^2$.\n\nBut the quadrilateral is tangential, so $a + c = b + d$. Squaring this equality and using the equalities obtained above we find that $2ac = 2bd$, or $ac = bd$. Plugging in $a = b + d - c$ we get $c^2 - (b + d)c + bd = 0$, which can be factored as $(c - b)(c - d) = 0$.\n\nSo, $c = b$ and $a = d$ or $c = d$ and $a = b$. In both cases the quadrilateral is a deltoid.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41644,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the origin of the coordinate system. The point $A(\\frac{5}{2}, -\\frac{5\\sqrt{3}}{2})$ is rotated around $O$ by $2013\\pi$ into the point $B$. The point $B$ is reflected across the bisector of the odd quadrants into the point $C$. Find the size of the angle $\\angle AOC$.",
    "options": [],
    "answer": "30°",
    "solution": "The coordinates of the point $B$ are $(-\\frac{5}{2}, \\frac{5\\sqrt{3}}{2})$ and the coordinates of $C$ are $(\\frac{5\\sqrt{3}}{2}, -\\frac{5}{2})$. All three points lie on a circle with the centre at $O$ and radius $5$. From the values of the trigonometric functions $\\sin(-30^\\circ) = -\\frac{1}{2}$ and $\\cos(-30^\\circ) = \\frac{\\sqrt{3}}{2}$ we conclude that the angle between $OC$ and the positive ray of the $x$-axis is $30^\\circ$. Similarly, $\\sin(-60^\\circ) = -\\frac{1}{2}$ and $\\cos(-60^\\circ) = \\frac{\\sqrt{3}}{2}$ imply that the angle between $OA$ and the positive ray of the $x$-axis is $60^\\circ$. From here we conclude that $\\angle AOC = 30^\\circ$.\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41645,
    "subject": "Mathematics (Multi-modal)",
    "question": "A rectangle has been divided into four parts by three line segments, as shown in the picture. After that, the four shapes obtained have been rearranged to form a square. What is the perimeter of this square?\n![](attached_image_1.png)",
    "options": [],
    "answer": "48",
    "solution": "Let us use the notation suggested in the figure. By Pythagoras' theorem we have $y = \\sqrt{15^2 - 9^2} = \\sqrt{144} = 12$. The two right triangles on the left side are similar, so $\\frac{x}{5} = \\frac{y}{15}$, or $x = \\frac{y}{3} = 4$. Hence, the sides of the rectangle measure $9$ and $16$, and its area is $144$. The area of the square is therefore also equal to $144$, so its side has the length $12$, and its perimeter is $48$.\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41646,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $x$ that satisfy the equation\n$$\n\\log_{\\sin x} \\left( \\frac{1}{2} \\sin 2x \\right) = 2.\n$$",
    "options": [],
    "answer": "x = pi/4 + 2k*pi, where k is any integer",
    "solution": "The base of the logarithm has to be positive and different from $1$, so $\\sin x > 0$ and $\\sin x \\ne 1$. Also, $\\frac{1}{2} \\sin 2x > 0$ since the logarithm is only defined for positive numbers. When all this holds we may use the definition of the logarithm to transform the equation into the equivalent equation $\\sin^2 x = \\frac{1}{2} \\sin 2x = \\sin x \\cos x$, or $\\sin x(\\sin x - \\cos x) = 0$. Since $\\sin x > 0$, we have $\\sin x - \\cos x = 0$. If $\\cos x = 0$, we would get $\\sin x = 0$, but this is not possible. So, we may divide by $\\cos x$ and get\n\n$$\n\\tan x = 1.\n$$\n\nFrom here we obtain the solutions $x = \\frac{\\pi}{4} + 2k\\pi$ and $x = \\frac{5\\pi}{4} + 2k\\pi$, where $k$ is an integer. But $\\sin(\\frac{5\\pi}{4} + 2k\\pi) = \\sin \\frac{5\\pi}{4} = -\\frac{\\sqrt{2}}{2} < 0$, so the only solutions are $x = \\frac{\\pi}{4} + 2k\\pi$, where $k$ is an integer.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41647,
    "subject": "Mathematics (Multi-modal)",
    "question": "When the third grade pupil Benjamin calculated the sum $1+2+3+\\ldots+2012$, he forgot to add some terms, and he got an incorrect sum that was divisible by $2011$. When Anika calculated the sum $A = 1+2+3+\\ldots+2013$, she forgot to add the same terms as Benjamin, and she got an incorrect sum $N$ that was divisible by $2014$. What is the ratio $N/A$ of the two sums?",
    "options": [],
    "answer": "2/3",
    "solution": "Let us denote the sum of the terms omitted by Benjamin by $x$. Since $1 + 2 + 3 + \\ldots + 2012 = \\frac{2012 \\cdot 2013}{2} = 1006 \\cdot 2013$, Benjamin's result was equal to $1006 \\cdot 2013 - x$. So, there exists a non-negative integer $m$, such that $1006 \\cdot 2013 - x = 2011m$.\n\nSince $A = 1 + 2 + 3 + \\ldots + 2013 = \\frac{2013 \\cdot 2014}{2} = 2013 \\cdot 1007$, Anika's result was equal to $N = 2013 \\cdot 1007 - x$. So, there exists a non-negative integer $n$, such that $2013 \\cdot 1007 - x = 2014n$.\n\nExpressing $x$ from both equalities and comparing the results we get $1006 \\cdot 2013 - 2011m = 2013 \\cdot 1007 - 2014n$, or $2014n - 2011m - 2013 = 0$. This last equality can be rearranged to $2011(n-m) = 2013-3n$.\n\nSince $2014n = 2013 \\cdot 1007 - x \\le 2013 \\cdot 1007$, we get $n \\le \\frac{2013 \\cdot 1007}{2014} < 1007$. So, $-1008 < 2013 - 3n \\le 2013$. At the same time $2013-3n$ is divisible by $3$ and the equality implies it is also divisible by $2011$. This is only possible for $2013 - 3n = 0$ or $n = 671$.\n\nHence, $\\frac{N}{A} = \\frac{2014n}{2013 \\cdot 1007} = \\frac{2014 \\cdot 671}{2013 \\cdot 1007} = \\frac{2}{3}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41648,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEF$ be a regular hexagon, let $P$ be the midpoint of the side $AB$ and let $R$ be the midpoint of the side $EF$, as shown in the figure. What is the ratio of the area of the quadrilateral $APRF$ to the area of the quadrilateral $BCDP$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "5/8",
    "solution": "Denote the length of the side of the regular hexagon by $a$. The quadrilateral $APRF$ is a trapezium with the altitude $\\frac{1}{2} \\cdot \\frac{a\\sqrt{3}}{2} = \\frac{a\\sqrt{3}}{4}$, so its area equals\n$$\nP_{APRF} = \\frac{a\\sqrt{3}}{4} \\cdot \\frac{|PR|+|AF|}{2} = \\frac{a\\sqrt{3}(\\frac{3}{2}a+a)}{8} = \\frac{5\\sqrt{3}a^2}{16}.\n$$\nThe triangle $PBD$ has a side of length $|BD| = 2 \\cdot \\frac{a\\sqrt{3}}{2} = \\sqrt{3}a$ and the altitude $|PB| = \\frac{a}{2}$, so its area equals\n$$\nP_{PBD} = \\frac{a}{2} \\cdot \\frac{\\sqrt{3}a}{2} = \\frac{\\sqrt{3}a^2}{4}.\n$$\nThe triangle $BCD$ has a side of length $|BD| = \\sqrt{3}a$ and the altitude $\\frac{a}{2}$, so its area also equals\n$$\nP_{BCD} = \\frac{\\sqrt{3}a^2}{4}.\n$$\nThe area of the quadrilateral $BCDP$ is then\n$$\nP_{BCDP} = P_{PBD} + P_{BCD} = 2 \\cdot \\frac{\\sqrt{3}a^2}{4} = \\frac{\\sqrt{3}a^2}{2}.\n$$\nThe ratio of the areas is\n$$\n\\frac{P_{APRF}}{P_{BCDP}} = \\frac{5}{8}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41649,
    "subject": "Mathematics (Multi-modal)",
    "question": "Žan wrote a sequence of four positive real numbers. The first term in the sequence was the number $3$, and the last term was the number $9$. The first three terms formed a geometric sequence, and the last three terms formed an arithmetic sequence. Determine all four terms of Žan's sequence.",
    "options": [],
    "answer": "3, 9/2, 27/4, 9",
    "solution": "Denote the second and the third term of Žan's sequence by $x$ and $y$. Then $3$, $x$, $y$ is a geometric sequence and $x^2 = 3y$. On the other hand, $x$, $y$, $9$ is an arithmetic sequence, so $2y = x + 9$. We may rewrite the second equation as $y = \\frac{x+9}{2}$. Plugging this into the first equation and rearranging we get $2x^2 - 3x - 27 = 0$, or $(2x - 9)(x + 3) = 0$. Since $x$ is positive we have $x = \\frac{9}{2}$. From here we get $y = \\frac{27}{4}$. Žan's sequence is $3$, $\\frac{9}{2}$, $\\frac{27}{4}$, $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41650,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that arbitrary real numbers $a$ and $b$ satisfy the inequality\n$$\n(a + ab - b^2)^2 + ab^2(a + 2) \\geq 0.\n$$\nWhen does the equality hold?",
    "options": [],
    "answer": "Equality holds if and only if a = b = 0 or a = b = -1.",
    "solution": "Expanding the left-hand side of the inequality we get $a^2 + a^2b^2 + b^4 + 2a^2b - 2ab^3 + a^2b^2$. This can be rearranged into $a^2(1+b)^2 + b^2(b-a)^2$, and the desired inequality now follows. At the same time we see that the equality holds if and only if $a = b = 0$ or $a = b = -1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41651,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1$ and $x_2$ be two different roots of the polynomial $p(x) = x^2 + a x + b$, and let $x_1^2 - \\frac{1}{2}$ and $x_2^2 - \\frac{1}{2}$ be the roots of $q(x) = x^2 + (a^2 - \\frac{1}{2}) x + b^2 - \\frac{1}{2}$. Find $a$ and $b$.",
    "options": [],
    "answer": "a = 0, b = -3/4",
    "solution": "The roots of the polynomial $p(x) = x^2 + a x + b$ are $x_1 = \\frac{-a + \\sqrt{a^2 - 4b}}{2}$ and $x_2 = \\frac{-a - \\sqrt{a^2 - 4b}}{2}$. They have to be different, so $a^2 - 4b \\neq 0$. From here we get\n$$\nx_1^2 - \\frac{1}{2} = \\frac{(a^2 - 2b - 1) - a \\sqrt{a^2 - 4b}}{2}\n$$\nand\n$$\nx_2^2 - \\frac{1}{2} = \\frac{(a^2 - 2b - 1) + a \\sqrt{a^2 - 4b}}{2}.\n$$\nSince these are the roots of the polynomial $q(x)$, they have to satisfy the equation $x^2 + (a^2 - \\frac{1}{2}) x + b^2 - \\frac{1}{2} = 0$. Inserting both into this equation and rearranging we get\n$$\n(4a^4 - 12a^2b + 8b^2 - 5a^2 + 6b) - (4a^3 - 4ab - 3a)\\sqrt{a^2 - 4b} = 0\n$$\nand\n$$\n(4a^4 - 12a^2b + 8b^2 - 5a^2 + 6b) + (4a^3 - 4ab - 3a)\\sqrt{a^2 - 4b} = 0.\n$$\nSo,\n$$\n4a^4 - 12a^2b + 8b^2 - 5a^2 + 6b = 0\n$$\nand\n$$\n(4a^3 - 4ab - 3a)\\sqrt{a^2 - 4b} = 0.\n$$\nSince $a^2-4b \\neq 0$ the second equation implies that $(4a^3-4ab-3a) = a(4a^2-4b-3) = 0$. If $a=0$, then the first equation implies that $8b^2+6b = 2b(4b+3) = 0$. Since $a^2-4b \\neq 0$ we have $b \\neq 0$, so $b = -\\frac{3}{4}$. If $a \\neq 0$, then $4a^2 - 4b - 3 = 0$, or $b = a^2 - \\frac{3}{4}$. Using this in the first equation we get $-2a^2 = 0$, or $a = 0$, which is a contradiction. So, $a = 0$ and $b = -\\frac{3}{4}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41652,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to line up the numbers $1, 2, 3, \\ldots, 2013$ so that the arithmetic mean of any two of the numbers is never located between them?",
    "options": [],
    "answer": "Yes",
    "solution": "Let us show that the statement is true for any $n \\in \\mathbb{N}$. We claim that the numbers $1, 2, \\dots, n$ can be lined up so that for any pair their arithmetic mean does not lie somewhere in between.\n\nFirst, we will show that this is true for $n = 2^m$ for all $m \\in \\mathbb{N}$. We will use induction on $m$.\n\nIn the base case $m = 1$ this is obvious.\n\nNow, let us assume that for some $m$ the numbers $1, 2, \\dots, 2^m$ can be arranged into the sequence\n$(a_1, a_2, \\dots, a_{2^m})$ so that for any pair their arithmetic mean does not lie in between the two numbers.\n\nWe notice that the sequence\n$$\n(b_1, b_2, \\dots, b_{2^{m+1}}) = (2a_1 - 1, 2a_2 - 1, \\dots, 2a_{2^m} - 1, 2a_1, 2a_2, \\dots, 2a_{2^m})\n$$\nis a permutation of $1, 2, \\dots, 2^{m+1}$ which satisfies the condition of the problem. Indeed, by the induction hypothesis the arithmetic mean of $b_i$ and $b_j$, where either $1 \\le i < j \\le 2^m$ or $2^m + 1 \\le i < j \\le 2^{m+1}$, does not lie in between them and if $1 \\le i \\le 2^m < j \\le 2^{m+1}$, then the arithmetic mean of $b_i$ and $b_j$ is not an integer at all. We have thus proven the statement for $n = 2^m$.\n\nFinally, if the positive integer $n$ is not a power of $2$, then there exists $m \\in \\mathbb{N}$ such that $n < 2^m$. In this case we can first arrange the numbers $1, 2, \\dots, 2^m$ into the sequence that satisfies the condition and then simply remove any numbers greater than $n$. The new sequence obtained in this way will obviously still satisfy the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41653,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIzračunaj vrednosti koeficientov polinoma $p$ s predpisom $p(x) = a x^{3} + b x^{2} + c x + d$, če je $0$ njegova ničla in velja $p(1) = \\frac{16}{3}$. Ničli odvoda polinoma $p$ sta $-1$ in $-3$.",
    "options": [],
    "answer": "a = 1/3, b = 2, c = 3, d = 0",
    "solution": "Solution:\n\nKer je $0$ ničla polinoma $p$, velja $p(0) = 0$.\n\n$$\np(0) = a \\cdot 0^3 + b \\cdot 0^2 + c \\cdot 0 + d = d = 0\n$$\nTorej je $d = 0$.\n\nPolinom je torej oblike:\n$$\np(x) = a x^3 + b x^2 + c x\n$$\n\nPodano je še $p(1) = \\frac{16}{3}$:\n$$\np(1) = a \\cdot 1^3 + b \\cdot 1^2 + c \\cdot 1 = a + b + c = \\frac{16}{3}\n$$\n\nOdvod polinoma je:\n$$\np'(x) = 3a x^2 + 2b x + c\n$$\n\nNičli odvoda sta $-1$ in $-3$:\n$$\np'(-1) = 3a (-1)^2 + 2b (-1) + c = 3a - 2b + c = 0\n$$\np'(-3) = 3a (-3)^2 + 2b (-3) + c = 27a - 6b + c = 0\n$$\n\nDobimo sistem enačb:\n\n1) $a + b + c = \\frac{16}{3}$\n\n2) $3a - 2b + c = 0$\n\n3) $27a - 6b + c = 0$\n\nOdštejmo drugo enačbo od tretje:\n$$\n(27a - 6b + c) - (3a - 2b + c) = 0\n$$\n$$\n24a - 4b = 0 \\implies 6a - b = 0 \\implies b = 6a\n$$\n\nVstavimo $b = 6a$ v drugo enačbo:\n$$\n3a - 2(6a) + c = 0\n$$\n$$\n3a - 12a + c = 0\n$$\n$$\n-9a + c = 0 \\implies c = 9a\n$$\n\nVstavimo $b = 6a$ in $c = 9a$ v prvo enačbo:\n$$\na + 6a + 9a = \\frac{16}{3}\n$$\n$$\n16a = \\frac{16}{3}\n$$\n$$\na = \\frac{1}{3}\n$$\n\nTorej:\n$$\nb = 6a = 2\n$$\n$$\nc = 9a = 3\n$$\n\nKoeficienti so:\n$$\na = \\frac{1}{3}, \\quad b = 2, \\quad c = 3, \\quad d = 0\n$$\n\nTorej je polinom:\n$$\np(x) = \\frac{1}{3} x^3 + 2 x^2 + 3 x\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41654,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that\n$$\n(n^2 + 11n - 4) \\cdot n! + 33 \\cdot 13^n + 4\n$$\nis a perfect square.",
    "options": [],
    "answer": "n = 1, 2",
    "solution": "For reasons of readability let us write $A_n = (n^2 + 11n - 4) \\cdot n! + 33 \\cdot 13^n + 4$. First, consider the value of $A_n$ modulo $8$ for $n \\ge 4$. We have $8 \\mid 1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot \\dots \\cdot n = n!$, so\n$$\nA_n = (n^2 + 11n - 4) \\cdot n! + 33 \\cdot 13^n + 4 \\equiv 0 + 1 \\cdot 5^n + 4 \\pmod{8}.\n$$\nSince $5^2 \\equiv 1 \\pmod{8}$, we have $5^{2k} \\equiv 1 \\pmod{8}$ and $5^{2k-1} \\equiv 5 \\pmod{8}$ for all $k \\in \\mathbb{N}$. For even $n$ we get $A_n \\equiv 5 \\pmod{8}$ and for odd $n$ we have $A_n \\equiv 1 \\pmod{8}$. On the other hand a perfect square can only give the remainder of $0, 1$ or $4$ when divided by $8$, so all even $n \\ge 4$ are out of consideration.\n\nNow, consider the value of $A_n$ modulo $7$ for $n \\ge 7$. We have $7 \\mid n!$, so\n$$\nA_n = (n^2 + 11n - 4) \\cdot n! + 33 \\cdot 13^n + 4 \\equiv 0 + 5 \\cdot (-1)^n + 4 \\pmod{7}.\n$$\nSince $(-1)^2 = 1 \\pmod{7}$, we have $(-1)^{2k} = 1 \\pmod{7}$ and $(-1)^{2k-1} = -1 \\pmod{7}$ for all $k \\in \\mathbb{N}$. For even $n$ we get $A_n \\equiv 2 \\pmod{7}$, for odd $n$ we get $A_n \\equiv 6 \\pmod{7}$. A perfect square can only give the remainders of $0, 1, 2$ or $4$ when divided by $7$, so we can exclude all odd $n \\ge 7$.\n\nThe remaining options are $n = 1, 2, 3$ and $5$.\nIf $n = 3$ we have\n$$\nA_3 \\equiv 3 \\cdot 1 + 3 \\cdot 3^3 + 4 \\equiv 3 \\pmod{5}.\n$$\nA perfect square can only give the remainder of $0, 1$ or $4$ when divided by $5$, so $A_3$ is not a perfect square.\nA similar reasoning helps us see that $n = 5$ does not work, since\n$$\nA_5 \\equiv (-4) \\cdot 0 + 3 \\cdot 3^5 + 4 \\equiv 3 \\pmod{5}.\n$$\nIf $n = 1$, then $A_1 = 441 = 21^2$. If $n = 2$, then $A_2 = 5625 = 75^2$. Thus, $n = 1$ and $n = 2$ are the only solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41655,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $p$ with real coefficients, such that\n$$\np(p(x)) = (x^2 + x + 1)p(x)\n$$\nfor all $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "p(x) = 0 and p(x) = x^2 + x",
    "solution": "The zero polynomial is obviously a valid solution. Let $p$ be a non-zero polynomial and write $p(x) = a_nx^n + a_{n-1}x^{n-1} + \\dots + a_0$, where $a_n \\ne 0$. The leading term on the left-hand side of the equality is equal to $a_n(a_nx^n)^n = a_n^{n+1}x^{n^2}$, and the leading term on the right-hand side is $x^2 \\cdot a_nx^n = a_nx^{n+2}$. The degrees of the two terms must be equal, so $n^2 = n + 2$, or $(n-2)(n+1) = 0$. Since $n$ is a non-negative integer, we conclude that $n=2$. Hence, $p(x) = ax^2 + bx + c$, where $a \\ne 0$. We have:\n$$\np(p(x)) = a^3 x^4 + 2a^2 bx^3 + (ab^2 + 2a^2 c + ab)x^2 + (2abc + b^2)x + (ac^2 + bc + c) \\quad (1)\n$$\n$$\n(x^2 + x + 1)p(x) = ax^4 + (b+a)x^3 + (c+b+a)x^2 + (c+b)x + c. \\quad (2)\n$$\n\nFrom the given equality we obtain the system of equations\n$$\n\\begin{align*}\na^3 &= a, \\\\\n2a^2b &= b+a, \\\\\nab^2 + 2a^2c + ab &= c+b+a, \\\\\n2abc + b^2 &= c+b, \\\\\nac^2 + bc + c &= c.\n\\end{align*}\n$$\nSince $a \\neq 0$, the first equation implies $a = 1$ or $a = -1$. If $a = 1$ the second equation becomes $2b = b + 1$, so $b = 1$. Inserting both into the third equation we find that $2c + 2 = c + 2$, so $c = 0$. It is easy to check that these three numbers also satisfy the final two equations. If $a = -1$, then the system has no solutions. There are two solutions to the problem, namely $p(x) = 0$ and $p(x) = x^2 + x$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41656,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bosta $x_{1}$ in $x_{2}$ različni ničli polinoma $p(x)=x^{2}+a x+b$, $x_{1}^{2}-\\frac{1}{2}$ in $x_{2}^{2}-\\frac{1}{2}$ pa naj bosta ničli polinoma $q(x)=x^{2}+(a^{2}-\\frac{1}{2}) x+b^{2}-\\frac{1}{2}$. Določi $a$ in $b$.",
    "options": [],
    "answer": "a = 0, b = -3/4",
    "solution": "Solution:\n\nNičli polinoma $p(x)=x^{2}+a x+b$ sta $x_{1}=\\frac{-a+\\sqrt{a^{2}-4 b}}{2}$ in $x_{2}=\\frac{-a-\\sqrt{a^{2}-4 b}}{2}$. Ker morata biti različni, je $a^{2}-4 b \\neq 0$. Od tod izračunamo\n$$\nx_{1}^{2}-\\frac{1}{2}=\\frac{\\left(a^{2}-2 b-1\\right)-a \\sqrt{a^{2}-4 b}}{2}\n$$\nin\n$$\nx_{2}^{2}-\\frac{1}{2}=\\frac{\\left(a^{2}-2 b-1\\right)+a \\sqrt{a^{2}-4 b}}{2}\n$$\nKer sta to ničli polinoma $q(x)$, morata ustrezati enačbi $x^{2}+(a^{2}-\\frac{1}{2}) x+b^{2}-\\frac{1}{2}=0$. Ko oboje vstavimo v to enačbo in preuredimo, dobimo\n$$\n\\left(4 a^{4}-12 a^{2} b+8 b^{2}-5 a^{2}+6 b\\right)-\\left(4 a^{3}-4 a b-3 a\\right) \\sqrt{a^{2}-4 b}=0\n$$\nin\n$$\n\\left(4 a^{4}-12 a^{2} b+8 b^{2}-5 a^{2}+6 b\\right)+\\left(4 a^{3}-4 a b-3 a\\right) \\sqrt{a^{2}-4 b}=0\n$$\nTorej mora veljati\n$$\n4 a^{4}-12 a^{2} b+8 b^{2}-5 a^{2}+6 b=0\n$$\nin\n$$\n\\left(4 a^{3}-4 a b-3 a\\right) \\sqrt{a^{2}-4 b}=0\n$$\nKer je $a^{2}-4 b \\neq 0$, iz druge enačbe sledi $\\left(4 a^{3}-4 a b-3 a\\right)=a\\left(4 a^{2}-4 b-3\\right)=0$. Če je $a=0$, potem iz prve enačbe sledi $8 b^{2}+6 b=2 b(4 b+3)=0$. Ker je $a^{2}-4 b \\neq 0$, mora biti $b \\neq 0$, torej je $b=-\\frac{3}{4}$. Če pa je $a \\neq 0$, potem mora biti $4 a^{2}-4 b-3=0$ oziroma $b=a^{2}-\\frac{3}{4}$. Ko slednje vstavimo v prvo enačbo, dobimo $-2 a^{2}=0$ oziroma $a=0$, kar je protislovje. Torej je $a=0$ in $b=-\\frac{3}{4}$.\n\n\n2. način. Po Vietovih pravilih je $x_{1}+x_{2}=-a, x_{1} x_{2}=b, x_{1}^{2}+x_{2}^{2}-1=\\frac{3}{2}-a^{2}$ in $\\left(x_{1}^{2}-\\frac{1}{2}\\right)\\left(x_{2}^{2}-\\frac{1}{2}\\right)=b^{2}-\\frac{1}{2}$. Najprej preoblikujmo tretjo enačbo, pri čemer upoštevamo prvi dve,\n$$\n\\frac{3}{2}-a^{2}=x_{1}^{2}+x_{2}^{2}=(x_{1}+x_{2})^{2}-2 x_{1} x_{2}=a^{2}-2 b\n$$\nod koder dobimo $b=a^{2}-\\frac{3}{4}$. Preoblikujmo še četrto enačbo,\n$$\nb^{2}-\\frac{1}{2}=\\left(x_{1}^{2}-\\frac{1}{2}\\right)\\left(x_{2}^{2}-\\frac{1}{2}\\right)=x_{1}^{2} x_{2}^{2}-\\frac{1}{2}(x_{1}^{2}+x_{2}^{2})+\\frac{1}{4}=b^{2}-\\frac{1}{2}\\left(\\frac{3}{2}-a^{2}\\right)+\\frac{1}{4}\n$$\nod koder sledi $a^{2}=0$ in zato $a=0$ ter $b=-\\frac{3}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41657,
    "subject": "Mathematics (Multi-modal)",
    "question": "One day $n$ students participated at a sporting competition with several different disciplines. Exactly $k$ students competed in each of the disciplines where $3 \\le k \\le n$. Each pair of students had exactly three disciplines in common and each triple has competed in exactly two common disciplines. Determine all possible values of $n$ and $k$.",
    "options": [],
    "answer": "n = 5, k = 4",
    "solution": "The 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$$\n3n(n-1) = mk(k-1). \\qquad (11)\n$$\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$$\n2n(n-1)(n-2) = mk(k-1)(k-2).\n$$\nSince $k, n \\ge 3$, we can divide one equation by the other and get\n$$\n\\frac{2(n-2)}{3} = k-2.\n$$\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$$\nm = \\frac{3n(n-1)}{k(k-1)} = \\frac{3(3l+2)(3l+1)}{(2l+2)(2l+1)}.\n$$\nHowever, $D(3l+1, 2l+1) = D(2l+1, l) = 1$ and $D(3l+2, 2l+1) = D(2l+1, l+1) = D(l+1, l) = 1$. So, $2l+1$ and $(3l+2)(3l+1)$ are relatively prime and $2l+1$ divides 3. This is only possible when $l=1$ and consequently $n=5$, $k=4$ and $m=5$.\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 given in the problem:\n$\\{A_1, A_2, A_3, A_4\\}, \\{A_1, A_2, A_3, A_5\\}, \\{A_1, A_2, A_4, A_5\\}, \\{A_1, A_3, A_4, A_5\\}, \\{A_2, A_3, A_4, A_5\\}.$\nThus, the only solution is $n=5, k=4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41658,
    "subject": "Mathematics (Multi-modal)",
    "question": "For real numbers $a$ and $b$, such that $|a| \\neq |b|$ and $a \\neq 0$, we have\n$$\n\\frac{a-b}{a^2+ab} + \\frac{a+b}{a^2-ab} = \\frac{3a-b}{a^2-b^2}.\n$$\nDetermine the value of the expression $\\frac{b}{a}$.",
    "options": [],
    "answer": "1/2",
    "solution": "Multiplying the equation by $a(a+b)(a-b)$ we get\n$$\n(a-b)^2 + (a+b)^2 = a(3a-b).\n$$\nAfter expanding the terms and moving all the terms to the right-hand side we get\n$$\n0 = a^2 - ab - 2b^2 = (a - 2b)(a + b).\n$$\nSince $a \\neq -b$, we get $a - 2b = 0$ or $a = 2b$. Since $a$ is non-zero, we get $\\frac{b}{a} = \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41659,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIzračunaj vrednosti parametra $n$, tako da bo graf funkcije $f$ s predpisom $f(x)=2x+n$ tvoril s koordinatnima osema trikotnik s ploščino 25.",
    "options": [],
    "answer": "n = 10 or n = -10",
    "solution": "Solution:\n\nGraf funkcije $f(x) = 2x + n$ seka $y$-os pri $x = 0$, torej v točki $(0, n)$.\n\nZa presečišče z $x$-osjo postavimo $f(x) = 0$:\n$$\n2x + n = 0 \\implies x = -\\frac{n}{2}\n$$\nTorej je presečišče z $x$-osjo v točki $\\left(-\\frac{n}{2}, 0\\right)$.\n\nTrikotnik, ki ga graf tvori s koordinatnima osema, ima oglišča v točkah $(0, 0)$, $(0, n)$ in $\\left(-\\frac{n}{2}, 0\\right)$.\n\nOsnova trikotnika je razdalja med $(0, 0)$ in $\\left(-\\frac{n}{2}, 0\\right)$, torej $\\left| -\\frac{n}{2} - 0 \\right| = \\frac{|n|}{2}$.\n\nVišina trikotnika je razdalja med $(0, 0)$ in $(0, n)$, torej $|n|$.\n\nPloščina trikotnika je:\n$$\nP = \\frac{\\text{osnova} \\times \\text{višina}}{2} = \\frac{\\frac{|n|}{2} \\cdot |n|}{2} = \\frac{|n|^2}{4}\n$$\n\nZahtevamo $P = 25$:\n$$\n\\frac{|n|^2}{4} = 25 \\implies |n|^2 = 100 \\implies |n| = 10\n$$\nTorej $n = 10$ ali $n = -10$.\n\nOdgovor: $n = 10$ ali $n = -10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41660,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $x > 0$. Reši enačbo $\\left(\\frac{2}{5}\\right)^{\\log^2 x + 1} = \\left(\\frac{25}{4}\\right)^{2 - \\log x^3}$.",
    "options": [],
    "answer": "10 and 100000",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41661,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse polinome $p$ z realnimi koeficienti, za katere velja\n$$\np(p(x)) = \\left(x^{2} + x + 1\\right) p(x)\n$$\nza vse $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "p(x) = 0 or p(x) = x^2 + x",
    "solution": "Solution:\n\nNičelni polinom je očitno rešitev naloge. Naj bo $p$ neničeln polinom in pišimo $p(x) = a_{n} x^{n} + a_{n-1} x^{n-1} + \\cdots + a_{0}$, kjer je $a_{n} \\neq 0$. Vodilni člen na levi strani enakosti je tedaj enak $a_{n} (a_{n} x^{n})^{n} = a_{n}^{n+1} x^{n^{2}}$, vodilni člen na desni strani pa je enak $x^{2} \\cdot a_{n} x^{n} = a_{n} x^{n+2}$. Ker morata biti ta dva člena iste stopnje, sledi $n^{2} = n + 2$ oziroma $(n-2)(n+1) = 0$. Ker je $n$ nenegativno celo število, mora biti $n = 2$. Pišimo torej $p(x) = a x^{2} + b x + c$, kjer je $a \\neq 0$. Tedaj velja\n$$\n\\begin{gathered}\np(p(x)) = a^{3} x^{4} + 2 a^{2} b x^{3} + \\left(a b^{2} + 2 a^{2} c + a b\\right) x^{2} + \\left(2 a b c + b^{2}\\right) x + \\left(a c^{2} + b c + c\\right) \\\\\n\\left(x^{2} + x + 1\\right) p(x) = a x^{4} + (b + a) x^{3} + (c + b + a) x^{2} + (c + b) x + c\n\\end{gathered}\n$$\nIz dane enakosti tako dobimo sistem enačb\n$$\n\\begin{aligned}\na^{3} & = a \\\\\n2 a^{2} b & = b + a \\\\\na b^{2} + 2 a^{2} c + a b & = c + b + a \\\\\n2 a b c + b^{2} & = c + b \\\\\na c^{2} + b c + c & = c\n\\end{aligned}\n$$\nKer je $a \\neq 0$, iz prve enakosti sledi $a = 1$ ali $a = -1$. Za $a = 1$ v drugi enačbi dobimo $2 b = b + 1$, torej $b = 1$. Če oboje vstavimo v tretjo enačbo, dobimo še $2 c + 2 = c + 2$ oziroma $c = 0$. Preverimo lahko, da ta tri števil ustrezajo tudi zadnjima dvema enakostima. Za $a = -1$ dobimo protisloven sistem. Rešitvi naloge sta torej dve, $p(x) = 0$ in $p(x) = x^{2} + x$.\n\nRešitev $p(x) = 0$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41662,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples of real numbers $(x, y, z)$ that satisfy the system of equations\n$$\n\\begin{aligned}\nx^2 + y^2 + 4z^2 &= 6y - 4, \\\\\n2xy - 4xz + 4yz &= y^2 + 5.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "(1, 3, 1) and (2, 3, 1/2)",
    "solution": "Subtracting the second equality from the first we notice that the left side is a square of a trinomial. Indeed, we get\n$$\n(x - y + 2z)^2 = -y^2 + 6y - 9,\n$$\nwhich can be rearranged into\n$$\n(x - y + 2z)^2 + (y - 3)^2 = 0.\n$$\nSince $x$, $y$ and $z$ are real numbers the expressions in the brackets must be 0. So, $y = 3$ and $x = y - 2z = 3 - 2z$. Plugging this into the first equation we get $2z^2 - 3z + 1 = 0$, or $(2z - 1)(z - 1) = 0$. We obtain the same by plugging $y = 3$ and $x = 3 - 2z$ into the second equation. So, $z = 1$ or $z = \\frac{1}{2}$. The solutions are $(1, 3, 1)$ and $(2, 3, \\frac{1}{2})$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41663,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV podjetju izdelujejo vijake s šestimi enakimi stroji. V devetih urah bi s temi stroji izdelali 900 vijakov. V koliko urah bodo izdelali 900 vijakov, če se po petih urah pokvarita dva stroja?",
    "options": [],
    "answer": "11 hours",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41664,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa realni števili $a$ in $b$, kjer je $|a| \\neq |b|$ in $a \\neq 0$, velja\n$$\n\\frac{a-b}{a^{2}+a b}+\\frac{a+b}{a^{2}-a b}=\\frac{3 a-b}{a^{2}-b^{2}}\n$$\nDoloči vrednost izraza $\\frac{b}{a}$.",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\n\nEnačbo pomnožimo z $a(a+b)(a-b)$ in dobimo\n$$\n(a-b)^{2}+(a+b)^{2}=a(3 a-b)\n$$\nKo odpravimo oklepaje in vse člene nesemo na desno stran, dobimo\n$$\n0=a^{2}-a b-2 b^{2}=(a-2 b)(a+b)\n$$\nKer je $a \\neq -b$, mora biti $a-2 b=0$ oziroma $a=2 b$. Ker $a$ ni enak 0 je torej $\\frac{b}{a}=\\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41665,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of primes $p$ and $q$, such that $p+q$ and $p+4q$ are perfect squares.",
    "options": [],
    "answer": "(5, 11), (13, 3)",
    "solution": "Let $p+q = x^2$ and $p+4q = y^2$ for some positive integers $x$ and $y$. Subtracting the equations we get $3q = y^2 - x^2 = (y-x)(y+x)$. Since $q$ is a prime and $x+y \\ge 2$ we have the following possibilities:\n\n1. $y - x = 1$ and $y + x = 3q$. From the first equation we get $y = x + 1$. Plugging this into the second equation we see that $2x + 1 = 3q$. This implies that $q$ is odd, so we may write $q = 2m + 1$ for some positive integer $m$. Now, $x = 3m + 1$ and $p = x^2 - q = 9m^2 + 4m = m(9m + 4)$. Since $p$ is a prime, we conclude that $m = 1$. Thus, $p = 13$ and $q = 3$, which are indeed prime.\n\n2. $y - x = 3$ and $y + x = q$. From the first equation we get $y = x + 3$. Plugging this into the second equation we see that $2x + 3 = q$. The initial equality then implies $p = x^2 - p = x^2 - q = x^2 - 2x - 3 = (x + 1)(x - 3)$. Since $p$ is a prime we have $x = 4$. Thus, $p = 5$ and $q = 11$, which are indeed prime.\n\n3. $y - x = q$ and $y + x = 3$. Since $x$ and $y$ are positive integers and $y > x$, we get $y = 2$ and $x = 1$, which is a contradiction since it immediately follows that $q = 1$.\n\nThe two solutions are $(5, 11)$ and $(13, 3)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41666,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa vrhu $75~\\mathrm{m}$ dolgega klanca je oddajnik, ki ga od vznožja vidimo pod kotom $10^\\circ$ (glej sliko). Kako visok je oddajnik (njegova višina je na sliki označena s $h$), če je naklon klanca $20^\\circ$? Rezultat zaokroži na dve mesti natančno.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "15.04",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41667,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a real number $x$ let $[x]$ denote the greatest integer not greater than $x$.\n\na. Prove that for all positive integers $a$, $b$ and $c$ we have\n$$\n\\left\\lfloor \\frac{\\left\\lfloor \\frac{c}{a} \\right\\rfloor}{b} \\right\\rfloor \\le \\left\\lfloor \\frac{c}{ab} \\right\\rfloor.\n$$\n\nb. Find an example showing that the above equality does not hold for all positive real numbers $a$, $b$ and $c$.",
    "options": [],
    "answer": "Example: a = 2, b = 1/2, c = 1",
    "solution": "a. The number $c$ can be written in the form $c = kab + r$, where $k$ is a non-negative integer and $r < ab$ is the remainder of $c$ when divided by $ab$. The number $r$ can be further written as $r = ma + n$, where $m$ is a non-negative integer and $n < a$ is the remainder of $r$ when divided by $a$. From this it follows that $m < b$, or we would have $r \\ge ab$. We see that\n$$\n\\left[ \\frac{c}{ab} \\right] = \\left[ k + \\frac{r}{ab} \\right] = k,\n$$\nbecause $0 \\le \\frac{r}{ab} < 1$, and\n$$\n\\left[ \\frac{\\left[ \\frac{c}{a} \\right]}{b} \\right] = \\left[ \\frac{\\left[ kb + m + \\frac{n}{a} \\right]}{b} \\right] = \\left[ \\frac{kb + m}{b} \\right] = \\left[ k + \\frac{m}{b} \\right] = k,\n$$\nbecause $0 \\le \\frac{n}{a} < 1$ and $0 \\le \\frac{m}{b} < 1$. From here we obtain the desired result.\n\nb. Let $a = 2$, $b = \\frac{1}{2}$ and $c = 1$. Then $\\left[ \\frac{c}{ab} \\right] = 1$ and $\\left[ \\frac{[\\frac{c}{a}]}{b} \\right] = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41668,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathcal{K}$ be a circle with centre $O$ and radius $r$. What is the upper bound for the length of $|OT|$ given that $T$ is the point in the plane such that there exists an equilateral triangle $ABT$, where $A$ and $B$ lie on the circle $\\mathcal{K}$?",
    "options": [],
    "answer": "2r",
    "solution": "Let $M$ be the intersection of the segments $OT$ and $AB$. Then $|OT| = |OM| + |MT|$. Write $|AB| = |AT| = |BT| = a$. We know that $AB \\perp OT$. By Pythagoras' theorem we have $|OM| = \\sqrt{r^2 - |BM|^2} = \\sqrt{r^2 - \\frac{a^2}{4}}$ and $|MT| = \\sqrt{a^2 - \\frac{a^2}{4}} = \\frac{\\sqrt{3}a}{2}$. We can now find an upper bound for $|OT|$ by using the inequality between the arithmetic and the quadratic mean (the A-G inequality):\n$$\n\\begin{aligned}\n|OT| &= |OM| + |MT| \\\\\n&= \\sqrt{r^2 - \\frac{a^2}{4}} + \\frac{\\sqrt{3}a}{2} \\\\\n&= \\sqrt{r^2 - \\frac{a^2}{4}} + \\frac{\\sqrt{3}a}{6} + \\frac{\\sqrt{3}a}{6} + \\frac{\\sqrt{3}a}{6} \\\\\n&\\le 4\\sqrt{\\frac{(r^2 - \\frac{a^2}{4}) + \\frac{a^2}{12} + \\frac{a^2}{12} + \\frac{a^2}{12}}{4}} \\\\\n&= \\sqrt{4r^2} = 2r.\n\\end{aligned}\n$$\nWe have shown that $|OT| \\le 2r$, so now we must only check if the equality can hold. The equality in the A-G inequality holds when all four terms are the same, i.e. when $\\sqrt{r^2 - \\frac{a^2}{4}} = \\frac{\\sqrt{3}a}{6} \\iff a = r\\sqrt{3}$. Since $a = |AB|$ can reach any value from the interval $(0, 2r]$, the equality holds exactly when $TA$ and $TB$ are tangents to the circle $\\mathcal{K}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41669,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{N} \\to \\mathbb{N}$ such that:\n* for any positive integers $a$ and $b$ which are not relatively prime we have $f(a)f(b) = f(ab)$,\n* for any positive integers $a$ and $b$ there exists a non-degenerate triangle with the sides of lengths $f(a)$, $f(b)$ and $f(a + b - 1)$.\n(Note: A triangle is non-degenerate if its three vertices do not lie on a common line.)",
    "options": [],
    "answer": "Either f(n) = 1 for all positive integers n; or f(n) = n for all n ≥ 2 with f(1) ∈ {1, 2, 3}.",
    "solution": "The second condition implies that for all $a, b \\in \\mathbb{N}$ the triangle inequalities hold:\n$$\n\\begin{aligned}\nf(a) + f(b) &> f(a+b-1), \\\\\nf(a) + f(a+b-1) &> f(b), \\\\\nf(b) + f(a+b-1) &> f(a).\n\\end{aligned}\n$$\nFirst, consider $a = b = 2$. It then follows from the first condition that $f(4) = f(2)^2$, and the second condition implies that $2f(2) > f(3)$.\n\nNow, let $a = 3$, $b = 2$. The second condition then implies that $f(2) + f(3) > f(4)$. So,\n$$\n\\begin{aligned}\nf(2)^2 &= f(4) < f(2) + f(3) < f(2) + 2f(2) = 3f(2) \\\\\n&\\implies f(2)(3 - f(2)) > 0.\n\\end{aligned}\n$$\nSince $f(2)$ is a positive integer there are only two possibilities: $f(2) = 1$ and $f(2) = 2$. We consider the two cases separately:\n\n* 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$$\nf(n+1) < f(n) + f(2) = 2 \\implies f(n+1) = 1.\n$$\nThis gives us the solution $f(n) = 1$ for all $n \\in \\mathbb{N}$.\n\n* If $f(2) = 2$, then $f(4) = f(2)^2 = 4$. What is more, for all $k \\in \\mathbb{N}$ the induction on the first condition implies that\n$$\nf(2^k) = f(2)f(2^{k-1}) = \\dots = f(2)^k = 2^k.\n$$\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 $2, 3, \\dots, n-1$. Using $a = n-1$, $b = 2$ we get\n$$\nf(n) < f(n-1) + f(2) = n+1 \\implies f(n) \\le n.\n$$\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$$\n\\begin{align*}\nf(n) + f(2^r - s + 1) &> f(2^r + s + 2^r - s + 1 - 1) = f(2^{r+1}) \\\\\n\\implies f(n) &> f(2^{r+1}) - f(2^r - s + 1) = 2^r + s - 1 = n - 1 \\\\\n\\implies f(n) &\\ge n.\n\\end{align*}\n$$\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": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41670,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circles $K_1$ and $K_2$ intersect in two distinct points $A$ and $B$. The tangents to $K_1$ through $A$ and $B$ intersect at $T$. Let $M$ be an arbitrary point on the circle $K_1$ distinct from $A$ and $B$. The line $MT$ meets the circle $K_1$ again at $C$, the line $MA$ meets the circle $K_2$ again at $K$ and the line $AC$ meets the circle $K_2$ again at $L$. Prove that the line $MC$ contains the midpoint of the segment $KL$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the intersection of the lines $MT$ and $KL$.\nBy Menelaus' theorem for the triangle $ALK$ and the colinear points $C, P$ and $M$ we have\n$$\n\\frac{AC}{CL} \\cdot \\frac{LP}{PK} \\cdot \\frac{KM}{MA} = -1.\n$$\nIn order to show that $|PL| = |PK|$ it suffices to see that\n$$\n\\frac{|AC|}{|CL|} = \\frac{|MA|}{|MK|}. \\qquad (12)\n$$\n*(Note: We will be using directed angles from this point on.)*\nThe points $B, C, A$ and $M$ are concyclic, so $\\nparallel BCL = -\\nparallel ACB = \\nparallel BMK$. The points $B, L, K$ and $A$ are also concyclic, so $\\nparallel CLB = \\nparallel ALB = \\nparallel MKB$. The triangles $BCL$ and $BMK$ have two angles in common, so they are similar. It follows that\n$$\n\\frac{|CL|}{|MK|} = \\frac{|BC|}{|BM|}. \\qquad (13)\n$$\nBy the tangent-chord angle theorem we have $\\nparallel TBC = \\nparallel BMC$. The triangles $TBC$ and $TMB$ have two angles in common and are similar. Using the tangent-chord angle theorem again we see that $\\nparallel CAT = \\nparallel CMA$. So, the triangles $TAC$ and $TMA$ have two angles in common and are similar. From here we find the ratios $\\frac{|BC|}{|BM|} = \\frac{|TB|}{|TM|}$ and $\\frac{|AC|}{|AM|} = \\frac{|TA|}{|TM|}$. Combining the two with the equality between the tangent segments $|TA| = |TB|$ we get\n$$\n\\frac{|BC|}{|BM|} = \\frac{|AC|}{|AM|}. \\qquad (14)\n$$\n\n(Note: The property (14) or $|AM| \\cdot |BC| = |AC| \\cdot |BM|$ is called the harmonic property of the cyclic quadrilateral.)\nIf we now combine the equalities (13) and (14), we get (12), which was our goal. From here we conclude that $P$ is the midpoint of $KL$ and so the midpoint of $KL$ does indeed lie on $MC$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41671,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVelikosti notranjih kotov trikotnika sestavljajo aritmetično zaporedje, vsota njihovih sinusov pa je enaka $\\frac{3+\\sqrt{3}}{2}$. Izračunaj velikosti notranjih kotov trikotnika.",
    "options": [],
    "answer": "30°, 60°, 90°",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41672,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa realno število $x$ označimo $[x]$ največje celo število, ki ni večje od $x$.\n\na. Dokaži, da za vsa naravna števila $a$, $b$ in $c$ velja\n$$\n\\left[\\frac{\\left[\\frac{c}{a}\\right]}{b}\\right]=\\left[\\frac{c}{ab}\\right]\n$$\n\nb. S primerom pokaži, da gornja enakost ne velja za vsa pozitivna realna števila $a$, $b$ in $c$.",
    "options": [],
    "answer": "The identity holds for all natural numbers. A counterexample for positive reals is a equals 2, b equals one half, c equals 1.",
    "solution": "Solution:\n\na.\nŠtevilo $c$ lahko zapišemo v obliki $c = k a b + r$, kjer je $k$ neko nenegativno celo število, $r < a b$ pa ostanek števila $c$ pri deljenju z $a b$. Število $r$ lahko nadalje zapišemo v obliki $r = m a + n$, kjer je $m$ neko nenegativno celo število, $n < a$ pa ostanek števila $r$ pri deljenju z $a$. Med drugim od tod sledi $m < b$, saj bi sicer bilo $r \\geq a b$. Od tod izračunamo\n$$\n\\left[\\frac{c}{a b}\\right] = \\left[k + \\frac{r}{a b}\\right] = k\n$$\nsaj je $0 \\leq \\frac{r}{a b} < 1$, in\n$$\n\\left[\\frac{\\left[\\frac{c}{a}\\right]}{b}\\right] = \\left[\\frac{k b + m + \\frac{n}{a}}{b}\\right] = \\left[\\frac{k b + m}{b}\\right] = \\left[k + \\frac{m}{b}\\right] = k\n$$\nsaj je $0 \\leq \\frac{n}{a} < 1$ in $0 \\leq \\frac{m}{b} < 1$. Od tod sledi enakost iz naloge.\n\nb.\nTaka števila so na primer $a = 2$, $b = \\frac{1}{2}$ in $c = 1$, saj je v tem primeru $\\left[\\frac{c}{a b}\\right] = 1$ in $\\left[\\frac{\\left[\\frac{c}{a}\\right]}{b}\\right] = 0$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41673,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči najmanjše naravno število, ki ga lahko zapišemo v obliki $3 a^{2} - a b^{2} - 2 b - 4$, kjer sta $a$ in $b$ naravni števili.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nOdgovor je $2$. Če vzamemo $a = 4$ in $b = 3$, dobimo $3 a^{2} - a b^{2} - 2 b - 4 = 2$. Dovolj je torej pokazati, da enačba $3 a^{2} - a b^{2} - 2 b - 4 = 1$ nima rešitev v naravnih številih.\n\nEnačbo preuredimo v $3 a^{2} - a b^{2} = 2 b + 5$. Ker je desna stran liha, mora biti tudi leva stran liha, torej mora biti $a$ liho število in $b$ sodo število. Število $b^{2}$ je zato deljivo s $4$, ostanek števila $a^{2}$ pri deljenju s $4$ pa je enak $1$. Leva stran ima torej pri deljenju s $4$ ostanek $3$, desna stran pa ostanek $1$, saj je $2 b$ deljivo s $4$.\n\nPrišli smo do protislovja, torej enačba nima rešitev v naravnih številih.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41674,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $n$ naravno število. Poenostavi izraz\n$$\n\\left(\\frac{a+a^{-1} b^{2}}{a-a^{-1} b^{2}}-1\\right)^{-1}\\left(b^{n}(a-b)^{-1}-b^{n}(a+b)^{-1}\\right)\n$$\nZa katere vrednosti $a$ in $b$ izraz nima pomena?",
    "options": [],
    "answer": "Simplified expression: b^{n-1}. Undefined when a = 0, or b = 0, or a^2 = b^2 (i.e., a = b or a = −b).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41675,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be positive integers such that $5m + n$ divides $5n + m$. Prove that $m$ divides $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "There exists a positive integer $k$ such that $5n + m = k(5m + n)$, or $(5 - k)n = (5k - 1)m$. The right-hand side is strictly positive, so the left-hand side must be strictly positive as well, which implies $5 - k > 0$ and so $k \\in \\{1, 2, 3, 4\\}$. If $k = 1$, then $4n = 4m$, so $n = m$. If $k = 2$, then $3n = 9m$, so $n = 3m$. If $k = 3$, then $2n = 14m$, so $n = 7m$. If $k = 4$, then $n = 19m$. In each case $m$ divides $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41676,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer of the form $3a^2 - ab^2 - 2b - 4$, where $a$ and $b$ are some positive integers.",
    "options": [],
    "answer": "2",
    "solution": "The answer is $2$. If $a = 4$ and $b = 3$, then $3a^2 - ab^2 - 2b - 4 = 2$.\n\nIt therefore suffices to show that the equation $3a^2 - ab^2 - 2b - 4 = 1$ has no solutions in positive integers. We can rewrite the equation as $3a^2 - ab^2 = 2b + 5$. The right-hand side is odd, so the left-hand side must be odd as well. So, $a$ must be odd and $b$ must be even, which implies that $b^2$ is divisible by $4$, and $a^2$ gives a remainder of $1$ when divided by $4$. It follows that the left-hand side gives the remainder of $3$, and the right-hand side gives the remainder of $1$ because $2b$ is divisible by $4$. This contradiction shows that the equation has no solution in positive integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41677,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be a point on the side $AB$ of the triangle $ABC$. The circumcircles of the triangles $BCD$ and $ACD$ meet the lines $AC$ and $BC$ again at points $E$ and $F$, respectively. The bisector of the segment $EF$ intersects the line $AB$ at $M$ and meets the altitude to $AB$ from $D$ at $N$. Let $T$ be the intersection of the lines $AB$ and $EF$, and let $U$ be the intersection of the line $CT$ and the circumcircle of the triangle $CDM$. Prove that $|NC| = |NU|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, we will show that the points $E$, $D$, $M$, $F$ and $N$ are concyclic.\n\nLet the circumcircle of the triangle $EDF$ intersect the line $AB$ at $M_1$. The quadrilaterals $BCED$ and $AFDC$ are cyclic, so $\\angle M_1EF = \\angle M_1DF = \\angle ACB = \\angle EDA = \\angle EFM_1$. The triangle $EM_1F$ is isosceles with the apex at $M_1$, so $M_1$ lies on the bisector of the segment $EF$. This implies that $M_1 = M$ and the points $E$, $D$, $M$ and $F$ are concyclic.\n\nThe bisector of the chord $EF$ is the diameter of the circle passing through the points $E$, $D$, $M$ and $F$. We have $\\angle MDN = \\pi/2$ and $N$ lies on the bisector, so by Thales' theorem $N$ must also lie on the circle. This means that $E$, $D$, $M$, $F$ and $N$ are concyclic.\n\nNow, $\\angle ENF = \\pi - \\angle FME = \\pi - \\angle FDE = \\pi - \\angle FDA - \\angle BDE + \\pi = 2 \\angle ECF$. Since $N$ lies on the bisector of the segment $EF$ and $\\angle ENF = 2 \\angle ECF$, we see that $N$ is the centre of the circumcircle of the triangle $EFC$.\n\nThe quadrilaterals $EDMF$ and $UDMC$ are cyclic and by the power of a point theorem we have $\\overrightarrow{TE} \\cdot \\overrightarrow{TF} = \\overrightarrow{TD} \\cdot \\overrightarrow{TM} = \\overrightarrow{TU} \\cdot \\overrightarrow{TC}$. We conclude that $UEFC$ is also a cyclic quadrilateral. Thus, the points $C$ and $U$ both lie on the circle centred at $N$ and $|NC| = |NU|$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41678,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDokaži, da je za poljubno liho naravno število $n$ izraz $(n+2)^3 + 3(n+2)^2 - n - 5$ deljiv z $48$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNaj bo $n$ liho naravno število.\n\nRazpišimo izraz:\n\n$$(n+2)^3 + 3(n+2)^2 - n - 5$$\n\nNajprej razvijemo $(n+2)^3$ in $(n+2)^2$:\n\n$(n+2)^3 = n^3 + 3n^2 \\cdot 2 + 3n \\cdot 2^2 + 2^3 = n^3 + 6n^2 + 12n + 8$\n\n$3(n+2)^2 = 3(n^2 + 4n + 4) = 3n^2 + 12n + 12$\n\nSkupaj:\n\n$n^3 + 6n^2 + 12n + 8 + 3n^2 + 12n + 12 - n - 5$\n\nZdružimo podobne člen:\n\n$n^3 + (6n^2 + 3n^2) + (12n + 12n - n) + (8 + 12 - 5)$\n\n$= n^3 + 9n^2 + 23n + 15$\n\nTorej izraz je:\n\n$$n^3 + 9n^2 + 23n + 15$$\n\nPokažimo, da je ta izraz deljiv z $48$ za liho $n$.\n\nKer je $48 = 16 \\cdot 3$, preverimo deljivost z $16$ in $3$ posebej.\n\n**Deljivost z $3$:**\n\nOpazimo, da je $n^3 + 9n^2 + 23n + 15 \\equiv n^3 + 2n + 0 \\pmod{3}$ (ker $9n^2 \\equiv 0$, $23n \\equiv 2n$, $15 \\equiv 0$).\n\nZa liho $n$:\n\n- Če $n \\equiv 1 \\pmod{3}$: $n^3 \\equiv 1$, $2n \\equiv 2$, vsota $1+2=3 \\equiv 0$\n- Če $n \\equiv 2 \\pmod{3}$: $n^3 \\equiv 8 \\equiv 2$, $2n \\equiv 4 \\equiv 1$, vsota $2+1=3 \\equiv 0$\n- Če $n \\equiv 0 \\pmod{3}$: $n^3 \\equiv 0$, $2n \\equiv 0$, vsota $0+0=0$\n\nTorej je izraz vedno deljiv s $3$.\n\n**Deljivost z $16$:**\n\nKer je $n$ liho, zapišimo $n = 2k+1$.\n\nIzraz postane:\n\n$$(2k+1)^3 + 9(2k+1)^2 + 23(2k+1) + 15$$\n\nIzračunajmo:\n\n$(2k+1)^3 = 8k^3 + 12k^2 + 6k + 1$\n\n$9(2k+1)^2 = 9(4k^2 + 4k + 1) = 36k^2 + 36k + 9$\n\n$23(2k+1) = 46k + 23$\n\nSkupaj:\n\n$8k^3 + 12k^2 + 6k + 1 + 36k^2 + 36k + 9 + 46k + 23 + 15$\n\nZdružimo podobne člen:\n\n$8k^3 + (12k^2 + 36k^2) + (6k + 36k + 46k) + (1 + 9 + 23 + 15)$\n\n$= 8k^3 + 48k^2 + 88k + 48$\n\n$= 8k^3 + 48k^2 + 88k + 48$\n\nOpazimo, da je $8k^3$ deljivo z $8$, $48k^2$ in $48$ sta deljiva z $16$, $88k = 16 \\cdot 5.5k$.\n\n$88k = 16 \\cdot 5k + 8k$\n\nTorej izraz je:\n\n$8k^3 + 48k^2 + 88k + 48 = 16(3k^2 + 5k + 3) + 8k^3 + 8k$\n\n$= 16(3k^2 + 5k + 3) + 8k(k^2 + 1)$\n\nKer je $k$ celo število, $8k(k^2 + 1)$ je vedno deljivo z $16$ za liho $k$ ali $k$ sodo, ker $k^2 + 1$ je sodo za liho $k$ in obratno.\n\nLahko pa preverimo nekaj konkretnih primerov:\n\nZa $k=0$ ($n=1$):\n\n$n^3 + 9n^2 + 23n + 15 = 1 + 9 + 23 + 15 = 48$\n\nZa $k=1$ ($n=3$):\n\n$3^3 + 9 \\cdot 9 + 23 \\cdot 3 + 15 = 27 + 81 + 69 + 15 = 192$\n\nOba izraza sta deljiva z $48$.\n\nTorej je izraz za liho $n$ vedno deljiv z $48$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41679,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana je funkcija $f$ s predpisom $f(x) = -2x^{2} + 8x - d$. Izračunaj vrednosti parametra $d$ tako,\n\na) da se bo graf funkcije $f$ dotikal osi $x$,\n\nb) da bo maksimalna vrednost funkcije $f$ enaka $6$,\n\nc) da bo funkcija $f$ povsod pozitivna,\n\nali pa utemelji, da taka vrednost parametra $d$ ne obstaja.",
    "options": [],
    "answer": "a) d = 8; b) d = 2; c) no such value exists",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41680,
    "subject": "Mathematics (Multi-modal)",
    "question": "Exactly $p-1$ distinct positive integers are written on the blackboard for some prime number $p$. The number $p$ is among these $p-1$ numbers. For any pair of the numbers the absolute value of their difference is also on the board. Prove that all the numbers on the blackboard are divisible by $p$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the numbers on the board by $a_1, a_2, \\dots, a_{p-1}$.\nWithout loss of generality we may assume that $a_1 < a_2 < \\dots < a_{p-1}$. Then the numbers $a_2 - a_1 < a_3 - a_1 < \\dots < a_{p-1} - a_1$ are also written on the blackboard. There are $p-2$ of them and they are all distinct. This is only possible when\n$$\n\\begin{aligned}\na_{p-2} &= a_{p-1} - a_1 \\\\\n&\\vdots \\\\\na_2 &= a_3 - a_1 \\\\\na_1 &= a_2 - a_1\n\\end{aligned}\n$$\nThus, $a_2 = 2a_1, a_3 = 3a_1, \\dots, a_{p-2} = (p-2)a_1$ and $a_{p-1} = (p-1)a_1$. Since $p$ is a prime and it appears on the blackboard we must have $a_1 = p$. We conclude that all the numbers on the blackboard are divisible by $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41681,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSimetrala notranjega kota $\\angle A C B$ ostrokotnega trikotnika $A B C$ seka stranico $A B$ v točki $D$. Očrtana krožnica trikotnika $A D C$ seka stranico $B C$ v različnih točkah $C$ in $E$. Premica skozi točko $B$, vzporedna premici $A E$, seka premico $C D$ v točki $F$. Dokaži, da je trikotnik $A F B$ enakokrak.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOznačimo $\\angle A C B=\\gamma$. Potem je $\\angle A C F=\\angle F C B=\\frac{\\gamma}{2}$, saj je $C F$ simetrala kota $\\angle A C B$. Zaradi koncikličnosti točk $A, D, E$ in $C$ sledi $\\angle D A E=\\angle D C E=\\frac{\\gamma}{2}$. Ker je premica $B F$ vzporedna premici $A E$, velja $\\angle A B F=\\angle D A E=\\frac{\\gamma}{2}$. Zaradi $\\angle A C F=\\frac{\\gamma}{2}=\\angle A B F$ so torej točke $A, F, B$ in $C$ konciklične. Od tod sledi $\\angle F A B=\\angle F C B=\\frac{\\gamma}{2}=\\angle A B F$, torej je trikotnik $A F B$ enakokrak z vrhom pri $F$.\n\nSklep $\\angle D C E=\\angle D A E$ ali smiselna uporaba koncikličnosti točk $A, D, E, C$ pri premetavanju kotov 1 točka\nSklep $\\angle D A E=\\angle A B F$ ali smiselna uporaba vzporednosti $A E$ in $B F$ pri premetavanju kotov 1 točka\nTetivnost $A F B C$ ali podobnost trikotnikov $F D A$ in $B D C$ 3 točke\nSklep $\\angle F A B=\\angle F C B$ in zato trikotnik $A F B$ enakokrak 2 točki",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41682,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $AB$ najdaljša stranica trikotnika $ABC$. Z $M$ in $N$ označimo taki točki na stranici $AB$, da velja $|AM| = |AC|$ ter $|BN| = |BC|$. Razpolovišči daljic $MC$ in $NC$ naj bosta $P$ in $R$, trikotniku $ABC$ včrtana krožnica pa naj se stranic $BC$ in $AC$ dotika v točkah $D$ in $E$. Dokaži, da so točke $P$, $R$, $D$ in $E$ konciklične.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nNaj bo $I$ središče trikotniku $ABC$ včrtane krožnice. Ker je $AMC$ enakokrak trikotnik z vrhom pri $A$, je $AP$ višina na osnovnico in hkrati simetrala kota $\\angle MAC$. Zato $I$ leži na premici $AP$. Podobno $I$ leži tudi na premici $BR$. Sledi $\\angle CPI = \\angle CPA = \\frac{\\pi}{2}$ in $\\angle IRC = \\angle BRC = \\frac{\\pi}{2}$, torej točki $P$ in $R$ ležita na krožnici s premerom $CI$. Ker sta $D$ in $E$ dotikališči včrtane krožnice, velja $\\angle CDI = \\angle IEC = \\frac{\\pi}{2}$, zato tudi točki $D$ in $E$ ležita na krožnici s premerom $CI$. Točke $P$, $R$, $D$ in $E$ so torej konciklične.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41683,
    "subject": "Mathematics (Multi-modal)",
    "question": "The bisector of the inner angle $\\angle ACB$ in an acute triangle $ABC$ intersects the side $AB$ at $D$. The circumcircle of the triangle $ADC$ intersects the side $BC$ in two distinct points, $C$ and $E$. The line through $B$ parallel to the line $AE$ intersects the line $CD$ at $F$. Prove that $AFB$ is an isosceles triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Write $\\angle ACB = \\gamma$. Then $\\angle ACF = \\angle FCB = \\frac{\\gamma}{2}$, since $CF$ is the bisector of the angle $\\angle ACB$. The points $A$, $D$, $E$ and $C$ are concyclic, so $\\angle DAE = \\angle DCE = \\frac{\\gamma}{2}$. The line $BF$ is parallel to the line $AE$, so $\\angle ABF = \\angle DAE = \\frac{\\gamma}{2}$. We have $\\angle ACF = \\frac{\\gamma}{2} = \\angle ABF$, so the points $A$, $F$, $B$ and $C$ are concyclic. From here we get $\\angle FAB = \\angle FCB = \\frac{\\gamma}{2} = \\angle ABF$ and so $AFB$ is an isosceles triangle with the apex at $F$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41684,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the parallelogram $ABCD$ we have $|AB| = |BD|$. Let $K$ be a point on the line $AB$ different from $A$, such that $|KD| = |AD|$. Denote the reflection of the point $C$ over $K$ by $M$, and the reflection of $B$ over $A$ by $N$. Prove that $MDN$ is an isosceles triangle with the apex at $D$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Triangle $ADK$ is isosceles with the apex at $D$. So, $\\angle BKD = 180^\\circ - \\angle DKA = 180^\\circ - \\angle KAD = \\angle CBK$ and $|DK| = |DA| = |BC|$. The triangles $DKB$ and $CBK$ are congruent because they have a common side $KB$, congruent angles $\\angle BKD = \\angle CBK$ and $|DK| = |BC|$. Thus, $\\angle DCK = \\angle BKC = \\angle DBK$. Since $M$ and $N$ are reflections of $C$ and $B$, we have $|CK| = |KM|$ and $|NA| = |AB|$. So,\n$$|CM| = 2|CK| = 2|DB| = 2|AB| = |NB|.$$\nThis implies that the triangles $CDM$ and $BDN$ are also congruent since $|DC| = |DB|$, $|CM| = |NB|$ and $\\angle DCM = \\angle DBN$. Thus, $|DN| = |DM|$ and the triangle $DMN$ is isosceles with the apex at $D$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41685,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana je funkcija $f$ s predpisom $f(x)=\\frac{12}{x^{2}+x-6}-2$. Zapiši ničle, pole, enačbo asimptote in začetno vrednost funkcije $f$ ter nariši graf funkcije $f$. Nariši še graf funkcije $g$ s predpisom $g(x)=|f(x)|$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Zeros: x = 3 and x = −4; Poles: x = 2 and x = −3; Asymptote: y = −2; Initial value: f(0) = −4.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41686,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathcal{K}$ be the circumcircle of the acute triangle $ABC$ with $|AB| < |AC|$. Let $p$ be the reflection of the line $BC$ over the line $AB$. The line $p$ intersects the circle $\\mathcal{K}$ at $B$ and $E$. The tangent to $\\mathcal{K}$ at $A$ intersects the line $p$ at $D$. Let $F$ be the reflection of the point $D$ over the point $A$. The line $CF$ intersects the circle $\\mathcal{K}$ at $C$ and $G$. Prove that the lines $CE$ and $GB$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the tangent-chord angle theorem we have $\\angle CAF = \\angle CBA$. Since $p$ is the reflection of the line $BC$ over the line $AB$, we have $\\angle CBA = \\angle ABD$. The points $A, B, E, C$ are concyclic, so $\\angle ABD = \\angle ACE$. Hence, $\\angle CAF = \\angle ACE$, and the line $CE$ is parallel to the line $FD$.\n\nWe would like to show that the line $GB$ is also parallel to $FD$. Using the tangent-chord angle theorem one more time we see that $\\angle DAB = \\angle ACB$. Also, $\\angle CBA = \\angle ABD$, so the triangles $BAD$ and $BCA$ are similar, having two congruent angles. This implies that $\\frac{|AD|}{|AB|} = \\frac{|CA|}{|CB|}$. Since $F$ is the reflection of the point $D$ over the point $A$, we have $|AD| = |FA|$. So, $\\frac{|FA|}{|AB|} = \\frac{|CA|}{|CB|}$, or $\\frac{|FA|}{|CA|} = \\frac{|AB|}{|CB|}$. We also have $\\angle CAF = \\angle CBA$, so the triangles $FAC$ and $ABC$ are similar as well. This implies that $\\angle AFC = \\angle BAC = \\angle BGC$ and the line $GB$ is parallel to $FD$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41687,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB$ be the longest side of the triangle $ABC$. Let $M$ and $N$ denote the points on the side $AB$, such that $|AM| = |AC|$ and $|BN| = |BC|$. Denote the midpoints of the segments $MC$ and $NC$ by $P$ and $R$. The incircle of the triangle $ABC$ touches the sides $BC$ and $AC$ at $D$ and $E$. Prove that the points $P, R, D$ and $E$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $I$ be the incentre of the triangle $ABC$.\n\n![](attached_image_1.png)\n\nTriangle *AMC* is isosceles with the apex at *A*, so the line *AP* is the altitude to the base and at the same time the bisector of the angle $\\angle MAC$. It follows that $I$ lies on $AP$. Similarly, $I$ lies on the line $BR$. This implies that $\\angle CPI = \\angle CPA = \\pi/2$ and $\\angle IRC = \\angle BRC = \\pi/2$, so the points $P$ and $R$ lie on the circle with diameter $CI$. Since $D$ and $E$ are the points where the incircle touches the sides of the triangle, we have $\\angle CDI = \\angle IEC = \\pi/2$, and $D$ and $E$ therefore lie on the circle with diameter $CI$. We have shown that the points $P$, $R$, $D$ and $E$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41688,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBrez uporabe računala izračunaj vrednost izraza $\\sqrt{|2 \\sqrt{5}-6|}-\\sqrt{2 \\sqrt{5}+6}$.",
    "options": [],
    "answer": "-2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41689,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDana je pokončna prizma, katere osnovna ploskev je enakostranični trikotnik s stranico, dolgo $12~\\mathrm{cm}$. Izračunaj višino pokončne piramide, ki ima enako višino, enako osnovno ploskev in enako površino plašča kot dana pokončna prizma.",
    "options": [],
    "answer": "2 cm",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41690,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are three piles of tokens on the table: the first contains $a$, the second $b$ and the third one $c$ tokens, and $a \\ge b \\ge c > 0$. Two players $A$ and $B$ take turns moving the tokens around. Player $A$ goes first. For each move, the active player first chooses two piles and then moves at least one token from the pile with fewer tokens into the pile containing more tokens. If the numbers of tokens in the two chosen piles are equal, then the player moves at least one token from any of them onto the other. A player wins if there is only one pile remaining after their move. Who has the winning strategy for different values of $a, b$ and $c$?",
    "options": [],
    "answer": "If b = c, then player B has a winning strategy; otherwise player A has a winning strategy.",
    "solution": "If $b = c$, then player $B$ has the winning strategy, otherwise player $A$ has the winning strategy.\n\nFirst, assume that $b = c$. Then the two piles with the smallest number of tokens contain the same amount. In this case, player $B$ can ensure that the situation remains like that every time he makes his move, whereas every time $A$ makes his move the pile with the least number of tokens contains less tokens than the remaining piles. Indeed, player $A$ always has to move at least one token from one of the smallest two piles onto one of the other piles. Player $B$ can then choose the biggest two piles which necessarily contain more tokens than the smallest pile, and distribute the tokens between the two so that the smallest two piles after his turn contain the same number of tokens. The game will end when the smallest two piles are emptied, which can only happen after $B$ completes his turn. Hence, $B$ can always win.\n\nNow, assume that $b > c$. In this case $A$ should move $b-c > 0$ tokens from the pile with $b$ tokens onto the pile with $a$ tokens. After his move the smallest two piles will contain $c$ tokens each. The situation is now the same as above, except this time $B$ is the first player to continue. Hence, $A$ has the winning strategy in this case.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41691,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDano je neskončno zaporedje, katerega prvi štirje členi so $2, 6, 18$ in $54$. Zapiši splošni člen $a_{n}$ danega zaporedja, če je količnik med poljubnima zaporednima členoma konstanten. Izračunaj še vsoto prvih $1000$ členov zaporedja s splošnim členom $b_{n} = \\log_{3}\\left(a_{n}\\right)$.",
    "options": [],
    "answer": "a_n = 2·3^{n−1};  S_1000 = 1000·log_3 2 + 499500",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41692,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $n$ is written on the blackboard. In each step we replace the number by the sum of any two positive integers whose product is equal to the number on the board. Determine the smallest number that can be obtained after a finite number of steps in terms of the initial number $n$.",
    "options": [],
    "answer": "min(n, 5)",
    "solution": "First, let us show the following: if the number on the blackboard has decreased after the change, then the new number is greater than or equal to $5$. We cannot obtain the number $1$ because the sum of two positive integers can never be equal to $1$. The number $2 = 1 + 1$ can only be obtained from $1$, the number $3 = 1 + 2$ can only be obtained from $2$, and the number $4 = 1 + 3 = 2 + 2$ can only be obtained from $3$ or $4$. So, if the number obtained is less than $5$, then it has not decreased during the change. If the starting number was $n < 5$, then we cannot get a smaller number.\n\nNow, let us show that if a number is greater than $5$, then it can be reduced. Assume that the number $m > 5$ is on the blackboard at a given time. If $m$ is even, i.e. of the form $m = 2k$, where $k > 2$ is a positive integer, then it can be replaced by $k + 2$. In this way we have obtained a number smaller than $m$, since the condition $k + 2 < 2k$ is equivalent to the assumption $k > 2$. If $m$ is odd, then we can express it as $m = 2k - 1$, where $k > 3$ is a positive integer. In this case, we can first replace it by $(2k - 1) + 1 = 2k$ and then with $k + 2$. Once again we have obtained a number smaller than $m$, because the condition $k + 2 < 2k - 1$ is equivalent to our assumption $k > 3$. Therefore, if the starting number is $n \\ge 5$, we can reduce it to the number $5$, using a finite number of steps, and according to the reasoning above this is the smallest number that can be obtained.\n\nThe smallest number we can obtain after a finite number of steps is equal to $5$ if $n \\ge 5$ and is equal to $n < 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41693,
    "subject": "Mathematics (Multi-modal)",
    "question": "Eight boxes numbered $1$ to $8$ are lined up next to an empty bag. Each box contains one token. Miha has a lot of extra tokens and he plays a game where the following two actions are allowed:\n* removing $1$ token from box $i$ ($i < 8$) and adding $2$ tokens into box $(i + 1)$,\n* removing $1$ token from box $i$ ($i < 8$) and moving $1$ token from box $(i + 1)$ into the bag.\n\nThe game ends when there are no possible moves left. At least how many tokens are in the bag at the end?",
    "options": [],
    "answer": "85",
    "solution": "Let $A_i$ and $B_i$ denote the first and second move taken on the boxes marked $i$ and $i+1$. Let $i \\le 6$. If Miha chooses to play $B_i$, then the number of tokens in the boxes marked $i$ and $i+1$ is reduced by $1$, the number of tokens in the box $i+2$ remains the same, and the number of tokens in the bag increases by $1$. If Miha chooses to play the sequence of moves $A_i$, $A_{i+1}$, $B_{i+1}$, then the number of tokens in the box $i$ decreases by $1$, the number of tokens in the box $i+1$ remains the same, and the number of tokens in the box $i+2$ and in the bag is increased by $1$. The end result is better in the second case since the only difference is that two boxes contain an extra token each. We conclude that the moves $B_i$ for $i \\le 6$ are not profitable.\n\nAs for other moves, only $B_7$ increases the number of tokens in the bag, so Miha should make that move as many times as possible. Without loss of generality we may assume that this move is only taken at the very end. Before that, Miha uses $A_i$, $i \\le 6$, to move the tokens into the seventh box which then contains $1 + 2 + 2^2 + \\dots + 2^6 = 2^7 - 1 = 127$ tokens, while the final box contains $1$ token.\n\nLet $x$ denote the number of tokens in the seventh box at a certain point in time, and let $y$ be the number of tokens in the last box at the same time. In order for Miha to make as many moves $B_7$ as possible, $\\min\\{x, y\\}$ has to be as large as possible. Since $A_7$ decreases the number of tokens in the seventh box by $1$ and increases the number of tokens in the last box by $2$, this minimum will be the greatest when we have $x \\le y$ for the first time, so when $y = x$.\n\nAfter $k$ moves $A_7$ we have $x = 127 - k$ and $y = 1 + 2k$. Thus, we are looking for the smallest $k$ such that $127 - k \\le 1 + 2k$, or $k \\ge \\frac{127-1}{3} = 42$. The smallest such $k$ is $k = 42$.\n\nAfter $42$ moves of type $A_7$ Miha can use the $B_7$ moves to move $127 - 42 = 85$ tokens into the bag, and this is the greatest possible number of tokens that he can have in the bag at the end.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41694,
    "subject": "Mathematics (Multi-modal)",
    "question": "The diagonals of a convex quadrilateral $ABCD$ intersect at $P$. Let $L, M$ and $N$ be the points on the segments $AB, BC$ and $CD$, respectively, such that\n$$\n\\frac{|AL|}{|LB|} = \\frac{|BM|}{|MC|} = \\frac{|CN|}{|ND|}.\n$$\nSuppose that the line $LM$ is tangent to the circumcircle of the triangle $MCN$ and the line $NM$ is tangent to the circumcircle of the triangle $MBL$. Prove that $\\angle DPA = \\angle NML$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the circumcircle of $\\triangle MBL$ by $K_1$ and the circumcircle of $\\triangle MCN$ by $K_2$. The line $NM$ is tangent to $K_1$, so $\\angle NML = \\angle MBL$ (the tangent-chord angle theorem). Similarly, $LM$ is tangent to $K_2$, so $\\angle NML = \\angle NCM$. Hence,\n$$\n\\angle NCM = \\angle NML = \\angle MBL \\qquad (3)\n$$\n$$\n\\angle DCB = \\angle CBA = \\angle NML \\qquad (4)\n$$\nWe have $\\angle LMB = \\angle(LM, MC) = \\angle MNC$ and (3), so the triangles $\\triangle MBL$ and $\\triangle NCM$ are similar. This implies\n$$\n\\frac{|LB|}{|MC|} = \\frac{|LM|}{|MN|} = \\frac{|BM|}{|CN|}. \\qquad (5)\n$$\nSince\n$$\n\\frac{|AL|}{|LB|} = \\frac{|BM|}{|MC|} = \\frac{|CN|}{|ND|},\n$$\nwe have\n$$\n\\frac{|AL|}{|AB|} = \\frac{|BM|}{|BC|} = \\frac{|CN|}{|CD|}.\n$$\nThus\n$$\n\\frac{|AL|}{|BM|} = \\frac{|AB|}{|BC|} \\qquad (6)\n$$\n$$\n\\frac{|BM|}{|CN|} = \\frac{|BC|}{|CD|} \\qquad (7)\n$$\nFrom (4), (5), (6) and (7) we conclude that $\\triangle ABC$, $\\triangle LMN$ and $\\triangle BCD$ are similar. Now, denote $\\angle BDC = \\alpha$, $\\angle CBD = \\beta$ and $\\angle DCB = \\gamma$. Then\n$$\n\\begin{aligned} \n\\angle DPA &= \\pi - \\angle CPD \\\\\n&= \\pi - (\\pi - \\angle PDC - \\angle DCP) \\\\\n&= \\pi - (\\pi - \\angle BDC - \\angle DCA) \\\\\n&= \\pi - (\\pi - \\angle BDC - (\\angle DCB - \\angle ACB)) \\\\\n&= \\pi - (\\pi - \\alpha - (\\gamma - \\alpha)) \\\\\n&= \\pi - (\\pi - \\alpha - \\gamma + \\alpha) \\\\\n&= \\pi - (\\pi - \\gamma) \\\\\n&= \\gamma \\\\\n&= \\angle DCB = \\angle NML.\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41695,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa mizi so trije kupčki žetonov: eden z $a$ žetoni, eden z $b$ žetoni in eden s $c$ žetoni, pri čemer velja $a \\geq b \\geq c > 0$. Igralca $A$ in $B$ izmenično prestavljata žetone. Začne igralec $A$. V vsaki potezi igralec najprej izbere dva kupčka in nato s tistega z manj žetoni prestavi vsaj en žeton na tistega z več žetoni. Če imata izbrana kupčka enako žetonov, prestavi vsaj en žeton s kateregakoli izmed njiju na drugega. Zmaga tisti igralec, po čigar potezi so vsi žetoni na enem kupčku. Določi, kdo ima zmagovalno strategijo, in sicer v odvisnosti od $a, b$ in $c$.",
    "options": [],
    "answer": "Player B wins when b = c; otherwise Player A wins.",
    "solution": "Solution:\n\nČe je $b = c$, potem ima zmagovalno strategijo igralec $B$, sicer pa ima zmagovalno strategijo igralec $A$.\n\nDenimo najprej, da je $b = c$. Imamo torej situacijo, ko je na dveh kupčkih z najmanj žetoni enako število žetonov. Igralec $B$ lahko v tem primeru poskrbi, da je situacija po vsaki njegovi potezi spet taka, medtem ko po vsaki potezi igralca $A$ kupček z najmanj žetoni vsebuje strogo manj žetonov kot preostala dva kupčka. Igralec $A$ mora namreč v svoji potezi nujno iz enega izmed kupčkov z najmanj žetoni prestaviti nekaj žetonov na nek drug kupček. Igralec $B$ potem izbere trenutno najvišja kupčka, ki imata zaradi poteze igralca $A$ sedaj strogo več žetonov kot najnižji kupček, in med njima premakne žetone tako, da bosta po njegovi potezi najnižja kupčka imela enako žetonov. Ko bosta kupčka z najmanj žetoni prazna, bodo vsi žetoni na enem kupčku in igra bo končana. To se lahko zgodi le po potezi igralca $B$, zato igralec $B$ zmaga.\n\nDenimo sedaj, da je $b > c$. Tedaj lahko igralec $A$ s kupčka z $b$ žetoni prestavi $b - c > 0$ žetonov na kupček z $a$ žetoni. Po njegovi potezi bosta zato kupčka z najmanj žetoni oba imela $c$ žetonov. Torej bo situacija taka kot zgoraj, le da bo na potezi igralec $B$. Zato ima v tem primeru zmagovalno strategijo igralec $A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41696,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a regular $n$-gon where $n > 1$ is an odd integer. At most how many vertices can be coloured red so that the centre of the $n$-gon does not lie inside a polygon, determined by the red vertices?",
    "options": [],
    "answer": "(n+1)/2",
    "solution": "The greatest number of vertices that can be painted red is $\\frac{n+1}{2}$. If we paint $\\frac{n+1}{2}$ consecutive vertices of the $n$-gon, then the condition is satisfied.\n\nNow, assume that we have painted at least $\\frac{n+1}{2}+1$ vertices. Let us denote the vertices of the $n$-gon in order using positive integers from $1$ to $2k+1=n$. We have assumed that at least $k+2$ vertices are red. Consider the pairs $(1, k+1)$, $(2, k+2)$, $(3, k+3)$, ..., $(k, 2k)$, $(k+1, 2k+1)$, which determine some of the longest diagonals of the $n$-gon. There are precisely $k+1$ pairs and each vertex is contained in one of them, so there exists at least one pair in which both vertices are red.\n\nWithout loss of generality we may assume that the red pair is $(1, k+1)$, renumbering the vertices if necessary. There are at least $k$ red vertices remaining, but there are only $k-1$ numbers $2, 3, ..., k$. So, at least one of the vertices numbered $k+2, k+3, ..., 2k+1$ must be red. Assume that the red vertex is $k+m$ where $2 \\le m \\le k+1$. The vertex $k+m$ lies on the same side of the line through $1$ and $k+1$ as the centre of the $n$-gon. We conclude that the centre of the $n$-gon lies inside the triangle with vertices $1, k+1, k+m$, since the segment with endpoints $1, k+1$ is one of the longest diagonals of the $n$-gon. But the vertices $1, k+1, k+m$ are all painted red, so the centre of the $n$-gon lies inside a triangle with red vertices.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41697,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIzračunaj vrednost parametra $a$ tako, da bosta premici z enačbama $y+3=a x-2 x+a^{2}$ in $\\frac{x}{2 a-5}-\\frac{y}{3 a-6}=1$ vzporedni.",
    "options": [],
    "answer": "4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41698,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa tabli je napisano neko naravno število $n$. Na vsakem koraku lahko število na tabli nadomestimo z vsoto dveh naravnih števil, katerih zmnožek je enak številu na tabli. Določi najmanjše število, ki je lahko po končno korakih zapisano na tabli, in sicer v odvisnosti od začetnega števila $n$.",
    "options": [],
    "answer": "5 if n >= 5; otherwise n",
    "solution": "Solution:\n\nPokažimo najprej, da če se število na tabli pri zamenjavi zmanjša, potem je novo število večje ali enako $5$. Števila $1$ po zamenjavi ne moremo dobiti, saj vsota dveh naravnih števil ni nikoli enaka $1$. Število $2=1+1$ lahko dobimo le iz števila $1$, število $3=1+2$ le iz števila $2$, število $4=1+3=2+2$ pa le iz števil $3$ ali $4$. Torej, če po zamenjavi dobimo število manjše od $5$, potem se pri zamenjavi število ni zmanjšalo. To pomeni, da če je začetno število $n<5$, potem v nobenem koraku ne bomo dobili manjšega števila.\n\nDokažimo sedaj, da lahko število, ki je večje od $5$, vedno zmanjšamo. Denimo, da imamo na nekem koraku na tabli število $m>5$. Če je $m$ sod, torej oblike $m=2k$, kjer je $k>2$ naravno število, potem ga lahko zamenjamo s številom $k+2$. S tem smo dobili število, ki je manjše od $m$, saj je pogoj $k+2<2k$ ekvivalenten pogoju $k>2$. Če je $m$ lih, torej oblike $m=2k-1$, kjer je $k>3$ naravno število, potem ga lahko najprej zamenjamo s številom $(2k-1)+1=2k$ in nato s številom $k+2$. Na ta način spet dobimo število, ki je manjše od $m$, saj je pogoj $k+2<2k-1$ ekvivalenten pogoju $k>3$. To pomeni, da če je začetno število $n \\geq 5$, ga lahko po končno korakih vedno zmanjšamo do števila $5$, manjšega števila pa po zgornjem razmisleku ne moremo dobiti.\n\nNajmanjše število, ki je lahko po končno korakih zapisano na tabli, je torej enako $5$ v primeru $n \\geq 5$ oziroma $n$ v primeru $n<5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41699,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV vrsti stoji 8 škatel, oštevilčenih s števili od 1 do 8, in prazna vreča. V vsaki škatli je 1 žeton. Miha, ki ima veliko žetonov, se igra igro, v kateri sta dovoljeni naslednji dve potezi:\n- odstrani 1 žeton iz škatle, oštevilčene z $i$ ($i < 8$), in doda 2 žetona v škatlo, oštevilčeno z $(i+1)$,\n- odstrani 1 žeton iz škatle, oštevilčene z $i$ ($i < 8$), in premakne 1 žeton iz škatle, oštevilčene z $(i+1)$, v vrečo.\nIgra se konča, ko ni več možno izvesti nobene poteze. Največ koliko žetonov je lahko na koncu v vreči?",
    "options": [],
    "answer": "85",
    "solution": "Solution:\n\nNaj $A_{i}$ oziroma $B_{i}$ označuje prvo oziroma drugo potezo, izvedeno na škatlah $i$ in $i+1$. Naj bo $i \\leq 6$. Če Miha izvede potezo $B_{i}$, potem se število žetonov v škatlah $i$ in $i+1$ zmanjša za 1, število žetonov v škatli $i+2$ ostane enako, število žetonov v vreči pa se poveča za 1. Če pa Miha izvede zaporedje potez $A_{i}$, $A_{i+1}$, $B_{i+1}$, potem se število žetonov v škatli $i$ zmanjša za 1, število žetonov v škatli $i+1$ ostane enako, število žetonov v škatli $i+2$ in v vreči pa se poveča za 1. V drugem primeru je torej rezultat boljši, saj je edina razlika to, da ima v dveh škatlah po en žeton več. Ta razmislek pokaže, da se Mihu potez $B_{i}$ za $i \\leq 6$ ne splača izvajati. Od preostalih potez le poteza $B_{7}$ poveča število žetonov v vreči, zato jo mora Miha izvesti čim večkrat. Brez škode lahko predpostavimo, da jo izvaja šele na koncu. Pred tem s potezami $A_{i}$, $i \\leq 6$, prestavi žetone v predzadnjo škatlo, ki potem vsebuje $1+2+2^{2}+\\ldots+2^{6}=2^{7}-1=127$ žetonov, zadnja škatla pa vsebuje 1 žeton.\n\nNaj $x$ označuje število žetonov v predzadnji škatli ob nekem času, $y$ pa število žetonov v zadnji škatli ob istem času. Da bo lahko Miha izvedel čim več potez $B_{7}$, mora biti $\\min\\{x, y\\}$ čim večji. Ker se s potezo $A_{7}$ število žetonov v predzadnji škatli zmanjša za 1, v zadnji pa poveča za 2, bo minimum največji takrat, ko bo prvič veljalo $x \\leq y$ in bo zato enak $x$. Po $k$ potezah $A_{7}$ bo $x=127-k$ in $y=1+2k$. Iščemo torej najmanjši $k$, pri katerem bo $127-k \\leq 1+2k$ oziroma $k \\geq \\frac{127-1}{3}=42$. Najmanjši tak $k$ je enak $k=42$. Po 42 potezah $A_{7}$ bo s potezami $B_{7}$ Miha v vrečo premaknil $127-42=85$ žetonov, kar je torej največje število žetonov, ki jih lahko ima Miha na koncu v vreči.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41700,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCena vstopnice za ogled nogometne tekme je bila $9$ evrov. Zaradi nezanimanja za nakup vstopnic so ceno znižali. Število kupcev se je povečalo za $50\\%$, izkupiček od prodanih vstopnic pa za $25\\%$. Koliko je bila cena vstopnice po znižanju? Za koliko odstotkov je bila nova cena vstopnice nižja od prvotne cene?",
    "options": [],
    "answer": "7.5 euros; decreased by 16 2/3%",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41701,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n \\ge 2$ let $\\tau(n)$ denote the number of positive divisors of $n$, and let $\\sigma(n)$ be the sum of these divisors. Prove that\n$$\n\\frac{n+1}{2} \\ge \\frac{\\sigma(n)}{\\tau(n)} \\ge \\frac{3}{2\\sqrt{2}}\\sqrt{n}.\n$$\nFor each of these inequalities also determine when the equality holds.",
    "options": [],
    "answer": "Right inequality: equality holds only for n equal to two. Left inequality: equality holds exactly when n is prime.",
    "solution": "First consider the right inequality. 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). Then\n$$\n\\tau(n) = (k_1 + 1)(k_2 + 1) \\cdots (k_s + 1)\n$$\n$$\n\\begin{align*}\n\\sigma(n) &= \\left(1 + p_1 + \\cdots + p_1^{k_1}\\right) \\left(1 + p_2 + \\cdots + p_2^{k_2}\\right) \\cdots \\left(1 + p_s + \\cdots + p_s^{k_s}\\right) \\\\\n&= \\sigma(p_1^{k_1} \\cdots p_s^{k_s}).\n\\end{align*}\n$$\nSo\n$$\n\\frac{\\sigma(n)}{\\tau(n)\\sqrt{n}} = \\frac{1+p_1+\\cdots+p_1^{k_1}}{(k_1+1)p_1^{\\frac{k_1}{2}}} \\cdot \\frac{1+p_2+\\cdots+p_2^{k_2}}{(k_2+1)p_2^{\\frac{k_2}{2}}} \\cdots \\frac{1+p_s+\\cdots+p_s^{k_s}}{(k_s+1)p_s^{\\frac{k_s}{2}}}.\n$$\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 would like to prove that\n$$\n\\frac{1 + p + \\cdots + p^k}{(k + 1) p^{\\frac{k}{2}}} \\geq \\frac{3}{2\\sqrt{2}}\n$$\nfor all $p \\ge 2$ and $k \\ge 1$ (the left-hand side is the smallest when $p = 2, k = 1$).\nLet $x = \\sqrt{p}$. For every positive integer $k$ the function\n$$\nf_k(x) = \\frac{1 + x^2 + x^4 + \\dots + x^{2k}}{x^k}\n$$\nis increasing when $x > 1$.\nWhen $k = 1$ and $x > y > 1$ we have\n$$\nf_1(x) - f_1(y) = (x - y) + \\frac{y - x}{xy} = (x - y)\\left(1 - \\frac{1}{xy}\\right) > 0.\n$$\nIn general\n$$\nf_{2l}(x) = 1 + f_1(x^2) + \\cdots + f_1(x^{2l}), \\quad (8)\n$$\n$$\nf_{2l+1}(x) = f_1(x^3) + \\cdots + f_1(x^{2l+1}). \\tag{9}\n$$\nThe functions $f_{2l}(x)$ and $f_{2l+1}(x)$ are obviously increasing. So, $f_k(x)$ is increasing in $x$.\n$$\n\\begin{align*}\n\\Rightarrow \\frac{1+p+p^2+\\cdots+p^k}{p^{\\frac{k}{2}}} &\\ge \\frac{1+2+\\cdots+2^k}{2^{\\frac{k}{2}}} \\\\\n\\Rightarrow \\frac{1+p+p^2+\\cdots+p^k}{(k+1)p^{\\frac{k}{2}}} &\\ge \\frac{1+2+\\cdots+2^k}{(k+1)2^{\\frac{k}{2}}}\n\\end{align*}\n$$\nDefine $g(k) = \\frac{1+2+\\cdots+2^k}{(k+1)2^{\\frac{k}{2}}}$, $k \\in \\mathbb{N}$. Let us show that $g(k)$ is increasing.\n$$\n\\begin{align*}\ng(k+1) - g(k) &= \\frac{1 + 2(1 + 2 + \\cdots + 2^k)}{(k+2)2^{\\frac{k}{2}}\\sqrt{2}} - \\frac{1 + 2 + \\cdots + 2^k}{(k+1)2^{\\frac{k}{2}}} \\\\\n&= \\frac{(k+1) + (1 + 2 + \\cdots + 2^k)(2k + 2 - \\sqrt{2}k - 2\\sqrt{2})}{(k+1)(k+2)2^{\\frac{k}{2}}\\sqrt{2}} > 0\n\\end{align*}\n$$\nSo,\n$$\n\\frac{1+p+p^2+\\dots+p^k}{(k+1)p^{\\frac{k}{2}}} \\ge \\frac{1+2+\\dots+2^k}{(k+1)2^{\\frac{k}{2}}} \\ge \\frac{3}{2\\sqrt{2}}\n$$\nWe conclude that\n$$\n\\frac{\\sigma(n)}{\\tau(n)\\sqrt{n}} \\ge \\left( \\frac{3}{2\\sqrt{2}} \\right)^s \\ge \\frac{3}{2\\sqrt{2}}\n$$\nThe equality holds when $n = 2$.\n\nFinally, consider the left inequality. Denote by $1 = d_1 < d_2 < \\dots < d_{\\tau(n)} = n$ 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 precisely the divisors of $n$, we also have $\\sigma(n) = \\sum_{i=1}^{\\tau(n)} \\frac{n}{d_i}$.\n\nFor each divisor $d_i$ we also have\n$$\nd_i + \\frac{n}{d_i} \\le n + 1 \\iff d_i^2 - d_i(n + 1) + n \\le 0 \\iff (n - d_i)(1 - d_i) \\le 0, \\quad (10)\n$$\nand the equality holds if and only if either $d_i = 1$ or $d_i = n$. Now, we can estimate\n$$\n\\begin{aligned} \\frac{\\sigma(n)}{\\tau(n)} &= \\frac{\\sigma(n) + \\sigma(n)}{2\\tau(n)} = \\frac{\\sum_{i=1}^{\\tau(n)} d_i + \\sum_{i=1}^{\\tau(n)} \\frac{n}{d_i}}{2\\tau(n)} = \\frac{\\sum_{i=1}^{\\tau(n)} \\left(d_i + \\frac{n}{d_i}\\right)}{2\\tau(n)} \\le \\\\ &\\le \\frac{\\sum_{i=1}^{\\tau(n)} (n+1)}{2\\tau(n)} = \\frac{\\tau(n) \\cdot (n+1)}{2\\tau(n)} = \\frac{n+1}{2}. \\end{aligned}\n$$\nThe equality holds if and only if it holds for each (10). This is only the case when $\\tau(n) \\le 2$. So, the equality holds when either $n = 1$ or $n$ is a prime.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41702,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a zoo rabbits, parrots and snakes have 24 heads, 14 wings and 62 legs. How many snakes are there in this zoo?\n(A) 5 (B) 6 (C) 7 (D) 8 (E) 9",
    "options": [],
    "answer": "A",
    "solution": "Rabbits and snakes have no wings and parrots have two wings each, so there are $14 \\div 2 = 7$ parrots in the zoo. Thus, the parrots account for $7 \\cdot 2 = 14$ legs. Snakes have no legs, so the rabbits have $62 - 14 = 48$ legs altogether. Each rabbit has $4$ legs, so there are $48 \\div 4 = 12$ rabbits in the zoo, as well as $24 - 12 - 7 = 5$ snakes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41703,
    "subject": "Mathematics (Multi-modal)",
    "question": "Samo wrote down a 3-digit odd positive integer on a piece of paper. He then told Peter what the last digit of this number was. Peter immediately concluded that the number Samo wrote down is not prime. What was this last digit?\n(A) 1\n(B) 3\n(C) 5\n(D) 7\n(E) 9",
    "options": [],
    "answer": "C",
    "solution": "Peter can only arrive at this conclusion if the last digit is even (and the number is divisible by $2$) or equal to $5$ (and the number is divisible by $5$). All other digits can be the last digit of a three-digit prime, e.g. $101$, $103$, $107$ and $109$ are prime. Samo's number was odd, so the last digit must have been $5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41704,
    "subject": "Mathematics (Multi-modal)",
    "question": "Which digit of the 7-digit number $2345678$ should we delete to get a 6-digit number divisible by $9$?\n(A) $8$   (B) $7$   (C) $6$   (D) $5$   (E) $4$",
    "options": [],
    "answer": "A",
    "solution": "A positive integer is divisible by $9$ if and only if the sum of its digits is divisible by $9$. The sum of the digits of $2345678$ is equal to $35 = 3 \\cdot 9 + 8$, so we should delete the digit $8$. We obtain the 6-digit number $234567$. The sum of its digits is equal to $27 = 3 \\cdot 9$, so it must be divisible by $9$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41705,
    "subject": "Mathematics (Multi-modal)",
    "question": "Miha has 4 boxes. One of the boxes contains only 20 cent coins, another contains only 10 cent coins, one contains only 2 cent coins and another contains only 1 cent coins. Miha can take the coins from 3 different boxes, taking 1 coin from one of the boxes, 2 coins from another and 3 coins from the third box. At most how many cents can Miha take from the boxes?\n(A) 61 (B) 62 (C) 82 (D) 92 (E) 96",
    "options": [],
    "answer": "C",
    "solution": "Miha will get the greatest amount by taking three 20 cents coins, two 10 cents coins and one 2 cents coin. The total in this case equals 82 cents.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41706,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be two different real numbers. For which $x$ does the equality $\\frac{x-a}{x-b} = \\frac{x-b}{x-a}$ hold?\n(A) $\\frac{a-b}{2}$\n(B) $\\frac{a^2+b^2}{a+b}$\n(C) $\\frac{a^2+b^2}{2(a+b)}$\n(D) $a+b$\n(E) $\\frac{a+b}{2}$",
    "options": [],
    "answer": "E",
    "solution": "Multiplying the given equality by $(x-a)(x-b)$, we get $(x-a)^2 = (x-b)^2$. Squaring both sides and subtracting $x^2$ we get $-2a x + a^2 = -2b x + b^2$, which can be further rearranged into $2(b-a)x = (b-a)(b+a)$. Since $b-a \\neq 0$, we can divide both sides of the equality by $b-a$ to get $2x = b+a$. From here we get $x = \\frac{a+b}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41707,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 17 girls and 12 boys on the playground. How many children, at least, would have to join them, so that they could be divided into two groups of equal sizes both containing equal numbers of girls and boys?\n\n(A) 1   (B) 3   (C) 5   (D) 7   (E) 9",
    "options": [],
    "answer": "D",
    "solution": "If the children form 2 groups of equal sizes such that each group contains $n$ boys and $n$ girls, then there must be $2n$ girls on the playground at that time. This is an even number that must also be equal to the number of boys on the playground. In order for the number of girls to be even and equal to the number of boys, at least one girl has to join them, as well as at least 6 boys. Therefore, at least 7 children have to join.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 41708,
    "subject": "Mathematics (Multi-modal)",
    "question": "During the course of a crime investigation 4 suspects have been arrested. Each of them gave a statement.\n\nŽan: \"Of all the suspects only I am innocent.\"\nAlen: \"Of all the suspects only I am guilty.\"\nZala: \"We are all innocent.\"\nBeno: \"At least 2 of the suspects are guilty.\"\n\nFurther investigation has shown that at least one of the suspects was guilty and the innocent were telling the truth while the guilty were lying. How many of the suspects were guilty?\n\n(A) 1 (B) 2 (C) 3 (D) 4\n(E) Impossible to say.",
    "options": [],
    "answer": "C",
    "solution": "Alen cannot be innocent since that would imply that he is telling the truth and would be, according to his own statement, guilty. Hence, Alen is guilty. From here we conclude that Zala was lying and must, therefore, also be guilty. Now, we see that Beno was telling the truth and is innocent. Finally, Žan must have been lying, so he must be guilty as well. Precisely three of the suspects were guilty, namely Žan, Alen and Zala.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 41709,
    "subject": "Mathematics (Multi-modal)",
    "question": "The segment $AB$ measures $20$ cm. The polygonal curve connecting the points $A$ and $B$ together with the segment $AB$ forms $7$ equilateral triangles. No segment of the polygonal curve lies on the segment $AB$. How long, in centimeters, is this polygonal curve?\n\n![](attached_image_1.png)\n\n(A) $40\\sqrt{3}$\n\n(B) $40$\n\n(C) $20 + 20\\sqrt{3}$\n\n(D) $20 + 20\\sqrt{2}$\n\n(E) Impossible to determine.",
    "options": [],
    "answer": "B",
    "solution": "Denote the lengths of the sides of these $7$ equilateral triangles by $a_1, a_2, \\dots, a_7$. The length of the segment $AB$ is then $a_1 + a_2 + \\dots + a_7$, and the length of the polygonal curve from $A$ to $B$ is $2a_1 + 2a_2 + \\dots + 2a_7 = 2(a_1 + a_2 + \\dots + a_7)$. Since this length is twice the length of the segment it must be equal to $40$ cm.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41710,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many two-digit numbers are there, such that the predecessor and the successor are a prime and a perfect square, not necessarily in this order?\n(A) 1 (B) 2 (C) 3 (D) 4 (E) 5",
    "options": [],
    "answer": "E",
    "solution": "Perfect squares that are also predecessors or successors of two-digit numbers are $9$, $16$, $25$, $36$, $49$, $64$, $81$ and $100$. Two-digit numbers that have a perfect square as their successors or predecessors are therefore $10$, $15$, $17$, $24$, $26$, $35$, $37$, $48$, $50$, $63$, $65$, $80$, $82$ and $99$. But a two-digit number that is preceded or succeeded by a prime must be even, so the only options are $10$, $24$, $26$, $48$, $50$, $80$ and $82$. Out of these, the only ones with a prime neighbour are $10$, $24$, $48$, $80$ and $82$. All five of these satisfy the condition. Thus, the correct answer is $5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41711,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $(m, n)$, such that\n$$\nm + 3n - 5 = 2v - 11d.\n$$\nHere $v$ is the least common multiple of the numbers $m$ and $n$, and $d$ is the greatest common divisor of $m$ and $n$.",
    "options": [],
    "answer": "(m, n) = (9, 1) and (65, 5)",
    "solution": "Let us write $m = d a$ and $n = d b$. Here $a$ and $b$ are relatively prime integers and $v = d a b$. Plugging this into the given equation we get $d a + 3 d b - 5 = 2 d a b - 11 d$, so\n\n$$\nd a + 3 d b - 5 = 2 d a b - 11 d\n$$\n\nwhich rearranges to\n\n$$\nd a + 3 d b - 2 d a b + 11 d = 5\n$$\n\nor\n\n$$\nd(a + 3b - 2ab + 11) = 5\n$$\n\nSince $d$ is a positive integer, $d$ must divide $5$. Thus, $d = 1$ or $d = 5$.\n\nFirst, let us assume that $d = 1$. Then the equation can be rearranged into\n\n$$\na + 3b - 5 = 2ab - 11\n$$\n\nor\n\n$$\na + 3b + 6 = 2ab - 11 + 6 + 5 = 2ab\n$$\n\nso\n\n$$\na + 3b + 6 = 2ab\n$$\n\nor\n\n$$\n2ab - a - 3b - 6 = 0\n$$\n\nor\n\n$$\na(2b - 1) = 3b + 6\n$$\n\nFrom here we conclude that $2b - 1$ divides $3b + 6$, so it must also divide $2(3b + 6) - 3(2b - 1) = 6b + 12 - 6b + 3 = 15$.\n\nSince $2b - 1$ is a positive integer there are four possibilities:\n\n- If $2b - 1 = 1$, then $b = 1$ and $a = 9$.\n- If $2b - 1 = 3$, then $b = 2$ and $a = 4$, which leads to a contradiction, since $a$ and $b$ should be relatively prime.\n- If $2b - 1 = 5$, then $b = 3$ and $a = 3$, which is again a contradiction.\n- If $2b - 1 = 15$, then $b = 8$ and $a = 2$, and once more this is not possible.\n\nHence, in this case we only have one solution, $(9, 1)$.\n\nNow, assume that $d = 5$. Then we can divide the equation by $5$ and rearrange it to get\n\n$$\na + 3b - 1 = 2ab - 11\n$$\n\nor\n\n$$\na + 3b + 10 = 2ab\n$$\n\nso\n\n$$\na(2b - 1) = 3b + 10\n$$\n\nThis implies that $2b - 1$ divides $3b + 10$, so it must also divide $2(3b + 10) - 3(2b - 1) = 6b + 20 - 6b + 3 = 23$.\n\nThere are two possibilities:\n\n- If $2b - 1 = 1$, then $b = 1$ and $a = 13$.\n- If $2b - 1 = 23$, then $b = 12$ and $a = 2$, a contradiction once more.\n\nIn this case we get the solution $(65, 5)$.\n\nTherefore, the solutions are $(m, n) = (9, 1)$ and $(65, 5)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41712,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be positive integers such that $a^2 + b^2 + c^2$ is divisible by $7$. Prove that $a^4 + b^4 + c^4$ is also divisible by $7$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "When divided by $7$ a perfect square can give the remainder $0$, $1$, $2$ or $4$. For $a^2 + b^2 + c^2$ to be divisible by $7$, the numbers $a^2$, $b^2$ and $c^2$ must either all give the remainder $0$, or they must give three different remainders, $1$, $2$ and $4$.\n\nIn the first case the numbers $a$, $b$, $c$ are divisible by $7$, so $a^4 + b^4 + c^4$ is divisible by $7$ as well.\n\nIn the second case we may assume that $a^2 = 7k + 1$, $b^2 = 7m + 2$ and $c^2 = 7n + 4$ for some integers $k$, $m$ and $n$. In this case\n$$\na^4 + b^4 + c^4 = (7k+1)^2 + (7m+2)^2 + (7n+4)^2 = 49(k^2+m^2+n^2)+14(k+m+n)+21,\n$$\nwhich is again divisible by $7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41713,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which there exists a rectangle, such that the lengths of its sides are positive integers while its perimeter equals $n$ and is equal to the area of the rectangle.",
    "options": [],
    "answer": "16, 18",
    "solution": "Let $a$ and $b$ be the lengths of the sides of a rectangle that satisfies the conditions. Then $2(a+b) = n = ab$. This equality can be rewritten as $a(b-2) = 2b$. Now we can express $a$ in terms of $b$:\n$$\na = \\frac{2b}{b-2} = \\frac{2(b-2)}{b-2} + \\frac{4}{b-2} = 2 + \\frac{4}{b-2}.\n$$\n\nSince $a$ and $b$ are positive integers, the number $b-2$ must divide $4$. On the other hand $b-2$ is greater than $-2$, so we conclude that it can only be equal to $-1, 1, 2$ or $4$. Thus, $b$ equals $1, 3, 4$ or $6$ and $a$, respectively, equals $-2, 6, 4$ or $3$. The first case is not a valid solution since $-2$ is not a positive integer. In the second and fourth case we get $n=18$, in the third case $n=16$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41714,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $a \\ge 0$ for which the equation $2|x-a| + 3|x+a| = 1$ has at least one real solution.",
    "options": [],
    "answer": "[0, 1/4]",
    "solution": "We need to consider several possibilities.\n\nIf $x < -a$, then $-2(x - a) - 3(x + a) = 1$, which implies $x = -\\frac{a+1}{5}$. For this to be a valid solution, we must have\n\n$-\\frac{a+1}{5} < -a$, or $a < \\frac{1}{4}$.\n\nIf $-a \\le x \\le a$, then $-2(x-a) + 3(x+a) = 1$, so $x = 1-5a$. This solution is valid only when $-a \\le 1-5a \\le a$, which is equivalent to $\\frac{1}{6} \\le a \\le \\frac{1}{4}$.\n\nIf $x > a$, then $2(x-a) + 3(x+a) = 1$, or $x = \\frac{1-a}{5}$. For this solution to be valid we must have $\\frac{1-a}{5} > a$, or $a < \\frac{1}{6}$.\n\nThe equation has at least one real solution if and only if $a \\le \\frac{1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41715,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$, such that\n$$\nf(xy) = x f(y) + 3 f(x) + 3\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = x/2 - 3/2",
    "solution": "Plugging $x = 0$ into the equation, we get $f(0) = 3 f(0) + 3$, which implies that $f(0) = -\\frac{3}{2}$.\n\nNow plug in $y = 0$ to get $f(0) = x f(0) + 3 f(x) + 3$. From here we can obtain\n$$\nf(x) = -\\frac{1}{3} x f(0) + \\frac{1}{3} f(0) - 1 = \\frac{1}{2} x - \\frac{3}{2}.\n$$\nIt is easy to verify that the function $f(x) = \\frac{1}{2} x - \\frac{3}{2}$ satisfies the given equation, and this is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41716,
    "subject": "Mathematics (Multi-modal)",
    "question": "A deltoid is inscribed into a circle with radius $r$. One of the sides of the deltoid is twice as long as the other. Find the ratio of the area of the deltoid to the area of the circle.",
    "options": [],
    "answer": "8/(5π)",
    "solution": "Denote the lengths of the sides of the inscribed deltoid by $a$ and $2a$. Due to symmetry, the longer of the two diagonals passes through the centre of the circle. By Thales' theorem there is a right angle between the sides of length $a$ and $2a$. By Pythagoras' theorem the length of the longer diagonal is equal to $\\sqrt{a^2 + (2a)^2} = \\sqrt{5a^2} = a\\sqrt{5}$, which implies that $r = \\frac{a\\sqrt{5}}{2}$. The area of the deltoid is equal to $2 \\cdot \\frac{a \\cdot 2a}{2} = 2a^2$ and the area of the circle is $\\pi r^2 = \\frac{5\\pi a^2}{4}$. The ratio is $\\frac{8}{5\\pi}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41717,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does the equation $\\cos(\\sin x) = \\sin(\\cos x)$ have at least one real solution?",
    "options": [],
    "answer": "No",
    "solution": "Rewrite the equation as $\\cos(\\sin x) = \\cos\\left(\\frac{\\pi}{2} - \\cos x\\right)$. This implies that $\\frac{\\pi}{2} - \\cos x = \\pm \\sin x + 2k\\pi$ for some $k \\in \\mathbb{Z}$, so $\\cos x \\pm \\sin x = \\frac{\\pi}{2} - 2k\\pi$. Since $|\\cos x \\pm \\sin x| \\le |\\cos x| + |\\sin x| \\le 2$, we have $k=0$, and so $\\cos x \\pm \\sin x = \\frac{\\pi}{2}$. Squaring both sides of the equation we get $\\cos^2 x \\pm 2\\cos x\\sin x + \\sin^2 x = \\frac{\\pi^2}{4}$ or $\\pm \\sin 2x = \\frac{\\pi^2}{4} - 1$. Since $\\frac{\\pi^2}{4} - 1 > \\frac{9}{4} - 1 > 1$, this equation has no real solutions, so the initial equation does not have any real solutions either.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41718,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right triangle with the right angle at $C$, such that $|BC| = a$ and $|AC| = b$. Let $D$ be a point on the opposite side of the line $AC$ from $B$, such that the triangle $ACD$ is similar to the triangle $ABC$. Let $E$ be a point on the line $CD$, such that $\\angle EBC$ is a right angle. Find the area of the quadrilateral $ABED$ in terms of $a$ and $b$.",
    "options": [],
    "answer": "(2ab^4 + 2a^3b^2 + a^5) / (2b(a^2 + b^2))",
    "solution": "The triangles $ACD$ and $ABC$ are similar, so $\\angle CAD = \\angle BAC$. We notice that the point $E$ does not lie on the same side of the line $BC$ as $A$, so $\\angle BCE = 180^\\circ - \\angle ACB - \\angle DCA = 90^\\circ - \\angle DCA = \\angle CAD = \\angle BAC$. We conclude that the triangles $ABC$ and $CEB$ are also similar since they have two congruent angles. By Pythagoras' theorem we have $|AB| = \\sqrt{a^2 + b^2}$.\n\nSimilarity of the triangles $ABC$ and $ACD$ implies\n$$\n\\frac{|AD|}{|AC|} = \\frac{|AC|}{|AB|} \\quad \\text{and} \\quad \\frac{|DC|}{|AC|} = \\frac{|CB|}{|AB|}\n$$\n![](attached_image_1.png)\nFrom the first equality we get $|AD| = \\frac{b^2}{\\sqrt{a^2+b^2}}$, and from the second $|DC| = \\frac{ab}{\\sqrt{a^2+b^2}}$. Since the triangles $ABC$ and $CEB$ are also similar, we have\n$$\n\\frac{|BE|}{|BC|} = \\frac{|CB|}{|CA|}\n$$\nwhich implies $|BE| = \\frac{a^2}{b}$.\n\nThus, the area of the quadrilateral $ABED$ equals\n$$\n\\begin{align*}\np &= \\frac{|AD| \\cdot |DC|}{2} + \\frac{|AC| \\cdot |CB|}{2} + \\frac{|CB| \\cdot |BE|}{2} \\\\\n&= \\frac{ab^3}{2(a^2 + b^2)} + \\frac{ab}{2} + \\frac{a^3}{2b} = \\frac{ab^4 + ab^2(a^2 + b^2) + a^3(a^2 + b^2)}{2b(a^2 + b^2)} \\\\\n&= \\frac{2ab^4 + 2a^3b^2 + a^5}{2b(a^2 + b^2)}.\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41719,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sides and one of the diagonals of a rhombus measure $60$ cm. Inside the rhombus there are $9$ points. Do there exist two of the points which are not more than $30$ cm apart? Justify your answer.",
    "options": [],
    "answer": "Yes",
    "solution": "The answer is yes. The shorter diagonal divides the rhombus into two equilateral triangles with the sides of length $60$ cm. Each of the two triangles can be further divided into $4$ equilateral triangles with the sides of length $30$ cm. We have thus divided the rhombus into $8$ equilateral triangles with the sides measuring $30$ cm. Since there are $9$ points inside the rhombus, at least one of these equilateral triangles must contain at least two points. Let us prove that two points contained in the same triangle are at most $30$ cm apart. Denote the points by $E$ and $F$ and denote the vertices of the equilateral triangle containing them by $A$, $B$ and $C$. Consider the triangle $AEF$. The inner angle at $A$ measures $60^\\circ$ or less, so one of the remaining two angles must measure at least $60^\\circ$. From here we conclude that $EF$ is not the longest side in the triangle $AEF$. The other two sides measure at most $30$ cm since the points $E$ and $F$ lie inside a circle centered at $A$ with radius $30$ cm. Thus, the distance between $E$ and $F$ is at most $30$ cm.\n\n![](attached_image_1.png)\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41720,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of real numbers $x$ and $y$ which satisfy the equations\n$$\nx + \\frac{1}{y-x} = 1,\n$$\n$$\ny + \\frac{1}{x-y} = 2.\n$$",
    "options": [],
    "answer": "(2, 1) and (1/2, 5/2)",
    "solution": "We add up the equations and get $x + y = 3$. From this we express $y = 3 - x$ and insert it in the first equation to get $x + \\frac{1}{3-2x} = 1$. We rearrange to get $2x^2 - 5x + 2 = 0$. We can factor the left side of this equation $2(x-2)(x - \\frac{1}{2}) = 0$. It follows $x = 2$ or $x = \\frac{1}{2}$. In the first case we have $y = 1$, and in the second $y = \\frac{5}{2}$. Since $x - y \\neq 0$ in both cases, $(2, 1)$ and $(\\frac{1}{2}, \\frac{5}{2})$ are indeed solutions of the given system of equations.\nWe subtract the equations and get $y - x - \\frac{2}{y-x} = 1$. Introducing a new variable $a = y - x$ and rearranging we get $a^2 - a - 2 = 0$. We can factor the left side of this equation $(a+1)(a-2) = 0$. It follows $a = -1$ or $a = 2$. In each case we insert $a = y - x$ in the first equation to get $x = 2$ and $x = \\frac{1}{2}$, respectively. It then follows from $a = y - x$ that $y = 1$ and $y = \\frac{5}{2}$, respectively. Since $x - y \\neq 0$ in both cases, $(2, 1)$ and $(\\frac{1}{2}, \\frac{5}{2})$ are indeed solutions of the given system of equations.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41721,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $a$ and $b$ which satisfy $2a^b = ab + 3$.",
    "options": [],
    "answer": "a = 3, b = 1",
    "solution": "Since $b$ is a positive integer, $a$ divides $2a^b$ and $ab$, and hence also $3$. Since $3$ is a prime number, we have $a = 1$ or $a = 3$.\n\nIf $a = 1$ we get equation $2 = b + 3$, which does not have solutions in positive integers.\n\nTherefore $a = 3$ and we get equation $2 \\cdot 3^b = 3b + 3$ which gives $2 \\cdot 3^{b-1} = b + 1$.\n\nOne solution of this equation is $b = 1$.\n\nLet's prove that there are no other solutions in positive integers. We shall use induction with respect to $b$ and prove that $2 \\cdot 3^{b-1} > b + 1$ for all $b \\ge 2$.\n\nFor $b = 2$ this is true since $2 \\cdot 3^1 = 6 > 3 = 2 + 1$.\n\nSuppose that $2 \\cdot 3^{b-1} > b + 1$ for some $b \\ge 2$. Then $2 \\cdot 3^b = 3 \\cdot 2 \\cdot 3^{b-1} > 3(b+1) > b + 2$, and induction is finished.\n\nThus we indeed have that $2 \\cdot 3^{b-1} > b + 1$ for all $b \\ge 2$, which means that $b = 1$ is the only solution of the equation $2 \\cdot 3^{b-1} = b + 1$ in positive integers.\n\nThe only solution to the given equation is the pair $a = 3$ and $b = 1$.\nAs in Solution 1 we deduce that $a = 3$ and $2 \\cdot 3^b = 3b + 3$. Let's take a look at the functions $f(x) = 3x + 3$ and $g(x) = 2 \\cdot 3^x$. Real solutions of the equation $2 \\cdot 3^b = 3b + 3$ are those points at which the graphs of functions $f$ and $g$ intersect. The function $f$ is a linear function, the function $g$ is a scalar multiple of the exponent function. The graphs of such two functions intersect at most twice. Since $f(1) = 6 = g(1)$ one intersection is at $x = 1$. The second intersection lies on the interval between $-1$ and $0$ since $f(-1) = 0 < \\frac{2}{3} = g(-1)$ and $f(0) = 3 > 2 = g(0)$. The equation $2 \\cdot 3^b = 3b + 3$ thus has two real solutions, only one of which is a positive integer.\n\nThe only solution to the given equation is the pair $a = 3$ and $b = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41722,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIzračunaj tak $x$, da bo zaporedje $x+2, 3x, x^{2}+5$ geometrijsko in bodo členi zaporedja cela števila. Izračunaj količnik in zapiši zaporedje.",
    "options": [],
    "answer": "x = 2; common ratio = 3/2; sequence: 4, 6, 9",
    "solution": "Solution:\n\nNaj bo zaporedje $a_1 = x+2$, $a_2 = 3x$, $a_3 = x^2 + 5$.\n\nKer je zaporedje geometrijsko, velja:\n$$\n\\frac{a_2}{a_1} = \\frac{a_3}{a_2}\n$$\n\nVstavimo izraze:\n$$\n\\frac{3x}{x+2} = \\frac{x^2 + 5}{3x}\n$$\n\nKrižno pomnožimo:\n$$\n(3x)^2 = (x+2)(x^2 + 5)\n$$\n$$\n9x^2 = (x+2)(x^2 + 5)\n$$\n$$\n9x^2 = x^3 + 2x^2 + 5x + 10\n$$\n$$\n0 = x^3 - 7x^2 + 5x + 10\n$$\n\nPoiščimo celoštevilske rešitve. Preizkusimo nekaj vrednosti:\n\nZa $x = -1$:\n$$\n(-1)^3 - 7(-1)^2 + 5(-1) + 10 = -1 - 7 + (-5) + 10 = -1 - 7 - 5 + 10 = -13 + 10 = -3\n$$\nZa $x = 1$:\n$$\n1 - 7 + 5 + 10 = 1 - 7 + 15 = -6 + 15 = 9\n$$\nZa $x = -2$:\n$$\n(-2)^3 - 7(-2)^2 + 5(-2) + 10 = -8 - 28 - 10 + 10 = -8 - 28 = -36\n$$\nZa $x = 2$:\n$$\n8 - 28 + 10 + 10 = 8 - 28 + 20 = -20 + 20 = 0\n$$\n\nTorej je $x = 2$ rešitev.\n\nPreverimo, če so vsi členi cela števila:\n$$\na_1 = 2 + 2 = 4\n$$\n$$\na_2 = 3 \\cdot 2 = 6\n$$\n$$\na_3 = 2^2 + 5 = 4 + 5 = 9\n$$\n\nVsi členi so cela števila.\n\nIzračunajmo količnik:\n$$\nk = \\frac{a_2}{a_1} = \\frac{6}{4} = \\frac{3}{2}\n$$\n\nZaporedje je $4, 6, 9$ s količnikom $\\frac{3}{2}$.\n\nOdgovor:\n$x = 2$, zaporedje je $4, 6, 9$, količnik je $\\frac{3}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41723,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ and $Q$ be two points in the interior of the triangle $ABC$ such that $\\angle PAC = \\angle BAQ$ and $\\angle CBP = \\angle QBA$. Denote by $P_A, P_B$, and $P_C$ the orthogonal projections of the point $P$ onto the sides $BC, CA$, and $AB$, and by $Q_A, Q_B$, and $Q_C$ the orthogonal projections of the point $Q$ onto the sides $BC, CA$, and $AB$. Prove that the points $P_A, P_B, P_C, Q_A, Q_B$, and $Q_C$ lie on the same circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\angle PAC = \\angle BAQ$ the right-angle triangles $PAP_B$ and $QAQ_C$ have equal angles and hence are similar. It follows $\\frac{|AP|}{|AQ|} = \\frac{|AP_B|}{|AQ_C|}$. We also have $\\angle BAP = \\angle BAC - \\angle PAC = \\angle BAC - \\angle BAQ = \\angle QAC$. Hence the right-angle triangles $PAP_C$ and $QAQ_B$ also have equal angles and are similar. It follows $\\frac{|AP|}{|AQ|} = \\frac{|AP_C|}{|AQ_B|}$. Combining both ratios we get\n$$\n|AP_B| \\cdot |AQ_B| = |AP_C| \\cdot |AQ_C| \\text{ oziroma } A\\vec{P}_B \\cdot A\\vec{Q}_B = A\\vec{P}_C \\cdot A\\vec{Q}_C,\n$$\nsince the points $P_B, Q_B, P_C$, and $Q_C$ all lie on the sides of the triangle. By the Power of a Point Theorem the points $P_B, Q_B, P_C$, and $Q_C$ are concyclic. Denote their common circle by $\\mathcal{K}_1$.\n\nSimilarly, since $\\angle CBP = \\angle QBA$ the right-angle triangles $PBP_A$ and $QBQ_C$ are similar. Since $\\angle ABP = \\angle QBC$ the right-angle triangles $PBP_C$ and $QBQ_A$ are also similar. It follows $\\frac{|BP_A|}{|BQ_C|} = \\frac{|BP_C|}{|BQ_A|}$. Therefore\n$$\nB\\vec{P}_A \\cdot B\\vec{Q}_A = B\\vec{P}_C \\cdot B\\vec{Q}_C,\n$$\n\nLet $M$ be the midpoint of the segment $PQ$. The center of the circle $\\mathcal{K}_1$ lies on the intersection of bisectors of the segments $P_BQ_B$ and $P_CQ_C$. Since both bisectors pass through $M$, $M$ is the center of the circle $\\mathcal{K}_1$. Similarly, the center of the circle $\\mathcal{K}_2$ lies on the intersection of bisectors of the segments $P_AQ_A$ and $P_CQ_C$. These bisectors also pass through $M$, hence $M$ is the center of the circle $\\mathcal{K}_2$. Therefore the circles $\\mathcal{K}_1$ and $\\mathcal{K}_2$ have a common center, and both pass through the points $P_C$ and $Q_C$, hence they are the same. It follows that the points $P_A, P_B, P_C, Q_A, Q_B$, and $Q_C$ all lie on a common circle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41724,
    "subject": "Mathematics (Multi-modal)",
    "question": "For how many positive integers $n$, $n \\le 2015$ is the fraction $\\frac{3n-1}{2n^2+1}$ reducible?",
    "options": [],
    "answer": "183",
    "solution": "Suppose that the fraction $\\frac{3n-1}{2n^2+1}$ is reducible. Then there exists a positive integer $a$ different from $1$ which divides $3n-1$ and $2n^2+1$. It follows that $a$ divides also $3(2n^2+1)-2n(3n-1) = 2n+3$ and hence also $3(2n+3)-2(3n-1) = 11$. Since $11$ is a prime number it follows $a=11$, hence $11$ divides $3n-1$ and $2n^2+1$. Therefore there exists an integer $k$ such that $3n-1=11k$. From this we express $n = \\frac{11k+1}{3}$. For this to be an integer, $3$ must divide $11k+1$ and hence $2k+1$, which can happen if and only if $k$ is of the form $k = 3m+1$ for some integer $m$. In this case we have $n = 11m+4$ and the number $2n^2+1 = 2(11m+4)^2+1 = 2 \\cdot 11^2m^2 + 4 \\cdot 11m + 33$ is also divisible by $11$. Because of $1 \\le n \\le 2015$ we have $0 \\le m \\le 182$. The given fraction is thus reducible for $183$ positive integers $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41725,
    "subject": "Mathematics (Multi-modal)",
    "question": "Eva, Igor, Marko, and Maruša each wrote a positive integer on a piece of paper. If we would erase the last digit of Eva's number we would get Igor's number. If we would erase the last digit of Igor's number we would get Marko's number. If we would erase the last digit of Marko's number we would get Maruša's number. The sum of all four written numbers was $3838$. Which numbers did Eva, Igor, Marko, and Maruša write?",
    "options": [],
    "answer": "Eva 3456, Igor 345, Marko 34, Maruša 3",
    "solution": "Eva's number must be at least four-digit so that we can erase the last digit three times in a row and still get a positive integer. Eva's number also can not be more than four-digit since the sum of all four numbers is a four-digit number. Denote Eva's number by $abcd$ where $a, b, c$, and $d$ are the digits. Then Igor's number is $abc$, Marko's number is $ab$, and Maruša's number is $a$. The sum of these four numbers is equal to\n$$\n1000a + 100(a + b) + 10(a + b + c) + (a + b + c + d) = 3838.\n$$\nSince $a, b, c$, and $d$ are digits we have $a + b + c + d \\le 36$. However, since ones of the number $3838$ equal $8$ we must have $a + b + c + d \\le 28$. Thus, in the calculation at most $2$ carries over from ones to tens. Since $a + b + c + 2 \\le 29$ at most $2$ carries over from tens to hundreds. Since $a + b + 2 \\le 20$ and hundreds of the number $3838$ equal $3$ at most $1$ carries over from hundreds to thousands. It follows that $a$ equals $2$ or $3$. The case $a = 2$ is not possible since we would have $10 \\le a + b = 2 + b \\le 11$ and at least $7$ would have to carry over from tens to hundreds. Thus $a = 3$.\n\nSince at most $2$ carries over from tens to hundreds $b$ must equal $3, 4$, or $5$. The case $b = 5$ is not possible since we would have $8 \\le 8 + c = a + b + c \\le 9$ and at least $4$ would have to carry over from ones to tens. The case $b = 3$ is also impossible. In this case exactly $2$ would have to carry over from ones to tens, which would mean $a+b+c+2=8+c<20$, and thus at most $1$ would carry over from tens to hundreds, which yields a contradiction. Therefore $b = 4$.\nSince at most $2$ carries over from ones to tens $c$ equals $4, 5$, or $6$. The case $c = 6$ is not possible since we would have $13 + c = a + b + c + d \\le 9$, a contradiction. If we had $c = 4$ exactly $2$ would carry over from ones to tens. In this case we would have $a + b + c + d = 11 + d$ and hence $d = 9$. But then ones of the sum would equal $0$, a contradiction. Thus $c = 5$ and hence $d = 6$.\n\nEva, Igor, Marko, and Maruša wrote numbers $3456$, $345$, $34$, and $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41726,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse pare realnih števil $x$ in $y$, ki zadoščajo enačbama\n$$\n\\begin{aligned}\n& x+\\frac{1}{y-x}=1 \\\\\n& y+\\frac{1}{x-y}=2\n\\end{aligned}\n$$",
    "options": [],
    "answer": "(2, 1) and (1/2, 5/2)",
    "solution": "Solution:\n\n1. način. Enačbi seštejemo in dobimo $x+y=3$. Od tod izrazimo $y=3-x$ in vstavimo v prvo enačbo, da dobimo $x+\\frac{1}{3-2x}=1$. Odpravimo ulomke in preoblikujemo do $2x^{2}-5x+2=0$. Levo stran te enačbe lahko razstavimo $2(x-2)\\left(x-\\frac{1}{2}\\right)=0$. Sledi $x=2$ ali $x=\\frac{1}{2}$. V prvem primeru je $y=1$, v drugem pa $y=\\frac{5}{2}$. Ker je v obeh primerih $x-y \\neq 0$, sta $(2,1)$ in $\\left(\\frac{1}{2}, \\frac{5}{2}\\right)$ res rešitvi danega sistema enačb.\n\n\n2. način. Enačbi odštejemo in dobimo $y-x-\\frac{2}{y-x}=1$. Z uvedbo spremenljivke $a=y-x$ preoblikujemo do $a^{2}-a-2=0$. Levo stran te enačbe lahko razstavimo $(a+1)(a-2)=0$. Sledi $a=-1$ in $a=2$. Za vsak primer $a$ vstavimo v prvo enačbo in dobimo $x=2$ in $x=\\frac{1}{2}$. V prvem primeru iz $a=y-x$ sledi $y=1$, v drugem pa $y=\\frac{5}{2}$. Ker je v obeh primerih $y-x \\neq 0$, sta $(2,1)$ in $\\left(\\frac{1}{2}, \\frac{5}{2}\\right)$ res rešitvi danega sistema enačb.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41727,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMarko se je odločil, da bo privarčeval 600 evrov, ki jih bo porabil za nakup novega računalnika. Po štirih mesecih varčevanja je ugotovil, da so njegovi mesečni prihranki glede na vrstni red mesecev varčevanja v razmerju $4: 3: 2: 2$. Za nakup novega računalnika mu tako manjka le še polovica najmanjšega mesečnega prihranka. Koliko je Marko privarčeval prvi mesec?",
    "options": [],
    "answer": "200",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41728,
    "subject": "Mathematics (Multi-modal)",
    "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$$\n\\frac{D(a,b) + v(a,b)}{a+b} \\le \\frac{a+b}{4}.\n$$",
    "options": [],
    "answer": "Detailed 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 co-prime. We have $v(a, b) = a_1 b_1 d$. We insert this in the condition of the problem to get\n$$\na_1 d + b_1 d \\mid d + a_1 b_1 d \\Rightarrow a_1 + b_1 \\mid 1 + a_1 b_1,\n$$\nand we can rearrange the inequality to\n$$\n\\frac{d + a_1 b_1 d}{a_1 d + b_1 d} \\le \\frac{a_1 d + b_1 d}{4} \\Leftrightarrow 1 + a_1 b_1 \\le d \\frac{(a_1 + b_1)^2}{4} \\Leftrightarrow 4 \\le d(a_1^2 + b_1^2) + (2d - 4)a_1 b_1.\n$$\nIf $d \\ge 2$ we have\n$$\nd(a_1^2 + b_1^2) + (2d - 4)a_1 b_1 \\ge 2(a_1^2 + b_1^2) + (2 \\cdot 2 - 4)a_1 b_1 = 2(a_1^2 + b_1^2) \\ge 2(1^2 + 1^2) = 4,\n$$\nLets now check the case when $d = 1$. We must prove the inequality\n$$\n4 \\le a_1^2 + b_1^2 - 2a_1 b_1 \\Leftrightarrow 4 \\le (a_1 - b_1)^2 \\Leftrightarrow 2 \\le |a_1 - b_1|.\n$$\nThis means we have to prove that the numbers $a_1$ and $b_1$ differ by at least 2. Hence it suffices to check if that they do not differ for less.\nIf $a_1 = b_1$ we get $a_1 = b_1 = 1$ due to co-primality and hence $a = b = 1$, which is impossible.\n\nIf the numbers differ by 1 we may assume that $b_1 = a_1 + 1$ since the other case is symmetric. We insert this to the condition of the problem to get\n$$\n\\begin{align*}\na_1 + (a_1 + 1) \\mid 1 + a_1(a_1 + 1) & = a_1^2 + a_1 + 1 \\\\\n\\Rightarrow \\quad 2a_1 + 1 \\mid 2(a_1^2 + a_1 + 1) & = (2a_1 + 1)a_1 + a_1 + 2 \\\\\n\\Rightarrow \\quad 2a_1 + 1 \\mid a_1 + 2\n\\end{align*}\n$$\nHowever, since $2a_1 + 1 \\ge a_1 + 2$ this is possible only if $a_1 = 1$. Since $d = 1$ we again get $a = 1$, which is impossible since $a, b > 1$.\nThus we proved that the numbers $a_1$ and $b_1$ differ by at least 2 which proves the inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41729,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse pare naravnih števil $a$ in $b$, za katere je $2 a^{b}=a b+3$.",
    "options": [],
    "answer": "a=3, b=1",
    "solution": "Solution:\n\n1. način. Ker je $b$ naravno število, $a$ deli $2 a^{b}$ in $a b$, torej $a$ deli tudi $3$. Ker je $3$ praštevilo, je $a=1$ ali $a=3$. Če je $a=1$, dobimo enačbo $2 = b + 3$, ki pa nima rešitev v naravnih številih. Torej je $a=3$, od koder sledi $2 \\cdot 3^{b} = 3b + 3$ oziroma $2 \\cdot 3^{b-1} = b + 1$. Ena rešitev te enačbe je $b=1$. Dokažimo, da drugih rešitev v naravnih številih ni. To bomo storili tako, da bomo z indukcijo na $b$ dokazali, da velja $2 \\cdot 3^{b-1} > b + 1$ za vse $b \\geq 2$.\n\nZa $b=2$ je to res, saj je $2 \\cdot 3^{1} = 6 > 3 = 2 + 1$. Predpostavimo, da velja $2 \\cdot 3^{b-1} > b + 1$ za nek $b \\geq 2$. Potem je $2 \\cdot 3^{b} = 3 \\cdot 2 \\cdot 3^{b-1} > 3(b + 1) > b + 2$. Indukcija je s tem končana. Torej res velja $2 \\cdot 3^{b-1} > b + 1$ za vse $b \\geq 2$, kar pomeni, da je $b=1$ edina rešitev enačbe $2 \\cdot 3^{b-1} = b + 1$ v naravnih številih.\n\nDano enačbo reši le par $a=3$ in $b=1$.\n\n\n2. način. Kot v prvi rešitvi sklepamo, da velja $a=3$ in $2 \\cdot 3^{b} = 3b + 3$. Oglejmo si funkciji $f(x) = 3x + 3$ in $g(x) = 2 \\cdot 3^{x}$. Realne rešitve enačbe $2 \\cdot 3^{b} = 3b + 3$ so tiste točke, pri katerih se grafa funkcij $f$ in $g$ sekata. Funkcija $f$ je linearna funkcija, funkcija $g$ pa je skalarni večkratnik eksponentne funkcije. Grafa takih dveh funkcij se sekata kvečjemu dvakrat. Ker je $f(1) = 6 = g(1)$, je eno presečišče pri $x=1$. Drugo presečišče leži na intervalu med $-1$ in $0$, saj velja $f(-1) = 0 < \\frac{2}{3} = g(-1)$ in $f(0) = 3 > 2 = g(0)$. Enačba $2 \\cdot 3^{b} = 3b + 3$ ima torej dve realni rešitvi, od katerih pa je le $b=1$ naravno število.\n\nDano enačbo reši torej le par $a=3$ in $b=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41730,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa koliko naravnih števil $n$, $n \\leq 2015$, ulomek $\\frac{3 n-1}{2 n^{2}+1}$ ni okrajšan?",
    "options": [],
    "answer": "183",
    "solution": "Solution:\n\nDenimo, da ulomek $\\frac{3 n-1}{2 n^{2}+1}$ ni okrajšan. Potem obstaja naravno število $a$, različno od $1$, ki deli $3 n-1$ in $2 n^{2}+1$. Sledi, da $a$ deli tudi $3\\left(2 n^{2}+1\\right)-2 n(3 n-1)=2 n+3$ in zato tudi $3(2 n+3)-2(3 n-1)=11$. Ker je $11$ praštevilo, sledi $a=11$, torej $11$ deli $3 n-1$ in $2 n^{2}+1$. To pomeni, da obstaja celo število $k$, da je $3 n-1=11 k$. Od tod izrazimo $n=\\frac{11 k+1}{3}$. Da bo to celo število, mora $3$ deliti $11 k+1$ oziroma $2 k+1$, kar pa se zgodi natanko tedaj, ko je $k$ oblike $k=3 m+1$ za neko celo število $m$. V tem primeru je $n=11 m+4$ in tudi število $2 n^{2}+1=2(11 m+4)^{2}+1=2 \\cdot 11^{2} m^{2}+4 \\cdot 11 m+33$ je deljivo z $11$. Ker mora biti $1 \\leq n \\leq 2015$, je $0 \\leq m \\leq 182$. Dan ulomek torej ni okrajšan za $183$ naravnih števil $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41731,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Naj bo $a > 0$ in $b > 0$. Poenostavi izraz\n$$\n7\\left(\\frac{a^{5}}{2 b^{4}}\\right)^{5} \\cdot \\left(\\frac{2 b^{2}}{a^{4}}\\right)^{5} - 3\\left(\\frac{a}{b^{2}}\\right)^{3+2x} \\cdot \\left(\\frac{a^{2}}{b^{4}}\\right)^{1-x} + \\left(4 a^{5} b\\right)^{5} : \\left(2 a^{4} b^{3}\\right)^{5}\n$$\nb) Izračunaj vrednost izraza za $a = 5^{\\frac{1}{5}}$ in $b = \\sqrt[5]{2}$.",
    "options": [],
    "answer": "a) 36 a^5 / b^{10}; b) 45",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41732,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $p$ of odd degree with real coefficients which satisfy\n$$\np(p(x)) \\le (p(x))^3\n$$\nfor all $x \\in \\mathbb{R}$ and have the coefficient at $x^2$ equal 0.",
    "options": [],
    "answer": "p(x) = x^3 + b with b ≤ 0",
    "solution": "Let $p(x) = a_nx^n + a_{n-1}x^{n-1} + \\dots + a_0$, where $a_n \\neq 0$. The leading term of the polynomial $p(p(x))$ equals $a_n(a_nx^n)^n = a_n^{n+1}x^{n^2}$, and the leading term of the polynomial $p(x)^3$ equals $(a_nx^n)^3 = a_n^3x^{3n}$, they are both of odd degree. Since the polynomial $p(x)^3 - p(p(x))$ has only non-negative values it must be of even degree, hence the leading terms must cancel each other out. Therefore $a_n^{n+1}x^{n^2} = a_n^3x^{3n}$, i.e. $n^2 = 3n$ and $a_n^{n+1} = a_n^3$. It follows $n=3$ and $a_n=1$ since $a_n \\neq 0$, i.e. the polynomial $p$ is of degree 3 with a leading coefficient 1. Since the coefficient of the polynomial $p$ at $x^2$ is equal to the polynomial $p(x) = x^3+ax+b$. We use this in the given inequality and simplifying it to get $ax^3+a^2x+ab+b \\le 0$ for all $x \\in \\mathbb{R}$. This is possible only if $a=0$, since the polynomial of odd degree always has positive values at some points. The inequality thus simplifies to $b \\le 0$. The polynomials that satisfy the conditions of the problem are exactly those of the form $p(x) = x^3+b$, where $b \\le 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41733,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa realni števili $x$ in $y$ velja\n$$\nx^{3}+x^{2}+x y+x+y+2=0 \\quad \\text{in} \\quad y^{3}-y^{2}+3 y-x=0\n$$\nDoloči vrednost izraza $x-y$.",
    "options": [],
    "answer": "-1",
    "solution": "Solution:\n\n1. način. Enačbi odštejemo, da dobimo\n$$\nx^{3}-y^{3}+x^{2}+y^{2}+x y+2 x-2 y+2=0\n$$\nLevo stran preoblikujemo\n$$\n\\begin{aligned}\n& x^{3}-y^{3}+x^{2}+y^{2}+x y+2 x-2 y+2= \\\\\n& =(x-y)\\left(x^{2}+x y+y^{2}\\right)+\\left(x^{2}+x y+y^{2}\\right)+2(x-y+1)= \\\\\n& =(x-y+1)\\left(x^{2}+x y+y^{2}\\right)+2(x-y+1)= \\\\\n& =(x-y+1)\\left(x^{2}+x y+y^{2}+2\\right)\n\\end{aligned}\n$$\nKer je $x^{2}+x y+y^{2}+2=\\frac{1}{2}\\left(x^{2}+(x+y)^{2}+y^{2}\\right)+2>0$, mora biti $x-y+1=0$. Vrednost izraza $x-y$ je enaka $-1$.\n\n\n2. način. Iz druge enačbe izrazimo $x=y^{3}-y^{2}+3 y$ in vstavimo v prvo enačbo ter poenostavimo, da dobimo\n$$\ny^{9}-3 y^{8}+12 y^{7}-18 y^{6}+34 y^{5}-19 y^{4}+21 y^{3}+11 y^{2}+4 y+2=0\n$$\nS precej spretnosti lahko levo stran enačbe razstavimo\n$$\n\\left(y^{3}-y^{2}+2 y+1\\right)\\left(y^{6}-2 y^{5}+8 y^{4}-7 y^{3}+13 y^{2}+2\\right)=0\n$$\nKer je\n$$\n\\begin{aligned}\ny^{6}-2 y^{5}+8 y^{4}-7 y^{3}+13 y^{2}+2 & =y^{4}\\left(y^{2}-2 y+1\\right)+7 y^{2}\\left(y^{2}-y+\\frac{1}{4}\\right)+\\frac{45}{4} y^{2}+2= \\\\\n& =y^{4}(y-1)^{2}+7 y^{2}\\left(y-\\frac{1}{2}\\right)^{2}+\\frac{45}{4} y^{2}+2>0\n\\end{aligned}\n$$\nmora biti $y^{3}-y^{2}+2 y+1=0$ oziroma $y=y^{3}-y^{2}+3 y+1$. Torej je\n$$\nx-y=\\left(y^{3}-y^{2}+3 y\\right)-\\left(y^{3}-y^{2}+3 y+1\\right)=-1 \\text{.}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41734,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, a_3, \\dots$ be a sequence of non-zero real numbers which satisfies $a_n^2 = -a_{n+1}a_{n-1}$ for all positive integers $n$, $n \\ge 2$. Prove that the sequence $a_2, a_4, a_6, \\dots$ is geometric.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For each $n \\ge 2$ we can rearrange the given recursive formula to $\\frac{a_{n+1}}{a_n} = -\\frac{a_n}{a_{n-1}}$. Using this we calculate\n$$\n\\frac{a_{2n+2}}{a_{2n}} = \\frac{a_{2n+2}}{a_{2n+1}} \\cdot \\frac{a_{2n+1}}{a_{2n}} = \\left(-\\frac{a_{2n+1}}{a_{2n}}\\right) \\cdot \\left(-\\frac{a_{2n}}{a_{2n-1}}\\right) = \\frac{a_{2n+1}}{a_{2n-1}}.\n$$\nUsing this equation twice in a row we get\n$$\n\\frac{a_{2n+2}}{a_{2n}} = \\frac{a_{2n+1}}{a_{2n-1}} = \\frac{a_{2n}}{a_{2n-2}},\n$$\nwhich means that the sequence $a_2, a_4, a_6, \\dots$ is geometric.\nAfter we rearrange the formula to $\\frac{a_{n+1}}{a_n} = -\\frac{a_n}{a_{n-1}}$ we see that the sequence of positive real numbers $|a_1|, |a_2|, |a_3|, \\dots$ is geometric, hence there exist positive real numbers $a$ and $q$ so that $|a_n| = aq^{n-1}$ for all positive integers $n$. It follows that for every $n$ we have either $a_n = aq^{n-1}$ or $a_n = -aq^{n-1}$. If we insert $2n+1$ instead of $n$ in the equation that the original sequence is defined with we get $a_{2n+2}a_{2n} = -a_{2n+1}^2 \\le 0$, hence the neighbouring terms in the sequence $a_2, a_4, a_6, \\dots$ are of opposite signs. We deduce that either\n$$\na_{2n} = (-1)^n a q^{2n-1} = -a q (-q^2)^{n-1}\n$$\nfor all $n$ or\n$$\na_{2n} = a q (-q^2)^{n-1}\n$$\nfor all $n$. In both cases the sequence $a_2, a_4, a_6, \\dots$ is geometric.\nWe square the equality $a_n^2 = -a_{n+1}a_{n-1}$ to get $(a_n^2)^2 = a_{n+1}^2 a_{n-1}^2$. We rearrange this to $\\frac{a_{n+1}^2}{a_n^2} = \\frac{a_n^2}{a_{n-1}^2}$ and notice that the sequence of positive real numbers $a_1^2, a_2^2, a_3^2, \\dots$ is geometric. Therefore there exist positive real numbers $a$ and $q$ such that $a_n^2 = aq^{n-1}$ for all positive integers $n$. It follows that for each $n$ we have either $a_n = \\sqrt{a}\\sqrt{q^{n-1}}$ or $a_n = -\\sqrt{a}\\sqrt{q^{n-1}}$. Similarly as in Solution 2 we get that either\n$$\na_{2n} = (-1)^n \\sqrt{a} \\sqrt{q^{2n-1}} = (-1)^n \\sqrt{a} \\sqrt{q} \\sqrt{q^{2n-2}} = (-1)^n \\sqrt{a q q^{n-1}} = -\\sqrt{a q} (-q)^{n-1}\n$$\nfor all $n$ or\n$$\na_{2n} = \\sqrt{a q} (-q)^{n-1}\n$$\nfor all $n$. In both cases the sequence $a_2, a_4, a_6, \\dots$ is geometric.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41735,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse pare naravnih števil $a$ in $b$, za katere je $a-b=101$ in je $a b$ popoln kvadrat.",
    "options": [],
    "answer": "a = 2601, b = 2500",
    "solution": "Solution:\n\n1. način. Naj bo $d$ največji skupni delitelj števil $a$ in $b$. Torej je $a=d m$ in $b=d n$, kjer sta $m$ in $n$ tuji naravni števili. Ker je $a b=d^{2} m n$ popoln kvadrat in sta $m$ in $n$ tuji števili, sta tudi $m$ in $n$ popolna kvadrata. Pišimo $m=x^{2}$ in $n=y^{2}$, kjer sta $x$ in $y$ naravni števili. Sledi $101=a-b=d(x+y)(x-y)$. Ker je $101$ praštevilo in je $x+y \\geq 2$, mora veljati $d=x-y=1$ in $x+y=101$. Od tod izračunamo $x=51$ in $y=50$, torej je $a=51^{2}=2601$ in $b=50^{2}=2500$.\n\n\n2. način. Naj bo $d$ največji skupni delitelj števil $a$ in $b$. Ker $d$ deli $101$ in je $101$ praštevilo, je lahko le $d=1$ ali $d=101$.\nČe je $d=1$, sta si $a$ in $b$ tuji števili. Ker je $a b$ popoln kvadrat, sledi, da sta tudi $a$ in $b$ popolna kvadrata. Pišimo $a=x^{2}$ in $b=y^{2}$, kjer sta $x$ in $y$ naravni števili. Tedaj je $101=a-b=x^{2}-y^{2}=(x+y)(x-y)$, od koder sledi $x-y=1$ in $x+y=101$, saj je $101$ praštevilo in $x+y \\geq 2$. Iz teh dveh enačb zlahka izračunamo $x=51$ in $y=50$, torej je $a=51^{2}=2601$ in $b=50^{2}=2500$.\nČe je $d=101$, lahko zapišemo $a=101 m$ in $b=101 n$, kjer sta $m$ in $n$ tuji naravni števili. Ker je $a b=101^{2} m n$ popoln kvadrat in sta si $m$ in $n$ tuji števili, sledi da sta tudi $m$ in $n$ popolna kvadrata. Pišimo $m=u^{2}$ in $n=v^{2}$, kjer sta $u$ in $v$ naravni števili. Tedaj je $101=a-b=101(m-n)$ oziroma $1=m-n=(u+v)(u-v)$, kar pa je protislovje, saj je $u+v \\geq 2$.\nEdina rešitev naloge je torej par $a=2601$ in $b=2500$.\n\n\n3. način. Zapišimo $a=b+101$ in izračunamo $a b=b(b+101)=n^{2}$ za neko naravno število $n$. Od tod izrazimo $n^{2}-b^{2}=(n-b)(n+b)=101 b$. Ker je $101$ praštevilo, velja $101 \\mid(n-b)$ ali $101 \\mid(n+b)$. Obravnavamo obe možnosti.\nČe $101 \\mid(n-b)$, lahko zapišemo $n-b=101 k$ za neko naravno število $k$. Iz enačbe $(n-b)(n+b)=101 b$ sledi $k n=b(1-k)$, kar je protislovje, ker je leva stran enačbe strogo pozitivna, desna pa manjša ali enaka nič.\nČe $101 \\mid(n+b)$, lahko zapišemo $n+b=101 l$ za neko naravno število $l$. Iz enačbe $(n-b)(n+b)=101 b$ sledi $l(101 l-2 b)=b$, oziroma $101 l^{2}=b(2 l+1)$, od koder izrazimo $b=\\frac{101 l^{2}}{2 l+1}$, kar mora biti naravno število. Torej $2 l+1$ deli $101 l^{2}$. Ker je $101$ praštevilo, mora veljati $(2 l+1) \\mid 101$, zato lahko zapišemo $101=(2 l+1) c$, kjer je $c \\in \\mathbb{N}$. Ker je $2 l+1>1$, mora veljati $c=1$ in $2 l+1=101$. Od tod izračunamo $l=50$ in $b=50^{2}=2500$. Nato izračunamo še $a=2601$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41736,
    "subject": "Mathematics (Multi-modal)",
    "question": "Real numbers $x$ and $y$ satisfy $x^3 + x^2 + xy + x + y + 2 = 0$ and $y^3 - y^2 + 3y - x = 0$. Determine the value of the expression $x - y$.",
    "options": [],
    "answer": "-1",
    "solution": "We subtract the equations to get\n$$\nx^3 - y^3 + x^2 + y^2 + xy + 2x - 2y + 2 = 0.\n$$\nWe rearrange the left side\n$$\n\\begin{align*}\nx^3 - y^3 + x^2 + y^2 + xy + 2x - 2y + 2 &= \\\\\n= (x - y)(x^2 + xy + y^2) + (x^2 + xy + y^2) + 2(x - y + 1) = \\\\\n= (x - y + 1)(x^2 + xy + y^2) + 2(x - y + 1) = \\\\\n= (x - y + 1)(x^2 + xy + y^2 + 2).\n\\end{align*}\n$$\nSince $x^2 + xy + y^2 + 2 = \\frac{1}{2}(x^2 + (x+y)^2 + y^2) + 2 > 0$ we must have $x - y + 1 = 0$. The value of the expression $x - y$ equals $-1$.\nFrom the second equation we express $x = y^3 - y^2 + 3y$, insert it in the second equation, and simplify to get\n$$\ny^{9} - 3y^{8} + 12y^{7} - 18y^{6} + 34y^{5} - 19y^{4} + 21y^{3} + 11y^{2} + 4y + 2 = 0.\n$$\nWith quite a lot of skill we can factor the left side of the equation to\n$$\n(y^{3} - y^{2} + 2y + 1)(y^{6} - 2y^{5} + 8y^{4} - 7y^{3} + 13y^{2} + 2) = 0.\n$$\nSince\n$$\n\\begin{align*}\ny^{6} - 2y^{5} + 8y^{4} - 7y^{3} + 13y^{2} + 2 &= y^{4}(y^{2} - 2y + 1) + 7y^{2}\\left(y^{2} - y + \\frac{1}{4}\\right) + \\frac{45}{4}y^{2} + 2 = \\\\\n= y^{4}(y - 1)^{2} + 7y^{2}\\left(y - \\frac{1}{2}\\right)^{2} + \\frac{45}{4}y^{2} + 2 > 0,\n\\end{align*}\n$$\nwe must have $y^3 - y^2 + 2y + 1 = 0$ which gives $y = y^3 - y^2 + 3y + 1$. Therefore\n$$\nx - y = (y^3 - y^2 + 3y) - (y^3 - y^2 + 3y + 1) = -1.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41737,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse polinome $p$ lihe stopnje $z$ realnimi koeficienti, za katere velja\n$$\np(p(x)) \\leq (p(x))^{3}\n$$\nza vse $x \\in \\mathbb{R}$ in ki imajo koeficient pri $x^{2}$ enak 0.",
    "options": [],
    "answer": "p(x) = x^3 + b with b ≤ 0",
    "solution": "Solution:\n\nPišimo $p(x) = a_{n} x^{n} + a_{n-1} x^{n-1} + \\cdots + a_{0}$, kjer je $a_{n} \\neq 0$. Vodilni člen polinoma $p(p(x))$ je enak $a_{n} (a_{n} x^{n})^{n} = a_{n}^{n+1} x^{n^{2}}$, vodilni člen polinoma $p(x)^{3}$ pa $(a_{n} x^{n})^{3} = a_{n}^{3} x^{3n}$, oba sta lihe stopnje. Ker je polinom $p(x)^{3} - p(p(x))$ povsod nenegativen, mora biti sode stopnje, zato se morata omenjena vodilna člena pokrajšati. Torej je $a_{n}^{n+1} x^{n^{2}} = a_{n}^{3} x^{3n}$ oziroma $n^{2} = 3n$ in $a_{n}^{n+1} = a_{n}^{3}$. Sledi $n = 3$ in $a_{n} = 1$, saj je $a_{n} \\neq 0$, tj. $p$ je polinom stopnje $3$ z vodilnim koeficientom $1$.\n\nKer je koeficient polinoma $p$ pri $x^{2}$ enak $0$, je polinom $p$ oblike $p(x) = x^{3} + a x + b$. Ko slednje upoštevamo v dani neenakosti in neenakost poenostavimo, dobimo $a x^{3} + a^{2} x + a b + b \\leq 0$ za vsak $x \\in \\mathbb{R}$. Slednje je mogoče le, če je $a = 0$, saj polinom lihe stopnje vedno zavzame tudi pozitivne vrednosti. Neenakost se v tem primeru poenostavi do $b \\leq 0$.\n\nPolinomi, ki zadoščajo pogojem naloge, so natanko polinomi oblike $p(x) = x^{3} + b$, kjer je $b \\leq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41738,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana je funkcija $f$ s predpisom $f(x)=A \\cdot \\cos \\left(x+\\frac{\\pi}{6}\\right)$. Določi $A \\in \\mathbb{R}$ tako, da bo veljalo $f\\left(-\\frac{\\pi}{3}\\right)=-\\sqrt{3}$. Za tako določeno funkcijo $f$ zapiši njeno zalogo vrednosti in nariši njen graf.",
    "options": [],
    "answer": "A = -2; range: [-2, 2]",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41739,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $a_{1}, a_{2}, a_{3}, \\ldots$ zaporedje neničelnih realnih števil, za katerega velja $a_{n}^{2} = -a_{n+1} a_{n-1}$ za vsa naravna števila $n$, $n \\geq 2$. Dokaži, da je zaporedje $a_{2}, a_{4}, a_{6}, \\ldots$ geometrijsko.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n1. način. Za vsak $n \\geq 2$ lahko dano rekurzivno formulo preoblikujemo v $\\frac{a_{n+1}}{a_{n}} = -\\frac{a_{n}}{a_{n-1}}$. S pomočjo te formule izračunamo\n$$\n\\frac{a_{2n+2}}{a_{2n}} = \\frac{a_{2n+2}}{a_{2n+1}} \\cdot \\frac{a_{2n+1}}{a_{2n}} = \\left(-\\frac{a_{2n+1}}{a_{2n}}\\right) \\cdot \\left(-\\frac{a_{2n}}{a_{2n-1}}\\right) = \\frac{a_{2n+1}}{a_{2n-1}}\n$$\nZ dvakratno zaporedno uporabo te enakosti dobimo\n$$\n\\frac{a_{2n+2}}{a_{2n}} = \\frac{a_{2n+1}}{a_{2n-1}} = \\frac{a_{2n}}{a_{2n-2}}\n$$\nkar pomeni, da je zaporedje $a_{2}, a_{4}, a_{6}, \\ldots$ geometrijsko.\n\n\n2. način. Potem, ko formulo preoblikujemo v $\\frac{a_{n+1}}{a_{n}} = -\\frac{a_{n}}{a_{n-1}}$, vidimo, da je zaporedje pozitivnih realnih števil $|a_{1}|, |a_{2}|, |a_{3}|, \\ldots$ geometrijsko, torej obstajata pozitivni realni števili $a$ in $q$, da je $|a_{n}| = a q^{n-1}$ za vsa naravna števila $n$. Sledi, da je za vsak $n$ bodisi $a_{n} = a q^{n-1}$ bodisi $a_{n} = -a q^{n-1}$. Če v enačbo, s katero je prvotno zaporedje definirano, vstavimo $2n+1$ namesto $n$, dobimo $a_{2n+2} a_{2n} = -a_{2n+1}^{2} \\leq 0$, zato sta v zaporedju $a_{2}, a_{4}, a_{6}, \\ldots$ sosednja člena nasprotnega predznaka. Od tod sklepamo, da je bodisi\n$$\na_{2n} = (-1)^{n} a q^{2n-1} = -a q (-q^{2})^{n-1}\n$$\nza vse $n$ bodisi\n$$\na_{2n} = a q (-q^{2})^{n-1}\n$$\nza vse $n$. V obeh primerih je $a_{2}, a_{4}, a_{6}, \\ldots$ geometrijsko zaporedje.\n\n\n3. način. Enakost $a_{n}^{2} = -a_{n+1} a_{n-1}$ kvadriramo, da dobimo $\\left(a_{n}^{2}\\right)^{2} = a_{n+1}^{2} a_{n-1}^{2}$. To formulo preoblikujemo v $\\frac{a_{n+1}^{2}}{a_{n}^{2}} = \\frac{a_{n}^{2}}{a_{n-1}^{2}}$ in vidimo, da je zaporedje pozitivnih realnih števil $a_{1}^{2}, a_{2}^{2}, a_{3}^{2}, \\ldots$ geometrijsko, torej obstajata pozitivni realni števili $a$ in $q$, da je $a_{n}^{2} = a q^{n-1}$ za vsa naravna števila $n$. Sledi, da je za vsak $n$ bodisi $a_{n} = \\sqrt{a} \\sqrt{q^{n-1}}$ bodisi $a_{n} = -\\sqrt{a} \\sqrt{q^{n-1}}$. Podobno kot pri 2. načinu dobimo, da je bodisi\n![](attached_image_1.png)\nza vse $n$ bodisi\n$$\na_{2n} = -(-1)^{n} \\sqrt{a} \\sqrt{q^{2n-1}} = \\sqrt{a q} (-q)^{n-1}\n$$\nza vse $n$. V obeh primerih je $a_{2}, a_{4}, a_{6}, \\ldots$ geometrijsko zaporedje.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41740,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{Z} \\to \\mathbb{Z}$ be an injective map such that $|f(m) - f(n)| \\le 2015$ holds for arbitrary integers $m$ and $n$ which satisfy $|m - n| \\le 2015$. Prove that\n$$\n|f(m) - f(n)| = |m - n|\n$$\nholds for all $m, n \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n$ be an arbitrary integer and\n$$\nS_n = \\{n - 2015, n - 2014, \\dots, n, \\dots, n + 2015\\}\n$$\nbe the set of all integers that differ from $n$ by at most $2015$.\nWe know that $S_{f(n)} = \\{f(n)-2015, f(n)-2014, \\dots, f(n)+2015\\}$.\nFrom the condition of the problem it follows that the numbers $f(n-2015), f(n-2014), \\dots, f(n+2015)$ are also elements of the set $S_{f(n)}$, since they differ from $f(n)$ by at most $2015$. Because $f$ is injective all these numbers are different. There are $4031$ of them, which is also the size of the set $S_{f(n)}$. Thus\n$$\nS_{f(n)} = \\{f(n) - 2015, \\dots, f(n) + 2015\\} = \\{f(n - 2015), \\dots, f(n + 2015)\\}\n$$\n\nLet us now look at the intersection of the sets $S_{f(n)}$ and $S_{f(n+1)}$. On one hand we have\n$$\n\\begin{aligned}\nS_{f(n)} \\cap S_{f(n+1)} &= \\{f(n - 2015), \\dots, f(n + 2015)\\} \\cap \\{f(n - 2014), \\dots, f(n + 2016)\\} \\\\\n&= \\{f(n - 2014), \\dots, f(n + 2015)\\}.\n\\end{aligned}\n$$\nSo the intersection has $4030$ elements. On the other hand we have\n$$\nS_{f(n)} \\cap S_{f(n+1)} = \\{f(n) - 2015, \\dots, f(n) + 2015\\} \\cap \\{f(n+1) - 2015, \\dots, f(n+1) + 2015\\}.\n$$\nThis set can have $4030$ elements if and only if $f(n)$ and $f(n+1)$ differ by $1$. Thus we proved that $|f(n+1) - f(n)| = 1$ for every integer $n$.\nIf for some $n$ we have $f(n+1) - f(n) = 1$ then by induction we have $f(n+k) - f(n) = k$ for every integer $k$. Similarly, if for some $n$ we have $f(n+1) - f(n) = -1$ then we have $f(n+k) - f(n) = -k$ for every integer $k$. In both cases we get $|f(m) - f(n)| = |m-n|$ for all integers $m$ and $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41741,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $a$ and $b$ such that $a - b = 101$ holds and $ab$ is a perfect square.",
    "options": [],
    "answer": "a=2601, b=2500",
    "solution": "Let $d$ be the greatest common divisor of the numbers $a$ and $b$. Then we have $a = dm$ and $b = dn$, where $m$ and $n$ are co-prime positive integers. Since $ab = d^2mn$ is a perfect square and integers $m$ and $n$ are co-prime, they are also perfect squares. Let $m = x^2$ and $n = y^2$, where $x$ and $y$ are positive integers. It follows $101 = a - b = d(x + y)(x - y)$. Since 101 is a prime number and we have $x + y \\ge 2$ it must be $d = x - y = 1$ and $x + y = 101$. From this we deduce $x = 51$ and $y = 50$, hence $a = 51^2 = 2601$ and $b = 50^2 = 2500$.\nLet $d$ be the greatest common divisor of the numbers $a$ and $b$. Since $d$ divides 101 which is a prime number we can only have $d = 1$ or $d = 101$.\nIf $d = 1$ the numbers $a$ and $b$ are co-prime. Since $ab$ is a perfect square it follows that $a$ and $b$ are also perfect squares. Let $a = x^2$ and $b = y^2$, where $x$ and $y$ are positive integers. Then $101 = a - b = x^2 - y^2 = (x+y)(x-y)$, which implies $x - y = 1$ and $x + y = 101$ since 101 is a prime number and $x + y \\ge 2$. From this we deduce $x = 51$ and $y = 50$, hence $a = 51^2 = 2601$ and $b = 50^2 = 2500$.\nIf $d = 101$ we can write $a = 101m$ and $b = 101n$, where $m$ and $n$ are co-prime positive integers. Since $ab = 101^2mn$ is a perfect square and integers $m$ and $n$ are co-prime, they are also perfect squares. Let $m = u^2$ and $n = v^2$, where $u$ and $v$ are positive integers. Then $101 = a - b = 101(m - n)$ which gives $1 = m - n = (u + v)(u - v)$. This is a contradiction since $u + v \\ge 2$.\nThus the only solution of the problem is the pair $a = 2601$ and $b = 2500$.\nWrite $a = b + 101$ and calculate $ab = b(b + 101) = n^2$ for some positive integer $n$. From this we express $n^2 - b^2 = (n - b)(n + b) = 101b$. Since 101 is a prime number we have $101|(n - b)$ or $101|(n + b)$. We consider each case separately.\nIf $101|(n - b)$ we can write $n - b = 101k$ for some positive integer $k$. From the equation $(n - b)(n + b) = 101b$ it follows $kn = b(1 - k)$, which is a contradiction since the left side of the equation is strictly positive and the right side is non-positive.\nIf $101|(n + b)$ we can write $n + b = 101l$ for some positive integer $l$. From the equation $(n - b)(n + b) = 101b$ it follows $l(101l - 2b) = b$ which gives $101l^2 = b(2l + 1)$\n\nand thus $b = \\frac{101l^2}{2l+1}$ must be a positive integer. Hence $2l+1$ divides $101l^2$. Since $101$ is a prime number we have $(2l+1)|101$ and we can write $101 = (2l+1)c$, where $c \\in \\mathbb{N}$. Since $2l+1 > 1$ we must have $c=1$ and $2l+1 = 101$. From this we deduce $l=50$ and $b = 50^2 = 2500$, and also $a = 2601$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41742,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoloči $m \\in \\mathbb{R}$ tako, da bo najmanjša vrednost funkcije $f(x) = (m-1)x^{2} + (1-3m)x + 2m + 1$ enaka $-1$.",
    "options": [],
    "answer": "3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41743,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIzračunaj naravni števili, katerih vsota je $168$, največji skupni delitelj pa $24$. Zapiši vse možne rešitve.",
    "options": [],
    "answer": "(24, 144), (48, 120), (72, 96)",
    "solution": "Solution:\n\nNaj bosta iskani števili $a$ in $b$. Podano je:\n\n- $a + b = 168$\n- $\\gcd(a, b) = 24$\n\nKer je njun največji skupni delitelj $24$, lahko zapišemo $a = 24m$, $b = 24n$, kjer sta $m$ in $n$ tuji števili ($\\gcd(m, n) = 1$).\n\nTorej:\n$$\na + b = 24m + 24n = 24(m + n) = 168\n$$\n$$\nm + n = \\frac{168}{24} = 7\n$$\n\nPoiščimo naravna števila $m$ in $n$, ki sta tuji in katerih vsota je $7$.\nMožne možnosti so:\n- $m = 1$, $n = 6$\n- $m = 2$, $n = 5$\n- $m = 3$, $n = 4$\n\nPreverimo, ali so pari tuji števili:\n- $\\gcd(1, 6) = 1$ (OK)\n- $\\gcd(2, 5) = 1$ (OK)\n- $\\gcd(3, 4) = 1$ (OK)\n\nTorej so vsi pari ustrezni.\n\nIzračunamo ustrezne vrednosti $a$ in $b$:\n\n1. $m = 1$, $n = 6$: $a = 24 \\cdot 1 = 24$, $b = 24 \\cdot 6 = 144$\n2. $m = 2$, $n = 5$: $a = 24 \\cdot 2 = 48$, $b = 24 \\cdot 5 = 120$\n3. $m = 3$, $n = 4$: $a = 24 \\cdot 3 = 72$, $b = 24 \\cdot 4 = 96$\n\nKer vrstni red ni pomemben, so vse rešitve:\n\n- $24$ in $144$\n- $48$ in $120$\n- $72$ in $96$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41744,
    "subject": "Mathematics (Multi-modal)",
    "question": "We write numbers $1, 2, \\dots, 2014$ in some order and denote them with $a_1, a_2, \\dots, a_{2014}$. We then rewrite them in some new order and denote them with $b_1, b_2, \\dots, b_{2014}$. Find the largest positive integer $k$ such that, regardless of the chosen orders, the number $3^k$ will certainly divide the number\n$$\n(a_1^2 - b_1^2)(a_2^2 - b_2^2)\\dots(a_{2014}^2 - b_{2014}^2).\n$$",
    "options": [],
    "answer": "672",
    "solution": "Let's prove that the answer is $n = 672$.\n\nWe can check that the number $a_i^2 - b_i^2$ is divisible by $3$ if and only if the numbers $a_i$ and $b_i$ are either both divisible by $3$ or both co-prime to $3$. Among the numbers $1, 2, \\dots, 2014$ there are exactly $671$ of them divisible by $3$, $672$ of them have remainder $1$, and $671$ of them have remainder $2$ by division with $3$. Thus we can form at most $2 \\cdot 671$ pairs $(a_i, b_i)$ where one of the numbers is divisible by $3$ and the other is not. We are left with at least $672$ pairs for which this does not hold. For each such pair $(a_i, b_i)$ we have $3 \\mid a_i^2 - b_i^2$. Hence $3^{672}$ divides $(a_1^2 - b_1^2)(a_2^2 - b_2^2) \\dots (a_{2014}^2 - b_{2014}^2)$ and we proved $n \\ge 672$.\n\nWe now construct an example where $3^{673}$ does not divide $(a_1^2 - b_1^2)(a_2^2 - b_2^2) \\dots (a_{2014}^2 - b_{2014}^2)$. For $i = 1, 2, \\dots, 671$ let $(a_i, b_i) = (3i, 3i + 2)$, and for $i = 672, 673, \\dots, 2 \\cdot 671$ let $(a_i, b_i) = (b_{i-671}, a_{i-671})$. In all these pairs $3$ does not divide $a_i^2 - b_i^2$. We are left to choose $672$ pairs of numbers where both numbers in the pair have remainder $1$ by division with $3$. Let this pair be $(a_{1343}, b_{1343}) = (1, 4), (a_{1344}, b_{1344}) = (4, 7), \\dots, (a_{2013}, b_{2013}) = (2010, 2013)$ and $(a_{2014}, b_{2014}) = (2013, 1)$. For these pairs $a_i^2 - b_i^2$ is divisible by $3$ but not by $9$. Therefore $(a_1^2 - b_1^2)(a_2^2 - b_2^2) \\dots (a_{2014}^2 - b_{2014}^2)$ is divisible by $3^{672}$ but not by $3^{673}$. Hence $n \\le 672$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41745,
    "subject": "Mathematics (Multi-modal)",
    "question": "At the beginning we have a triple of pairwise distinct positive integers $(a, b, c)$ which satisfies $a + b + c = 2015$. Then in each step we replace the current triple of numbers $(x, y, z)$ with the triple $(y + z - x, z + x - y, x + y - z)$. At least how many steps must we make so that starting with a triple $(a, b, c)$ we will certainly get a triple which has at least one negative number?",
    "options": [],
    "answer": "9",
    "solution": "Let's look at the sum of the triple. This does not change after each step since $(y+z-x)+(z+x-y)+(x+y-z) = x+y+z$. Moreover, the values of the new triple do not depend on the order of the numbers in the original triple, since the terms are symmetrical in all three variables. Hence we may assume $a < b < c$ and we can change the order of the numbers in the triple.\n\n**Solution 1:** After the first step we get the triple $(a + b - c, a + c - b, b + c - a)$ and we have\n$$\na + b - c < a + c - b < b + c - a,\n$$\nafter the second step we get the triple $(3a - b - c, 3b - a - c, 3c - a - b)$ and we have\n$$\n3a - b - c < 3b - a - c < 3c - a - b.\n$$\n\nLet's look at the differences between individual numbers of the triple. By induction we check that after $n$ steps we have the following:\n* The difference between the smallest and the largest number in the triple is equal to $2^n(c-a)$.\n* The difference between the smallest and the middle number in the triple is equal to $2^n(b-a)$ if $n$ is even and $2^n(c-b)$ if $n$ is odd.\n* The difference between the middle and the largest number in the triple is equal to $2^n(c-b)$ if $n$ is even and $2^n(b-a)$ if $n$ is odd.\n\nAfter $n$ steps we have the triple of the form $(t, t + 2^n(b-a), t + 2^n(c-a))$ if $n$ is even and $(t, t + 2^n(c-b), t + 2^n(c-a))$ if $n$ is odd. Here $t$ is some integer which can be calculated from the condition on the sum. We have\n$$\n\\begin{aligned}\n2015 &= t + (t + 2^n(b-a)) + (t + 2^n(c-a)) = 3t + 2^n(b+c-2a) \\\\\n&\\Rightarrow t = \\frac{2015 - 2^n(b+c-2a)}{3}, \\text{ if } n \\text{ even, and}\n\\end{aligned}\n$$\n$$\n\\begin{aligned}\n2015 &= t + (t + 2^n(c-b)) + (t + 2^n(c-a)) = 3t + 2^n(2c-a-b) \\\\\n&\\Rightarrow t = \\frac{2015 - 2^n(2c-a-b)}{3}, \\text{ if } n \\text{ odd.}\n\\end{aligned}\n$$\nIf any of the numbers in the triple will be negative, it will surely be $t$. Since the sum of three consecutive numbers cannot equal $2015$, the difference between the largest and the smallest number in the initial triple is at least $3$. This gives $c-a \\ge 3$. Thus we can estimate $b+c-2a = (c-a)+(b-a) \\ge 4$ and $2c-a-b = (c-a)+(c-b) \\ge 4$. It follows\n$$\nt \\le \\frac{2015 - 4 \\cdot 2^n}{3}.\n$$\n\nWith the initial triple $(670, 672, 673)$ we get $t = \\frac{2015-2^8(673+672-2 \\cdot 670)}{3} = 245$ after $8$ steps. Therefore this triple does not yield a negative number after $8$ steps.\nSo we have to make at least $9$ steps to get a triple with at least one negative number.\n\n\n**Solution 2:** 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 $x \\mapsto 2015-2x$.\nWe can check by induction that if we apply this transformation $n$ times to the number $x$ we get\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$$\n\\begin{aligned} 2^n x + \\frac{1-2^n}{3} \\cdot 2015 &\\ge 0 \\\\ \\Leftrightarrow \\frac{1}{\\frac{2015}{3}-x} \\cdot \\frac{2015}{3} &\\ge 2^n. \\end{aligned}\n$$\n\nFor odd $n$ we must have\n$$\n\\begin{aligned} -2^n x + \\frac{1+2^n}{3} \\cdot 2015 &\\ge 0 \\\\ \\Leftrightarrow \\frac{1}{x - \\frac{2015}{3}} \\cdot \\frac{2015}{3} &\\ge 2^n. \\end{aligned}\n$$\n\nSince $2015$ is not a sum of three consecutive positive integers, we must have $a \\le 670$ and $c \\ge 673$. Inserting this in the above inequalities we get\n$$\n2^n \\le \\frac{1}{\\frac{2015}{3} - a} \\cdot \\frac{2015}{3} \\le \\frac{1}{\\frac{2015}{3} - 670} \\cdot \\frac{2015}{3} = 403 \\Rightarrow n \\le 8,\n$$\n$$\n2^n \\le \\frac{1}{c - \\frac{2015}{3}} \\cdot \\frac{2015}{3} \\le \\frac{1}{673 - \\frac{2015}{3}} \\cdot \\frac{2015}{3} = \\frac{2015}{4} \\Rightarrow n \\le 8.\n$$\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": 41746,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ and $E$ be points on the sides $BC$ and $CA$ of the triangle $ABC$ respectively. The circumcircle of the triangle $CDE$ and the line through $C$, which is parallel to $AB$, intersect again in a point $L$. The line $DL$ intersects the side $AB$ in a point $M$. Denote by $N$ the point on the line $AB$ such that $\\angle NDA = \\angle MEB$ and the point $A$ lies between the points $N$ and $B$. Let $T$ be the intersection of the lines $AD$ and $BE$. Assuming $|DM| = |EN|$, prove that $CT$ is the angle bisector of $\\angle ACB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the intersection of the lines $CT$ and $AB$ by $U$. We want to prove that\n$$\n\\frac{|AU|}{|UB|} = \\frac{|AC|}{|CB|}, \\qquad (2)\n$$\nsince it will follow that $CT$ is the angle bisector of $\\angle ACB$.\n\nSince the points $C, E, D$, and $L$ are concyclic, and the lines $AB$ and $CL$ are parallel, we have $\\angle AED = \\pi - \\angle DEC = \\angle CLD = \\pi - \\angle DMA$. Thus the points $A, M, D$, and $E$ are concyclic.\nMoreover, we have $\\angle EDN = \\angle EDA - \\angle NDA = \\angle EMA - \\angle MEB = \\pi - \\angle BME =$\n\n$\\prec MEB = \\prec EBN$. Hence the points $N, B, D$, and $E$ are also concyclic.\nThis gives $\\prec NED = \\pi - \\prec DBN = \\pi - \\prec DCL = \\pi - \\prec DEL$. Hence the points $N, E$, and $L$ are collinear.\n\nSince the lines $AB$ and $CL$ are parallel the triangles $MBD$ and $LCD$ have the same angles. Therefore they are similar and we have\n$$\n\\frac{|MD|}{|BD|} = \\frac{|DL|}{|DC|}. \\qquad (3)\n$$\nFor the same reason the triangles $NAE$ and $LCE$ also have the same angles. Therefore they are similar and we have\n$$\n\\frac{|NE|}{|AE|} = \\frac{|EL|}{|EC|}. \\qquad (4)\n$$\nSince the lines $AD, BE$, and $CU$ intersect in the same point the Ceva's theorem gives\n$$\n\\frac{|AU|}{|UB|} \\frac{|BD|}{|DC|} \\frac{|CE|}{|EA|} = 1.\n$$\nWe also have $\\prec ACB = \\prec ECD = \\prec ELD$ and $\\prec DEL = \\prec DCL = \\prec CBA$. Therefore the triangles $ABC$ and $DEL$ are similar and we get\n$$\n\\frac{|LE|}{|DL|} = \\frac{|CB|}{|AC|}.\n$$\nInserting (3) and (4) into the Ceva's theorem equation we get\n$$\n\\frac{|AU|}{|UB|} \\frac{|LE|}{|DL|} = 1. \\qquad (5)\n$$\nInserting this into (5) we get exactly (2) which we set out to prove. Therefore the line $CT$ is indeed the angle bisector of $\\prec ACB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41747,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNaj bo $x \\in \\mathbb{R}$. Izračunaj vrednost izraza\n$$\n\\frac{2^{2014} + 2^{2015} - 2^{2016} - 2^{2017}}{(-3)^{2014} - (-3)^{2016}} \\cdot \\left(\\frac{3}{2}\\right)^{2013} \\cdot \\frac{4^{2x+2}}{(-2)^{4x}}\n$$\nRezultat naj bo točen.",
    "options": [],
    "answer": "12",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41748,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $P$ razpolovišče stranice $AB$ trikotnika $ABC$. Zrcalna slika poltraka $PC$ pri zrcaljenju čez premico $AB$ seka trikotniku $ABC$ očrtano krožnico v točki $D$. Naj bo $E$ drugo presečišče premice $CP$ s trikotniku $ABC$ očrtano krožnico. Dokaži, da je $|AE| = |BD|$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n1. način. Naj bo $\\mathcal{K}$ trikotniku $ABC$ očrtana krožnica. Ker je poltrak $PD$ zrcalna slika poltraka $PC$ pri zrcaljenju čez premico $AB$, velja $\\Varangle DPB = \\Varangle BPC = \\Varangle APE$. Oglejmo si zrcaljenje preko simetrale stranice $AB$. Točka $A$ se prezrcali v točko $B$. Ker je $P$ središče stranice $AB$, se pri zrcaljenju ohrani. Iz zgornje enakosti kotov zato sledi, da se poltrak $PE$ prezrcali v poltrak $PD$. Ker simetrala stranice $AB$ poteka skozi središče krožnice $\\mathcal{K}$, se krožnica $\\mathcal{K}$ prezrcali sama vase. Torej se presečišče poltraka $PE$ in krožnice $\\mathcal{K}$, tj. točka $E$, prezrcali v presečišče poltraka $PD$ in krožnice $\\mathcal{K}$, tj. točko $D$. Pokazali smo, da se daljica $AE$ prezrcali v daljico $BD$. Ker zrcaljenje ohranja razdalje, je $|AE| = |BD|$.\n\n\n2. način. Naj bo $\\mathcal{K}$ trikotniku $ABC$ očrtana krožnica. Ker je poltrak $PD$ zrcalna slika poltraka $PC$ pri zrcaljenju čez premico $AB$, velja $\\Varangle DPB = \\Varangle BPC = \\Varangle APE$. Velja tudi $|AP| = |PB|$. Pokazati želimo še $|PE| = |PD|$. Ker točka $P$ leži na simetrali $AB$ in gre simetrala skozi središče krožnice $\\mathcal{K}$ (označimo ga z $O$), velja: $|OE| = |OD|$, $\\Varangle APO = \\Varangle BPO$ in $\\Varangle EPA = \\Varangle DPB$. Ker imata trikotnika $EPO$ in $DPO$ še skupno stranico $OP$, sta skladna in zato je tudi $|PE| = |PD|$. Torej sta si trikotnika $AEP$ in $BDP$ skladna in je $|AE| = |BD|$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41749,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoloči parameter $a \\in \\mathbb{R}$ tako, da bo graf funkcije $f(x)=\\frac{3-2a}{a+5} x+\\frac{2a-1}{3-a}$ sekal ordinatno os nad koordinatnim izhodiščem in da bo funkcija $f$ padajoča.",
    "options": [],
    "answer": "(3/2, 3)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41750,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIz črk besede LJUBEZNIVOST sestavljamo besede dolžine 4. Črke se ne smejo ponavljati.\n\na) Koliko različnih besed lahko sestavimo?\n\nb) Koliko različnih besed lahko sestavimo, če uporabimo le samoglasnike?\n\nc) Izračunaj verjetnost, da iz naključno izbranih črk sestavimo besedo NEBO.\n\nd) Izračunaj verjetnost, da iz naključno izbranih črk sestavimo besedo, ki se začne in konča s soglasnikom.\n\n(8 točk)",
    "options": [],
    "answer": "a) 11880; b) 24; c) 1/11880; d) 14/33",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41751,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bosta $D$ in $E$ zaporedoma razpolovišči stranic $B C$ in $C A$ trikotnika $A B C$. Premici $A D$ in $B E$ sekata trikotniku $A B C$ očrtano krožnico zaporedoma še v točkah $P$ in $Q$. Denimo, da je $|D P|=|E Q|$. Dokaži, da je trikotnik $A B C$ enakokrak z vrhom $C$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n1. način. Ker sta $D$ in $E$ razpolovišči daljic $B C$ in $A C$, sta premici $D E$ in $A B$ vzporedni. Od tod sledi $\\angle E D A = \\angle B A D$, po izreku o obodnih kotih pa velja $\\angle B A D = \\angle B A P = \\angle B Q P$. Torej je $\\angle E Q P = \\angle E D A = \\pi - \\angle P D E$, kar pomeni, da je štirikotnik $E D P Q$ tetiven, trikotnika $T E D$ in $T P Q$ pa sta si podobna. Naj bo $T$ težišče trikotnika $A B C$ in $x = |D P| = |E Q|$. Iz podobnosti trikotnikov $T E D$ in $T P Q$ izpeljemo\n$$\n\\frac{|T D| + x}{|T E|} = \\frac{|T P|}{|T E|} = \\frac{|T Q|}{|T D|} = \\frac{|T E| + x}{|T D|}\n$$\nTo enakost lahko preoblikujemo do\n$$\n(|T D| - |T E|)(|T D| + |T E| + x) = 0\n$$\nod koder sledi $|T E| = |T D|$, saj je $|T D| + |T E| + x > 0$. Trikotnik $D E T$ je torej enakokrak z vrhom pri $T$. Zaradi vzporednosti premic $A D$ in $B E$, je tudi trikotnik $A B T$ enakokrak z vrhom pri $T$, torej velja $|B E| = |B T| + |T E| = |A T| + |T D| = |A D|$. Od tod sledi, da sta trikotnika $A B E$ in $B A D$ skladna, saj imata dva para enako dolgih stranic in enak kot med njima $\\angle E B A = \\angle B A D$. Torej je $|A E| = |B D|$ in zato $|A C| = |B C|$.\n\n\n2. način. Kot v prvi rešitvi dokažemo, da sta premici $D E$ in $A B$ vzporedni in da je štirikotnik $E D P Q$ tetiven. Zato je $\\angle E P D = \\angle E Q D$. Ker sta tetivi $D P$ in $E Q$ enako dolgi, pa je $\\angle D E P = \\angle Q D E$. Zato je $\\angle P D E = \\angle D E Q$, kar pomeni, da je štirikotnik $E D P Q$ enakokrak trapez. Torej je $\\angle T E D = \\angle E D T = \\angle B A T$ oziroma $\\angle B E D = \\angle B A D$. To pomeni, da je štirikotnik $A B D E$ tetiven. Ker pa sta premici $A B$ in $E D$ vzporedni, je tudi štirikotnik $A B D E$ enakokrak trapez. Sledi $|A E| = |B D|$ in zato $|A C| = |B C|$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41752,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ and $E$ be the midpoints of sides $BC$ and $CA$ of the triangle $ABC$ respectively. The lines $AD$ and $BE$ intersect the circumcircle of the triangle $ABC$ additionally in points $P$ and $Q$ respectively. Suppose that $|DP| = |EQ|$. Prove that the triangle $ABC$ is isosceles with apex $C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nSolution:\n\nSince $D$ 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\n\n$\\angle EQP = \\angle EDA = \\pi - \\angle PDE$ which means that the quadrilateral $EDPQ$ is cyclic, and the triangles $TED$ and $TPQ$ are similar. Let $T$ be the centroid of the triangle $ABC$ and $x = |DP| = |EQ|$. From the similarity of the triangles $TED$ and $TPQ$ we deduce\n$$\n\\frac{|TD| + x}{|TE|} = \\frac{|TP|}{|TE|} = \\frac{|TQ|}{|TD|} = \\frac{|TE| + x}{|TD|}\n$$\nWe rearrange this to\n$$\n(|TD| - |TE|)(|TD| + |TE| + x) = 0,\n$$\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 $|AC| = |BC|$.\nAs in the first solution we prove that the lines $DE$ and $AB$ are parallel and that the quadrilateral $EDPQ$ is cyclic. Therefore $\\angle EPD = \\angle EQD$. Since the chords $DP$ and $EQ$ are of equal length we have $\\angle DEP = \\angle QDE$. Hence $\\angle PDE = \\angle DEQ$ which means that the quadrilateral $EDPQ$ is an isosceles trapezoid. Thus $\\angle TED = \\angle EDT = \\angle BAT$ and so $\\angle BED = \\angle BAD$. This means that the quadrilateral $ABDE$ is cyclic. Since the lines $AB$ and $ED$ are parallel the quadrilateral $ABDE$ is also isosceles trapezoid. It follows $|AE| = |BD|$ and hence $|AC| = |BC|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41753,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral which satisfies $\\angle BAC = \\angle ACB = 20^\\circ$, $\\angle DCA = 30^\\circ$ and $\\angle CAD = 40^\\circ$. Determine the size of the angle $\\angle CBD$.",
    "options": [],
    "answer": "80°",
    "solution": "![](attached_image_1.png)\n\n**Solution 1.** Notice that the triangle $ABC$ is isosceles with apex at vertex $B$. Let $E$ be such a point on the line $CD$ that $\\angle CAE = 30^\\circ$, so that the triangle $ACE$ is also isosceles with apex at vertex $E$. Since $\\angle CAD = 40^\\circ$ the point $E$ lies between $C$ and $D$. Thus the quadrilateral $ABCE$ is a deltoid and its diagonals $AC$ and $BE$ intersect at a right-angle. Therefore we have $\\angle BEC = 90^\\circ - 30^\\circ = 60^\\circ$ and hence $\\angle BAD = 60^\\circ = \\angle BEC = 180^\\circ - \\angle DEB$. This means that the points $A, B, E$, and $D$ are concyclic. By Angles Subtended by Same Arc Theorem we have $\\angle EBD = \\angle EAD = 40^\\circ - 30^\\circ = 10^\\circ$, hence $\\angle CBD = \\angle CBE + \\angle EBD = (90^\\circ - 20^\\circ) + 10^\\circ = 80^\\circ$.\n\n\n**Solution 2.** Notice that the triangle $ABC$ is isosceles with apex at vertex $B$. Let $\\mathcal{K}$ be the circle with center $B$ which passes through the points $A$ and $C$. Let's prove that the point $D$ also lies on the circle $\\mathcal{K}$. From the data in the problem we calculate $\\angle ADC = 180^\\circ - 40^\\circ - 30^\\circ = 110^\\circ$ and $\\angle CBA = 180^\\circ - 20^\\circ - 20^\\circ = 140^\\circ$. Therefore the central angle subtended by the arc $\\widehat{AC}$ equals $\\angle ABC = 360^\\circ - \\angle CBA = 220^\\circ$ and is double the angle $\\angle ADC$. By the Angle at the Center Theorem the angle $\\angle ADC$ is the inscribed angle subtended by the arc $\\widehat{AC}$, which means that the point $D$ lies on the circle $\\mathcal{K}$. From this it follows that the triangle $CBD$ is isosceles with apex at vertex $B$ and hence $\\angle CBD = 180^\\circ - 2\\angle DCB = 180^\\circ - 100^\\circ = 80^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41754,
    "subject": "Mathematics (Multi-modal)",
    "question": "A square $ABCD$ and two points $E$ and $F$ outside of this square are given so that the triangles $BEC$ and $CFD$ are equilateral. Prove that the triangle $AEF$ is also equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $ABCD$ is a square and the triangles $BEC$ and $CFD$ are equilateral we have $|EB| = |BA| = |AD| = |DF|$. Therefore the triangles $EBA$ and $ADF$ are isosceles with apexes at vertexes $B$ and $D$, and they have legs of the same length. We also have $\\angle EBA = \\angle ADF = 90^\\circ + 60^\\circ = 150^\\circ$, hence this two triangles are congruent. It follows $|AE| = |AF|$. We also have $\\angle FAD = \\angle BAE = \\frac{1}{2}(180^\\circ - \\angle EBA) = 15^\\circ$, hence $\\angle EAF = 90^\\circ - \\angle BAE - \\angle FAD = 60^\\circ$. Thus the triangle $EAF$ is isosceles with apex at vertex $A$ and apex angle of $60^\\circ$, which meant it is in fact equilateral.\n\n![](attached_image_1.png)\nSince $ABCD$ is a square and the triangles $BEC$ and $CFD$ are equilateral we have $|EB| = |BA| = |AD| = |DF| = |CE| = |CF|$. Thus the triangles $EBA$, $ADF$, and $ECF$ are isosceles. We have $\\angle EBA = \\angle ADF = 90^\\circ + 60^\\circ = 150^\\circ$ and $\\angle ECF = 360^\\circ - 90^\\circ - 2 \\cdot 60^\\circ = 150^\\circ$. Therefore the triangles $EBA$, $ADF$, and $ECF$ have two sides and the angle between them congruent and are hence congruent themselves. It follows that $|EA| = |AF| = |EF|$ and thus the triangle $AEF$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41755,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be the midpoint of the side $AB$ of the triangle $ABC$. The mirror image of the ray $PC$ by reflection across the line $AB$ intersects the circumcircle of the triangle $ABC$ in the point $D$. Let $E$ be the other intersection of the line $CP$ with the circumcircle of the triangle $ABC$. Prove that $|AE| = |BD|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\mathcal{K}$ be the circumcircle of the triangle $ABC$. Since the ray $PD$ is the mirror image of the ray $PC$ by reflection across the line $AB$ we have $\\angle DPB = \\angle BPC = \\angle APE$. Let's look at the reflection across the bisector of the side $AB$. The point $A$ is reflected to the point $B$. Since $P$ is the midpoint of the side $AB$ it is the fixed point of the reflection. From the above equality of angles it thus follows that the ray $PE$ is reflected to the ray $PD$. Since the bisector of the side $AB$ contains the center of the circle $\\mathcal{K}$, this circle is invariant under reflection. Therefore the point $E$, which is the intersection of the ray $PE$ and the circle $\\mathcal{K}$, is reflected to the point $D$, which is the intersection of the ray $PD$ and the circle $\\mathcal{K}$. We thus proved that the segment $AE$ is reflected to the segment $BD$. Since reflection preserves distances we have $|AE| = |BD|$.\nLet $\\mathcal{K}$ be the circumcircle of the triangle $ABC$. Since the ray $PD$ is the mirror image of the ray $PC$ by reflection across the line $AB$ we have $\\angle DPB = \\angle BPC = \\angle APE$. We also have $|AP| = |PB|$. We want to prove that $|PE| = |PD|$ also holds. Since the point $P$ lies on the bisector of the side $AB$ which contains the center of the circle $\\mathcal{K}$, denote it by $O$, we have $|OE| = |OD|$, $\\angle APO = \\angle BPO$, and $\\angle EPA = \\angle DPB$. Since the triangles $EPO$ and $DPO$ also have a common side $OP$ they are congruent and we have $|PE| = |PD|$. Thus the triangles $AEP$ and $BDP$, and we have $|AE| = |BD|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41756,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je kvadrat $A B C D$ in taki točki $E$ in $F$ izven kvadrata, da sta trikotnika $B E C$ in $C F D$ enakostranična. Dokaži, da je tudi trikotnik $A E F$ enakostraničen.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n1. način. Ker je $A B C D$ kvadrat in sta trikotnika $B E C$ in $C F D$ enakostranična, je $|E B|=|B A|=|A D|=|D F|$. Torej sta trikotnika $E B A$ in $A D F$ enakokraka z vrhoma pri $B$ in $D$ in imata enako dolge krake. Hkrati velja $\\Varangle E B A=\\Varangle A D F=90^{\\circ}+60^{\\circ}=150^{\\circ}$, zato sta omenjena trikotnika skladna. Sledi $|A E|=|A F|$. Poleg tega velja $\\Varangle F A D=\\Varangle B A E=\\frac{1}{2}\\left(180^{\\circ}-\\Varangle E B A\\right)=15^{\\circ}$, zato je $\\Varangle E A F=90^{\\circ}-\\Varangle B A E-\\Varangle F A D=60^{\\circ}$. Torej je trikotnik $E A F$ enakokrak s kotom $60^{\\circ}$ pri vrhu $A$, od koder sledi, da je enakostraničen.\n\n\n2. način. Ker je $A B C D$ kvadrat in sta trikotnika $B E C$ in $C F D$ enakostranična, je $|E B|=|B A|=|A D|=|D F|=|C E|=|C F|$. Trikotniki $E B A, A D F$ in $E C F$ so torej enakokraki. Velja $\\Varangle E B A=\\Varangle A D F=90^{\\circ}+60^{\\circ}=150^{\\circ}$ ter $\\Varangle E C F=360^{\\circ}-90^{\\circ}-2 \\cdot 60^{\\circ}=150^{\\circ}$. Trikotniki $E B A, A D F$ in $E C F$ se torej ujemajo v dveh stranicah in kotu med njima, zato so skladni. Sledi $|E A|=|A F|=|E F|$, zato je trikotnik $A E F$ enakostraničen.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41757,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples of positive integers $(a, b, p)$ such that $p$ is a prime and\n$$\n\\frac{5^a + 2^b p}{5^a - 2^b p}\n$$\nis a perfect square of a positive integer.",
    "options": [],
    "answer": "(1, 1, 2), (2, 3, 3), (2, 2, 5)",
    "solution": "Let $\\frac{5^a + 2^b p}{5^a - 2^b p} = n^2$. Since $5^a - 2^b p$ is odd, we also have $5^a + 2^b p$ is even. Thus $5^a - 2^b p$ and $5^a + 2^b p$ are both positive integers. However, the number $5^a - 2^b p$ is odd, hence $5^a - 2^b p$ is odd. Thus we can write $5^a - 2^b p = 2 \\cdot 5^a$. However, the number $5^a - 2^b p$ is odd, hence $5^a - 2^b p$ is odd. We consider two cases:\n* If $m \\ge 1$, we have $2^b p = 5^a - 5^m$, hence $2^b p$ is divisible by $5$. It follows $p=5$ and we get equation $2^b = 5^{a-1} - 5^{m-1}$. Since $2^b$ is not divisible by $5$ the same holds for $5^{m-1}$. Thus the only possibility is $m=1$ and we get equation\n$$\n5^{a-1} - 2^b = 1. \\qquad (1)\n$$\nObviously, $b=1$ is not a solution, and if $b=2$ we get a solution $a=2$. For $b \\ge 3$ view the equation mod $8$. We have $2^b \\equiv 0 \\pmod 8$. Hence $5^{a-1} \\equiv 1 \\pmod 8$, which means that $a-1=2k$ for some $k \\in \\mathbb{N}$. We can rearrange the equation (1) to $(5^k-1)(5^k+1) = 2^b$. Therefore the numbers $5^k-1$ and $5^k+1$ are both powers of $2$. The only powers of $2$ that differ by $2$ are $2^1$ and $2^2$. This would mean that $5^k=3$ which is not possible.\n\nThe only solution of the equation (1) is thus $a=2$ and $b=2$. Inserting this in the original equation we get $n=3$. Thus we get a solution $(2,2,5)$.\n\n$$\n5^a - 2^b = 1.\n$$\nIf $b=1$ we get $5^a - 1 = 2p$. Since $4|5^a - 1$ the only possibility is $p=2$ and $a=1$. Inserting this in the original equation we get $n=3$. Thus $(1,1,2)$ is a solution.\nIf $b = 2$ we get $5^a - 1 = 4p$. From the original equation then follows $n^2 = 8p+1$ which gives $(n+1)(n-1) = 8p$. The number $n$ must be odd and bigger than $1$, hence we write $n = 2s + 1$. We get $(s+1)s = 2p$. The only solution of this equation is $s = 2$ and $p = 3$. However, this would mean that $5^a = 13$, which has no solutions in positive integers.\nWe are left with the case $b \\ge 3$. In this case we have $2^b p \\equiv 0 \\pmod 8$ and hence $5^a \\equiv 1 \\pmod 8$. Thus $a = 2k$. However, we then get $5^{2k} \\equiv 1 \\pmod 3$, hence $2^b p \\equiv 0 \\pmod 3$ must hold, which implies $p = 3$. It follows that $(5^k - 1)(5^k + 1) = 3 \\cdot 2^b$. However $5^k + 1 \\equiv 2 \\pmod 4$, hence $5^k + 1$ must have some odd divisor. The only possibility is thus $5^k + 1 = 2 \\cdot 3$. This gives $a = 2$ and $b = 3$. Inserting these values in the original equation we get $n = 7$. Thus $(2, 3, 3)$ is a solution.\n\nTherefore the solutions are $(1, 1, 2)$, $(2, 3, 3)$, and $(2, 2, 5)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41758,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $ABCD$ štirikotnik, za katerega velja $\\Varangle BAC = \\Varangle ACB = 20^\\circ$, $\\Varangle DCA = 30^\\circ$ in $\\Varangle CAD = 40^\\circ$. Določi velikost kota $\\Varangle CBD$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "80°",
    "solution": "Solution:\n\n1. način. Opazimo da je trikotnik $ABC$ enakokrak z vrhom pri $B$. Naj bo $E$ taka točka na premici $CD$, da je $\\Varangle CAE = 30^\\circ$, tako da bo tudi trikotnik $ACE$ enakokrak z vrhom pri $E$. Ker je $\\Varangle CAD = 40^\\circ$, točka $E$ leži med $C$ in $D$. Potem je štirikotnik $ABCE$ deltoid, zato se njegovi diagonali $AC$ in $BE$ sekata pod pravim kotom. Torej velja $\\Varangle BEC = 90^\\circ - 30^\\circ = 60^\\circ$ in zato $\\Varangle BAD = 60^\\circ = \\Varangle BEC = 180^\\circ - \\Varangle DEB$. To pomeni, da so točke $A, B, E$ in $D$ konciklične. Po izreku o obodnih kotih velja $\\Varangle EBD = \\Varangle EAD = 40^\\circ - 30^\\circ = 10^\\circ$, zato je $\\Varangle CBD = \\Varangle CBE + \\Varangle EBD = (90^\\circ - 20^\\circ) + 10^\\circ = 80^\\circ$.\n\n![](attached_image_2.png)\n\n2. način. Opazimo da je trikotnik $ABC$ enakokrak z vrhom pri $B$. Naj bo $\\mathcal{K}$ krožnica s središčem $B$, ki gre skozi točki $A$ in $C$. Dokažimo, da tudi točka $D$ leži na krožnici $\\mathcal{K}$. Iz podatkov naloge izračunamo $\\Varangle ADC = 180^\\circ - 40^\\circ - 30^\\circ = 110^\\circ$ in $\\Varangle CBA = 180^\\circ - 20^\\circ - 20^\\circ = 140^\\circ$. Torej je središčni kot nad lokom $\\widehat{AC}$ enak $\\Varangle ABC = 360^\\circ - \\Varangle CBA = 220^\\circ$ in je dvakratnik kota $\\Varangle ADC$. Po izreku o središčnem in obodnem kotu je torej $\\Varangle ADC$ obodni kot nad lokom $\\widehat{AC}$, kar pomeni, da točka $D$ leži na krožnici $\\mathcal{K}$. Od tod sledi, da je trikotnik $CBD$ enakokrak z vrhom pri $B$, zato je $\\Varangle CBD = 180^\\circ - 2 \\Varangle DCB = 180^\\circ - 100^\\circ = 80^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41759,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nReši enačbo $\\frac{4^{\\log_{3} x}}{6} = \\frac{2^{\\log_{3}(x+1)}}{3}$.",
    "options": [],
    "answer": "(3 + sqrt(21)) / 2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41760,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana je pravokotna mreža velikosti $7 \\times 9$ (glej sliko). V spodnjem levem vozlišču mreže je kolonija mravelj, v zgornjem desnem vozlišču pa je njihovo mravljišče. V vseh ostalih vozliččih mreže je po eno zrno riža. Vsaka mravlja iz kolonije se na poti do mravljišča sprehaja po povezavah mreže, vendar le v smereh desno ali navzgor. Na svoji poti pobere vsa zrna riža, na katera naleti, in ko pride do mravljišča, tam tudi ostane. Najmanj koliko mravelj bi moralo biti v koloniji, da bi lahko pobrale vsa zrna riža?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nKolonija mora vsebovati najmanj 8 mravelj.\n\nDokažimo najprej, da 8 mravelj lahko pobere vsa zrna riža. To lahko storijo na primer tako, da se $i$-ta od teh 8 mravelj sprehodi najprej navpično navzgor do $i$-te vrstice mreže, nato po $i$-ti vrstici v desno do skrajnega desnega roba mreže in nazadnje navzgor do mravljišča. Tako teh 8 mravelj pobere vsa zrna riža, saj $i$-ta mravlja izprazni celotno $i$-to vrstico mreže.\n\nSedaj pa dokažimo, da 7 mravelj ne more pobrati vseh zrn riža. Oglejmo si zrna riža, ki ležijo na označenih vozliščih mreže.\n\n![](attached_image_2.png)\n\nKer se mravlje lahko sprehajajo le v desno ali navzgor, nobena mravlja ne more pobrati dveh izmed teh zrn riža. Torej 7 mravelj ne more pobrati vseh teh 8 zrn riža.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41761,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV podjetju, ki ga vodi več direktorjev, imajo sef, ki je zaklenjen s šestimi ključavnicami. Vsak direktor ima tri ključe, s katerimi lahko odklene tri različne ključavnice. Z vsakim ključem lahko odklene natanko eno ključavnico.\n\nNobena dva direktorja ne moreta odkleniti istih treh ključavnic in nobena dva direktorja skupaj ne moreta odpreti sefa. Največ koliko direktorjev vodi to podjetje?",
    "options": [],
    "answer": "10",
    "solution": "Solution:\n\n1. način. Podjetje vodi največ 10 direktorjev.\n\nOznačimo ključavnice oziroma pripadajoče ključe s številkami od 1 do 6. Vsak direktor ima set 3 ključev od 3 različnih ključavnic in nobena dva direktorja nimata enakih setov. Zato najprej preštejmo, koliko različnih setov 3 ključev obstaja. Za posamezen set ključev imamo na izbiro 6 ključavnic, med katerimi moramo izbrati 3. Gre torej za kombinacije, pri katerih izmed 6 elementov izberemo 3. Takih kombinacij je $\\binom{6}{3}=20$ (preštejemo jih lahko tudi tako, da si izpišemo vse možnosti). Vsakemu setu 3 ključev pripada komplementarni set 3 ključev, to je set, v katerem so ključi preostalih 3 ključavnic. Tako na primer setu ključev $\\{2,3,5\\}$ pripada komplementarni set ključev $\\{1,4,6\\}$. Množico vseh različnih setov ključev razbijemo na 10 parov komplementarnih si setov ključev. Ker vsak tak par ključev skupaj odklene sef, nobena dva direktorja ne moreta imeti komplementarnih si setov ključev. Torej je direktorjev lahko največ 10, saj iz vsakega para komplementarnih si setov ključev kvečjemu en set pripada nekemu direktorju.\n\nPokažimo, da 10 direktorjev res lahko vodi podjetje, torej da obstaja 10 setov ključev, ki ustrezajo pogojem naloge. Vse kar moramo storiti je, da iz vsakega para komplementarnih si setov ključev izberemo po en set (na primer tistega, ki vsebuje ključ 1). Tako ima lahko 10 direktorjev na primer naslednje sete ključev: $\\{1,2,3\\},\\{1,2,4\\},\\{1,2,5\\},\\{1,2,6\\},\\{1,3,4\\}$, $\\{1,3,5\\},\\{1,3,6\\},\\{1,4,5\\},\\{1,4,6\\},\\{1,5,6\\}$.\n\n\n2. način. Označimo ključavnice oziroma pripadajoče ključe s številkami od 1 do 6. Zapišimo vse možne različne kombinacije treh ključev in sicer v parih:\n- $\\{1,2,3\\}$ in $\\{4,5,6\\}$,\n- $\\{1,2,4\\}$ in $\\{3,5,6\\}$,\n- $\\{1,2,5\\}$ in $\\{3,4,6\\}$,\n- $\\{3,4,5\\}$ in $\\{1,2,6\\}$,\n- $\\{1,3,4\\}$ in $\\{2,5,6\\}$,\n- $\\{1,3,5\\}$ in $\\{2,4,6\\}$,\n- $\\{2,4,5\\}$ in $\\{1,3,6\\}$,\n- $\\{1,4,5\\}$ in $\\{2,3,6\\}$,\n- $\\{2,3,5\\}$ in $\\{1,4,6\\}$,\n- $\\{2,3,4\\}$ in $\\{1,5,6\\}$.\n\nOpazimo, da vsak od teh parov kombinacij ključev vsebuje vseh 6 ključev, torej kvečjemu ena kombinacija iz vsakega para pripada nekemu direktorju. Ker je teh parov 10, podjetje vodi največ 10 direktorjev.\n\nVidimo tudi, da 10 direktorjev res lahko vodi podjetje, saj imajo lahko ti direktorji na primer prve kombinacije ključev iz vsakega para. Nobeni dve kombinaciji iz tega izbora ne vsebujeta vseh šestih ključev, saj pri vsaki od njih manjka ključ 6.\n\n\n3. način. Naj predstavlja $\\mathcal{K}=\\{1,2,3,4,5,6\\}$ množico vseh ključev. Vsak direktor ima 3 različne ključe, torej neko podmnožico množice $\\mathcal{K}$, ki ima natanko 3 elemente. Označimo vse take podmnožice s $\\binom{\\mathcal{K}}{3}$. Ker nobena dva direktorja nimata enakega kompleta ključev, vsak element $\\binom{\\mathcal{K}}{3}$ predstavlja set ključev enega potencialnega direktorja. Naj bosta $A, B \\subset \\binom{\\mathcal{K}}{3}$ kompleta ključev dveh različnih direktorjev. Če bi veljalo $A \\cup B=\\mathcal{K}$, bi lahko ta dva direktorja skupaj odprla sef. Naj bo $G$ graf, kjer je\n$$\nV(G)=\\binom{\\mathcal{K}}{3} \\quad \\text{ in } \\quad \\{A, B\\} \\in E(G) \\Longleftrightarrow A \\cup B=\\mathcal{K}\n$$\nVsaka neodvisna množica tega grafa ustreza neki skupini direktorjev, ki zadošča pogojem naloge. Nas zanima največja neodvisna množica. Naj $\\alpha(G)$ označuje moč največje neodvisne množice v grafu $G$. Graf $G$ je prikazan na spodnji sliki (vozlišča smo zaradi preglednosti označili kar s števili, katerih števke so elementi množic, ki ustrezajo vozliščem grafa):\n\n| 123 | 124 | 125 | 126 | 134 | 135 | 136 | 145 | 146 | 156 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 0 | 0 | 9 | 9 | 9 | 0 | 9 | 0 | 9 | 0 |\n| 456 | 356 | 346 | 345 | 256 | 246 | 245 | 236 | 235 | 234 |\n\nČe je $G=G_{1}+G_{2}$, tj. disjunkna unija dveh grafov, potem velja $\\alpha(G)=\\alpha\\left(G_{1}\\right)+\\alpha\\left(G_{2}\\right)$. Ker je $\\alpha\\left(K_{2}\\right)=1$, je $\\alpha(G)=10$, saj je $G \\cong \\underbrace{K_{2}+K_{2}+\\cdots+K_{2}}_{10}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41762,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a company led by several directors they have a safe which is locked by six locks. Each director has three keys which unlock three different locks. With each key he can unlock exactly one lock. No two directors can unlock the same three locks and no two directors together can unlock the safe. At most how many directors is leading this company?",
    "options": [],
    "answer": "10",
    "solution": "The company is led by at most **10** directors.\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). To 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 to $10$ pairs of complementary sets of keys. Since every such pair of keys unlocks the safe 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\nLets 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: $\\{1, 2, 3\\}$, $\\{1, 2, 4\\}$, $\\{1, 2, 5\\}$, $\\{1, 2, 6\\}$, $\\{1, 3, 4\\}$, $\\{1, 3, 5\\}$, $\\{1, 3, 6\\}$, $\\{1, 4, 5\\}$, $\\{1, 4, 6\\}$, $\\{1, 5, 6\\}$.\nDenote the locks and the corresponding keys with numbers $1$ through $6$. Let's write down all possible different combinations of three keys in pairs:\n\n* $\\{1, 2, 3\\}$ and $\\{4, 5, 6\\}$,\n* $\\{1, 2, 4\\}$ and $\\{3, 5, 6\\}$,\n* $\\{1, 2, 5\\}$ and $\\{3, 4, 6\\}$,\n* $\\{3, 4, 5\\}$ and $\\{1, 2, 6\\}$,\n* $\\{1, 3, 4\\}$ and $\\{2, 5, 6\\}$,\n* $\\{1, 3, 5\\}$ and $\\{2, 4, 6\\}$,\n* $\\{2, 4, 5\\}$ and $\\{1, 3, 6\\}$,\n* $\\{1, 4, 5\\}$ and $\\{2, 3, 6\\}$,\n* $\\{2, 3, 5\\}$ and $\\{1, 4, 6\\}$,\n* $\\{2, 3, 4\\}$ and $\\{1, 5, 6\\}$.\n\nNotice that each of these pairs of sets of keys contains all $6$ keys so at most one set of keys from each pair belongs to some director. Since there are $10$ of these pairs the company is led by at most $10$ directors.\nLet $\\mathcal{K} = \\{1, 2, 3, 4, 5, 6\\}$ represent the set of all keys. Each director has $3$ different keys, which is a subset of $\\mathcal{K}$ with $3$ elements. Denote the set of all such subsets by $\\binom{\\mathcal{K}}{3}$. Since no two directors have the same set of keys each element of $\\binom{\\mathcal{K}}{3}$ represents a set of keys of one potential director. Let $A, B \\subset \\binom{\\mathcal{K}}{3}$ be the sets of keys of two different directors. If we had $A \\cup B = \\mathcal{K}$ these two directors together would be able to open the safe. Let $G$ be the graph with\n$$\nV(G) = \\binom{\\mathcal{K}}{3} \\quad \\text{and} \\quad \\{A, B\\} \\in E(G) \\iff A \\cup B = \\mathcal{K}.\n$$\nEvery independent set of this graph corresponds to some group of directors which satisfies the conditions of the problem. We are interested in the biggest independent set. Let $\\alpha(G)$ denote the power of this set. The graph is shown in the picture below (for simplicity we denote each vertex with an integer whose digits are the elements of the set represented by the vertex):\n\n![](attached_image_1.png)\n\nIf $G = G_1 + G_2$, i.e. the disjoint union of two graphs then $\\alpha(G) = \\alpha(G_1) + \\alpha(G_2)$. Since $\\alpha(K_2) = 1$ and $G \\cong K_2 + K_2 + \\dots + K_2$ we have $\\alpha(G) = 10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41763,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKonstruiraj trikotnik $ABC$ s podatki: $a = 4~\\mathrm{cm}$, $\\beta = 75^\\circ$, $t_c = 5~\\mathrm{cm}$. Kot konstruiraj s šestilom in ravnilom.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41764,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLuka se je po ravni cesti zapeljal čez gorski prelaz. Na oddaljenosti $x$ metrov v vodoravni smeri od vznožja prelaza se je nahajal na nadmorski višini $h(x) = -\\frac{x^{2}}{20000} + \\frac{3x}{5} + 560$ metrov (glej sliko).\n\n![](attached_image_1.png)\n\na) Na kateri nadmorski višini se nahaja najvišja točka prelaza?\nb) Izračunaj naklon ceste na minuto natančno v točki, ko je bil Luka od vznožja prelaza oddaljen 4000 metrov v vodoravni smeri.\n\n(8 točk)",
    "options": [],
    "answer": "a) 2360 m; b) 11 degrees 19 minutes",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41765,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV trgovini z oblačili imajo dva tedna akcijsko razprodajo. Prvi teden je kupec pri nakupu treh oblačil dobil najcenejši kos zastonj. Janez je prvi teden kupil jakno, hlače in pulover. Za jakno in hlače je plačal 115,01 evra ter zaradi te akcije prihranil $17,85 \\%$ vrednosti oblačil v redni prodaji. Drugi teden imajo akcijo, ki ponuja popust na vse jakne v višini $20 \\%$. Za isti nakup kot v prvem tednu bi Janez v drugem tednu plačal 125 evrov. Izračunaj ceno puloverja, hlač in jakne v redni prodaji.",
    "options": [],
    "answer": "Sweater: 24.99 euros, Pants: 40.01 euros, Jacket: 75.00 euros",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41766,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV vrsti stoji $n$ luči, $n \\geq 3$, ki so oštevilčene s števili od $1$ do $n$. Na začetku je vsaka liha luč v vrsti prižgana, vsaka soda luč pa ugasnjena. V vsaki potezi lahko hkrati zamenjamo stanje treh zaporednih luči v vrsti (ugasnjene prižgemo, prižgane pa ugasnemo).\n\na. Dokaži, da vrstni red izvajanja potez za končno stanje luči ni pomemben.\n\nb. Za katera števila $n$ lahko v končno mnogo potezah pridemo do stanja, v katerem bo vsaka liha luč v vrsti ugasnjena, vsaka soda luč pa prižgana?",
    "options": [],
    "answer": "Exactly those n that are multiples of 3",
    "solution": "Solution:\n\na.\nOglejmo si, kaj se dogaja s stanjem ene luči pri izvajanju potez. Vsaka poteza stanje luči bodisi zamenja bodisi ohrani. Končno stanje neke luči je zato odvisno le od števila izvedenih potez, ki stanje te luči zamenjajo, ni pa odvisno od vrstnega izvajanja potez. Torej tudi končno stanje vseh luči ni odvisno od vrstnega reda izvajanja potez, ampak le od tega, katere poteze izvedemo.\n\nb.\nTo je mogoče natanko za tiste $n$, ki so deljivi s $3$.\n\nDenimo najprej, da je $n$ deljiv s $3$ in pokažimo, da v tem primeru res lahko pridemo do željenega stanja. Naj bo $P_{i}$ poteza, pri kateri zamenjamo stanja $i$-te, $(i+1)$-ve in $(i+2)$-ge luči v vrsti. Ker je $n$ deljiv s $3$ in vsaka poteza spremeni stanje $3$ luči, lahko v tem primeru z $\\frac{n}{3}$ potezami, to so $P_{1}, P_{4}, P_{7}, \\ldots, P_{n-2}$, spremenimo stanje vseh luči v vrsti in s tem pridemo do željenega stanja.\n\nSedaj pa pokažimo, da je do željenega stanja mogoče priti le v primeru, če je $n$ deljiv s $3$. Ker po točki (a) vrstni red potez ni pomemben, lahko predpostavimo, da poteze $P_{i}$ izvajamo po vrsti, najprej tiste z manjšim $i$. Če potezo $P_{i}$ izvedemo dvakrat, se stanje luči ohrani. Torej lahko nadalje predpostavimo, da vsako potezo $P_{i}$ izvedemo kvečjemu enkrat. Gremo torej po vrsti po potezah $P_{1}, P_{2}, P_{3}, \\ldots$ in se za vsako odločimo, ali jo izvedemo ali ne. Ob koncu mora biti vsaka liha luč ugasnjena, vsaka soda pa prižgana. Ker moramo stanje $1.$ luči spremeniti, spremeni pa ga lahko le poteza $P_{1}$, moramo potezo $P_{1}$ nujno izvesti. Po tej potezi so prve tri luči že v pravem stanju. Ker od preostalih potez stanje $2.$ luči spremeni le poteza $P_{2}$, te poteze ne smemo izvesti. Med potezami $P_{3}, P_{4}, P_{5}, \\ldots$ stanje $3.$ luči spremeni le poteza $P_{3}$, zato tudi te poteze ne smemo izvesti. Sedaj smo pri potezi $P_{4}$ in razmišljamo podobno. Stanje $4.$ luči moramo spremeniti, zato moramo potezo $P_{4}$ izvesti. Po tej potezi so $4.$, $5.$ in $6.$ luč v pravem stanju, zato poteze $P_{5}$ ne smemo izvesti in poteze $P_{6}$ tudi ne. Nadaljujemo v enakem smislu, potezo $P_{7}$ moramo izvesti, potez $P_{8}$ in $P_{9}$ pa ne, in tako dalje. S tem smo ugotovili, da moramo nujno izvesti zaporedje potez $P_{1}, P_{4}, P_{7}, \\ldots$ Da bo to zaporedje spremenilo stanje vseh luči v vrsti, mora biti $n$ deljiv s $3$. Če ima namreč $n$ ostanek $1$ pri deljenju s $3$, potem stanja zadnje luči ne bomo spremenili, saj poteza $P_{n}$ ni dopustna, ker bi spremenila stanje le ene same luči. Podobno, če ima $n$ ostanek $2$ pri deljenju s $3$, potem stanja zadnjih dveh luči ne bomo spremenili, saj potezi $P_{n-1}$ in $P_{n}$ nista dopustni.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41767,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ lights in a line, $n \\ge 3$, numbered with numbers 1 through $n$. At the beginning each odd light in the line is turned on and each even light is turned off. In each move we may simultaneously change the state of three consecutive lights (turn on or turn off).\n\na. Prove that the order of the moves we make is not important for the final state of the lights.\n\nb. For which numbers $n$ can we in a finite number of moves reach a state when each odd light in the line is turned off and each even light is turned on?",
    "options": [],
    "answer": "Exactly those n that are divisible by 3",
    "solution": "a. Let's look at what happens with the state of one light when executing a move. Each move either changes or preserves the state of the light. Thus the final state of some light only depends on the number of moves made that change the state of that light, and not on the order of the moves made. Therefore the final state of all the lights also does not depend on the order of the moves made.\n\nb. This is possible for exactly those $n$ which are divisible by 3.\nSuppose first that $n$ is divisible by 3 and let's prove that in this case we can indeed reach the desired state. Let $P_i$ be the move where we change the states of the $i$-th, $(i+1)$-st, and $(i+2)$-nd light in the line. Since $n$ is divisible by 3 and each move changes the state of 3 lights we can change the states of all the lights in the row by making $\\frac{n}{3}$ moves $P_1, P_4, P_7, \\dots, P_{n-2}$, and thus reach the desired state.\n\nNow let's prove that we can reach the desired state only if $n$ is divisible by 3. We already know that the order of the moves is not important, so we may assume that we make moves $P_i$ in the increasing order determined by $i$. If we make the move $P_i$ twice the state of the light is preserved. Therefore we may also assume that we make each move $P_i$ at most one time. This basically means that we only need to decide for each move $P_i$ if we make it or not. At the end each odd light must be turned off and each even light turned on. Since we need to change the state of the 1st light, which can only be done by the move $P_1$, we must make the move $P_1$. After this move the first three lights are already in the desired state. Since $P_2$ is the only other move that changes the state of the 2nd light we must not make this move. Among the moves $P_3, P_4, P_5, \\dots$ the move $P_3$ is the only one that changes the state of the $3\\text{rd}$ light, thus we must also not make this move. Now we are at the move $P_4$ and we make a similar argument. We need to change the state of the $4\\text{th}$ light, hence we must make the move $P_4$. After this move the $4\\text{th}$, $5\\text{th}$, and $6\\text{th}$ light are in the desired state, so we must not make the move $P_5$ nor the move $P_6$. We continue in a similar matter, we make move $P_7$ and don't make moves $P_8$ and $P_9$, and so on. This means that we must necessarily make the sequence of moves $P_1, P_4, P_7, \\dots$. For this sequence to change the state of all the lights in the line, $n$ must be divisible by 3. Indeed, if $n$ has remainder 1 when divided by 3, then we will not change the state of the last light since the move $P_n$ is not allowed as it only changes the state of one light. Similarly, if $n$ has remainder 2 when divided by 3, then we will not change the state of the last two lights since the moves $P_{n-1}$ and $P_n$ are not allowed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41768,
    "subject": "Mathematics (Multi-modal)",
    "question": "First $n$ positive integers are written on the whiteboard. Ana and Bine are playing a game. In each move one of the players must erase 3 numbers whose average value is an integer. Ana starts and then they take turns after each move. The player that can not make a move loses. Determine who has the winning strategy and justify the answer.",
    "options": [],
    "answer": "Ana wins if n ≡ 3, 4, 5 (mod 6) and also when n = 7; otherwise Bine wins.",
    "solution": "In each step average value of chosen three numbers will be integer if and only if their sum will be divisible by 3. Therefore it is enough to know only residues of numbers modulo 3. Let us denote with $x_0, x_1$ and $x_2$ how many numbers on board give residue 0, 1 or 2 modulo 3, respectively.\nWe can present current state on the board with triple of these nonnegative integers $(x_0, x_1, x_2)$. Possible moves in this state are\n$$\n\\begin{array}{lcl}\n(x_0, x_1, x_2) & \\rightarrow & (x_0 - 1, x_1 - 1, x_2 - 1), \\quad \\text{if } x_0, x_1, x_2 \\ge 1, \\\\\n(x_0, x_1, x_2) & \\rightarrow & (x_0 - 3, x_1, x_2), \\quad \\text{if } x_0 \\ge 3, \\\\\n(x_0, x_1, x_2) & \\rightarrow & (x_0, x_1 - 3, x_2), \\quad \\text{if } x_1 \\ge 3, \\\\\n(x_0, x_1, x_2) & \\rightarrow & (x_0, x_1, x_2 - 3), \\quad \\text{if } x_2 \\ge 3.\n\\end{array}\n$$\n\nBecause of symmetry of these moves order of integers in triple is not important. Therefore we will present each state rather with ordered triple $(y_0, y_1, y_2)$, where $(y_0, y_1, y_2)$ is a permutation of $(x_0, x_1, x_2)$ with $y_0 \\ge y_1 \\ge y_2$.\nThe sum $y_0 + y_1 + y_2$ modulo 3 is invariant after each move. We denote $(\\hat{y}_0, \\hat{y}_1, \\hat{y}_2)$ to be final state of the game where no moves are possible anymore. Because $\\hat{y}_0 \\ge \\hat{y}_1 \\ge \\hat{y}_2$ it must hold $\\hat{y}_2 = 0$ and $2 \\ge \\hat{y}_0 \\ge \\hat{y}_1 \\ge 0$. With respect to $n$ we have following possibilities:\n\n* $n \\equiv 2 \\pmod{3}$\nFinal state $(\\hat{y}_0, \\hat{y}_1, \\hat{y}_2)$ is (2,0,0) or (1,1,0). In either case players had to make $\\frac{n-2}{3}$ moves together to reach this state. If this number is even Bine wins else Ana wins. Therefore if $n \\equiv 2 \\pmod{6}$ Bine wins and if $n \\equiv 5 \\pmod{6}$ Ana wins.\n\n* $n \\equiv 0 \\pmod{3}$\nAt the beginning of game we have $y_0 = y_1 = y_2$ and after each move property $y_0 \\equiv y_1 \\equiv y_2 \\pmod{3}$ is invariant. Therefore final state will be $(\\hat{y}_0, \\hat{y}_1, \\hat{y}_2) = (0,0,0)$. To reach it $\\frac{n}{3}$ will have to be played. So if $n \\equiv 0 \\pmod{6}$ Bine wins and if $n \\equiv 3 \\pmod{6}$ Ana wins.\n\n• $n \\equiv 1 \\pmod{6}$\nIn this case final states are (2,2,0) and (1,0,0). State (2,2,0) can be reached after odd number of moves and therefore is winning for Ana. State (1,0,0) can be reached after even number of moves and therefore is winning for Bine. For $n = 7$ Ana can make first move (3, 2, 2) $\\rightarrow$ (2, 2, 0) and wins.\nFor $n > 7$ we will prove that Bine has winning strategy. There exist two numbers in the triplet such that in each turn they are congruent modulo 3. Bine's strategy is to decrease of these two numbers below 2. If he succeeds it will be also possible to decrease the other number below 2 and therefore final state will be (1,0,0). Starting triple is $(\\frac{n-1}{3} + 1, \\frac{n-1}{3}, \\frac{n-1}{3})$. Bine chooses one of the numbers $\\frac{n-1}{3}$ and tries to decrease it below 2 in as few turns as possible. Ana can prevent him to do so only if at the end Bine gets number 2 and Ana in the meantime decreases number $\\frac{n-1}{3} + 1$ on 0. Because Bine is decreasing his number in each turn by 3 that can happen only if $\\frac{n-1}{3} \\equiv 2 \\pmod{3}$. That can happen because then $\\frac{n-1}{3} + 1 \\equiv 0 \\pmod{3}$ holds and Ana needs less turns to reach 0 with decreasing by 3. In this case Bine has to adjust his strategy in his penultimate move. For $n > 7$ the state before that move is $(3, 5, \\frac{n-1}{3})$. Then Bine must not decrease 5 by 3 but rather decrease all three numbers by 1 to reach state $(2, 4, \\frac{n-1}{3} - 1)$. In the next move Ana cannot decrease number 2 to 0 therefore Bine will be able to decrease number 4 below number 2. So Bine can definitely reach state (1,0,0) and win.\n\n• $n \\equiv 4 \\pmod{6}$\nFor $n = 4$ Ana makes move (2, 1, 1) $\\rightarrow$ (1, 0, 0) and wins. For $n = 10$ Ana makes move (4, 3, 3) $\\rightarrow$ (4, 3, 0). Then regardless of the move Bine makes Ana can make another move and again wins by reaching state (1, 0, 0).\nFor $n > 10$ Ana can in the first move erase biggest three numbers from the board. By that the game transforms into the previous case with $n \\equiv 1 \\pmod{6}$ and $n > 7$. Now Ana is a second player and she has a winning strategy.\n\nWe conclude that Ana wins if $n \\equiv 3, 4, 5 \\pmod{6}$ and if $n = 7$. Otherwise Bine wins.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41769,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a football tournament participated $n$ teams. Each team played exactly one match with each other team. There was a total of $2015n$ matches played in the tournament. How many teams participated in the tournament?\n(A) 2015 (B) 4029 (C) 4030 (D) 4031\n(E) It is impossible to determine.",
    "options": [],
    "answer": "D",
    "solution": "There was a total of $\\binom{n}{2}$ matches played in the tournament, thus $\\binom{n}{2} = \\frac{n(n-1)}{2} = 2015n$. From this we deduce $n = 4031$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41770,
    "subject": "Mathematics (Multi-modal)",
    "question": "The figure shows three concentric semicircles with radii $1$, $2$, and $4$. Regions denoted by $X$ have area $x$ and regions denoted by $Y$ have area $y$. What is the ratio between areas $x$ and $y$?\n\n![](attached_image_1.png)\n\n(A) $1 : 3$ (B) $1 : 2$ (C) $2 : 3$ (D) $3 : 8$ (E) $4 : 9$",
    "options": [],
    "answer": "D",
    "solution": "Areas of the three semicircles are equal to $\\frac{\\pi}{2}$, $\\frac{4\\pi}{2}$, and $\\frac{16\\pi}{2}$. Therefore, the area $x$ is equal to $\\frac{4\\pi}{2} - \\frac{\\pi}{2} = \\frac{3\\pi}{2}$ and the area $y$ is equal to $\\frac{16\\pi}{2} - \\frac{4\\pi}{2} = 6\\pi$. The ratio we are looking for is $\\frac{3\\pi}{2} : 6\\pi = 3 : 8$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41771,
    "subject": "Mathematics (Multi-modal)",
    "question": "Using eight isosceles right-angled triangles we assemble a regular octagon with an octagonal hole, as shown in the figure. The side length of the regular octagon equals $1$ cm. How many centimetres is the side length of the octagonal hole?\n\n![](attached_image_1.png)\n\n(A) $\\frac{\\sqrt{2}}{2}$\n(B) $\\frac{1}{2}$\n(C) $2 - \\sqrt{2}$\n(D) $\\sqrt{2} - 1$\n(E) $\\sqrt{3}$",
    "options": [],
    "answer": "D",
    "solution": "The legs of each right-angled triangle are of length $1$ cm and its hypotenuse is of length $\\sqrt{2}$ cm. Thus, the side of the octagonal hole is of length $\\sqrt{2} - 1$ cm.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41772,
    "subject": "Mathematics (Multi-modal)",
    "question": "We cut off all five of the vertices of an equilateral pyramid with edge length of $5$ cm, so that all the edges of the cut off small pyramids are shorter than $1$ cm. Which of the listed polygons is not a face of the resulting polyhedron?\n(A) triangle\n(B) quadrilateral\n(C) pentagon\n(D) hexagon\n(E) octagon\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "C",
    "solution": "The face that was the base of the pyramid is now an octagon. Faces that were on the side of the pyramid are now hexagons. At the apex vertex of the pyramid the resulting face is a quadrilateral, at all the other vertices of the pyramid the resulting faces are triangles. None of the faces of the resulting polyhedron is a pentagon.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 41773,
    "subject": "Mathematics (Multi-modal)",
    "question": "Which integer $n$ satisfies\n$$(n - 1)(n - 3) \\dots (n - 2015) = n(n + 2)(n + 4) \\dots (n + 2014)$$\n(A) -4028 (B) -2014 (C) 2015 (D) 4030 (E) None.",
    "options": [],
    "answer": "E",
    "solution": "If $n$ is an odd integer the left side of the equality is even and the right side is odd. Thus the equality is not satisfied for odd integers. Similarly, if $n$ is an even integer the left side of the equality is odd and the right side is even. Therefore no integer satisfies the equality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41774,
    "subject": "Mathematics (Multi-modal)",
    "question": "Lili figured out that the average of digits of the year 2015 equals 2, since $\\frac{2+0+1+5}{4} = 2$. How many times in the 21st century after the year 2015 will a year have the same average of digits as the year 2015?\n(A) 1\n(B) 2\n(C) 5\n(D) 6\n(E) 9",
    "options": [],
    "answer": "C",
    "solution": "The sum of digits of such a year must be $8$. Since it is a year in 21st century the first two digits must be $2$ and $0$. The sum of the last two digits must be equal to $6$. There are five such years after the year $2015$, they are $2024$, $2033$, $2042$, $2051$, and $2060$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41775,
    "subject": "Mathematics (Multi-modal)",
    "question": "Sara is looking for a three-digit positive integer $xyz$ ($z$ are ones, $y$ are tens, and $x$ are hundreds) which satisfies $1 \\le x < y < z$ and the property that the sum of the numbers $xyz$, $yzx$, and $zxy$ is a three-digit number which has equal digits. At most\n\nhow many such three-digit numbers can Sara find?\n(A) 3 (B) 5 (C) 7 (D) 9 (E) 11",
    "options": [],
    "answer": "C",
    "solution": "The sum of the numbers $\\overline{xyz}$, $\\overline{yzx}$, and $\\overline{zxy}$ is equal to $100(x + y + z) + 10(y + z + x) + (z + x + y)$. Since this must be a three-digit number it must hold that $x + y + z < 10$ and therefore it also holds that all the digits of the sum are automatically equal. So we are looking for the number of triples of digits $(x, y, z)$ which satisfy $1 \\le x < y < z$ and $x + y + z < 10$. There are exactly 7 such triples, they are $(1, 2, 3)$, $(1, 2, 4)$, $(1, 2, 5)$, $(1, 2, 6)$, $(1, 3, 4)$, $(1, 3, 5)$, and $(2, 3, 4)$. Sara can find at most 7 such three-digit numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41776,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the value of the expression $\\sqrt{0.04^3}$?\n(A) $\\frac{1}{5}$\n(B) $\\frac{1}{25}$\n(C) $\\frac{1}{125}$\n(D) $0.04$\n(E) $0.016$",
    "options": [],
    "answer": "C",
    "solution": "We calculate $\\sqrt{0.04^3} = \\sqrt{\\left(\\frac{4}{100}\\right)^3} = \\left(\\sqrt{\\frac{4}{100}}\\right)^3 = \\left(\\frac{2}{10}\\right)^3 = \\left(\\frac{1}{5}\\right)^3 = \\frac{1}{125}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41777,
    "subject": "Mathematics (Multi-modal)",
    "question": "A polynomial $p(x) = 2015x^{2013} - 2$ and a real number $h$ are given so that $p(h) = -2015$. What is the value of $p(-h)$?\n(A) 2011 (B) 2012 (C) 2013 (D) 2014 (E) 2015",
    "options": [],
    "answer": "A",
    "solution": "From data it follows that $2015h^{2013} - 2 = -2015$ which means $2015h^{2013} = -2013$. Thus $p(-h) = 2015(-h)^{2013} - 2 = -2015h^{2013} - 2 = -(-2013) - 2 = 2011$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41778,
    "subject": "Mathematics (Multi-modal)",
    "question": "During the first school hour the ratio of the number of boys and girls in the classroom was $3 : 4$. After 4 more girls entered the classroom and 4 boys left it before the second school hour, the ratio became $2 : 5$. How many more girls than boys were in the classroom during the second school hour?\n\n(A) 5\n(B) 6\n(C) 7\n(D) 8\n(E) 12",
    "options": [],
    "answer": "E",
    "solution": "Denote the number of boys in the classroom during the first school hour by $f$ and the number of girls by $d$. Then $\\frac{f}{d} = \\frac{3}{4}$. During the second school hour there are $f - 4$ boys and $d + 4$ girls in the classroom, therefore the ratio equals $\\frac{f-4}{d+4} = \\frac{2}{5}$. From first equation we deduce $f = \\frac{3}{4}d$ and insert it into second equation to get\n$$\n\\frac{3}{4}d - 4 = \\frac{2}{5}d + 4\n$$\n\nRearranging the equation we get $\\frac{7}{4}d = 28$, therefore $d = 16$ and $f = 12$. During the second school hour there are 8 boys and 20 girls in the classroom, which means there are 12 girls more than boys.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 41779,
    "subject": "Mathematics (Multi-modal)",
    "question": "We inscribe a regular octagon in a square with side of length $a$, so that 4 sides of the octagon lie on the sides of the square. Express the side length of the inscribed octagon in terms of $a$.",
    "options": [],
    "answer": "a(√2 - 1)",
    "solution": "Denote by $x$ the side length of the inscribed octagon. The four triangles that are formed at the vertices of the square are isosceles right-angled triangles. Since their hypotenuse is of length $x$, their legs are of length $\\frac{x}{\\sqrt{2}}$. Thus\n$$\na = x + 2\\frac{x}{\\sqrt{2}} = x + x\\sqrt{2} = x(\\sqrt{2} + 1).\n$$\nFrom this we deduce\n\n![](attached_image_1.png)\n$$\nx = \\frac{a}{\\sqrt{2} + 1} = \\frac{a(\\sqrt{2} - 1)}{2 - 1} = \\sqrt{2}a - a.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41780,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circles $\\mathcal{K}_1$ and $\\mathcal{K}_2$ in the figure are the circumcircle and incircle of the equilateral triangle $ABC$. A square $DEFG$ is inscribed in the circle $\\mathcal{K}_2$ so that the point $D$ lies on the side $AB$. The circles $\\mathcal{K}_3$ and $\\mathcal{K}_4$ are of the same size and touch each other, each of them also touches two sides of the square $DEFG$. Determine the ratio of the radii of the circles $\\mathcal{K}_1$ and $\\mathcal{K}_4$.\n![](attached_image_1.png)",
    "options": [],
    "answer": "2 + 2*sqrt(2)",
    "solution": "![](attached_image_2.png)\n![](attached_image_3.png)\nDenote by $r_1, r_2, r_3$, and $r_4$ the radii of the circles $K_1, K_2, K_3$, and $K_4$. We know that $r_3 = r_4$. Since the triangle $ABC$ is equilateral the circles $K_1$ and $K_2$ have a common center which we denote by $S$. The triangle $BSD$ is a half of an equilateral triangle, thus $r_1 = |SB| = 2|SD| = 2r_2$. Since $GE$ is the diameter of the circle $K_2$ we have $|GE| = 2r_2 = r_1$. Let's express the length of $|GE|$ in another way in terms of $r_4$. Let $T$ be the point where the circles $K_3$ and $K_4$ touch. Denote the center of the circle $K_4$ by $R$ and its contact points with sides $DE$ and $EF$ of the square $DEFG$ by $U$ and $V$. Since the circles $K_3$ and $K_4$ are of equal size, $T$ is the center of the square $DEFG$ and thus $|GE| = 2|TE|$. The quadrilateral $RUEV$ is a square with side length $r_4$ therefore its diagonal is of length $|RE| = \\sqrt{2}r_4$. From this we deduce $|GE| = 2|TE| = 2(|TR| + |RE|) = 2(r_4 + \\sqrt{2}r_4) = (2 + 2\\sqrt{2})r_4$. It follows $r_1 = (2 + 2\\sqrt{2})r_4$ and thus $\\frac{r_1}{r_4} = (2 + 2\\sqrt{2})$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41781,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $x$ which satisfy the equation\n$$\n2 \\sin^2 2x \\geq 3 \\cos 2x.\n$$",
    "options": [],
    "answer": "x ∈ [π/6 + kπ, 5π/6 + kπ] for k ∈ ℤ",
    "solution": "We use the double angle formulas for trigonometric functions to get\n$$\n8 \\sin^2 x \\cos^2 x \\ge 3 \\cos^2 x - 3 \\sin^2 x.\n$$\nUsing the identity $\\cos^2 x = 1 - \\sin^2 x$ we rearrange the inequality to\n$$\n8 \\sin^4 x - 14 \\sin^2 x + 3 \\le 0.\n$$\nIntroducing a new variable $y = \\sin^2 x$ we get $8y^2 - 14y + 3 \\le 0$ which factors to $(4y - 1)(2y - 3) \\le 0$. Since $0 \\le y \\le 1$ the factor $(2y - 3)$ is automatically negative.\n\nThe given inequality is thus equivalent to $4y - 1 \\ge 0$, i.e. $\\sin^2 x \\ge \\frac{1}{4}$. The latter inequality holds when either $\\sin x \\ge \\frac{1}{2}$ which is equivalent to $\\frac{\\pi}{6} + 2k\\pi \\le x \\le \\frac{5\\pi}{6} + 2k\\pi$, $k \\in \\mathbb{Z}$, or when $\\sin x \\le -\\frac{1}{2}$ which is equivalent to $-\\frac{5\\pi}{6} + 2k\\pi \\le x \\le -\\frac{\\pi}{6} + 2k\\pi$, $k \\in \\mathbb{Z}$. The real numbers $x$ that satisfy the given inequality are therefore exactly those that satisfy $\\frac{\\pi}{6} + k\\pi \\le x \\le \\frac{5\\pi}{6} + k\\pi$ for some integer $k$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41782,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many quadruples of positive integers $(a, b, c, d)$ satisfy the inequalities\n$$\na > b > c > d \\quad \\text{and} \\quad \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} > 2?\n$$",
    "options": [],
    "answer": "2",
    "solution": "If $d \\ge 2$ we get $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} \\le 4 \\cdot \\frac{1}{2} = 2$ which is a contradiction. Thus $d = 1$.\n\nIf $c \\ge 3$ we get $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} \\le 3 \\cdot \\frac{1}{3} + 1 = 2$ which is again a contradiction. Thus $c = 2$.\n\nSimilarly $b = 3$ since we would otherwise get $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} \\le 2 \\cdot \\frac{1}{4} + \\frac{1}{2} + 1 = 2$.\n\nTherefore $\\frac{1}{a} + \\frac{1}{3} + \\frac{1}{2} + \\frac{1}{1} > 2$ which means $\\frac{1}{a} > \\frac{1}{6}$. Since $a > b = 3$ we can only have $a = 4$ or $a = 5$.\n\nInequalities are satisfied by two quadruples $(4, 3, 2, 1)$ and $(5, 3, 2, 1)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41783,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, a_3, \\dots$ be an arithmetic sequence. For an arbitrary positive integer $k$ denote $S_k = a_1 + a_2 + \\dots + a_k$. Prove that\n$$\nS_n - S_m = \\frac{n-m}{n+m} S_{n+m}\n$$\nholds for all positive integers $m$ and $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $d$ the difference of two consecutive terms of this arithmetic sequence. By the formula for the sum of an arithmetic sequence we have\n$$\nS_k = k a_1 + \\frac{k(k-1)}{2} d.\n$$\nUsing this we calculate\n$$\nS_{n+m} = (n+m)a_1 + \\frac{(n+m)(n+m-1)}{2} d = (n+m)\\left(a_1 + (n+m-1)\\frac{d}{2}\\right)\n$$\nand\n$$\nS_n - S_m = (n-m)a_1 + (n(n-1)-m(m-1))\\frac{d}{2} = (n-m)a_1 + (n-m)(n+m-1)\\frac{d}{2} = \\\\\n= (n-m)\\left(a_1 + (n+m-1)\\frac{d}{2}\\right).\n$$\nThe last term in both equations is equal, therefore\n$$\n(n+m)(S_n - S_m) = (n-m)S_{n+m}\n$$\nwhich proves the desired equality since $n + m \\ne 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41784,
    "subject": "Mathematics (Multi-modal)",
    "question": "Jure wrote positive integers 1 through 2015 on a whiteboard. Urška then inspected the written numbers from smallest to largest respectively and erased each number that was not divisible by 3. From the numbers still left on the whiteboard she then erased from smallest to largest each number that was not divisible by $3^2$. From the remaining numbers she then erased from smallest to largest each number that was not divisible by $3^3$, and so on. Which number did Urška erase last?",
    "options": [],
    "answer": "1458",
    "solution": "After Urška inspected all the written numbers for the first time only numbers divisible by 3 were left on the whiteboard. After her second inspection only numbers divisible by $3^2$ were left. After the third inspection only numbers divisible by $3^3$ were left, and so on. The largest power of the number 3 smaller than 2015 is $3^6 = 729$ since $3^7 = 2187$. Thus, after Urška inspected the numbers for the sixth time only the numbers divisible by 729 were left on the whiteboard, they are 729 in 1458. In the next step she erased both this numbers since they are not divisible by 2187. Urška erased the number 1458 last.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41785,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be real numbers which satisfy the equality $\\frac{a^2}{1+a^2} + \\frac{b^2}{1+b^2} = 1$. Determine all possible values of the expression\n$$\n(a + b) \\left( \\frac{a}{1 + a^2} + \\frac{b}{1 + b^2} \\right).\n$$",
    "options": [],
    "answer": "0, 2",
    "solution": "We multiply the given equation by $(1+a^2)(1+b^2) \\neq 0$ and simplify it to get the equivalent equation $a^2b^2 = 1$. Therefore $ab = \\pm 1$. We also simplify the given expression\n$$\n\\begin{aligned}\n(a+b) \\left( \\frac{a}{1+a^2} + \\frac{b}{1+b^2} \\right) &= (a+b) \\frac{a+ab^2+b+a^2b}{(1+a^2)(1+b^2)} = (a+b) \\frac{(a+b)(1+ab)}{(1+a^2)(1+b^2)} = \\\\\n&= \\frac{(a^2+b^2+2ab)(1+ab)}{1+a^2+b^2+a^2b^2}.\n\\end{aligned}\n$$\n\nIf $ab = -1$ the value of this expression equals $0$, if $ab = 1$ the value of this expression equals\n$$\n\\frac{(a^2 + b^2 + 2) \\cdot 2}{a^2 + b^2 + 2} = 2.\n$$\n\nWe can calculate that these two values are actually achieved. For example, the value $0$ is attained when $a = 1$ and $b = -1$, and the value $2$ is attained when $a = b = 1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41786,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $x$ which satisfy the equation\n$$\n\\sqrt[3]{2x+13} - \\sqrt[3]{2x-13} = 2.\n$$",
    "options": [],
    "answer": "{-7, 7}",
    "solution": "We raise the equation to the third power to get\n$$\n(2x + 13) - 3 \\left(\\sqrt[3]{2x + 13}\\right)^2 \\sqrt[3]{2x - 13} + 3 \\sqrt[3]{2x + 13} \\left(\\sqrt[3]{2x - 13}\\right)^2 - (2x - 13) = 8.\n$$\nWe then rearrange it\n$$\n26 - 3\\sqrt[3]{2x + 13}\\sqrt[3]{2x - 13}\\left(\\sqrt[3]{2x + 13} - \\sqrt[3]{2x - 13}\\right) = 8\n$$\nand use $\\sqrt[3]{2x + 13} - \\sqrt[3]{2x - 13} = 2$ to get\n$$\n26 - 6\\sqrt[3]{2x + 13}\\sqrt[3]{2x - 13} = 8.\n$$\nWe rearrange this to $\\sqrt[3]{2x + 13}\\sqrt[3]{2x - 13} = 3$ and raise it to the third power to get $(2x + 13)(2x - 13) = 27$ which means $4x^2 - 169 = 27$. It follows $x^2 = 49$ and thus $x = \\pm 7$.\nWe rearrange the equation to $\\sqrt[3]{2x + 13} = \\sqrt[3]{2x - 13} + 2$ and raise it to the third power to get\n$$\n2x + 13 = 2x - 13 + 6(\\sqrt[3]{2x - 13})^2 + 12\\sqrt[3]{2x - 13} + 8.\n$$\nWe simplify this to get\n$$\n6(\\sqrt[3]{2x - 13})^2 + 12\\sqrt[3]{2x - 13} - 18 = 0.\n$$\nIntroducing a new variable $y = \\sqrt[3]{2x - 13}$ and dividing by 6 we get a quadratic equation $y^2 + 2y - 3 = 0$ which factors to $(y + 3)(y - 1) = 0$ and thus has solutions $y = -3$ and $y = 1$. From the new variable equation we get $x = \\frac{y^3+13}{2}$. When $y = -3$ we get $x = -7$, when $y = 1$ we get $x = 7$. The initial equation thus has two solutions $x = -7$ and $x = 7$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41787,
    "subject": "Mathematics (Multi-modal)",
    "question": "The bisector of the angle at $A$, the altitude from $A$, and the median from $A$ divide the angle at $A$ into 4 equal parts. Determine the sizes of angles of the triangle $ABC$.",
    "options": [],
    "answer": "∠A = 90°, and ∠B, ∠C are 22.5° and 67.5° (in either order).",
    "solution": "Let $D$ be the midpoint of the side $BC$, $E$ be the foot of the altitude from vertex $A$, and $S$ be the intersection of the angle bisector at vertex $A$ with the side $BC$. Since the angle bisector divides the angle into two equal parts it must lie between the altitude and the median. Thus the point $S$ lies between the points $E$ and $D$. We have two possibilities. Either $D$ lies between $B$ and $S$, and $E$ lies between $S$ and $C$, or $E$ lies between $B$ and $S$, and $D$ lies between $S$ and $C$. Due to symmetry we may consider only the first case.\n\nDenote the angle $\\angle BAC$ by $4\\alpha$ where $0 < \\alpha < \\frac{\\pi}{4}$. Then\n$$\n|BE| = |AE| \\tan 3\\alpha, \\quad |DE| = |AE| \\tan 2\\alpha \\quad \\text{in} \\quad |CE| = |AE| \\tan \\alpha.\n$$\nFrom this we deduce\n$$\n|BD| = |BE| - |DE| = |AE|(\\tan 3\\alpha - \\tan 2\\alpha)\n$$\nand\n$$\n|CD| = |CE| + |DE| = |AE|(\\tan \\alpha + \\tan 2\\alpha).\n$$\nSince $D$ is a midpoint of the side $BC$ it follows $\\tan 3\\alpha - \\tan 2\\alpha = \\tan \\alpha + \\tan 2\\alpha$ and thus\n$$\n\\tan 3\\alpha - \\tan \\alpha - 2 \\tan 2\\alpha = 0.\n$$\nWe can rearrange the left side\n$$\n\\begin{align*}\n\\tan 3\\alpha - \\tan \\alpha - 2 \\tan 2\\alpha &= \\frac{\\sin 3\\alpha}{\\cos 3\\alpha} - \\frac{\\sin \\alpha}{\\cos \\alpha} - 2 \\frac{\\sin 2\\alpha}{\\cos 2\\alpha} = \\\\\n&= \\frac{\\sin 3\\alpha \\cos \\alpha - \\sin \\alpha \\cos 3\\alpha}{\\cos \\alpha \\cos 3\\alpha} - 2 \\frac{\\sin 2\\alpha}{\\cos 2\\alpha} = \\frac{\\sin 2\\alpha}{\\cos \\alpha \\cos 3\\alpha} - 2 \\frac{\\sin 2\\alpha}{\\cos 2\\alpha} = \\\\\n&= \\frac{\\sin 2\\alpha(\\cos 2\\alpha - 2 \\cos \\alpha \\cos 3\\alpha)}{\\cos \\alpha \\cos 2\\alpha \\cos 3\\alpha} = - \\frac{\\sin 2\\alpha \\cos 4\\alpha}{\\cos \\alpha \\cos 2\\alpha \\cos 3\\alpha}.\n\\end{align*}\n$$\nTo get the fourth equality we used the addition formula for sine\n$$\n\\sin 2\\alpha = \\sin(3\\alpha - \\alpha) = \\sin 3\\alpha \\cos \\alpha - \\sin \\alpha \\cos 3\\alpha\n$$\nand to get the last equality we used the product formula for cosine\n$$\n2 \\cos \\alpha \\cos 3\\alpha = \\cos(3\\alpha + \\alpha) + \\cos(3\\alpha - \\alpha) = \\cos 4\\alpha + \\cos 2\\alpha.\n$$\nIt follows that $\\sin 2\\alpha = 0$ or $\\cos 4\\alpha = 0$, thus $2\\alpha = k\\pi$ or $4\\alpha = \\frac{\\pi}{2} + k\\pi$, where $k \\in \\mathbb{Z}$. Due to $0 < \\alpha < \\frac{\\pi}{4}$ the only possibility is $\\alpha = \\frac{\\pi}{8}$. Thus $\\angle BAC = 4\\alpha = \\frac{\\pi}{2}$, $\\angle ACB = \\frac{\\pi}{2} - \\alpha = \\frac{3\\pi}{8}$, and $\\angle CBA = \\frac{\\pi}{8}$.\n\nTaking into account both symmetrical cases we conclude that the angles of the triangle $ABC$ are either $\\angle BAC = \\frac{\\pi}{2}$, $\\angle ACB = \\frac{3\\pi}{8}$, and $\\angle CBA = \\frac{\\pi}{8}$ or $\\angle BAC = \\frac{\\pi}{2}$, $\\angle ACB = \\frac{\\pi}{8}$, and $\\angle CBA = \\frac{3\\pi}{8}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41788,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathcal{K}$ be the circumcircle and $\\mathcal{I}$ the incenter of the triangle $ABC$. Denote by $D$ the midpoint of the arc $BC$ of the circle $\\mathcal{K}$ which does not contain the point $A$, and by $E$ the midpoint of the arc $CA$ of the circle $\\mathcal{K}$ which does not contain the point $B$. Let $F$ be the mirror image of the point $\\mathcal{I}$ by reflection across the side $AB$ and suppose that $F$ lies on the circle $\\mathcal{K}$. What is the size of the angle $\\angle DFE$?",
    "options": [],
    "answer": "pi/3",
    "solution": "![](attached_image_1.png)\nSince the points $D$ and $E$ are the midpoints of the corresponding arcs of the circle $\\mathcal{K}$ they lie on the angle bisectors of angles $\\angle BAC$ and $\\angle CBA$. Due to symmetry we\n\nhave $\\angle FAB = \\angle BAI = \\frac{1}{2}\\angle BAC$ and thus $\\angle FED = \\angle FAD = \\angle BAC$. Similarly we get $\\angle EDF = \\angle CBA$. It follows $\\angle DFE = \\angle ACB$. Let's calculate the size of the angle $\\angle ACB$. The points $F, B, C$, and $A$ are concyclic, therefore $\\angle ACB = \\pi - \\angle BFA$. Since $F$ is a mirror image of the point $I$ we have\n$$\n\\begin{align*} \n\\angle BFA &= \\angle AIB = \\pi - \\frac{1}{2}\\angle BAC - \\frac{1}{2}\\angle CBA = \\pi - \\frac{1}{2}(\\angle BAC + \\angle CBA) = \\\\ \n&= \\pi - \\frac{1}{2}(\\pi - \\angle ACB) = \\frac{\\pi}{2} + \\frac{1}{2}\\angle ACB. \n\\end{align*}\n$$\n\nTherefore $\\angle ACB = \\pi - (\\frac{\\pi}{2} + \\frac{1}{2}\\angle ACB) = \\frac{\\pi}{2} - \\frac{1}{2}\\angle ACB$ which implies $\\angle ACB = \\frac{\\pi}{3}$ and thus also $\\angle DFE = \\frac{\\pi}{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41789,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let\n$$\nN = 2^{15} \\cdot 2015.\n$$\nHow many divisors of $N^2$ are strictly smaller than $N$ and do not divide $N$?",
    "options": [],
    "answer": "291",
    "solution": "A number with prime factorization $p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}$ has\n$$\n\\tau(p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}) = (\\alpha_1 + 1)(\\alpha_2 + 1) \\dots (\\alpha_k + 1)\n$$\ndivisors. The number $N$ can be factored as $N = 2^{15} \\cdot 5 \\cdot 13 \\cdot 31$, so\n$$\n\\tau(N) = (15 + 1) \\cdot (1 + 1) \\cdot (1 + 1) \\cdot (1 + 1) = 2^7 = 128\n$$\n$$\n\\text{and } \\tau(N^2) = (2 \\cdot 15 + 1) \\cdot (2 \\cdot 1 + 1) \\cdot (2 \\cdot 1 + 1) \\cdot (2 \\cdot 1 + 1) = 31 \\cdot 3^3 = 837.\n$$\nIf $d$ divides $N^2$ then $\\frac{N^2}{d}$ also divides $N^2$, so all divisors can be sorted into pairs $(d, \\frac{N^2}{d})$, omitting the number $N$, which would otherwise be paired with itself. In each pair exactly one divisor is smaller than $N$ and one is greater than $N$. This implies that the number of divisors of $N^2$ which are strictly smaller than $N$ is equal to the number of pairs, which is\n$$\n\\frac{\\tau(N^2) - 1}{2}.\n$$\n\nEach divisor of $N$ also divides $N^2$, so we have to subtract the number of divisors of $N$ smaller than $N$. The number of divisors of $N^2$ which are smaller than $N$ and do not divide $N$ is therefore equal to\n$$\n\\frac{\\tau(N^2) - 1}{2} - (\\tau(N) - 1) = \\frac{836}{2} - 127 = 291.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41790,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$ and $Q(x)$ be quadratic polynomials with rational coefficients and assume that the roots of $P(x)$ are irrational numbers $a$ and $b$. If $a + 2016$ is one of the roots of $Q(x)$, find all possible values for the other root.",
    "options": [],
    "answer": "b + 2016",
    "solution": "Let $c$ denote the other root of $Q$. Since $P$ and $Q$ have rational coefficients the Vieta's formulas imply\n$$\n\\begin{aligned}\na+b &= p_1 \\in \\mathbb{Q} \\quad \\text{and} \\quad ab = p_2 \\in \\mathbb{Q}, \\\\\n(a+2016)+c &= q_1 \\in \\mathbb{Q} \\quad \\text{and} \\quad (a+2016)c = q_2 \\in \\mathbb{Q}.\n\\end{aligned}\n$$\nExpressing $b = p_1 - a$ and $c = q_1 - a - 2016$ from the equations on the left and inserting them into the equations on the right we get\n$$\n\\begin{aligned}\na(p_1 - a) &= p_2, \\\\\n(a + 2016)(q_1 - a - 2016) &= q_2.\n\\end{aligned}\n$$\nSubtracting one equation from the other we get\n$$\n\\begin{aligned}\na(p_1 - a) - (a + 2016)(q_1 - a - 2016) &= p_2 - q_2 \\\\\n\\Rightarrow a(p_1 - q_1 + 4032) &= p_2 - q_2 + 2016q_1 - 2016^2 \\in \\mathbb{Q}.\n\\end{aligned}\n$$\nIf $p_1 - q_1 + 4032 \\neq 0$, we can write\n$$\na = \\frac{p_2 - q_2 + 2016q_1 - 2016^2}{p_1 - q_1 + 4032} \\in \\mathbb{Q}.\n$$\nIt would follow that $a$ is rational, which is not the case. So, $p_1 - q_1 + 4032 = 0$. From here\n$$\n(a+b) - (a+2016+c) + 4032 = 0 \\Rightarrow c = b+2016.\n$$\n\nHence, the only candidate for the other root of $Q$ is the number $b + 2016$. We have to check that there indeed exists a polynomial $Q$ with rational coefficients such that $a + 2016$ and $b + 2016$ are its roots. An example of such a polynomial would be\n$$\nQ(x) = (x - a - 2016)(x - b - 2016) = x^2 - (a + b)x + ab + 2016(a + b) + 2016^2,\n$$\nsince $a + b \\in \\mathbb{Q}$ and $ab + 2016(a + b) + 2016^2 \\in \\mathbb{Q}$. The only possible value for the other root of the polynomial $Q$ is therefore $b + 2016$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41791,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume that $n \\ge 3$ people of different ages sit around a round table. We call any pair consisting of two people $A$ and $B$ a *newbie pair*, if\n(i) $A$ and $B$ do not sit next to one another and\n(ii) on at least one of the arcs connecting $A$ and $B$ along the table edge all people are older than $A$ and $B$.\nDetermine the minimal number of newbie pairs. (From Nordic MO 2008.)",
    "options": [],
    "answer": "n - 3",
    "solution": "Denote the people in the order of ascending age by $A_1, A_2, \\dots, A_n$. Using induction on $n$ we show that regardless of the seating arrangement there are exactly $n-3$ newbie pairs.\n\nIn the base case of $n=3$ every person is adjacent to every other person so there are no newbie pairs.\n\nFor the induction step assume that any seating arrangement of $n$ people contains exactly $n-3$ newbie pairs and consider an arbitrary arrangement of $n+1$ people. We wish to show that there are exactly $n-2$ newbie pairs in this arrangement.\n\nRemoving the oldest person $A_{n+1}$ from the table, the induction hypothesis tells us that there are exactly $n-3$ newbie pairs left. Now let $A_{n+1}$ sit back down. All $n-3$ newbie pairs will still remain such because person $A_{n+1}$ was the oldest and does not affect the other pairs. Also, $A_{n+1}$ will not form any new pairs with any of the other people. If $A_{n+1}$ and $A_i$ are not adjacent there is at least one person on each of the two arcs between them that is younger than $A_{n+1}$.\n\nFinally, consider the remaining pairs which were not newbie pairs before. Out of these only the pair seated next to $A_{n+1}$ turns into a newbie pair. Indeed, there is now one person between them that is older than both of them, namely $A_{n+1}$.\n\nTherefore, this arrangement of $n+1$ people contains exactly $n-2$ newbie pairs. This concludes the induction step.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 41792,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$, such that\n$$\nf\\left(\\frac{x}{f(y)}\\right) = \\frac{(f(x))^2}{y f(f(x))}\n$$\nfor all $x, y > 0$.",
    "options": [],
    "answer": "f(x) = k x for any constant k > 0",
    "solution": "Let us show that the function $f$ is surjective. Substituting $y \\mapsto \\frac{(f(x))^2}{y f(f(x))}$ in the initial equation we get\n$$\nf\\left(\\frac{x}{f\\left(\\frac{(f(x))^2}{y f(f(x))}\\right)}\\right) = y,\n$$\nwhich means that for any $y$ there exists a number which is mapped into $y$ by $f$. Hence, $f$ is surjective.\n\nSurjectivity implies the existence of $c \\in \\mathbb{R}^+$, such that $f(c) = 1$. Insert $y = c$ into the initial equation.\n$$\n\\begin{aligned}\n& f\\left(\\frac{x}{f(c)}\\right) = \\frac{(f(x))^2}{c f(f(x))} \\\\\n\\Rightarrow \\quad & f(x) = \\frac{(f(x))^2}{c f(f(x))} \\\\\n\\Rightarrow \\quad & c f(f(x)) = f(x),\n\\end{aligned}\n$$\nbut because $f$ is surjective we can substitute $f(x)$ for any $y \\in \\mathbb{R}^+$. This implies that\n$$\nf(y) = \\frac{1}{c} y \\quad \\text{for all } y \\in \\mathbb{R}^{+}.\n$$\nIt is easy to check that all functions of the form $f(x) = kx$, where $k \\in \\mathbb{R}^+$ is an arbitrary constant, satisfy the original equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41793,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral and let $\\mathcal{K}$ be its circumcircle. Denote the intersection of the lines $AB$ and $CD$ by $E$, so that $|AB| = |BE|$. Denote the intersection of the tangents to $\\mathcal{K}$ from $B$ and $D$ by $F$, so that $AB$ and $DF$ are parallel. Show that the points $A, C$ and $F$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us use Pascal's Theorem for the points $A, B, C$ and $D$ on the circle. Here we use the points $B$ and $D$ twice and denote their copies by $B_1$ and $D_1$. We have $AB \\cap CD = E$, $BB_1 \\cap DD_1 = F$ and the lines $AD_1$ and $B_1C$ either intersect or are parallel.\n\nIf $AD_1$ and $B_1C$ intersect, denote this intersection by $G$. The Pascal's Theorem implies that $E, F$ and $G$ are collinear. We now wish to apply Ceva's Theorem for the triangle $AEG$ and the points $B, F$ and $D$ to show that $AF, BG$ and $DE$ are concurrent. The lines $BG$ and $DE$ intersect at $C$, so it suffices to show that\n$$\n\\frac{AB \\cdot EF \\cdot GD}{BE \\cdot FG \\cdot DA} = 1. \\qquad (4)\n$$\nSince $AB \\parallel DF$ the similarity of the triangles $AEG$ and $DFG$ implies that $\\frac{EF}{FG} = \\frac{DA}{GD}$. The point $B$ is the midpoint of the segment $AE$, so we see that (4) does indeed hold. The lines $AF, BG$ and $DE$ intersect at $C$ and the points $A, C$ and $F$ are collinear.\n\n![](attached_image_1.png)\n\nIf $AD_1$ and $B_1C$ are parallel then by Pascal's Theorem the line $EF$ is parallel to both of them as well. So, $AEFD$ is a parallelogram and the point $C$ is the midpoint of the diagonal $DE$. The other diagonal also passes through $C$, so the points $A, C$ and $F$ are collinear.\nLet $T$ be the intersection of the lines $AC$ and $BD$, and let the point $T'$ be the intersection of $AE$ and $BD$. We wish to show that $T = T'$. Since $AE$ is a symmedian we have\n$$\n\\frac{|BT'|}{|T'D|} = \\frac{|AB|^2}{|AD|^2}.\n$$\nSince $T$ and $T'$ both lie on the segment $BD$, it suffices to show that for $T$ we also have\n$$\n\\frac{|BT|}{|TD|} = \\frac{|AB|^2}{|AD|^2}. \\qquad (5)\n$$\n*Remark: This part of the reasoning can also be done using Menelaus' Theorem and Power-of-a-Point Theorem instead of the Symmedian Theorem.*\n\nBy Menelaus' Theorem for the triangle $BED$ and the collinear points $A, T$ and $C$ we have\n$$\n\\frac{|BT|}{|TD|} = \\frac{|DC|}{|CE|} = \\frac{|EA|}{|AB|} = 1.\n$$\nSince $|EA| = 2|AB|$, we get\n$$\n\\frac{|BT|}{|TD|} = \\frac{|CE|}{2|DC|}.\n$$\nThe Power-of-a-Point Theorem for the point $E$ and the circle $K$ states that $|EC| \\cdot |ED| = |EB| \\cdot |EA| = 2|AB|^2$. So,\n$$\n\\frac{|BT|}{|TD|} = \\frac{2|AB|^2}{2|DC| \\cdot |ED|} = \\frac{|AB|^2}{|DC| \\cdot |ED|}.\n$$\nSince $AB \\parallel DF$ and $DF$ is tangent to $K$, $D$ must be the midpoint of the arc $\\widehat{BA}$. The triangle $ABD$ is therefore isosceles with the apex at $D$ and $|AD| = |BD|$. It follows that\n$$\n\\asymp DCB = \\pi - \\asymp BAD = \\pi - \\asymp DBA = \\asymp EBD.\n$$\nThis implies that the triangles $DBC$ and $DEB$ are similar and\n$$\n\\frac{|DC|}{|DB|} = \\frac{|DB|}{|DE|} \\Rightarrow |DC| \\cdot |ED| = |DB|^2 = |AD|^2.\n$$\nThus,\n$$\n\\frac{|BT|}{|TD|} = \\frac{|AB|^2}{|AD|^2}.\n$$\nTherefore, $T = T'$, so the points $A, C$ and $F$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41794,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for all positive integers $n \\ge 2$ we have\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": "Detailed solution",
    "solution": "We use proof by induction.\nFor $n = 2$ we have\n$$\n\\frac{1}{2} + \\sqrt{\\frac{1}{2}} < \\frac{4}{3} \\Leftrightarrow \\sqrt{\\frac{1}{2}} < \\frac{5}{6} \\Leftrightarrow \\frac{1}{2} < \\frac{25}{36},\n$$\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$$\n\\frac{1}{2} + \\sqrt{\\frac{1}{2}} + \\sqrt[3]{\\frac{2}{3}} + \\dots + \\sqrt[n]{\\frac{n-1}{n}} + \\sqrt[n+1]{\\frac{n}{n+1}} < \\frac{(n+1)^2}{n+2}. \\quad (1)\n$$\nBy the induction hypothesis we can estimate\n$$\n\\frac{1}{2} + \\sqrt{\\frac{1}{2}} + \\sqrt[3]{\\frac{2}{3}} + \\dots + \\sqrt[n]{\\frac{n-1}{n}} + \\sqrt[n+1]{\\frac{n}{n+1}} < \\frac{n^2}{n+1} + \\sqrt[n+1]{\\frac{n}{n+1}}.\n$$\nIt therefore suffices to show that\n$$\n\\begin{aligned}\n\\frac{n^2}{n+1} + \\sqrt[n+1]{\\frac{n}{n+1}} &\\le \\frac{(n+1)^2}{n+2} \\\\\n\\Leftrightarrow \\sqrt[n+1]{\\frac{n}{n+1}} &\\le \\frac{(n+1)^2}{n+2} - \\frac{n^2}{n+1} = \\frac{n^2 + 3n + 1}{(n+2)(n+1)} = 1 - \\frac{1}{(n+2)(n+1)}.\n\\end{aligned}\n$$\nUsing the Arithmetic-Geometric Mean Inequality we get\n$$\n\\sqrt[n+1]{\\frac{n}{n+1}} = \\sqrt[n+1]{\\frac{n}{n+1} \\cdot \\underbrace{1 \\cdot 1 \\cdots 1}_{n}} \\le \\frac{\\underbrace{\\frac{n}{n+1} + 1 + \\cdots + 1}_{n}}{n+1} = \\frac{n^2 + 2n}{(n+1)^2} = 1 - \\frac{1}{(n+1)^2}.\n$$\nFrom here it follows that\n$$\n\\sqrt[n+1]{\\frac{n}{n+1}} \\le 1 - \\frac{1}{(n+1)^2} < 1 - \\frac{1}{(n+2)(n+1)},\n$$\nwhich proves the inequality (1) and completes the induction step.\n\n*Remark: The inequality*\n$$\n\\sqrt[n+1]{\\frac{n}{n+1}} \\le 1 - \\frac{1}{(n+1)^2}\n$$\nalso follows directly from the Bernoulli inequality which states that\n$$\n(1+x)^r \\le 1+rx \\text{ for all } x > -1 \\text{ and } 0 \\le r \\le 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41795,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral such that $|AB| = |BC| = |CD|$. The lines $AB$ and $CD$ intersect at the point $E$, and the circumcircles of the triangles $ABC$ and $BDE$ intersect at $B$ and $F$. Let $P$ denote the intersection of the lines $AC$ and $BF$. Prove that $EP$ is the bisector of the angle $\\angle AED$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**II/3.** Due to different configurations we will proceed using directed angles.\nLet $T$, distinct from $E$, be the intersection of the circumcircles of the triangles $BDE$ and $ACE$. The power line (radical axis) of the circumcircles of the triangles $ABC$ and $BDE$ is $BF$, the power line of the circumcircles of the triangles $ABC$ and $ACE$ is $AC$, and the power line of the circumcircles of the triangles $ACE$ and $BDE$ is $ET$. The Power-of-a-Point theorem states that these three lines intersect at a single point. The lines $AC$ and $BF$ intersect at $P$, so $P$ must also lie on the line $ET$. Due to the concyclicity of the points $B, T, D$ and $E$ we have\n$$\n\\asymp TBA = \\asymp TBE = \\pi - \\asymp EDT = \\asymp TDC,\n$$\nand due to concyclicity of $A, T, C$ and $E$ we have\n$$\n\\asymp BAT = \\pi - \\asymp TAE = \\asymp ECT = \\asymp DCT.\n$$\nThe triangles $ABT$ and $CDT$ have three matching angles and $|AB| = |CD|$, so they are congruent. This implies that $|AT| = |CT|$ and $|BT| = |DT|$. Since $|AB| = |BC| = |CD|$, we see that the triangles $ABT$ and $CBT$ have three matching sides and are congruent and the same is true for the triangles $BCT$ and $DCT$. These two congruences imply that $\\asymp TBA = \\asymp CBT$ and $\\asymp TCB = \\asymp DCT$.\nIn the triangle $BCE$ the point $T$ lies on the intersection of the bisectors of the angles $\\asymp CBE$ and $\\asymp ECB$. So, $T$ is the incentre of the triangle $BCE$ and the bisector of the angle $\\asymp BEC$ also passes through $T$. The points $E, P$ and $T$ are collinear, so $EP$ is the bisector of $\\asymp AED$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41796,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a set of pairwise distinct positive integers. A division of $A$ into two disjoint non-empty subsets $A_1$ and $A_2$ is good, if the least common multiple of the elements of $A_1$ is the same as the greatest common divisor of the elements of $A_2$. Show that for a set with $3n + 2$ elements the maximal possible number of good divisions is equal to $2n$. (A simplification of the problem C3 from the IMO Shortlist 2015.)",
    "options": [],
    "answer": "2n",
    "solution": "Let $a_1 < a_2 < \\dots < a_{3n+2}$ be the positive integers in our set. For each good division of the set into subsets $A_1$ and $A_2$ the least common multiple of the numbers from $A_1$ is greater than or equal to all of the elements of $A_1$. Similarly, the greatest common divisor of the numbers in $A_2$ is smaller than or equal to all the numbers in $A_2$. So, all the numbers in $A_1$ have to be smaller than all of the numbers in $A_2$ and any good division can only be of the form\n$$\nA_1 = \\{a_1, a_2, \\dots, a_k\\} \\text{ in } A_2 = \\{a_{k+1}, a_{k+2}, \\dots, a_{3n+2}\\},\n$$\nwhere $1 \\le k \\le 3n+1$.\n\nFor each good division let us call the greatest element $a_k$ of $A_1$ the *dividing number*. For each $k$, $1 \\le k \\le 3n+1$, let $l_k$ denote the least common multiple of the numbers $a_1, \\dots, a_k$ and let $d_k$ denote the greatest common divisor of the numbers $a_{k+1}, \\dots, a_{3n+2}$.\n\nAt the moment we have $3n+1$ divisions which can potentially be good. We have to show that there are at most $2n$ good divisions. We observe the validity of the following propositions.\n\n**Proposition 1:** If $a_k$ and $a_{k+1}$ are both dividing numbers then we have $d_k = l_k = d_{k+1} = l_{k+1} = a_{k+1}$.\n\n*Proof:* Since $a_k$ and $a_{k+1}$ are both dividing we have $d_k = l_k$ and $d_{k+1} = l_{k+1}$. Also, $l_{k+1} = \\text{lcm}(a_{k+1}, l_k)$ and $d_k = \\text{gcd}(a_{k+1}, d_{k+1})$. But $l_k = d_k|a_{k+1}$, so $l_{k+1} = \\text{lcm}(a_{k+1}, l_k) = a_{k+1}$. Similarly, $a_{k+1} = l_{k+1} = d_{k+1}$, so $d_k = \\text{gcd}(a_{k+1}, d_{k+1}) = a_{k+1}$.\n\n**Proposition 2:** The numbers $a_{k-1}$, $a_k$ and $a_{k+1}$ cannot all be dividing numbers.\n\n*Proof:* If they were, then by Proposition 1 we would have $a_k = d_k = d_{k+1} = a_{k+1}$, which contradicts the assumption that the elements of the set $A$ are all distinct.\n\n**Proposition 3:** The numbers $a_1$ and $a_2$ cannot both be dividing numbers.\n\n*Proof:* If they were, then by Proposition 1 we would have $a_1 = l_1 = l_2 = a_2$, which contradicts the assumption that $a_1 < a_2$.\n\n**Proposition 4:** The numbers $a_{3n}$, $a_{3n+1}$ cannot both be dividing numbers.\n\n*Proof:* If they were, then by Proposition 1 we would have $a_{3n+1} = d_{3n+1} = a_{3n+2}$, which contradicts the assumption that $a_{3n+1} < a_{3n+2}$.\n\nThere are $3n+1$ candidates for the dividing numbers (i.e. $a_1, a_2, \\dots, a_{3n+1}$). However, out of any three subsequent numbers at most two can be dividing numbers. Also, the first two and the last two cannot both be dividing numbers. So, the number of good divisions is at most $\\frac{2((3n+2)-2)}{3} = 2n$.\n\nConsider the set\n$$\n\\begin{aligned}\nB &= \\{6^i, 2 \\cdot 6^i, 3 \\cdot 6^i \\mid 0 \\le i \\le n-1\\} \\cup \\{6^n, 2 \\cdot 6^n\\} \\\\\n&= \\{1, 2, 3, 6, \\dots, 3 \\cdot 6^{n-1}, 6^n, 2 \\cdot 6^n\\}.\n\\end{aligned}\n$$\nIn this set the dividing numbers are exactly $1, 6^1, \\dots, 6^{n-1}$ and $3, 3 \\cdot 6^1, \\dots, 3 \\cdot 6^{n-1}$. There are $2n$ of them, which shows that a set with $3n+2$ numbers that has $2n$ good divisions, exists.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41797,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an acute triangle $ABC$ let $D$ denote the foot of the altitude from $A$, and let $H$ be the orthocentre. Drop the tangent from the point $B$ to the circle centered at $H$ with radius $HD$ and let $P$, distinct from $D$, be the point where the tangent touches the circle. Drop the tangent from the point $C$ to the circle centered at $A$ with radius $AD$ and let $R$, distinct from $D$, be the point where the tangent touches the circle. Prove that the line $PR$ passes through two of the feet of the altitudes of the triangle $ABC$. (Inspired by a problem from Estonian Selection Exams 2015.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "There are two different configurations of this problem depending on whether the point $P$ lies inside the triangle $ABC$ or outside, so we use directed angles in our solution.\nLet $K_1$ denote the circle centered at $H$ with radius $HD$. Let $K_2$ denote the circle centered at $A$ with radius $AD$. Let $E$ and $F$ be the feet of the altitudes of the triangle $ABC$ from $B$ and $C$ respectively. We show that the points $P, R, E$ and $F$ are collinear.\nFrom $\\angle HFA = \\angle HEA = \\frac{\\pi}{2}$ it follows that the points $A, F, H$ and $E$ are concyclic.\nAlso, $\\angle AEB = \\angle ADB = \\frac{\\pi}{2}$ implies that $A, B, D$ and $E$ are concyclic.\nThe tangent $BP$ is perpendicular to the radius $HP$ of the circle $K_1$, so $\\angle BPH = \\frac{\\pi}{2}$. This implies that $\\angle BPH = \\angle BFH = \\angle BDH = \\frac{\\pi}{2}$, so the points $B, D, H, P$ and $F$ are concyclic. Similarly, the tangent $CR$ is perpendicular to the radius $AR$ of the circle $K_2$, so $\\angle ARC = \\angle AFC = \\angle ADC = \\frac{\\pi}{2}$ and the points $A, F, D, C$ and $R$ are concyclic as well.\nThe lengths of the tangents from their intersection to the point where they touch the circle are equal, so $|BP| = |BD|$, and the distance from the centre to points on the circle is constant, so $|HP| = |HD|$. The triangles $BPH$ and $BDH$ match in all three sides so they are congruent and $\\angle HBP = \\angle DBH$. From here and the concyclicities proven above it follows that\n$$\n\\angle HFP = \\angle HBP = \\angle DBH = \\angle DBE = \\angle DAE = \\angle HAE = \\angle HFE.\n$$\nThis implies that $E, F$ and $P$ are collinear.\nSimilarly, $|CR| = |CD|$ and $|AR| = |AD|$, so the triangles $ARC$ and $ADC$ are congruent and $\\angle DAC = \\angle CAR$. This implies that\n$$\n\\angle HFR = \\angle CFR = \\angle CAR = \\angle DAC = \\angle HAE = \\angle HFE.\n$$\nand the points $E, F$ and $R$ are collinear.\nWe have shown that the line $EF$ passes through the points $P$ and $R$, so $E, F, P$ and $R$ are collinear.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41798,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the greatest positive integer $m$ such that each square of the $m \\times m$ array can be painted either red or blue so that not all the squares at the intersection of any two rows and any two columns are the same colour. (From Finnish MO 2014.)",
    "options": [],
    "answer": "4",
    "solution": "For each row consider all pairs of squares from this row such that both squares are the same colour. From the condition of the problem it follows that no two rows can have any of these pairs in common. So, each pair can occur in at most one row. There are $\\binom{m}{2} = \\frac{m(m-1)}{2}$ pairs altogether and the squares in each pair are either both red or both blue. So there are $2 \\cdot \\frac{m(m-1)}{2} = m(m-1)$ possible pairs.\n\nOn the other hand we can estimate the lowest possible number of same-colour pairs in each row. Suppose that a row contains $k$ red and $m-k$ blue squares. Then the number of same-colour pairs is equal to $\\binom{k}{2} + \\binom{m-k}{2}$. This number can be bounded below by using the Quadratic-Arithmetic Means Inequality\n$$\n\\begin{aligned}\n\\binom{k}{2} + \\binom{m-k}{2} &= \\frac{k(k-1)}{2} + \\frac{(m-k)(m-k-1)}{2} \\\\\n&= \\frac{k^2 + (m-k)^2}{2} - \\frac{m}{2} \\\\\n&\\ge \\left( \\frac{k+(m-k)}{2} \\right)^2 - \\frac{m}{2} = \\frac{m(m-2)}{4}.\n\\end{aligned}\n$$\nSince the number of possible same-coloured pairs is at least as large as the total number of same-colour pairs in all the rows combined and we have bounded the number of same-colour pairs from below, we have\n$$\n\\begin{aligned}\nm(m-1) &\\ge m \\cdot \\left( \\frac{m(m-2)}{4} \\right) \\\\\n\\Rightarrow 0 &\\ge m^2 - 6m + 4.\n\\end{aligned}\n$$\nSolving the quadratic equation we see that $m \\le 3 + \\sqrt{5} < 6$. So, $m \\le 5$.\n\nWhen $m = 5$ the lower bound from above has the value of $\\frac{m(m-2)}{4} = \\frac{15}{4} > 3$. This means that any colouring will create at least 4 same-colour pairs in each row. Since the number of possible pairs is $m(m-1) = 20$ and there are 5 rows, each row has to contain exactly 4 same-colour pairs. This can only happen when the row contains 3 squares of one colour and 2 squares of the other colour. Altogether we have to use all of the 10 red pairs and all of the 10 blue pairs. But the number of rows is odd. So, the number of squares of one colour will have to be greater than the number of squares of the other colour and the same will be true for the number of red pairs and the number of blue pairs. Hence, the $5 \\times 5$ array cannot be coloured as desired.\n\n![](attached_image_1.png)\n\nWe have shown that the greatest positive integer with the desired property is $m = 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41799,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $a, b, c$ and $d$ such that\n$$\n4^a \\cdot 5^b - 3^c \\cdot 11^d = 1.\n$$",
    "options": [],
    "answer": "(1, 2, 2, 1)",
    "solution": "Consider the equation modulo $3$. We get $(-1)^b \\equiv 1 \\pmod{3}$, so $b$ must be even and we can write $b = 2b_1$. The equation can now be rearranged into\n$$\n2^{2a} \\cdot 5^{2b_1} - 1 = 3^c \\cdot 11^d \\Rightarrow (2^a \\cdot 5^{b_1} - 1)(2^a \\cdot 5^{b_1} + 1) = 3^c \\cdot 11^d.\n$$\nWe have $\\gcd(2^a \\cdot 5^{b_1} - 1, 2^a \\cdot 5^{b_1} + 1) = \\gcd(2^a \\cdot 5^{b_1} - 1, 2) = 1$, so the two factors are relatively prime. One of them must be equal to $3^c$ and the other to $11^d$. The case when $2^a \\cdot 5^{b_1} - 1 = 11^d$ has no solutions since reducing mod $5$ gives us $-1 \\equiv 1^d$ (mod $5$), which is not possible.\n\nThe only possibility left is\n$$\n\\begin{aligned}\n2^a \\cdot 5^{b_1} - 1 &= 3^c \\\\\n2^a \\cdot 5^{b_1} + 1 &= 11^d.\n\\end{aligned} \n\\qquad (2)\n$$\n\nSubtracting one of the equations from the other we get\n$$\n11^d - 3^c = 2. \\qquad (3)\n$$\nWhen $d = 1$ we get $c = 2$ and the solution $(a, b, c, d) = (1, 2, 2, 1)$. This is also the solution of the initial equation.\nWhen $d \\ge 2$ we have $c \\ge 4$ and the equation can be rearranged as\n$$\n\\begin{aligned}\n11^d - 3^c &= 2 = 11 - 3^2 \\\\\n\\Rightarrow 11^d - 11 &= 3^c - 3^2 \\\\\n\\Rightarrow 11 \\cdot (11^{d-1} - 1) &= 9 \\cdot (3^{c-2} - 1).\n\\end{aligned}\n$$\nThis implies $11|3^{c-2}-1$ or $3^{c-2} \\equiv 1 \\pmod{11}$. It can be checked that this holds if and only if $c - 2 = 5k$ for some $k \\in \\mathbb{N}$. So,\n$$\n\\begin{aligned}\n11 \\cdot (11^{d-1} - 1) &= 9 \\cdot (3^{5k} - 1) \\\\\n\\Rightarrow 11 \\cdot (11^{d-1} - 1) &= 9 \\cdot (3^5 - 1) \\cdot (3^{5(k-1)} + \\dots + 3^5 + 1)\n\\end{aligned}\n$$\nand $3^5 - 1 = 242 = 2 \\cdot 11^2 | 11 \\cdot (11^{d-1} - 1)$, but this is not possible since $11 \\nmid 11^{d-1} - 1$ for any $d \\ge 2$.\nThe only solution is therefore $(a, b, c, d) = (1, 2, 2, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41800,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPloščina pravilnega šestkotnika je $96 \\sqrt{3}~\\mathrm{cm}^{2}$. Koliko je obseg tega šestkotnika?\n\n(A) $48~\\mathrm{cm}$\n(B) $24~\\mathrm{cm}$\n(C) $96~\\mathrm{cm}$\n(D) $16~\\mathrm{cm}$\n(E) $20~\\mathrm{cm}$",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nUporabimo formulo za ploščino pravilnega šestkotnika $\\frac{6 a^{2} \\sqrt{3}}{4} = 96 \\sqrt{3}$ ter izračunamo $a = 8~\\mathrm{cm}$. Obseg pravilnega šestkotnika je $6a = 48~\\mathrm{cm}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41801,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSara in Klara imata skupaj 816 evrov. Če bi Sara porabila $\\frac{3}{5}$ svojega denarja in Klara $\\frac{3}{7}$ svojega denarja, bi obema ostalo enako. Koliko denarja ima Sara?\n\n(A) 408 evrov\n(B) 366 evrov\n(C) 336 evrov\n(D) 480 evrov\n(E) 816 evrov",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nOznačimo z $x$ količino Sarinega denarja in z $y$ količino Klarinega denarja. Potem je $x + y = 816$ in $\\frac{2}{5} x = \\frac{4}{7} y$. Iz druge enačbe izrazimo $x = \\frac{10}{7} y$ in vstavimo v prvo enačbo. Tako dobimo $\\frac{17}{7} y = 816$. Torej je $y = 336$ in $x = 480$. Sara ima 480 evrov.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41802,
    "subject": "Mathematics (Multi-modal)",
    "question": "Draw 4 distinct lines in the plane and let $n$ denote the number of intersections (if more than one line passes through the same point it still only counts as one intersection). Which of the following is the set of all possible values of $n$?\n(A) $\\{0, 2, 3, 4, 5, 6\\}$\n(B) $\\{0, 1, 3, 4, 5, 6\\}$\n(C) $\\{0, 1, 3, 4, 6\\}$\n(D) $\\{1, 3, 4, 5, 6\\}$\n(E) $\\{0, 1, 2, 3, 4, 6\\}$",
    "options": [],
    "answer": "B",
    "solution": "As shown in the figures, four distinct lines can intersect in $0$, $1$, $3$, $4$, $5$ or $6$ points.\n![](attached_image_1.png)\n\nAssume that there are exactly two intersections. At each of these either two or three lines meet. Suppose that at one of the intersections, denote it by $A$, three of the lines meet. One of these three lines, call it $p$, has to pass through the other intersection, denoted $B$. The remaining two lines through $A$ are denoted by $q$ and $r$, and the final line, which also passes through $B$, is denoted by $s$. Now, the line $s$ intersects at least one of $q$ and $r$ in a third intersection, which is a contradiction. So, exactly two lines should meet at each of the two intersections. But in this case the third intersection would again have to exist. The correct answer is (B).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41803,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three friends, Andrej, Blaž and Cene, were playing badminton. For each game, two of them were playing one against the other, and the third was free. After each game, the winner of the game played against the one that was free in the last game. Andrej played $17$ games and Blaž played $23$ games. At least how many games did Cene play?\n(A) $6$\n(B) $11$\n(C) $12$\n(D) $14$\n(E) $23$",
    "options": [],
    "answer": "D",
    "solution": "Let $n$ be the number of games played by Cene. Then the total number of games was $\\frac{17+23+n}{2}$, which implies that $n$ is even. The total number of games was at least $23$ since Blaž has played this many of them. Since Cene was free for at most one game in a row he had to have played at least $11$ games, but $n$ is even which means he played at least $12$ games. So the total number of games was at least $\\frac{17+23+12}{2} = 26$, which in turn implies that Cene has played at least $13$ games, or, due to parity, at least $14$. In this case Andrej and Blaž have played $\\frac{17+23-14}{2} = 13$ times, Andrej and Cene $\\frac{17+14-23}{2} = 4$ times and Blaž and Cene $\\frac{23+14-17}{2} = 10$ times. To show that this is indeed possible consider the following example: C-A, A-B, B-C, B-A, A-C, A-B, B-C, B-A, A-C, A-B, B-C, B-A, A-C, A-B, B-C, B-A, B-C, B-A, B-C, B-A, B-C, B-A, B-C, B-A, B-C, B-A, B-C, B-A, B-C. The correct answer is (D).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41804,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJure je na ravnini risal po 4 različne premice v različnih položajih in z $n$ označil število presečišč premic v posamezni legi. S katero množico so podane natanko vse možne vrednosti števila $n$?\n\n(A) $\\{0,2,3,4,5,6\\}$\n(B) $\\{0,1,3,4,5,6\\}$\n(C) $\\{0,1,3,4,6\\}$\n(D) $\\{1,3,4,5,6\\}$\n(E) $\\{0,1,2,3,4,6\\}$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nKot prikazujejo slike, se lahko štiri različne premice v ravnini sekajo v $0,1,3,4$, 5 ali 6 točkah.\n\n![](attached_image_1.png)\n\nDenimo, da sta presečišči natanko 2. V vsakem od teh presečišč se potem sekata bodisi 2 premici bodisi 3 premice. Denimo, da se v enem presešišču, označimo ga z $A$, sekajo 3 premice. Ena od teh treh premic, označimo jo s $p$, mora potem potekati tudi skozi drugo presečišče, ki ga označimo z $B$. Preostali premici, ki potekata skozi $A$ označimo s $q$ in $r$, zadnjo premico, ki poteka skozi $B$, pa označimo z $s$. Tedaj bo premica $s$ sekala vsaj eno izmed premic $q$ in $r$ v tretjem presečišču, kar je protislovje. Torej bi se morali v vsakem izmed dveh presečišč sekati natanko 2 premici. Toda tudi v tem primeru bi obstajalo med temi premicami vsaj še eno presečišče. Pravilen odgovor je (B).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41805,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKatera izmed navedenih enačb premic je enačba simetrale daljice s krajiščema $A(3,-3)$ in $B(2,-2)$ ?\n(A) $y=-x-3$\n(B) $x+y-1=0$\n(C) $x-y-5=0$\n(D) $3 x-2 y=0$\n(E) $\\frac{x}{3}-\\frac{y}{2}=1$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\nSimetrala daljice $AB$ poteka skozi razpolovišče $S\\left(\\frac{5}{2},-\\frac{5}{2}\\right)$ daljice $AB$ in je pravokotna na daljico $AB$. Smerni koeficient premice skozi točki $A$ in $B$ je $k_{1}=\\frac{-2-(-3)}{2-3}=-1$. Torej je smerni koeficient simetrale $k_{2}=-\\frac{1}{k_{1}}=1$. Upoštevamo $y=kx+n$ oziroma $-\\frac{5}{2}=1 \\cdot \\frac{5}{2}+n$ in dobimo $n=-5$. Torej je enačba simetrale daljice $AB$ enaka $y=x-5$ oziroma $x-y-5=0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41806,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAritmetična sredina dveh pozitivnih števil je 65, njuna geometrijska sredina pa 60. Koliko je absolutna vrednost razlike teh dveh števil?\n\n(A) 10\n(B) 20\n(C) 30\n(D) 40\n(E) 50",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nUpoštevamo $\\frac{a+b}{2}=65$ in $\\sqrt{a b}=60$. Preoblikujemo v sistem enačb $a+b=130$ in $a b=3600$. Iz prve enačbe izrazimo $b=130-a$ in vstavimo v drugo enačbo ter jo preoblikujemo v $a^{2}-130 a+3600=0$. Dobimo rešitvi $a_{1}=40$ in $b_{1}=90$ ter $a_{2}=90$ in $b_{2}=40$. Torej sledi $|a-b|=50$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41807,
    "subject": "Mathematics (Multi-modal)",
    "question": "Take an equilateral triangle $ABC$ with the side of length $a$. Draw the squares $BADE$ and $CBFG$ on the sides $AB$ and $BC$, respectively (see figure). How long is the line segment $DG$?\n![](attached_image_1.png)\n(A) $(\\sqrt{2} + 1)a$\n(B) $(2\\sqrt{2} + 1)a$\n(C) $(\\sqrt{3} + 1)a$\n(D) $\\sqrt{3}a$\n(E) 1",
    "options": [],
    "answer": "(\\sqrt{3} + 1)a",
    "solution": "Denote the intersections of the segment $DG$ with the lines $AB$ and $BC$ by $K$ and $L$ respectively. Due to symmetry the segment $DG$ is parallel to the segment $AC$, so $KBL$ is an equilateral triangle. The triangles $DKA$ and $LGC$ are one half of an equilateral triangle with altitudes of length $a$ and sides $|DK| = |GL| = \\frac{2a}{\\sqrt{3}}$. The side of the equilateral triangle $KBL$ therefore measures\n$$\n|KL| = a - |AK| = a - \\frac{|DK|}{2} = a - \\frac{a}{\\sqrt{3}}.\n$$\nThe length of the segment $DG$ is\n$$\n|DG| = 2|DK| + |KL| = \\frac{4a}{\\sqrt{3}} + \\left(a - \\frac{a}{\\sqrt{3}}\\right) = a + \\frac{3a}{\\sqrt{3}} = (1 + \\sqrt{3})a.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41808,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the equality of the three fractions $\\frac{2}{6} = \\frac{3}{9} = \\frac{58}{174}$ each digit from $1$ to $9$ occurs exactly once and the value of all three fractions is $\\frac{1}{3}$. Here is another example of such equality, $\\frac{*}{*} = \\frac{*}{*} = \\frac{7*}{15*}$, where some of the digits have been replaced by an asterisk. What is the value of all three fractions in this case?\n(A) $\\frac{1}{2}$ (B) $\\frac{1}{3}$ (C) $\\frac{2}{3}$ (D) $\\frac{3}{4}$ (E) $\\frac{3}{5}$",
    "options": [],
    "answer": "A",
    "solution": "Sorting the available answers we get $\\frac{1}{3} < \\frac{1}{2} < \\frac{3}{5} < \\frac{2}{3} < \\frac{3}{4}$. The last of the three fractions in the equality can be estimated as\n$$\n\\frac{1}{3} = \\frac{60}{180} < \\frac{7*}{15*} < \\frac{90}{150} = \\frac{3}{5}.\n$$\nSo, the only possible answer is (A). After some quick trial and error we get $\\frac{2}{4} = \\frac{3}{6} = \\frac{79}{158}$, so (A) is indeed the correct answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41809,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNad stranicama $AB$ in $BC$ enakostraničnega trikotnika $ABC$ s stranico dolžine $a$ načrtamo kvadrata $BAD E$ in $CBFG$ (glej sliko). Koliko je dolga daljica $DG$?\n(A) $(\\sqrt{2}+1)a$\n(B) $(2\\sqrt{2}+1)a$\n(C) $(\\sqrt{3}+1)a$\n(D) $\\sqrt{3}a$\n(E) 1\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "C",
    "solution": "Solution:\nOznačimo presečišči daljice $DG$ s stranicama $AB$ in $BC$ zaporedoma s $K$ in $L$. Zaradi simetrije je daljica $DG$ vzporedna stranici $AC$. Torej je trikotnik $KBL$ enakostraničen, trikotnika $DKA$ in $LGC$ pa sta polovici enakostraničnih trikotnikov z višinama dolžine $a$ in stranicama $|DK|=|GL|=\\frac{2a}{\\sqrt{3}}$. Stranica enakostraničnega trikotnika $KBL$ je zato dolga\n$$\n|KL|=a-|AK|=a-\\frac{|DK|}{2}=a-\\frac{a}{\\sqrt{3}}\n$$\nDolžina daljice $DG$ je enaka\n$$\n|DG|=2|DK|+|KL|=\\frac{4a}{\\sqrt{3}}+\\left(a-\\frac{a}{\\sqrt{3}}\\right)=a+\\frac{3a}{\\sqrt{3}}=(1+\\sqrt{3})a\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41810,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVrednost izraza $10^{2016} - 10^{15}$ je naravno število. Koliko je vsota števk tega naravnega števila?\n\n(A) 1\n(B) 17\n(C) 2001\n(D) 18000\n(E) 18009",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nNaravno število, ki predstavlja vrednost izraza $10^{2016} - 10^{15}$ ima 2016 števk, zadnjih 15 je enakih 0, ostale pa so enake 9. Vsota števk tega števila je torej $(2016 - 15) \\times 9 = 18009$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41811,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bodo $a$, $b$ in $c$ dolžine stranic trikotnika in naj velja $c^{2}=a^{2}+b^{2}+a b$. Katera izmed navedenih trditev velja za tak trikotnik?\n(A) Trikotnik je enakostraničen.\n(B) Trikotnik je ostrokoten.\n(C) Višinska točka leži izven trikotnika.\n(D) Trikotnik je pravokoten.\n(E) Središče trikotniku očrtane krožnice je razpolovišče najdaljše stranice.",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nIz kosinusnega izreka $c^{2}=a^{2}+b^{2}-2 a b \\cos \\gamma$ in enakosti $c^{2}=a^{2}+b^{2}+a b$ dobimo $a b=-2 a b \\cos \\gamma$. Od tod sledi $\\cos \\gamma=-\\frac{1}{2}$ oziroma $\\gamma=120^{\\circ}$. Trikotnik je topokoten, torej leži višinska točka izven trikotnika.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41812,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $f(x)=\\frac{1}{\\sqrt{a^{x}}}$, $a>0$ in $a \\neq 1$. Koliko je $f\\left(\\log _{a} 4\\right)$ ?\n\n(A) 2\n(B) $-\\frac{1}{2}$\n(C) $\\frac{1}{2}$\n(D) -4\n(E) -2",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nIzračunamo $f\\left(\\log _{a} 4\\right)=\\frac{1}{\\sqrt{a^{\\log _{a} 4}}}=\\frac{1}{\\sqrt{4}}=\\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41813,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEnakostranični trikotnik na sliki je s črtami, vzporednimi eni od stranic, razdeljen na 5 enako širokih pasov, od katerih so trije pobarvani. Kolikšen delež trikotnika je pobarvan?\n\n(A) $52 \\%$\n(B) $58 \\%$\n(C) $60 \\%$\n(D) $68 \\%$\n(E) $72 \\%$\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nEnakokrak trikotnik na sliki razdelimo na manjše skladne trikotnike, tako da dorišemo še črte vzporedne preostalima dvema stranicama. Potem je od 25 majhnih trikotnikov pobarvanih 15. Torej je pobarvanih $\\frac{15}{25}=\\frac{60}{100}$ trikotnika, kar je $60 \\%$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41814,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKoliko je $1+3,5+6+8,5+\\ldots+2501$, če je razlika med zaporednima seštevancema stalna?\n\n(A) 1249750\n(B) 1251000\n(C) 1251750,5\n(D) 1252251\n(E) 1253502,5",
    "options": [],
    "answer": "D",
    "solution": "Solution:\nUgotovimo, da moramo sešteti $n$ členov aritmetičnega zaporedja z diferenco $d=2,5$. Upoštevamo $a_n=a_1+(n-1)d$ oziroma $2501=1+(n-1) \\cdot 2,5$. Rešimo enačbo in ugotovimo, da moramo sešteti $1001$ členov. Upoštevamo $S_n=\\frac{n}{2}\\left(a_1+a_n\\right)$ ali $S_n=\\frac{n}{2}\\left(2a_1+(n-1)d\\right)$ in dobimo $S_{1001}=1252251$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41815,
    "subject": "Mathematics (Multi-modal)",
    "question": "The value of the expression $10^{2016} - 10^{15}$ is a positive integer. Determine the sum of its digits.\n\n(A) 1\n(B) 17\n(C) 2001\n(D) 18\\,000\n(E) 18\\,009",
    "options": [],
    "answer": "E",
    "solution": "The positive integer which represents the value of the expression $10^{2016} - 10^{15}$ has 2016 digits. The last 15 of them are 0 and the rest are equal to 9. The sum of the digits is therefore $(2016 - 15) \\cdot 9 = 18\\,009$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41816,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSmetana predstavlja $7 \\%$ mase mleka, iz $61 \\%$ mase smetane pa nastane maslo. Iz koliko mleka nastane $3 \\mathrm{~kg}$ masla?\n(A) manj kot $5 \\mathrm{~kg}$\n(B) $17 \\mathrm{~kg}$\n(C) $54 \\mathrm{~kg}$\n(D) $69 \\mathrm{~kg}$\n(E) več kot $70 \\mathrm{~kg}$",
    "options": [],
    "answer": "E",
    "solution": "Solution:\nOznačimo z $x$ maso mleka v kg. V $x$ kg mleka je $0{,}07 x \\mathrm{~kg}$ smetane. Iz te smetane nastane $0{,}61 \\cdot 0{,}07 x \\mathrm{~kg}$ masla. Velja $0{,}61 \\cdot 0{,}07 x=3$. Sledi, da je $x \\doteq 70{,}258 \\mathrm{~kg}$. $3 \\mathrm{~kg}$ masla nastane iz več kot $70 \\mathrm{~kg}$ mleka.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41817,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of every three consecutive terms of the sequence $a_1, a_2, a_3, a_4, \\dots$ equals $2016$. It also holds that $a_{667} = 667$ and $a_{1004} = 1004$. What is the value of $a_{2016}$?\n(A) 0\n(B) 345\n(C) 667\n(D) 1004\n(E) 2016",
    "options": [],
    "answer": "B",
    "solution": "Since the sum of any three consecutive terms is the same we have $a_{n+3} = a_n$ for all $n$. This implies that $a_{667} = a_{667+3 \\cdot 449} = a_{2014}$ and $a_{1004} = a_{1004+3 \\cdot 337} = a_{2015}$. From $a_{2014} + a_{2015} + a_{2016} = 2016$ we conclude that $a_{2016} = 2016 - 667 - 1004 = 345$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41818,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV zaporedju $a_{1}, a_{2}, a_{3}, a_{4}, \\ldots$ je vsota vsakih treh zaporednih členov enaka 2016. Velja še $a_{667}=667$ in $a_{1004}=1004$. Koliko je $a_{2016}$?\n\n(A) 0\n(B) 345\n(C) 667\n(D) 1004\n(E) 2016",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nKer je vsota vsakih treh zaporednih členov zaporedja enaka, velja $a_{n+3}=a_{n}$ za vsak $n$. Od tod sledi $a_{667}=a_{667+3 \\cdot 449}=a_{2014}$ in $a_{1004}=a_{1004+3 \\cdot 337}=a_{2015}$. Ker je $a_{2014}+a_{2015}+a_{2016}=2016$, je $a_{2016}=2016-667-1004=345$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41819,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEden izmed kotov izjemnega enakokrakega trikotnika je $10^{\\circ}$ večji od nekega drugega kota tega trikotnika. Najmanj koliko je velik najmanjši kot izjemnega enakokrakega trikotnika?\n\n(A) $52^{\\circ} 20^{\\prime}$\n(B) $53^{\\circ} 20^{\\prime}$\n(C) $56^{\\circ} 40^{\\prime}$\n(D) $63^{\\circ} 20^{\\prime}$\n(E) $66^{\\circ} 40^{\\prime}$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nOznačimo kote izjemnega enakokrakega trikotnika z $\\alpha, \\alpha$ in $\\gamma$. Tedaj je razlika med $\\alpha$ in $\\gamma$ enaka $10^{\\circ}$. Imamo dve možnosti.\n\nČe je $\\alpha>\\gamma$, je $\\alpha-\\gamma=10^{\\circ}$ in seveda $\\alpha+\\alpha+\\gamma=180^{\\circ}$. Od tod izračunamo $3 \\alpha=190^{\\circ}$, torej je $\\alpha=63^{\\circ} 20^{\\prime}$ in $\\gamma=53^{\\circ} 20^{\\prime}$.\n\nČe pa je $\\alpha<\\gamma$, je $\\gamma-\\alpha=10^{\\circ}$ in $\\alpha+\\alpha+\\gamma=180^{\\circ}$. Sledi $3 \\alpha=170^{\\circ}$, torej je $\\alpha=56^{\\circ} 40^{\\prime}$ in $\\gamma=66^{\\circ} 40^{\\prime}$.\n\nNajmanjša možna vrednost najmanjšega kota je torej $53^{\\circ} 20^{\\prime}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41820,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa neničelni realni števili $a$ in $b$, $a \\neq -1$ in $b \\neq -1$, velja\n$$\n\\frac{a}{b+1} + \\frac{b}{a+1} = 1.\n$$\nKatera trditev o izrazu\n$$\n\\frac{a}{b} + \\frac{b}{a} - \\frac{1}{ab}\n$$\nje pravilna?\n\n(A) Izraz lahko zavzame poljubno vrednost z intervala $(0,1]$.\n(B) Izraz lahko zavzame poljubno vrednost z intervala $[1,2)$.\n(C) Vrednost izraza je 1.\n(D) Vrednost izraza je 2.\n(E) O vrednosti izraza ne moremo povedati ničesar.",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nV dani enakosti odpravimo ulomke in jo poenostavimo do $a^{2} + b^{2} = ab + 1$. Dani izraz postavimo na skupni imenovalec, da dobimo\n$$\n\\frac{a^{2} + b^{2} - 1}{ab}.\n$$\nIz enakosti sledi, da je vrednost izraza enaka\n$$\n\\frac{ab}{ab} = 1.\n$$\nPravilen odgovor je (C).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41821,
    "subject": "Mathematics (Multi-modal)",
    "question": "The lines containing the chords $AC$ and $BD$ of a circle with radius $1$ intersect at an angle of $45^\\circ$. The length of the chord $AB$ equals $1$ (see figure).\n![](attached_image_1.png)\nWhat is the size of the angle $\\angle CBD$?\n\n(A) $45^\\circ$\n(B) $60^\\circ$\n(C) $65^\\circ$\n(D) $75^\\circ$\n(E) $85^\\circ$",
    "options": [],
    "answer": "D",
    "solution": "Let $S$ denote the centre of the circle. Then the triangle $SBA$ is equilateral since all of its sides have length $1$. So, $\\angle BSA = 60^\\circ$. By the Central Angle Theorem $\\angle BCA = \\frac{1}{2} \\angle BSA = 30^\\circ$. The outer angle of the triangle is always equal to the sum of the non-adjacent inner angles, which implies $\\angle CBD = 45^\\circ + 30^\\circ = 75^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41822,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKvadratni enačbi $x^{2} + a x + 2 = 0$ in $x^{2} + 2 x + a = 0$, pri čemer je $a$ realno število, imata realne rešitve. Vsota kvadratov rešitev prve enačbe je enaka vsoti kvadratov rešitev druge enačbe. Koliko je $a$?\n\n(A) $-4$\n(B) $-2$\n(C) $0$\n(D) $4$\n(E) Nič od naštetega.",
    "options": [],
    "answer": "(A)",
    "solution": "Solution:\n\nEnačba $x^{2} + a x + 2 = 0$ ima diskriminanto $D = a^{2} - 8 \\geq 0$ in rešitvi $x_{1,2} = \\frac{-a \\pm \\sqrt{a^{2} - 8}}{2}$, za kateri velja $x_{1}^{2} + x_{2}^{2} = a^{2} - 4$.\n\nEnačba $x^{2} + 2 x + a = 0$ ima diskriminanto $D = 4 - 4a \\geq 0$ in rešitvi $x_{1,2} = \\frac{-2 \\pm \\sqrt{4 - 4a}}{2}$, za kateri velja $x_{1}^{2} + x_{2}^{2} = 4 - 2a$.\n\nSledi $a^{2} - 4 = 4 - 2a$, kar lahko preoblikujemo v $(a - 2)(a + 4) = 0$.\n\nRešitev $a = 2$ ni prava, saj bi bila diskriminanta druge enačbe negativna, rešitev $a = -4$ pa ustreza vsem pogojem.\n\nPravilen odgovor je (A).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41823,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKoliko je vrednost izraza $\\left(\\frac{\\sqrt{5}-3}{2}\\right)^{2016} \\frac{\\sqrt{5}-3}{2}\\left(\\frac{\\sqrt{5}+3}{2}\\right)^{2016}$?\n(A) 1\n(B) $\\frac{\\sqrt{5}}{2}$\n(C) $\\frac{\\sqrt{5}-3}{2}$\n(D) $\\frac{3-\\sqrt{5}}{2}$\n(E) $\\frac{(\\sqrt{5})^{2016}+3^{2016}}{8}$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n$$\n\\begin{gathered}\n\\left(\\frac{\\sqrt{5}-3}{2}\\right)^{2016} \\frac{\\sqrt{5}-3}{2}\\left(\\frac{\\sqrt{5}+3}{2}\\right)^{2016}=\\left(\\frac{\\sqrt{5}-3}{2} \\cdot \\frac{\\sqrt{5}+3}{2}\\right)^{2016} \\frac{\\sqrt{5}-3}{2}= \\\\\n=\\left(\\frac{5-9}{4}\\right)^{2016} \\frac{\\sqrt{5}-3}{2}=1 \\cdot \\frac{\\sqrt{5}-3}{2}=\\frac{\\sqrt{5}-3}{2}\n\\end{gathered}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41824,
    "subject": "Mathematics (Multi-modal)",
    "question": "Which quadratic equation has roots $1 - \\sqrt{2}$ and $\\frac{1}{\\sqrt{2}-1}$?\n(A) $x^2 + 2x - 1 = 0$   (B) $x^2 + \\sqrt{2}x - 1 = 0$   (C) $x^2 + x - 1 = 0$\n(D) $x^2 - 2x - 1 = 0$   (E) $x^2 - x + 2 = 0$",
    "options": [],
    "answer": "D",
    "solution": "Rationalize the second root to get $\\frac{1}{\\sqrt{2}-1} = \\frac{\\sqrt{2}+1}{2-1} = \\sqrt{2} + 1$. These roots belong to the equation $(x - (1 - \\sqrt{2})) (x - (\\sqrt{2} + 1)) = 0$. Expanding the left side we get $x^2 - 2x - 1 = 0$. The correct answer is (D).",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41825,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKatera izmed navedenih trditev ne velja za funkcijo $f$ s predpisom $f(x)=\\frac{1}{2}-\\frac{1}{2} \\cos x$?\n(A) Zaloga vrednosti funkcije $f$ je $[0,1]$.\n(B) Osnovna perioda funkcije $f$ je $2 \\pi$.\n(C) Funkcija $f$ je soda.\n(D) Ničle funkcije $f$ so $x=k \\pi,\\ k \\in \\mathbb{Z}$.\n(E) Funkcija $f$ doseže največjo vrednost za $x=\\pi+2 k \\pi,\\ k \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "D",
    "solution": "Solution:\nNiçle funkcije $f$ so rešitve enačbe $\\frac{1}{2}-\\frac{1}{2} \\cos x=0$, torej $x=2 k \\pi,\\ k \\in \\mathbb{Z}$, in ne $x=k \\pi,\\ k \\in \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41826,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a real number for which the quadratic equations $x^2 + a x + 2 = 0$ and $x^2 + 2x + a = 0$ have real roots. The sum of the squares of the roots of the first equation equals the sum of the squares of the roots of the second equation. What is the value of $a$?\n\n(A) $-4$\n(B) $-2$\n(C) $0$\n(D) $4$\n(E) None of the listed values.",
    "options": [],
    "answer": "A",
    "solution": "The discriminant of the equation $x^2 + a x + 2 = 0$ is $D = a^2 - 8 \\ge 0$. Its roots $x_{1,2} = \\frac{-a \\pm \\sqrt{a^2-8}}{2}$ satisfy $x_1^2 + x_2^2 = a^2 - 4$.\n\nThe discriminant of the equation $x^2 + 2x + a = 0$ is $D = 4 - 4a \\ge 0$ and the roots $x_{1,2} = \\frac{-2 \\pm \\sqrt{4-4a}}{2}$ satisfy $x_1^2 + x_2^2 = 4 - 2a$.\n\nIt follows that $a^2 - 4 = 4 - 2a$, which can be rearranged as $(a-2)(a+4) = 0$.\n\nThe solution $a=2$ is not the correct one because it would result in the discriminant of the second equation being negative. The solution $a=-4$ satisfies all of the conditions.\n\nThe correct answer is (A).",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41827,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKatera izmed naštetih kvadratnih enačb ima rešitvi $1-\\sqrt{2}$ in $\\frac{1}{\\sqrt{2}-1}$?\n\n(A) $x^{2}+2x-1=0$\n(B) $x^{2}+\\sqrt{2}x-1=0$\n(C) $x^{2}+x-1=0$\n(D) $x^{2}-2x-1=0$\n(E) $x^{2}-x+2=0$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\nDrugo ničlo racionaliziramo $\\frac{1}{\\sqrt{2}-1}=\\frac{\\sqrt{2}+1}{2-1}=\\sqrt{2}+1$. Predpisani rešitvi ima torej kvadratna enačba $(x-(1-\\sqrt{2}))(x-(\\sqrt{2}+1))=0$. Ko levo stran zmnožimo, dobimo $x^{2}-2x-1=0$. Pravilen odgovor je (D).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41828,
    "subject": "Mathematics (Multi-modal)",
    "question": "Non-zero real numbers $a$ and $b$ satisfy $\\frac{a}{b+1} + \\frac{b}{a+1} = 1$. Which of the statements about the expression $\\frac{a}{b} + \\frac{b}{a} - \\frac{1}{ab}$ is correct?\n(A) The expression can take any value in the interval $(0, 1]$.\n(B) The expression can take any value in the interval $[1, 2)$.\n(C) The value of the expression is 1.\n(D) The value of the expression is 2.\n(E) We cannot say anything about the value of the expression.",
    "options": [],
    "answer": "C",
    "solution": "Remove the fractions and simplify the equation to get $a^2 + b^2 = ab + 1$. Change the fractions in the given expression to the common denominator to obtain $\\frac{a^2 + b^2 - 1}{ab}$. The equality now implies that the value of the expression is $\\frac{ab}{ab} = 1$. The correct answer is (C).",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 41829,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKocko z robom dolžine $a$ razrežemo na 8 malih skladnih kock. Površino ene take male kocke označimo s $P$. Koliko je $P$?\n\n(A) $\\frac{a^{2}}{2}$\n(B) $\\frac{3 a^{2}}{4}$\n(C) $\\frac{a^{2}}{8}$\n(D) $\\frac{3 a^{2}}{2}$\n(E) $a^{8}$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nOznačimo z $a_{1}$ dolžino roba male kocke. Upoštevamo, da je prostornina male kocke $a_{1}^{3} = \\frac{a^{3}}{8}$ in dobimo $a_{1} = \\frac{a}{2}$. Torej velja $P = 6 a_{1}^{2} = 6\\left(\\frac{a}{2}\\right)^{2} = \\frac{3 a^{2}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41830,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nV katerega izmed navedenih izrazov lahko preoblikujemo izraz $\\left(x+y+\\frac{1}{4}\\right)^{2}-\\left(x+y-\\frac{1}{4}\\right)^{2}$?\n(A) $4 x y$\n(B) $\\frac{1}{16}$\n(C) $\\frac{1}{8}$\n(D) 0\n(E) $x+y$",
    "options": [],
    "answer": "E",
    "solution": "Solution:\nDani izraz razstavimo po pravilu razlike kvadratov in dobimo\n$\\left(x+y+\\frac{1}{4}-\\left(x+y-\\frac{1}{4}\\right)\\right)\\left(x+y+\\frac{1}{4}+\\left(x+y-\\frac{1}{4}\\right)\\right)=\\frac{1}{2}(2 x+2 y)=x+y$. Dani izraz lahko preoblikujemo v izraz $x+y$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41831,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $x$ and $y$ that solve the system of equations\n$$\n\\log_3 x^2 + \\log_2 y^3 = 1,\n$$\n$$\n\\log_9 x^4 + \\log_4 y^9 = 2.\n$$",
    "options": [],
    "answer": "x = ±1/√3, y = 4^(1/3)",
    "solution": "Denote $a = \\log_3 x^2$ and $b = \\log_2 y^3$. Then\n$$\n\\log_9 x^4 = \\frac{\\log_3 x^4}{\\log_3 9} = \\frac{2 \\log_3 x^2}{2} = a \\quad \\text{and} \\quad \\log_4 y^9 = \\frac{\\log_2 y^9}{\\log_2 4} = \\frac{3 \\log_2 y^3}{2} = \\frac{3}{2}b.\n$$\nInserting this into the initial equations we get $a+b=1$ and $a+\\frac{3}{2}b=2$. Subtracting the first equation from the second yields $\\frac{1}{2}b=1$ or $b=2$. From the first equation we get $a=-1$. From $\\log_3 x^2 = -1$ it follows that $x^2 = \\frac{1}{3}$ or $x = \\pm\\frac{1}{\\sqrt{3}}$, and from $\\log_2 y^3 = 2$ we have $y^3 = 4$ or $y = \\sqrt[3]{4}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41832,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $a$, $b$, $c$, and $d$ that solve the system of equations\n$$\na^2 + b^2 + c^2 = d + 13,\n$$\n$$\na + 2b + 3c = \\frac{d}{2} + 13.\n$$",
    "options": [],
    "answer": "(0, 2, 3, 0), (2, 2, 3, 4), (1, 1, 3, -2), (1, 3, 3, 6), (1, 2, 2, -4), (1, 2, 4, 8)",
    "solution": "From the second equation we express $d = 2a + 4b + 6c - 26$ and insert it into the first equation to get\n$$\na^2 + b^2 + c^2 = 2a + 4b + 6c - 13.\n$$\nMoving all the terms to the left we have $a^2 + b^2 + c^2 - 2a - 4b - 6c + 13 = 0$, which we now rewrite as the sum of perfect squares\n$$\n(a - 1)^2 + (b - 2)^2 + (c - 3)^2 = 1.\n$$\nAll three perfect squares on the left are non-negative integers, so one of them is equal to 1 and the other two are 0. If $(a - 1)^2 = 1$, we have $a = 0$ or $a = 2$ as well as $b = 2$ and $c = 3$. If $(b - 2)^2 = 1$, we have $b = 1$ or $b = 3$ as well as $a = 1$ and $c = 3$. If $(c - 3)^2 = 1$, we have $c = 2$ or $c = 4$ as well as $a = 1$ and $b = 2$. In each case we can determine $d$ from the equation given above. The integer solutions $(a, b, c, d)$ of the given system of equations are therefore $(0, 2, 3, 0)$, $(2, 2, 3, 4)$, $(1, 1, 3, -2)$, $(1, 3, 3, 6)$, $(1, 2, 2, -4)$ and $(1, 2, 4, 8)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41833,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nPoišči vse četverice celih števil $a, b, c$ in $d$, ki rešijo sistem enačb\n$$\n\\begin{aligned}\n& a^{2}+b^{2}+c^{2}=d+13 \\\\\n& a+2 b+3 c=\\frac{d}{2}+13\n\\end{aligned}\n$$",
    "options": [],
    "answer": "[[0, 2, 3, 0], [2, 2, 3, 4], [1, 1, 3, -2], [1, 3, 3, 6], [1, 2, 2, -4], [1, 2, 4, 8]]",
    "solution": "Solution:\nIz druge enačbe izrazimo $d=2 a+4 b+6 c-26$ in vstavimo v prvo enačbo, da dobimo\n$$\na^{2}+b^{2}+c^{2}=2 a+4 b+6 c-13\n$$\nVse člene postavimo na levo stran $a^{2}+b^{2}+c^{2}-2 a-4 b-6 c+13=0$ in dopolnimo do popolnih kvadratov\n$$\n(a-1)^{2}+(b-2)^{2}+(c-3)^{2}=1\n$$\nKer so vsi trije popolni kvadrati nenegativna cela števila, je eden od njih enak 1, druga dva pa sta enaka 0. Če je $(a-1)^{2}=1$, sledi $a=0$ ali $a=2$ ter $b=2$ in $c=3$. Če je $(b-2)^{2}=1$, sledi $b=1$ ali $b=3$ ter $a=1$ in $c=3$. Če pa je $(c-3)^{2}=1$, sledi $c=2$ ali $c=4$ ter $a=1$ in $b=2$. V vsakem posameznem primeru iz zgornje enačbe izračunamo še $d$. Celoštevilske rešitve $(a, b, c, d)$ danega sistema so torej $(0,2,3,0),(2,2,3,4),(1,1,3,-2)$, $(1,3,3,6),(1,2,2,-4)$ in $(1,2,4,8)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41834,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all relatively prime integers $x$ and $y$ that solve the equation\n$$\n4x^3 + y^3 = 3xy^2.\n$$",
    "options": [],
    "answer": "(-1, 1), (1, -1), (1, 2), (-1, -2)",
    "solution": "Adding $3y^3$ to the equation and factoring both sides we obtain $4(x + y)(x^2 - xy + y^2) = 3y^2(x + y)$. Moving all terms to the left side we can bring out $(x + y)$, so $(x + y)(4x^2 - 4xy + y^2) = 0$. By factoring the second expression we get $(x + y)(2x - y)^2 = 0$. So, $y = -x$ or $y = 2x$. Since $x$ and $y$ are relatively prime, we conclude that $x$ can only be $1$ or $-1$. From this we get four pairs $(x, y)$ of relatively prime solutions to the equation, namely $(1, -1)$, $(-1, 1)$, $(1, 2)$ and $(-1, -2)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41835,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs of real numbers $a$ and $b$, which satisfy the inequality\n$$\na^{2}(2a - b) + b^{2}(2b - a) \\geq 0.\n$$",
    "options": [],
    "answer": "a + b ≥ 0",
    "solution": "Expanding the left side of the inequality we get $2a^3 - a^2b - ab^2 + 2b^3$ and factoring it back up again results in\n$$\n\\begin{aligned}\n2a^3 - a^2b - ab^2 + 2b^3 &= 2(a^3 + b^3) - ab(a + b) = 2(a + b)(a^2 - ab + b^2) - ab(a + b) = \\\\\n&= (a + b)(2a^2 - 3ab + 2b^2).\n\\end{aligned}\n$$\nThe second factor is non-negative since $2a^2 - 3ab + 2b^2 = 2(a^2 - \\frac{3}{2}ab + b^2) = 2((a - \\frac{3}{4}b)^2 + \\frac{7}{16}b^2) \\ge 0$. So, either $2((a - \\frac{3}{4}b)^2 + \\frac{7}{16}b^2) = 0$ and $a = b = 0$ or $a + b \\ge 0$. The inequality is satisfied by all pairs of reals $a$ and $b$ such that $a + b \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 41836,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nPoišči vse pare realnih števil $x$ in $y$, ki rešijo sistem enačb\n$$\n\\begin{aligned}\n& \\log_{3} x^{2} + \\log_{2} y^{3} = 1 \\\\\n& \\log_{9} x^{4} + \\log_{4} y^{9} = 2\n\\end{aligned}\n$$",
    "options": [],
    "answer": "(x, y) = (1/sqrt(3), 2^(2/3)) or (x, y) = (-1/sqrt(3), 2^(2/3))",
    "solution": "Solution:\nS prehodom na novo osnovo sistem enačb preoblikujemo do\n$$\n\\begin{aligned}\n& \\frac{\\log x^{2}}{\\log 3} + \\frac{\\log y^{3}}{\\log 2} = 1 \\\\\n& \\frac{\\log x^{4}}{\\log 9} + \\frac{\\log y^{9}}{\\log 4} = 2\n\\end{aligned}\n$$\nOpazimo, da mora biti število $y$ nujno pozitivno, zato je $\\log y^{3} = 3 \\log y$ in $\\log y^{9} = 9 \\log y$. Število $x$ pa je lahko tudi negativno, zato velja $\\log x^{2} = 2 \\log |x|$ in $\\log x^{4} = 4 \\log |x|$. Če upoštevamo še $\\log 4 = \\log 2^{2} = 2 \\log 2$ in $\\log 9 = \\log 3^{2} = 2 \\log 3$, lahko sistem enačb zapišemo kot\n$$\n\\begin{aligned}\n& \\frac{2 \\log |x|}{\\log 3} + \\frac{3 \\log y}{\\log 2} = 1 \\\\\n& \\frac{4 \\log |x|}{2 \\log 3} + \\frac{9 \\log y}{2 \\log 2} = 2\n\\end{aligned}\n$$\nSistem najlažje rešimo, če prvo enačbo odštejemo od druge, saj se člena z $\\log |x|$ pokrajšata in dobimo $\\frac{3 \\log y}{2 \\log 2} = 1$. Od tod izračunamo $\\log y = \\frac{2}{3} \\log 2 = \\log 2^{\\frac{2}{3}}$, torej je $y = 2^{\\frac{2}{3}} = \\sqrt[3]{4}$. Če $\\log y = \\frac{2}{3} \\log 2$ vstavimo v prvo enačbo, dobimo še $\\frac{2 \\log |x|}{\\log 3} = -1$, od koder izrazimo $\\log |x| = -\\frac{1}{2} \\log 3 = \\log 3^{-\\frac{1}{2}}$. Torej je $|x| = 3^{-\\frac{1}{2}} = \\frac{1}{\\sqrt{3}}$ oziroma $x = \\pm \\frac{1}{\\sqrt{3}}$.\n\n\n2. način. Označimo $a = \\log_{3} x^{2}$ in $b = \\log_{2} y^{3}$. Tedaj je\n$$\n\\log_{9} x^{4} = \\frac{\\log_{3} x^{4}}{\\log_{3} 9} = \\frac{2 \\log_{3} x^{2}}{2} = a \\quad \\text{in} \\quad \\log_{4} y^{9} = \\frac{\\log_{2} y^{9}}{\\log_{2} 4} = \\frac{3 \\log_{2} y^{3}}{2} = \\frac{3}{2} b\n$$\nKo to vstavimo v prvotni enačbi, dobimo enačbi $a + b = 1$ in $a + \\frac{3}{2} b = 2$. Prvo enačbo odštejemo od druge, da dobimo $\\frac{1}{2} b = 1$ oziroma $b = 2$. Iz prve enačbo izračunamo še $a = -1$. Iz $\\log_{3} x^{2} = -1$ sledi $x^{2} = \\frac{1}{3}$ oziroma $x = \\pm \\frac{1}{\\sqrt{3}}$, iz $\\log_{2} y^{3} = 2$ pa $y^{3} = 4$ oziroma $y = \\sqrt[3]{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41837,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse pare tujih celih števil $x$ in $y$, ki rešijo enačbo\n$$\n4 x^{3}+y^{3}=3 x y^{2}\n$$",
    "options": [],
    "answer": "(1, -1), (1, 2), (-1, 1), (-1, -2)",
    "solution": "Solution:\n\nŠtevilo $x$ deli $4 x^{3}$ in $3 x y^{2}$, zato mora deliti tudi $y^{3}$. Ker pa sta števili $x$ in $y$ tuji, je to mogoče le, če je $x=1$ ali $x=-1$.\n\nČe je $x=1$, dobimo enakost $4+y^{3}=3 y^{2}$, ki jo preoblikujemo v $4=y^{2}(3-y)$. Torej $y^{2}$ deli $4$, zato je bodisi $y^{2}=1$ ali $y^{2}=4$. V prvem primeru iz enakosti sledi $y=-1$, v drugem pa $y=2$. Če pa je $x=-1$, dobimo enakost $-4+y^{3}=-3 y^{2}$, ki jo preoblikujemo do $y^{2}(y+3)=4$. Sklepamo podobno kot v prvem primeru, da pridemo do rešitev $y=1$ in $y=-2$.\n\nPari $(x, y)$ tujih celih števil, ki rešijo enačbo, so torej $(1,-1),(1,2),(-1,1)$ in $(-1,-2)$.\n\n\n2. način. Dani enačbi prištejemo $3 y^{3}$ in nato obe strani razstavimo, da dobimo enačbo $4(x+y)\\left(x^{2}-x y+y^{2}\\right)=3 y^{2}(x+y)$. Če postavimo vse člene na levo stran enačbe, lahko izpostavimo $(x+y)$, torej $(x+y)\\left(4 x^{2}-4 x y+y^{2}\\right)=0$, nato pa razstavimo še drugi faktor, da dobimo enačbo $(x+y)(2 x-y)^{2}=0$. Torej je bodisi $y=-x$ ali pa $y=2 x$. Ker pa sta $x$ in $y$ tuji števili, je lahko $x$ enak le $1$ ali $-1$. Tako dobimo štiri pare $(x, y)$ tujih števil, ki rešijo enačbo, to so $(1,-1),(-1,1),(1,2)$ in $(-1,-2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41838,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoloči vse pare realnih števil $a$ in $b$, ki ustrezajo neenakosti\n$$\na^{2}(2a-b) + b^{2}(2b-a) \\geq 0\n$$",
    "options": [],
    "answer": "a + b ≥ 0",
    "solution": "Solution:\n\nLevo stran neenakosti zmnožimo, da dobimo $2a^{3} - a^{2}b - ab^{2} + 2b^{3}$, in jo razstavimo\n$$\n\\begin{aligned}\n2a^{3} - a^{2}b - ab^{2} + 2b^{3} &= 2(a^{3} + b^{3}) - ab(a + b) = 2(a + b)(a^{2} - ab + b^{2}) - ab(a + b) = \\\\\n&= (a + b)(2a^{2} - 3ab + 2b^{2})\n\\end{aligned}\n$$\nDrugi faktor je nenegativen, saj velja $2a^{2} - 3ab + 2b^{2} = 2(a^{2} - \\frac{3}{2}ab + b^{2}) = 2\\left(\\left(a - \\frac{3}{4}b\\right)^{2} + \\frac{7}{16}b^{2}\\right) \\geq 0$. Torej mora biti bodisi $2\\left(\\left(a - \\frac{3}{4}b\\right)^{2} + \\frac{7}{16}b^{2}\\right) = 0$ in zato $a = b = 0$ ali pa $a + b \\geq 0$. Neenakosti iz naloge ustrezajo vsi pari realnih števil $a$ in $b$, za katere velja $a + b \\geq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41839,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa trikotnik $ABC$ velja $|AB| = 2|AC|$, $D$ pa je taka točka na poltraku $CA$, da velja $|CD| = 3|AC|$. Dokaži, da pravokotnica iz točke $C$ na premico $BD$ razpolavlja daljico $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOznačimo presečišče pravokotnice iz točke $C$ na premico $BD$ in premice $BD$ z $E$, presečišče premic $CE$ in $AB$ pa z $M$. Velja $|AD| = 2|AC| = |AB|$, zato je trikotnik $DBA$ enakokrak z vrhom pri $A$, torej velja $\\angle BDA = \\angle ABD$. Sledi $\\angle ACM = 90^{\\circ} - \\angle EDC = 90^{\\circ} - \\angle BDA = 90^{\\circ} - \\angle ABD = \\angle EMB = \\angle CMA$. Torej je tudi trikotnik $CAM$ enakokrak z vrhom pri $A$. Od tod sledi $|AM| = |AC| = \\frac{1}{2}|AB|$, kar pomeni, da je $M$ razpolovišče daljice $AB$.\n\n\n2. način. Naj bosta točki $E$ in $M$ kot v prvi rešitvi in naj bo $F$ presečišče pravokotnice skozi točko $A$ na premico $DB$ in premice $DB$. Iz podobnosti trikotnikov $DFA$ in $DEC$ sledi $\\frac{|DF|}{|FE|} = \\frac{|DA|}{|AC|} = 2$ oziroma $|DF| = 2|FE|$. Kot v prvi rešitvi sklepamo, da je trikotnik $DBA$ enakokrak z vrhom pri $A$, torej velja $|DF| = |FB|$. Iz teh dveh enakosti sledi $|FE| = |EB|$. Iz podobnosti trikotnikov $FBA$ in $EBM$ sledi $\\frac{|AM|}{|MB|} = \\frac{|FE|}{|EB|} = 1$, torej je $M$ razpolovišče daljice $AB$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41840,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathcal{K}$ be a circle with centre $O$, and let $\\mathcal{K}'$ be a circle that goes through the point $O$ with a radius that is greater than twice the radius of the circle $\\mathcal{K}$. A common tangent of the circles $\\mathcal{K}$ and $\\mathcal{K}'$ touches the circle $\\mathcal{K}$ at point $A$, and it touches the circle $\\mathcal{K}'$ at point $B$. Let $C$ denote the mirror image of the point $B$ with respect to point $A$. The line $AO$ intersects the circle $\\mathcal{K}'$ at points $O$ and $D$, and line $CD$ intersects circle $\\mathcal{K}'$ at points $D$ and $E$. Prove that line $BE$ is a tangent of the circle $\\mathcal{K}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the tangent from $B$ to the circle $K$ (distinct from the tangent $AB$) touch the circle $K$ at $A'$. We show that the points $B$, $A'$ and $E$ are collinear.\n\nWe have $|AB| = |A'B|$ and $|AO| = |A'O|$, so the triangles $ABO$ and $A'BO$ have three equal sides and are therefore congruent. This implies $\\angle OBA' = \\angle ABO$. By the Tangent-Chord Theorem in the circle $K'$ we have $\\angle ABO = \\angle BDO$. The line $AB$ is tangent to $K$, so $\\angle OAB = 90^\\circ$. We have $|AB| = |AC|$, so the triangles $ABD$ and $ACD$ match in two sides and the angle between them. We conclude that they are congruent and\n\n![](attached_image_1.png)\n\n$\\angle BDO = \\angle BDA = \\angle ADC = \\angle ODE$. The angles over the same chord of $K'$ are equal, so $\\angle ODE = \\angle OBE$.\n\nWe have shown that $\\angle OBA' = \\angle OBE$, which implies that the points $B$, $A'$ and $E$ are colinear and the line $BE$ is tangent to $K$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41841,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $\\mathcal{K}$ krožnica s središčem $O$ in $\\mathcal{K}'$ krožnica, ki poteka skozi točko $O$ in ima polmer večji od dvakratnika polmera krožnice $\\mathcal{K}$. Skupna tangenta krožnic $\\mathcal{K}$ in $\\mathcal{K}'$ se dotika krožnice $\\mathcal{K}$ v točki $A$, krožnice $\\mathcal{K}'$ pa v točki $B$. Označimo s $C$ zrcalno sliko točke $B$ pri zrcaljenju čez točko $A$. Premica $A O$ seka krožnico $\\mathcal{K}'$ v točkah $O$ in $D$, premica $C D$ pa seka krožnico $\\mathcal{K}'$ v točkah $D$ in $E$. Dokaži, da je premica $B E$ tangenta na krožnico $\\mathcal{K}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\nTangenta iz točke $B$ na krožnico $\\mathcal{K}$ različna od tangente $A B$ naj se krožnice $\\mathcal{K}$ dotika v točki $A'$. Dokazali bomo, da so točke $B, A'$ in $E$ kolinearne.\nKer je $|A B|=|A' B|$ in $|A O|=|A' O|$, se trikotnika $A B O$ in $A' B O$ ujemata v vseh treh stranicah in sta skladna. Sledi $\\angle O B A' = \\angle A B O$. Po izreku o kotu med tetivo in tangento za krožnico $\\mathcal{K}'$ velja $\\angle A B O = \\angle B D O$. Premica $A B$ je tangentna na $\\mathcal{K}$, zato je $\\angle O A B = 90^\\circ$. Ker je $|A B|=|A C|$, se trikotnika $A B D$ in $A C D$ ujemata v dveh stranicah in kotu med njima. Zato sta skladna in sledi $\\angle B D O = \\angle B D A = \\angle A D C = \\angle O D E$. Zaradi ujemanja kotov nad istim lokom znotraj krožnice $\\mathcal{K}'$ velja $\\angle O D E = \\angle O B E$.\nDokazali smo $\\angle O B A' = \\angle O B E$, kar pomeni, da so točke $B, A'$ in $E$ kolinearne, zato je premica $B E$ tangenta na krožnico $\\mathcal{K}$.\n\n\n2. način.\n![](attached_image_2.png)\nNaj bo $O'$ središče krožnice $\\mathcal{K}'$ in $F$ presečišče premic $O O'$ in $B E$. Dokazali bomo, da je premica $B E$ pravokotna na $O O'$, daljica $O F$ pa radij krožnice $\\mathcal{K}$.\nEnako kot v prvi rešitvi velja, da sta trikotnika $A B D$ in $A C D$ skladna. Po izreku o središčnem in obodnem kotu za krožnico $\\mathcal{K}'$ velja $\\angle B O' O = 2 \\angle B D O = 2 \\angle O D E = \\angle O O' E$. Ker je $|O' B|=|O' O|=|O' E|$, se trikotnika $O' B O$ in $O' E O$ ujemata v dveh stranicah in kotu med njima. Zato sta skladna in velja $|B O|=|E O|$. S tem smo dokazali, da je $O E O' B$ deltoid. Ker se diagonali deltoida sekata pravokotno, sledi $\\angle B F O = 90^\\circ$.\nKer je $A B$ tangentna tudi na krožnico $\\mathcal{K}'$, trikotnik $O' B O$ pa je enakokrak, velja $\\angle B O A = 90^\\circ - \\angle A B O = \\angle O B O' = \\angle O' O B = \\angle F O B$. Velja tudi $\\angle O A B = 90^\\circ = \\angle B F O$. Torej se trikotnika $A B O$ in $F B O$ ujemata v kotih in skupni stranici $B O$. Zato sta skladna in sledi $|O F|=|O A|$.\nDokazali smo, da je $O F$ radij krožnice $\\mathcal{K}$, premica $B E$ pa je nanj pravokotna v točki $F$. Zato je $B E$ tangenta na krožnico $\\mathcal{K}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41842,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle and let $D$ be a point in the interior of the triangle, such that $\\angle BAD = \\angle DCB$ and $\\angle CBD = \\angle DAC$. Prove that the lines $AD$ and $BC$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "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$. The equality $\\angle BAD = \\angle DCB$ implies that the triangles $GAD$ and $ECD$ are similar since they have two common angles. So,\n$$\n\\frac{|GD|}{|AD|} = \\frac{|ED|}{|CD|}.\n$$\nSimilarly, the equality $\\angle CBD = \\angle DAC$ implies that the triangles $FAD$ and $EBD$ are similar, so\n![](attached_image_1.png)\n$$\n\\frac{|AD|}{|FD|} = \\frac{|BD|}{|ED|}.\n$$\nIf we multiply the above inequalities we get\n$$\n\\frac{|GD|}{|FD|} = \\frac{|BD|}{|CD|} \\quad \\text{or} \\quad \\frac{|GD|}{|BD|} = \\frac{|FD|}{|CD|}.\n$$\nSince $\\angle GDB = \\angle CDF$, we conclude that the triangles $GDB$ and $FDC$ are similar, so $\\angle DBG = \\angle FCD$ and $\\angle DBA = \\angle ACD$. This and the assumptions of the problem imply that\n$$\n\\angle BAD + \\angle CBD + \\angle DBA = \\frac{1}{2}(\\angle BAC + \\angle ACB + \\angle CBA) = 90^\\circ,\n$$\nso $\\angle AEB = 180^\\circ - (\\angle BAD + \\angle CBD + \\angle DBA) = 90^\\circ$, which is what we wanted to show.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41843,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTočke $A(2,-1)$, $B(-1,-5)$ in $C(-3,-13)$ so oglišča trikotnika $ABC$. Izračunaj dolžino težničnice na stranico $BC$. Rezultat naj bo natančen in delno korenjen. Točki $A$ in $B$ prezrcali čez abscisno os. Dobljeni točki označi z $A'$ in $B'$. Nariši štirikotnik $AA'B'B$ in izračunaj njegovo ploščino.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "median length = 4*sqrt(5); area = 18",
    "solution": "Solution:\n\nZapis koordinat razpolovišča stranice $BC$: $R(-2,-9)$\n\nUgotovitev, da je dolžina težiščnice enaka razdalji med točkama $A$ in $R$\n\nIzračun dolžine težiščnice:\n$$\nt_{BC} = d(A, R) = \\sqrt{(-2-2)^2 + (-9+1)^2}\n$$\n\nZapis rešitve:\n$$\nt_{BC} = 4\\sqrt{5}\n$$\n\nPrezrcaljeni točki $A$ in $B$ ter narisan štirikotnik $AA'B'B$\n\n1. način\nUporaba formule za ploščino trapeza:\n$$\nS = \\frac{a + c}{2} \\cdot v = \\frac{10 + 2}{2} \\cdot 3\n$$\n\nZapis rešitve:\n$$\nS = 18\n$$\n\n2. način\nUporaba formule za ploščino trikotnika ali za ploščino pravokotnika in pravokotnega trikotnika\n\nZapis rešitve:\n$$\nS = 18\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41844,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle such that $|AB| = 2|AC|$ and let $D$ be a point on the ray $CA$ such that $|CD| = 3|AC|$. Prove that the line from $C$ perpendicular to the line $BD$ bisects the segment $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the intersection of the line through $C$ perpendicular to $BD$ and the line $BD$ by $E$ and the intersection of the lines $CE$ and $AB$ by $M$.\n\nWe have $|AD| = 2|AC| = |AB|$, so the triangle $DBA$ is isosceles with the apex at $A$, and so $\\angle BDA = \\angle ABD$.\n\nThis implies $\\angle ACM = 90^\\circ - \\angle EDC = 90^\\circ - \\angle BDA = 90^\\circ - \\angle ABD = \\angle EMB = \\angle CMA$.\n\nSo, the triangle $CAM$ is also isosceles with the apex at $A$. From here we get $|AM| = |AC| = \\frac{1}{2}|AB|$, which implies that $M$ is the midpoint of the segment $AB$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41845,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo\n$$\nf(x) = \\ln x \\quad \\text{in} \\quad g(x) = \\frac{1}{3} x^{3} - x^{2} - 6x + 4\n$$\nter $p$ tangenta na graf funkcije $f$ v točki $S(a, b)$, $q$ pa tangenta na graf funkcije $g$ v točki $T(a, c)$. Tangenta $p$ je pravokotna na tangento $q$. Izračunaj vrednost $a$.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nOdvod funkcije $f$: $f'(x) = \\frac{1}{x}$\n\nOdvod funkcije $g$: $g'(x) = x^{2} - 2x - 6$\n\nUpoštevanje $k_1 = f'(a) = \\frac{1}{a}$ in $k_2 = g'(a) = a^{2} - 2a - 6$\n\nUpoštevanje $k_1 k_2 = -1$ oziroma $\\frac{a^{2} - 2a - 6}{a} = -1$\n\nPreoblikovanje enačbe v $a^{2} - a - 6 = 0$\n\nZapis rešitev kvadratne enačbe $a_1 = 3$ in $a_2 = -2$\n\nUgotovitev, da rešitev $a_2 = -2$ ne ustreza ($f(-2)$ ne obstaja)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41846,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with the angle $\\angle BAC = 30^\\circ$, and let $P$ be an arbitrary point inside this triangle. Denote the mirror images of the point $P$ with respect to the sides $BC$, $CA$, and $AB$ by $P_A$, $P_B$, and $P_C$, respectively. Let $P_A P_B P_C$ be an equilateral triangle. Prove that $\\angle BPC = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D$, $E$ and $F$ denote the intersections of the segments $PP_A$, $PP_B$ and $PP_C$ with the segments $BC$, $CA$ and $AB$ respectively. The points $D$, $E$ and $F$ are also the midpoints of $PP_A$, $PP_B$ and $PP_C$, so the triangles $DEF$ and $P_A P_B P_C$ are similar. We conclude that $DEF$ is an equilateral triangle and $\\angle EDF = 60^\\circ$.\n![](attached_image_1.png)\nFrom $\\angle PEC = 90^\\circ$ and $\\angle CDP = 90^\\circ$ it follows that the points $P$, $D$, $C$ and $E$ are concyclic, which implies $\\angle DPC = \\angle DEC$. Similarly the points $B$, $D$, $P$ and $F$ are concyclic and $\\angle BPD = \\angle BFD$. From here $\\angle BPC = \\angle BPD + \\angle DPC = \\angle BFD + \\angle DEC = (180^\\circ - \\angle DFA) + (180^\\circ - \\angle AED) = 360^\\circ - \\angle DFA - \\angle AED = \\angle EDF + \\angle FAE = 60^\\circ + 30^\\circ = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41847,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je ostrokoten trikotnik $ABC$ in taka točka $D$ v notranjosti tega trikotnika, da velja $\\Varangle BAD = \\Varangle DCB$ in $\\Varangle CBD = \\Varangle DAC$. Dokaži, da sta premici $AD$ in $BC$ pravokotni.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOznačimo z $E$ presečišče premic $AD$ in $BC$, z $F$ presečišče premic $BD$ in $CA$ ter z $G$ presečišče premic $CD$ in $AB$. Ker je $\\Varangle BAD = \\Varangle DCB$, sta trikotnika $GAD$ in $ECD$ podobna, saj imata dva skladna kota. Torej je\n$$\n\\frac{|GD|}{|AD|} = \\frac{|ED|}{|CD|}\n$$\nPodobno iz enakosti $\\Varangle CBD = \\Varangle DAC$ sledi, da sta tudi trikotnika $FAD$ in $EBD$ podobna, zato velja\n$$\n\\frac{|AD|}{|FD|} = \\frac{|BD|}{|ED|}\n$$\nČe zgornji dve enakosti zmnožimo, dobimo\n$$\n\\frac{|GD|}{|FD|} = \\frac{|BD|}{|CD|} \\quad \\text{oziroma} \\quad \\frac{|GD|}{|BD|} = \\frac{|FD|}{|CD|}\n$$\nKer je hkrati $\\Varangle GDB = \\Varangle CDF$, sledi, da sta tudi trikotnika $GDB$ in $FDC$ podobna, torej je $\\Varangle DBG = \\Varangle FCD$ oziroma $\\Varangle DBA = \\Varangle ACD$. Od tod in iz podatkov naloge sledi\n$$\n\\Varangle BAD + \\Varangle CBD + \\Varangle DBA = \\frac{1}{2}(\\Varangle BAC + \\Varangle ACB + \\Varangle CBA) = 90^{\\circ}\n$$\ntorej je $\\Varangle AEB = 180^{\\circ} - (\\Varangle BAD + \\Varangle CBD + \\Varangle DBA) = 90^{\\circ}$, kar je bilo potrebno pokazati.\n\n\nSolution 2:\n\n![](attached_image_2.png)\n\nNaj bo $C'$ zrcalna slika točke $C$ pri zrcaljenju preko premice $BD$. Torej je $\\Varangle BC'D = \\Varangle DCB = \\Varangle BAD$. Poleg tega točki $A$ in $C'$ ležita na istem bregu premice $BD$, saj točka $D$ leži znotraj trikotnika $ABC$. Od tod sledi, da so točke $A, B, D$ in $C'$ konciklične.\n\nLočimo dva primera. Če točki $C'$ in $D$ ležita na nasprotnih bregovih premice $AB$, potem velja $\\Varangle C'AD = 180^{\\circ} - \\Varangle DBC' = 180^{\\circ} - \\Varangle CBD = 180^{\\circ} - \\Varangle DAC$. V tem primeru so torej točke $C', A$ in $C$ kolinearne. Če pa točki $C'$ in $D$ ležita na istem bregu premice $AB$, potem iz dejstva, da točki $A$ in $C'$ ležita na istem bregu premice $BD$, sledi, da točki $A$ in $D$ ležita na nasprotnih bregovih premice $BC'$. Torej velja $\\Varangle BAC' = 180^{\\circ} - \\Varangle C'DB = 180^{\\circ} - \\Varangle BDC = \\Varangle DCB + \\Varangle CBD = \\Varangle BAD + \\Varangle DAC = \\Varangle BAC$. Tudi v tem primeru so točke $C', A$ in $C$ kolinearne.\n\nKer je premica $CC'$ po definiciji točke $C'$ pravokotna na premico $BD$, sklepamo, da je premica $BD$ višina trikotnika $ABC$. Zaradi simetrije lahko na enak način dokažemo, da je tudi premica $CD$ višina trikotnika $ABC$. To pa pomeni, da je $D$ višinska točka trikotnika $ABC$, zato je tudi premica $AD$ pravokotna na premico $BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41848,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $m$ realno število in $f(x) = (m-2)x^{2} - 2mx + 3m$.\n\na) Izračunaj vrednost parametra $m$ tako, da bo graf funkcije $f$ potekal skozi točko $T(-2,3)$. Za tako izračunano vrednost parametra $m$ zapiši predpis funkcije $f$.\n\nb) Izračunaj, za katero vrednost parametra $m$ graf funkcije $f$ ni parabola. Kaj je v tem primeru graf funkcije $f$? Za tako izračunano vrednost parametra $m$ zapiši predpisa funkcije $f$ in njene inverzne funkcije $f^{-1}$.",
    "options": [],
    "answer": "a) m = 1; f(x) = -x^2 - 2x + 3. b) m = 2; f(x) = -4x + 6; f^{-1}(x) = -1/4 x + 3/2.",
    "solution": "Solution:\n\na.\n\nVstavimo točko $T(-2, 3)$ v funkcijo $f(x)$:\n\n$$\nf(-2) = (m-2)(-2)^2 - 2m(-2) + 3m = 3\n$$\n\nIzračunamo:\n\n$$\n(m-2) \\cdot 4 + 4m + 3m = 3\n$$\n$$\n4m - 8 + 4m + 3m = 3\n$$\n$$\n(4m + 4m + 3m) - 8 = 3\n$$\n$$\n11m - 8 = 3\n$$\n$$\n11m = 11\n$$\n$$\nm = 1\n$$\n\nPredpis funkcije $f$ za $m = 1$ je:\n\n$$\nf(x) = (1-2)x^2 - 2 \\cdot 1 \\cdot x + 3 \\cdot 1 = -x^2 - 2x + 3\n$$\n\nb.\n\nGraf funkcije $f$ ni parabola, kadar je koeficient ob $x^2$ enak $0$:\n\n$$\nm-2 = 0 \\implies m = 2\n$$\n\nV tem primeru je funkcija $f$ linearna:\n\n$$\nf(x) = (2-2)x^2 - 2 \\cdot 2 x + 3 \\cdot 2 = 0 - 4x + 6 = -4x + 6\n$$\n\nInverzna funkcija $f^{-1}$:\n\nNaj bo $y = -4x + 6$.\n\n$$\ny = -4x + 6\n$$\n$$\ny - 6 = -4x\n$$\n$$\nx = -\\frac{1}{4}y + \\frac{3}{2}\n$$\n\nTorej:\n\n$$\nf^{-1}(x) = -\\frac{1}{4}x + \\frac{3}{2}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41849,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKot $\\Varangle BAC$ trikotnika $ABC$ je velik $30^\\circ$, $P$ pa je poljubna točka v notranjosti tega trikotnika. Zrcalne slike točke $P$ pri zrcaljenu čez stranice $BC$, $CA$ in $AB$ zaporedoma označimo s $P_A$, $P_B$ in $P_C$. Denimo, da je $P_A P_B P_C$ enakostraničen trikotnik. Dokaži, da je tedaj $\\Varangle BPC = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nNaj bodo $D, E$ in $F$ zaporedoma presečisča daljic $P P_A$, $P P_B$ in $P P_C$ s stranicami $BC$, $CA$ in $AB$. Točke $D, E$ in $F$ so hkrati tudi razpolovišča daljic $P P_A$, $P P_B$ in $P P_C$, zato sta si trikotnika $DEF$ in $P_A P_B P_C$ podobna. Torej je tudi trikotnik $DEF$ enakostraničen in velja $\\Varangle EDF = 60^\\circ$.\n\nKer je $\\Varangle PEC = 90^\\circ$ in $\\Varangle CDP = 90^\\circ$, so točke $P, D, C$ in $E$ konciklične. Torej velja $\\Varangle DPC = \\Varangle DEC$. Podobno so tudi točke $B, D, P$ in $F$ konciklične in velja $\\Varangle BPD = \\Varangle BFD$. Od tod sledi $\\Varangle BPC = \\Varangle BPD + \\Varangle DPC = \\Varangle BFD + \\Varangle DEC = \\left(180^\\circ - \\Varangle DFA\\right) + \\left(180^\\circ - \\Varangle AED\\right) = 360^\\circ - \\Varangle DFA - \\Varangle AED = \\Varangle EDF + \\Varangle FAE = 60^\\circ + 30^\\circ = 90^\\circ$.\n\n\n2. način. Kot v prvi rešitvi pokažemo, da so točke $P, D, C$ in $E$ konciklične. Ker sta $P_B$ in $P_C$ zrcalni sliki točke $P$ pri zrcaljenju preko premic $CA$ in $AB$, velja $|A P_B| = |A P| = |A P_C|$. Torej je trikotnik $A P_C P_B$ enakokrak z vrhom pri $A$. Poleg tega velja $\\Varangle PAC = \\Varangle CAP_B$ in $\\Varangle P_C AB = \\Varangle BAP$. Zato je $\\Varangle P_C A P_B = 2 \\Varangle BAC = 60^\\circ$. To pomeni, da je trikotnik $A P_C P_B$ enakostraničen. Ker pa je po predpostavki tudi trikotnik $P_A P_B P_C$ enakostraničen, je štirikotnik $A P_C P_A P_B$ romb, zato sta premici $A P_C$ in $P_A P_B$ vzporedni. Prav tako sta vzporedni tudi premici $DE$ in $P_A P_B$, saj sta točki $D$ in $E$ razpolovišči daljic $P P_A$ in $P P_B$. Torej sta tudi premici $DE$ in $A P_C$ vzporedni. Od tod in iz koncikličnosti točk $P, D, C$ in $E$ sledi $\\Varangle DPC = \\Varangle DEC = \\Varangle P_C AC$. Zaradi simetrije lahko na enak način pokažemo, da je tudi $\\Varangle BPD = \\Varangle BAP_B$. Z upoštevanjem obeh enakosti dobimo $\\Varangle BPC = \\Varangle BPD + \\Varangle DPC = \\Varangle BAP_B + \\Varangle P_C AC = \\Varangle P_C A P_B + \\Varangle BAC = 30^\\circ + 60^\\circ = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41850,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Nariši graf funkcije $f$ s predpisom\n$$\nf(x)=\\begin{cases}\n2x+1 & \\text{za } x>1 \\\\\n-x+5 & \\text{za } x \\leq 1\n\\end{cases}\n$$\n\nb) Naj bo $g(x)=2x+1$. Zapiši vse vrednosti spremenljivke $x$, za katere velja enakost $g^{-1}(x)=g\\left(x^{-1}\\right)$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "x = 4 or x = -1",
    "solution": "Solution:\n\na)\nNarisana premica z enačbo $y=2x+1$ oziroma del te premice.\nNarisana premica z enačbo $y=-x+5$ oziroma del te premice.\nPravilno označen graf funkcije $f$ v $x=1$.\n\nb)\nPostopek za izračun predpisa inverzne funkcije.\nZapis $g^{-1}(x)=\\frac{x-1}{2}$.\nZapis $g\\left(x^{-1}\\right)=2x^{-1}+1$.\nZapis enačbe $x^{2}-3x-4=0$.\nZapis rešitev enačbe $x_{1}=4$ in $x_{2}=-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41851,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoenostavi izraz\n$$\n\\frac{\\frac{2 x y}{x+y}-x}{\\frac{1}{y}+\\frac{1}{x-2 y}}+\\frac{\\left(x^{2}-x y+y^{2}\\right)\\left(x^{3}-x(x-y)^{2}\\right)}{x^{3}+y^{3}}\n$$\nZa katere realne vrednosti $x$ in $y$ izraz nima pomena?",
    "options": [],
    "answer": "The expression simplifies to xy. It is undefined for y = 0, x = -y, x = 2y, or x = y.",
    "solution": "Solution:\n\nRazširjanje ulomkov na skupni imenovalec\n$$\n\\frac{\\frac{2 x y}{x+y}-x}{\\frac{1}{y}+\\frac{1}{x-2 y}}=\\frac{\\frac{x y-x^{2}}{x+y}}{\\frac{x-y}{y(x-2 y)}}\n$$\n\nRazreševanje dvojnega ulomka\n\nPoenostavitev prvega člena izraza\n$$\n= -\\frac{x y(x-2 y)}{x+y}\n$$\n\nRazcep\n$$\nx^{3}+y^{3}=(x+y)\\left(x^{2}-x y+y^{2}\\right)\n$$\n\nPoenostavitev\n$$\n\\frac{\\left(x^{2}-x y+y^{2}\\right)\\left(x^{3}-x^{3}+2 x^{2} y-x y^{2}\\right)}{(x+y)\\left(x^{2}-x y+y^{2}\\right)}=\\frac{x y(2 x-y)}{x+y}\n$$\n\nPoenostavitev izraza\n$$\n\\frac{-x y(x-2 y)}{x+y}+\\frac{x y(2 x-y)}{x+y}=x y\n$$\n\nIzraz nima pomena za $y=0$, $x=-y$, $x=2 y$ in $x=y$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41852,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nObseg pravokotnika je $4~\\mathrm{cm}$, velikost kota med diagonalama pa $60^\\circ$. Nariši skico pravokotnika z označenim kotom med diagonalama. Izračunaj dolžini obeh stranic tega pravokotnika. Rezultat naj bo natančen in zapisan v obliki okrajšanih ulomkov z racionaliziranimi imenovalci. Nalogo reši brez uporabe žepnega računala.",
    "options": [],
    "answer": "a = 3 - sqrt(3) cm, b = sqrt(3) - 1 cm",
    "solution": "Solution:\n\nSkica pravokotnika z označenim kotom med diagonalama\n\n![](attached_image_1.png)\n\nUpoštevanje $2a + 2b = 4$\n\nZapis ali upoštevanje $\\tan 30^\\circ = \\frac{b}{a}$\n\nZapis ali upoštevanje $\\tan 30^\\circ = \\frac{\\sqrt{3}}{3}$\n\nReševanje sistema enačb:\n\n$2a + 2b = 4$\n\n$\\tan 30^\\circ = \\frac{b}{a}$\n\n$\\Rightarrow a + b = 2$\n\n$\\Rightarrow b = 2 - a$\n\n$\\tan 30^\\circ = \\frac{b}{a} = \\frac{2 - a}{a} = \\frac{\\sqrt{3}}{3}$\n\n$\\Rightarrow 3(2 - a) = a \\sqrt{3}$\n\n$6 - 3a = a \\sqrt{3}$\n\n$6 = 3a + a \\sqrt{3}$\n\n$6 = a(3 + \\sqrt{3})$\n\n$a = \\frac{6}{3 + \\sqrt{3}}$\n\nRacionalizacija imenovalca:\n\n$a = \\frac{6}{3 + \\sqrt{3}} \\cdot \\frac{3 - \\sqrt{3}}{3 - \\sqrt{3}} = \\frac{6(3 - \\sqrt{3})}{(3 + \\sqrt{3})(3 - \\sqrt{3})} = \\frac{6(3 - \\sqrt{3})}{9 - 3} = \\frac{6(3 - \\sqrt{3})}{6} = 3 - \\sqrt{3}$\n\n$a = (3 - \\sqrt{3})~\\mathrm{cm}$\n\n$b = 2 - a = 2 - (3 - \\sqrt{3}) = -1 + \\sqrt{3}$\n\n$b = (-1 + \\sqrt{3})~\\mathrm{cm}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41853,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIzbrati si moramo sedemmestno geslo, ki vsebuje vsaj eno črko in vsaj eno števko. Izbiramo lahko med znaki $A, B, C, G, J, M, R, Z$ in $3$.\n\na) Koliko je vseh možnih izbir za geslo, če se znaki ne smejo ponavljati?\n\nb) Koliko je vseh možnih izbir za geslo, če se znaki lahko ponavljajo?",
    "options": [],
    "answer": "a) 141120; b) 2685816",
    "solution": "Solution:\n\na.\nUpoštevanje, da je v geslu 6 črk in ena števka.\n\nUgotovitev, da je 7 možnih položajev za števko.\n\nIzračun števila možnih razporeditev črk v geslo $8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3$.\n\nIzračun števila možnih izbir za geslo $7 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 = 141120$.\n\nb.\nUgotovitev, da je vseh možnih izbir za geslo, če ne bi bilo omejitve, da mora vsebovati vsaj eno črko in vsaj eno številko, enako $9^{7}$ oziroma $4782969$.\n\nUgotovitev, da je vseh možnih izbir za geslo, ki bi vsebovalo samo črke, enako $8^{7}$ oziroma $2097152$.\n\nUgotovitev, da je samo eno geslo, ki je sestavljeno samo iz števke $3$.\n\nIzračun števila možnih izbir za geslo $9^{7} - 8^{7} - 1 = 2685816$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41854,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $n \\geq 2$ naravno število. Katja želi pobarvati vsako polje tabele velikosti $n \\times n$ bodisi črno bodisi zeleno, tako da bo izmed poljubnih štirih polj, ki jih lahko prekrijemo z domino oblike\n| $\\square$ |\n| :--- |\n, liho mnogo polj črnih. Na koliko načinov lahko to stori?",
    "options": [],
    "answer": "2^{2n-1}",
    "solution": "Solution:\n\nV prvi vrstici in prvem stolpcu tabele je skupaj $n + (n-1) = 2n - 1$ polj. Ta polja lahko Katja pobarva na $2^{2n-1}$ različnih načinov, saj ima za vsako polje na voljo dve barvi. Pokažimo, da lahko Katja, ne glede na to, kako pobarva prvo vrstico in prvi stolpec, potem pobarva preostanek tabele na en sam pravilen način. Denimo torej, da je Katja že pobarvala prvo vrstico in prvi stolpec tabele. Postavimo domino v zgornji levi kot tabele, tako da prekrije tri že pobarvana polja. Če je izmed teh treh polj liho mnogo črnih, potem mora Katja četrto polje, pokrito z domino, pobarvati zeleno, če pa je izmed teh treh polj sodo mnogo črnih, mora Katja četrto polje, pokrito z domino, pobarvati črno. V obeh primerih lahko četrto polje, to je 2. polje v 2. stolpcu, pobarva na en sam način. Denimo, da je tudi to polje že pobarvala, in pomaknimo domino eno vrstico nižje. Ker domina spet pokriva tri že pobarvana polja, lahko zgornji sklep ponovimo, in ugotovimo, da lahko Katja pobarva 3. polje v 2. stolpcu na en sam pravilen način. Podobno sklepamo, da lahko tudi 4. polje in nato vsako naslednje polje v 2. stolpcu pobarva na en sam način. Torej lahko Katja 2. stolpec pobarva na en sam način. Ko je enkrat 2. stolpec cel pobarvan, se pomaknemo za en stolpec v desno in ponovimo sklep na 3. stolpcu. Katja lahko torej 3. stolpec pobarva na en sam način. Podobno sklepamo, da lahko Katja tudi 4. stolpec in potem vsak naslednji stolpec pobarva na en sam način. Torej, če Katja pobarva prvo vrstico in prvi stolpec na poljuben način, potem lahko preostanek tabele pobarva na en sam način. Barvanje, ki ga tako dobi, res ustreza pogoju, saj smo se pri sklepanju pomikali po eno vrstico in po en stolpec, torej smo spotoma preverili vse možne postavitve domine na tabelo.\n\n2. način. Katja lahko $2 \\times 2$ kvadratek pobarva na 8 različnih načinov. Denimo, da je že pobarvala $(n-1) \\times (n-1)$ kvadrat. Sedaj bi rada dopolnila še eno vrstico in en stolpec. Ne glede na to, kako je pobarvan že pobarvan kvadrat, lahko pobarva $2 \\times 2$ kvadratek z dvema nepobarvanima poljema na dva načina; če sta obe že pobarvani polji črni, se lahko odloči, katero od nepobarvanih polj bo zeleno in katero črno, če pa je od pobarvanih polj eno zeleno in eno črno, lahko preostali polji pobarva obe zeleno ali obe črno. To lahko neodvisno naredi z $2 \\times 2$ kvadratkoma v nasprotnih kotih. Po enakem argumentu kot zgoraj so vsa ostala nepobarvana polja s tem določena, torej je možnosti za $n \\times n$ tabelo štirikrat toliko kot za $(n-1) \\times (n-1)$ tabelo. Barvanje je res pravilno po enakem argumentu kot zgoraj. Iz rekurzivne zveze in začetnega pogoja dobimo, da je možnosti za $n \\times n$ tabelo $2^{2n-1}$.\n\n3. način. Vsak $2 \\times 2$ kvadratek lahko pobarvamo na 8 načinov. Denimo, da smo že pobarvali zgornji levi koten $2 \\times 2$ kvadratek. Potem lahko $2 \\times 1$ domino desno od njega pobarvamo na 2 načina in tako nadaljujemo do konca prvega $2 \\times n$ pasu. Podobno lahko $1 \\times 2$ polje pod prvim $2 \\times 2$ kvadratkom pobarvamo na 2 načina. Polje na mestu $(3,3)$ je sedaj enolično določeno, kot so tudi vsa polja desno od njega, saj je četrto polje v $2 \\times 2$ kvadratku enolično določeno s preostalimi tremi. Postopek ponavljamo do konca vsakega pasu in v vsaki vrstici. Skupno dobimo $2^{2n-1}$ možnosti. Barvanje je tudi res pravilno, saj smo sproti preverili vse postavitve $2 \\times 2$ kvadratkov.\n\n4. način. Vsak $2 \\times 2$ kvadratek lahko pobarvamo na 8 načinov. Denimo, da smo že pobarvali zgornji levi koten $2 \\times 2$ kvadratek. Potem lahko $2 \\times 2$ kvadratek spodaj desno (po diagonali), ki se z že pobarvanim ujema v zgornjem levem kvadratku, pobarvamo na 4 različne načine. Ta postopek ponavljamo po diagonali navzdol, na vsakem koraku pridobimo 4 načine barvanja kvadratka. Diagonalo iz $2 \\times 2$ kvadratkov lahko torej pobarvamo na $8 \\cdot \\underbrace{4 \\cdot 4 \\cdots 4}_{n-2} = 2^{2n-1}$ načinov. Četrto polje v $2 \\times 2$ kvadratku je enolično določeno s preostalimi tremi. Nepobarvani del tabele lahko enolično pobarvamo tako, da vsakemu nepobarvanemu kvadratku določimo barvo preko $2 \\times 2$ kvadratka s tremi že pobarvanimi polji (začnemo pri diagonali in barvamo polja v ostale smeri). Tako dobljeno barvanje je pravilno, saj smo sproti preverili vse postavitve $2 \\times 2$ kvadratkov.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41855,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $n \\geq 3$ naravno število. Na vsako polje tabele velikosti $n \\times n$ želimo zapisati eno izmed števil $1,2$ ali $3$, tako da bodo na poljubnih treh poljih, ki jih lahko prekrijemo z domino oblike $\\square \\square$, pri čemer lahko domino tudi zavrtimo, zapisana različna števila. Na koliko načinov lahko to storimo?",
    "options": [],
    "answer": "12",
    "solution": "Solution:\n\nNajprej opazimo, da, če zapišemo števili v dve sosednji polji vrstice oziroma stolpca, potem so števila, ki jih moramo zapisati v preostala polja te vrstice oziroma stolpca, enolično določena.\nČe znamo tabelo izpolniti na pravilen način, potem bo ostala pravilno izpolnjena tudi, če števila $1$, $2$ in $3$ med sabo poljubno premešamo. Števila $1$, $2$, $3$ lahko med sabo premešamo na $6$ različnih načinov, to so $(1,2,3)$, $(1,3,2)$, $(2,1,3)$, $(2,3,1)$, $(3,1,2)$ in $(3,2,1)$. Torej je dovolj, da preštejemo le tiste pravilno izpolnjene tabele, ki imajo v prvih treh poljih prve vrstice po vrsti zapisana števila $1,2,3$, rezultat pa bomo potem pomnožili s $6$.\nČe v prva tri polja prve vrstice zapišemo po vrsti števila $1,2$ in $3$, potem imamo za prvo polje v drugi vrstici dve možnosti, $2$ ali $3$. V obeh primerih je preostanek zgornjega levega $3 \\times 3$ kvadrata tabele enolično določen\n\n| 1 | 2 | 3 |\n| :--- | :--- | :--- |\n| 2 | 3 | 1 |\n| 3 | 1 | 2 |\n\noziroma\n\n| 1 | 2 | 3 |\n| :---: | :---: | :---: |\n| 3 | 1 | 2 |\n| 2 | 3 | 1 |\n\nPo zgornji opazki je potem enolično določen tudi preostanek tabele. V prvem primeru bodo na vsaki diagonali, ki potekajo desno-navzgor, enaka števila, isto število pa se bo ponovilo vsake $3$ diagonale. V drugem primeru pa bodo na vsaki diagonali, ki poteka desno-navzdol enaka števila, isto število pa se bo ponovilo na vsake $3$ diagonale. V obeh primerih bo torej tabela pravilno izpolnjena, saj nobena domina ne more prekriti dveh enakih števil.\n\nTako lahko torej tabelo izpolnimo na dva različna načina, če pa upoštevamo še premešanja, je vseh možnih načinov $2 \\cdot 6=12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41856,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be a positive integer. We wish to inscribe into each square of the $n \\times n$ array one of the numbers $1$, $2$ or $3$, so that any three squares which can be covered using the following shape [ ] [ ] [ ] (which can also be rotated) will contain three different numbers. In how many different ways can this be done?",
    "options": [],
    "answer": "12",
    "solution": "First notice that choosing two of the numbers in adjacent squares completely determines the remaining numbers in that row or column.\nIf we can inscribe the numbers as required then any permutation of the labels $1$, $2$ and $3$ will also give us a correct solution. We can put the numbers $1$, $2$ and $3$ in $6$ different orders, namely $(1, 2, 3)$, $(1, 3, 2)$, $(2, 1, 3)$, $(2, 3, 1)$, $(3, 1, 2)$ and $(3, 2, 1)$.\nSo, we have to count the arrays which contain $1$, $2$, $3$, in this order, in the first three squares of the first row, and multiply the result by $6$.\nPlacing $1$, $2$ and $3$ in the first three squares of the first row gives us two possibilities for the first square of the second row, $2$ or $3$. In both cases the remainder of the upper left $3 \\times 3$ square is then uniquely determined and is\n\n| 1 | 2 | 3 |\n|---|---|---|\n| 2 | 3 | 1 |\n| 3 | 1 | 2 |\n\n| 1 | 2 | 3 |\n|---|---|---|\n| 3 | 1 | 2 |\n| 2 | 3 | 1 |\n\nor\nUsing the same argument we can show that the remainder of the array is now also uniquely determined. In the first case each of the right-up diagonals will contain the same number with each of the number repeating on every third diagonal. In the second case each of the right-down diagonals will contain the same number with each number repeating on every third diagonal. In both cases the array will be filled correctly since it will not be possible for the tile to cover the same number twice.\nWe conclude that there are two ways to fill the array in this way. Allowing for permutations, there are $2 \\cdot 6 = 12$ solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41857,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPremica poteka skozi presečišče premic $11 x+3 y-7=0, 12 x+y-19=0$ in razpolovišče daljice s krajiščema $A(3,-2)$ in $B(-1,6)$. Zapiši enačbo te premice $v$ vseh treh oblikah in jih poimenuj. Premico nariši.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Explicit (slope–intercept) form: y = -7x + 9; Implicit (general) form: 7x + y - 9 = 0; Intercept form: x/(9/7) + y/9 = 1",
    "solution": "Solution:\n\nPo reševanju sistema enačb s katerokoli metodo dobimo presečišče $P(2,-5)$. Določimo razpolovišče daljice $AB: S(1,2)$. Izračunamo smerni koeficient $k=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{2+5}{1-2}=-7$.\n\nVstavimo podatke v enačbo premice npr.: $y-y_{1}=k\\left(x-x_{1}\\right)$ in po ureditvi dobimo $y=-7x+9$. Preoblikujemo jo še v preostali dve obliki. Premico narišemo.\n\n![](attached_image_2.png)\n\n- Zapis eksplicitne oblike enačbe premice: $y=-7x+9$\n- Zapis implicitne oblike enačbe premice: $7x+y-9=0$\n- Zapis odsekovne oblike enačbe premice: $\\frac{x}{\\frac{9}{7}}+\\frac{y}{9}=1$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41858,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Reši neenačbo $x^{4}-2 x^{3}-4 x^{2}>6-5 x$.\n\nb) Reši enačbo $x^{6}-5 x^{3}=14$.",
    "options": [],
    "answer": "a) (-∞, -2) ∪ (3, ∞)\nb) x = ∛7 or x = -∛2",
    "solution": "Solution:\n\na) Uredimo neenačbo $x^{4}-2 x^{3}-4 x^{2}+5 x-6>0$. Rešimo enačbo $x^{4}-2 x^{3}-4 x^{2}+5 x-6=0$ in zapišemo realni rešitvi enačbe $x_{1}=-2$ in $x_{2}=3$. Zapišemo rešitev neenačbe $x \\in (-\\infty,-2) \\cup (3, \\infty)$.\n\nb) Uredimo enačbo $x^{6}-5 x^{3}-14=0$ in jo razcepimo $\\left(x^{3}-7\\right)\\left(x^{3}+2\\right)=0$. Zapišemo realni rešitvi enačbe $x_{1}=\\sqrt[3]{7}$ in $x_{2}=-\\sqrt[3]{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41859,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana je pravilna šeststrana piramida z dolžino osnovnega roba $a=10~\\mathrm{cm}$ in dolžino stranskega roba $s=13~\\mathrm{cm}$.\n\na) Natančno izračunaj površino piramide.\n\nb) Natančno izračunaj ploščino osnega preseka piramide, ki nastane, če piramido presekamo z ravnino, ki poteka skozi vrh piramide in dve oglišči osnovne ploskve.\n\n(10 točk)",
    "options": [],
    "answer": "a) 150√3 + 360; b) 10√69",
    "solution": "Solution:\n\na)\nS pomočjo Pitagorovega izreka izračunamo višino stranske ploskve piramide $v_{s}=12~\\mathrm{cm}$. S pomočjo formule za izračun ploščine osnovne ploskve $S_{op}=\\frac{6 a^{2} \\sqrt{3}}{4}=\\frac{6 \\cdot 10^{2} \\sqrt{3}}{4}=150 \\sqrt{3}\\ \\mathrm{cm}^{2}$ in ploščine plašča $S_{pl}=6 \\cdot \\frac{a \\cdot v_{s}}{2}=6 \\cdot \\frac{10 \\cdot 12}{2}=360~\\mathrm{cm}^{2}$ izračunamo površino piramide $P=(150 \\sqrt{3}+360)\\ \\mathrm{cm}^{2}$.\n\n\nb)\nUgotovimo, da je osni presek piramide enakokraki trikotnik s stranicami $20~\\mathrm{cm}, 13~\\mathrm{cm}$ in $13~\\mathrm{cm}$. S pomočjo Pitagorovega izreka izračunamo višino trikotnika, ki je osni presek piramide $v_{op}=\\sqrt{69}~\\mathrm{cm}$. Izračunamo ploščino osnega preseka $S_{op}=\\frac{2 a \\cdot v_{op}}{2}=\\frac{20 \\cdot \\sqrt{69}}{2}=10 \\sqrt{69}~\\mathrm{cm}^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41860,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNatančno izračunaj vrednost izraza $\\left(\\sin \\frac{38 \\pi}{3}\\right)^{5}: 4^{\\log _{2} 3-\\log _{16} 256}$.",
    "options": [],
    "answer": "sqrt(3)/2",
    "solution": "Solution:\n\n$$\n\\left(\\sin \\frac{38 \\pi}{3}\\right)^{5}=\\sin ^{5}\\left(12 \\pi+\\frac{2 \\pi}{3}\\right)=\\sin ^{5}\\left(\\frac{2 \\pi}{3}\\right)=\\left(\\frac{\\sqrt{3}}{2}\\right)^{5}=\\frac{9 \\sqrt{3}}{32}\n$$\nPoenostavimo in izračunamo drugi del izraza:\n$$\n4^{\\log _{2} 3-\\log _{16} 256}=4^{\\log _{2} 3}: 4^{\\log _{16} 256}=2^{\\log _{2} 9}: 4^{2}=\\frac{9}{16}\n$$\nIzračunamo vrednost danega izraza:\n$$\n\\frac{9 \\sqrt{3}}{32}: \\frac{9}{16}=\\frac{\\sqrt{3}}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41861,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMarko in France imata vsak svoj sadovnjak z jablanami in hruškami. France ima $20 \\%$ manj jablan, a 7 hrušk več kot Marko. Skupaj imata 218 dreves. Naslednje leto namerava France na novo posaditi še 22 jablan in 24 hrušk, da bo imel jablan 1,5-krat toliko kot hrušk. Koliko ima letos vsak od njiju jablan in koliko hrušk?\n\n(10 točk)",
    "options": [],
    "answer": "Marko: 85 apples and 29 pears; France: 68 apples and 36 pears.",
    "solution": "Solution:\n\nNaj ima Marko $x$ jablan in $y$ hrušk. Potem ima France $0,8 x$ jablan in $y+7$ hrušk. Skupaj imata $1,8 x+2 y+7=218$ dreves, dobimo enačbo $1,8 x+2 y=211$. Naslednje leto bo imel France $0,8 x+22$ jablan in $y+31$ hrušk. Ker je jablan 1,5-krat več kot hrušk, dobimo enačbo $0,8 x+22=1,5(y+31)$. Preoblikujemo v enačbo $1,6 x-3 y=49$. Rešitvi sistema dveh linearnih enačb z dvema neznankama ($1,8 x+2 y=211$ in $1,6 x-3 y=49$) sta $x=85$ in $y=29$. To pomeni, da ima Marko 85 jablan in 29 hrušk, France pa 68 jablan in 36 hrušk.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41862,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa realno število $x$, kjer $x \\notin\\left\\{0, \\frac{1}{2}, \\frac{2}{3}, 1,2\\right\\}$, poenostavi izraz\n$$\n\\left(\\frac{3 x^{-\\frac{1}{3}}}{x^{\\frac{2}{3}}-2 x^{-\\frac{1}{3}}}-\\frac{x^{\\frac{1}{3}}}{x^{\\frac{4}{3}}-x^{\\frac{1}{3}}}\\right)^{-1}-\\left(\\frac{1-2 x}{3 x-2}\\right)^{-1}\n$$\nin izračunaj njegovo vrednost za $x=\\sqrt{3}$. Končni rezultat okrajšaj in po potrebi racionaliziraj imenovalec.",
    "options": [],
    "answer": "(6*sqrt(3)+3)/11",
    "solution": "Solution:\n\nV imenovalcu ulomka izpostavimo potenco $z$ najmanjšim eksponentom:\n$$\n\\left(\\frac{3 x^{-\\frac{1}{3}}}{x^{-\\frac{1}{3}}(x-2)}-\\frac{x^{\\frac{1}{3}}}{x^{\\frac{1}{3}}(x-1)}\\right)^{-1}-\\left(\\frac{1-2 x}{3 x-2}\\right)^{-1}\n$$\nKrajšamo ulomka, razširimo ulomek v oklepaju na skupni imenovalec in zapišemo obratno vrednost ulomkov:\n$$\n\\left(\\frac{(x-2)(x-1)}{2 x-1}\\right)-\\left(\\frac{3 x-2}{1-2 x}\\right)\n$$\nZa tem zopet določimo skupni imenovalec in skrčimo izraz. Tako dobimo ulomek:\n$$\n\\frac{x^{2}}{2 x-1}\n$$\nNato vstavimo za $x=\\sqrt{3}$ in racionaliziramo imenovalec:\n$$\n\\frac{3}{2 \\sqrt{3}-1} \\cdot \\frac{2 \\sqrt{3}+1}{2 \\sqrt{3}+1} = \\frac{6 \\sqrt{3}+3}{11}\n$$\n\n---\n\n2. način.\n\n$$\n\\begin{aligned}\n& \\left(\\frac{3 x^{-\\frac{1}{3}}}{x^{\\frac{2}{3}}-2 x^{-\\frac{1}{3}}}-\\frac{x^{\\frac{1}{3}}}{x^{\\frac{4}{3}}-x^{\\frac{1}{3}}}\\right)^{-1}-\\left(\\frac{1-2 x}{3 x-2}\\right)^{-1} = \\left(\\frac{\\frac{3}{\\sqrt[3]{x}}}{(\\sqrt[3]{x})^{2}-\\frac{2}{\\sqrt[3]{x}}}-\\frac{\\sqrt[3]{x}}{x \\cdot \\sqrt[3]{x}-\\sqrt[3]{x}}\\right)^{-1}-\\frac{3 x-2}{1-2 x}\n\\end{aligned}\n$$\n\n![](attached_image_1.png)\n\n$$\n\\begin{aligned}\n& =\\left(\\frac{3}{x-2}-\\frac{1}{x-1}\\right)^{-1}-\\frac{3 x-2}{1-2 x} \\\\\n& =\\left(\\frac{2 x-1}{(x-2)(x-1)}\\right)^{-1}-\\frac{3 x-2}{1-2 x} \\\\\n& =\\frac{(x-2)(x-1)}{2 x-1}+\\frac{3 x-2}{2 x-1}=\\frac{x^{2}}{2 x-1} .\n\\end{aligned}\n$$\n\nVstavimo $x=\\sqrt{3}$:\n$$\n\\frac{(\\sqrt{3})^{2}}{2 \\sqrt{3}-1}=\\frac{3}{2 \\sqrt{3}-1}\n$$\nRacionaliziramo imenovalec:\n$$\n\\frac{3}{2 \\sqrt{3}-1} \\cdot \\frac{2 \\sqrt{3}+1}{2 \\sqrt{3}+1}=\\frac{6 \\sqrt{3}+3}{11}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41863,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $a_{1}, \\frac{1}{5}, a_{3}, \\frac{16}{125}, \\ldots$ geometrijsko zaporedje s samimi pozitivnimi členi. Kolikšna je vsota $a_{1}+a_{3}+a_{5}+\\ldots+a_{99}$?\n\n(A) $\\frac{5}{4}\\left(1-0,8^{50}\\right)$\n(B) $\\frac{5}{4} \\cdot 0,2^{50}$\n(C) $\\frac{5}{4}-0,2^{50}$\n(D) $\\frac{5}{4}\\left(1-8 \\cdot 10^{-50}\\right)$\n(E) $5 \\cdot 0,2^{48}$",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nIz $\\frac{a_{3}}{\\frac{1}{5}}=\\frac{\\frac{16}{125}}{a_{3}}$ sledi, da je $a_{3}=\\frac{4}{25}$. Potem je $q=\\frac{4}{5}$, $a_{1}=\\frac{1}{4}$. Vsota je $S=a_{1} \\cdot \\frac{q^{50}-1}{q-1}=\\frac{5}{4}\\left(1-0,8^{50}\\right)$. Pravilen odgovor je (A).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41864,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLik $A$ ima 2 oglišči več in 55 diagonal več kot lik $B$. Koliko oglišč ima lik $A$?\n\n(A) 25\n(B) 28\n(C) 30\n(D) 40\n(E) Ni možno določiti.",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nŠtevilo diagonal se izračuna z $d = \\frac{n(n-3)}{2}$. Če ima lik $B$ $n$ oglišč, jih ima lik $A$ $n+2$ in za število diagonal velja\n$$\n\\frac{(n+2)(n-1)}{2} = \\frac{n(n-3)}{2} + 55.\n$$\nDobimo $n = 28$. Iz tega sledi, da ima lik $A$ $30$ oblišč. Pravilen odgovor je (C).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41865,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKoliko je takih 6-mestnih števil, ki se ne začnejo z 0 in vsebujejo število 2017 kot strnjen podniz? Npr. število 820178 vsebuje strnjen podniz 2017, število 820817 pa ne.\n(A) 100\n(B) 190\n(C) 200\n(D) 280\n(E) 300",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n6-mestno število, ki vsebuje strnjen podniz 2017 je ene od treh oblik: $x y 2017$, $x 2017 y$ ali $2017 x y$, kjer sta $x$ in $y$ števki. Hitro opazimo, da nobeno naravno število ne more biti dveh oblik hkrati. Torej moramo le prešteti, koliko števil je posamezne oblike, saj s tem nobenega števila ne bomo šteli dvakrat.\n\nŠtevil oblike $x y 2017$ je $9 \\cdot 10 = 90$, saj $x$ ne sme biti enak $0$.\n\nŠtevil oblike $x 2017 y$ je $9 \\cdot 10 = 90$,\n\nštevil oblike $2017 x y$ pa $10 \\cdot 10 = 100$.\n\nVseh števil skupaj je $90 + 90 + 100 = 280$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41866,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV preglednici so podane vrednosti kvadratne funkcije $f$. Koliko je $f(4)$?\n\n| $x$ | -2 | 0 | 1 | 2 |\n| :---: | :---: | :---: | :---: | :---: |\n| $f(x)$ | 0 | $\\frac{8}{3}$ | 3 | $\\frac{8}{3}$ |\n\n(A) $\\frac{2}{3}$\n(B) $\\frac{5}{3}$\n(C) -3\n(D) 0\n(E) Nič od navedenega",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nEna ničla funkcije $f$ je $-2$, v točki $T(1,3)$ ima graf funkcije $f$ teme, torej je $4$ druga ničla funkcije $f$, zato je $f(4)=0$. Pravilen odgovor je (D).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41867,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nZa kateri $x$ bo imel izraz $\\frac{x^{3}-2 x^{2}-9 x+18}{x^{2}-9}$ vrednost 0?\n(A) 9\n(B) 3\n(C) -3\n(D) 0\n(E) 2",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n$\\frac{x^{3}-2 x^{2}-9 x+18}{x^{2}-9}=\\frac{x^{2}(x-2)-9(x-2)}{x^{2}-9}=\\frac{(x-2)\\left(x^{2}-9\\right)}{x^{2}-9}=x-2$\n\nin vrednost je enaka 0 za $x=2$. Pravilen odgovor je (E).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41868,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nPolona je razdelila pravokotnik na 9 manjših pravokotnikov in na tri izmed štirih vogalnih pravokotnikov zapisala njihove ploščine (glej sliko). Koliko je ploščina četrtega vogalnega pravokotnika?\n(A) 9\n(B) 13.5\n(C) 14\n(D) 15\n(E) 16\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "D",
    "solution": "Solution:\nOznačimo višini prve in tretje vrstice pravokotnikov z $x$ in $y$, širini prvega in tretjega stolpca pravokotnikov pa $z$ in $w$. Tedaj so ploščine treh vogalnih pravokotnikov s številkami enake $x z = 8$, $x w = 12$, $y z = 10$, ploščina četrtega vogalnega pravokotnika pa je $y w = \\frac{y z \\cdot x w}{x z} = \\frac{10 \\cdot 12}{8} = 15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41869,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKrogu očrtamo in včrtamo enakostranični trikotnik. Koliko je razmerje dolžin stranic krogu očrtanega in včrtanega trikotnika?\n(A) $1: 3$\n(B) $2: 1$\n(C) $4: 1$\n(D) $5: 2$\n(E) $\\sqrt{3}: 1$",
    "options": [],
    "answer": "(B)",
    "solution": "Solution:\n\nNaj bo $v$ dolžina višine krogu očrtanega enakostraničnega trikotnika, $v_1$ pa dolžina višine krogu včrtanega enakostraničnega trikotnika in $r$ polmer kroga. Za očrtani trikotnik velja $\\frac{1}{3} v = r$ in za včrtani trikotnik velja $\\frac{2}{3} v_1 = r$. Iz obeh enakosti sledi, da je razmerje višin očrtanega in včrtanega trikotnika enako $v : v_1 = 2 : 1$. V enakem razmerju sta tudi dolžini stranic trikotnikov. Pravilen odgovor je (B).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41870,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nS katerim izmed navedenih števil je deljivo število $5^{2017} + 5^{2016} + 5^{2015}$?\n\n(A) 15\n(B) 31\n(C) 2015\n(D) 39\n(E) 2017",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\n$5^{2017} + 5^{2016} + 5^{2015} = 5^{2015}(5^2 + 5 + 1) = 31 \\cdot 5^{2015}$. Število je deljivo z $31$. Pravilen odgovor je (B).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41871,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa koliko različnih načinov lahko sedejo v vrsto z 8 sedeži v gledališču 4 pari, če želi vsak posamezen par sedeti skupaj?\n\n(A) $4!$\n(B) $2 \\cdot 4!$\n(C) $24$\n(D) $384$\n(E) $256$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nNačinov razporeditev 4 parov v vrsto je $4! = 24$. Če želijo pari sedeti skupaj, se lahko posedejo na $4! \\cdot 2^4 = 384$ načinov. Pravilen odgovor je (D).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41872,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nZa linearno funkcijo $f$ velja $f(-1)+f(3)=-8$ in $f(1)+f(5)=4$. Koliko je $f(2)+f(7)$ ?\n(A) 12\n(B) 13\n(C) 14\n(D) 15\n(E) 16",
    "options": [],
    "answer": "B",
    "solution": "Solution:\nLinearna funkcija je oblike $f(x)=k \\cdot x+n$. Iz prve zveze dobimo $2k+2n=-8$, iz druge pa $6k+2n=4$. Iz tega sledi $k=3$ in $n=-7$, $f(x)=3x-7$. Tako velja $f(2)+f(7)=13$. Pravilen odgovor je (B).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41873,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPrvo število zaporedja šestih števil je enako $4$, zadnje pa $47$. Vsako število od vključno tretjega naprej je enako vsoti prejšnjih dveh števil. Naj bo $S$ vsota vseh šestih števil zaporedja. Tedaj $S$ leži na intervalu med\n(A) $51$ in $90$\n(B) $91$ in $100$\n(C) $101$ in $110$\n(D) $111$ in $120$\n(E) $121$ in $160$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nOznačimo drugo število v zaporedju z $a$. Tedaj je zaporedje števil enako $4, a, a+4, 2a+4, 3a+8, 5a+12$. Zadnje število je enako $5a+12=47$, od koder izračunamo $a=7$. Vsota vseh šestih števil je enaka $S=4+a+(a+4)+(2a+4)+(3a+8)+(5a+12)=12a+32=12 \\cdot 7+32=116$. Pravilen odgovor je (D).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41874,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLetalska družba potniku za prtljago ne zaračuna dodatnih stroškov, če masa prtljage ne preseže določene dovoljene mase. Vsak dodaten kilogram prtljage pa mora potnik doplačati. Gospod in gospa Kotnik imata skupno prtljago, zato se jima ne bi zaračunalo dodatnih stroškov, če masa skupne prtljage ne bi presegla dvakratnika določene dovoljene mase. Njuna skupna prtljaga tehta $60~\\mathrm{kg}$, doplačati pa sta morala 11 evrov. Prtljaga gospoda Novaka, ki potuje sam, prav tako tehta $60~\\mathrm{kg}$. Toda on je moral za prtljago doplačati 33 evrov. Največ koliko kilogramov prtljage lahko ima en potnik, da mu zanjo ni treba nič doplačati?\n\n(A) 11\n(B) 18\n(C) 20\n(D) 24\n(E) 39",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nOznačimo največjo dovoljeno težo za enega potnika z $d$. Gospod in gospa Kotnik imata skupaj $60-2d$ dodatnih kilogramov prtljage, za katere sta plačala 11 evrov. Torej sta za vsak dodaten kilogram plačala $\\frac{11}{60-2d}$ evrov.\n\nGospod Novak ima $60-d$ dodatnih kilogramov prtljage, za katere je plačal 33 evrov. Za vsak kilogram je torej plačal $\\frac{33}{60-d}$ evrov.\n\nKer morata biti tarifi enaki, je\n$$\n\\frac{11}{60-2d} = \\frac{33}{60-d}.\n$$\nEnačbo delimo z 11 in odpravimo ulomke, da dobimo\n$$\n60-d = 3(60-2d).\n$$\nOd tod izračunamo $d=24$.\n\nPravilen odgovor je torej (D).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41875,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKatero število je rešitev enačbe $2^{x} \\cdot 5^{x}=0,01 \\cdot\\left(10^{x-2}\\right)^{4}$ ?\n(A) $\\frac{2}{3}$\n(B) $\\frac{10}{3}$\n(C) $\\frac{1}{3}$\n(D) $-2$\n(E) $-\\frac{3}{2}$",
    "options": [],
    "answer": "(B)",
    "solution": "Solution:\nUpoštevamo pravila za računanje s potencami in preoblikujemo enačbo v obliko $10^{x}=10^{4 x-10}$. Rešitev enačbe je $x=\\frac{10}{3}$. Pravilen odgovor je (B).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41876,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $n$ naravno število. Kateri od spodaj navedenih izrazov je enakovreden izrazu\n$$\\sqrt[\\sqrt[n]{n^{n}} \\cdot n+\\sqrt[n]{n^{n}}]{\\sqrt[n]{n^{n+n}}}$$?\n\n(A) 1\n(B) $\\sqrt[n]{n}$\n(C) $n$\n(D) $\\sqrt{n}$\n(E) $n^{-\\frac{1}{2}}$",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\n$\\sqrt[n]{n^{n}} = n$, $\\sqrt[n+n]{n^{n}} = \\sqrt[2]{n}$, $\\sqrt[n]{n^{n+n}} = n^{2}$. Vrednost izraza\n$$\n\\sqrt[\\sqrt[n]{n^{n}} \\cdot \\sqrt[n+n]{n^{n}}]{\\sqrt[n^{n+n}]}\n$$\nje enaka\n$$\n\\frac{n \\cdot \\sqrt{n}}{n^{2}} = \\frac{1}{\\sqrt{n}} = n^{-\\frac{1}{2}}.\n$$\nPravilen odgovor je (E).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41877,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV kvadrat s stranico dolžine $a$ je včrtana krožnica s središčem v $O$. Štiri manjše krožnice s središči $O_{1}$, $O_{2}$, $O_{3}$ in $O_{4}$ se dotikajo večje krožnice in po dveh stranic kvadrata (glej sliko). Koliko je ploščina kvadrata $O_{1} O_{2} O_{3} O_{4}$?\n\n(A) $3 a^{2}(3-2 \\sqrt{2})$\n(B) $4 a^{2}(3-2 \\sqrt{2})$\n(C) $a^{2}(3-\\sqrt{2})$\n(D) $a^{2}(2-\\sqrt{2})$\n(E) $3 a^{2}(2-\\sqrt{2})$\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nOznačimo oglišča osnovnega kvadrata z $A_{1}$, $A_{2}$, $A_{3}$ in $A_{4}$, tako da je za vsak $i$ oglišče $A_{i}$ najbližje točki $O_{i}$. Polmer večje krožnice je $R=\\frac{a}{2}$, polmer manjših krožnic pa označimo z $r$. Tedaj je $|A_{1} O_{1}|=r \\sqrt{2}$, saj je to diagonala majhnega kvadratka s stranico dolžine $r$, podobno je $|A_{3} O_{3}|=r \\sqrt{2}$. Diagonala osnovnega kvadrata je zato dolga\n$$\n\\begin{aligned}\n|A_{1} A_{3}| & =|A_{1} O_{1}|+|O_{1} O|+|O O_{3}|+|O_{3} A_{3}|=r \\sqrt{2}+(r+R)+(R+r)+r \\sqrt{2}= \\\\\n& =2 R+2 r+2 r \\sqrt{2}=a+2 r(1+\\sqrt{2})\n\\end{aligned}\n$$\nKer pa je to diagonala osnovnega kvadrata, mora biti dolga $a \\sqrt{2}$. Sledi\n$$\na+2 r(1+\\sqrt{2})=a \\sqrt{2}\n$$\nod koder izrazimo $r=\\frac{a(\\sqrt{2}-1)}{2(\\sqrt{2}+1)}$. Ulomek racionaliziramo, da dobimo\n$$\nr=\\frac{a(\\sqrt{2}-1)^{2}}{2}=\\frac{3-2 \\sqrt{2}}{2} a\n$$\nPloščina kvadrata $O_{1} O_{2} O_{3} O_{4}$ je enaka\n$$\npl=(a-2 r)^{2}=(2 \\sqrt{2} a-2 a)^{2}=4 a^{2}(\\sqrt{2}-1)^{2}=4 a^{2}(3-2 \\sqrt{2})\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41878,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKenguruju Pitagori so všeč le tista naravna števila, ki so deljiva s $4$ in imajo vsoto števk enako $3$ ter imajo v zapisu natanko $5$ števk enakih $0$. Koliko naravnih števil je všeč kenguruju Pitagori?\n\n(A) $15$\n(B) $16$\n(C) $19$\n(D) $20$\n(E) $25$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nŠtevke, ki so enake $0$, vsoti števk ne prispevajo ničesar. Iz neničelnih števk pa lahko dobimo vsoto $3$ le na tri načine: $3=3$, $1+2=3$ ali $1+1+1=3$. Ker je natanko $5$ števk števila, ki je všeč kenguruju Pitagori, enakih $0$, imamo torej tri možnosti.\n\nNaravno število, ki ima števke $3,0,0,0,0,0$ in je deljivo s $4$, je samo $300000$. Ničle namreč ne smejo nastopati na začetku števila.\n\nNaravna števila s števkami $1,2,0,0,0,0,0$ so po vrsti $1000002$, $1000020$, $1000200$, $1002000$, $1020000$, $1200000$, $2000001$, $2000010$, $2000100$, $2001000$, $2010000$, $2100000$. Od teh števila $1000002$, $2000001$ in $2000010$ niso deljiva s $4$, ostala pa so.\n\nNaravna števila s števkami $1,1,1,0,0,0,0,0$, ki so deljiva s $4$, morajo imeti na zadnjih dveh mestih števko $0$. Taka števila (razvrščena po številu števk $0$ na desni strani) so $10001100$, $10010100$, $10100100$, $11000100$, $10011000$, $10101000$, $11001000$, $10110000$, $11010000$, $11100000$.\n\nSkupaj je kenguruju Pitagori všeč natanko $20$ naravnih števil.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41879,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa koliko \\% se spremeni vrednost ulomka, če števec povečamo za $20 \\%$, imenovalec pa zmanjšamo za $20 \\%$?\n\n(A) $0 \\%$\n(B) $20 \\%$\n(C) $40 \\%$\n(D) $50 \\%$\n(E) $60 \\%$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\n$a + 0,2a = 1,2a$, $b - 0,2b = 0,8b$ in $\\frac{1,2a}{0,8b} = 1,5 \\frac{a}{b}$. Vrednost ulomka se poveča za $50 \\%$. Pravilen odgovor je (D).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41880,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKateri od navedenih zapisov predstavlja definicijsko območje funkcije $f(x)=\\frac{2x}{|x|-2}$?\n\n(A) $\\mathbb{R}$\n(B) $(-\\infty, 2) \\cup (2, \\infty)$\n(C) $\\mathbb{R}^{+}$\n(D) $\\mathbb{R} \\setminus \\{-2, 2\\}$\n(E) $(-2, 2)$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nFunkcija ni definirana za $|x|-2=0$. Rešitvi enačbe sta $x_{1}=-2$, $x_{2}=2$. Definicijsko območje funkcije $f$ je $\\mathbb{R} \\setminus \\{-2, 2\\}$. Pravilen odgovor je (D).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41881,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDani sta funkciji $f, g: \\mathbb{R} \\rightarrow \\mathbb{R}$, podani s predpisoma $f(x)=6-4x-2x^{2}$ in $g(x)=-x+4$.\n\na) V istem koordinatnem sistemu nariši grafa obeh funkcij.\n\nb) Zapiši enačbo tangente na graf funkcije $f$, ki je vzporedna grafu funkcije $g$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "y = -x + 57/8",
    "solution": "Solution:\n\na) Ničli funkcije $f$ sta $-3$ in $1$, teme grafa funkcije $f$ je v točki $T(-1,8)$. Graf funkcije $g$ ima smerni koeficient $-1$, ordinatno os pa seka v točki $N(0,4)$. Narišemo oba grafa.\n\n![](attached_image_2.png)\n\nIzračun ničel in začetne vrednosti funkcije $f$ ..... 1 točka\nIzračun temena grafa funkcije $f$ in narisan graf funkcije $f$ ..... 1 točka\nNarisan graf funkcije $g$ ..... 1 točka\n\nb) Tangenta vzporedna grafu funkcije $g$ ima enačbo $y=-x+n$. Tangenta se dotika parabole, zato imata krivulji eno skupno točko. Enačbi krivulj izenačimo in dobimo kvadratno enačbo $2x^{2}+3x+(n-6)=0$. Kvadratna enačba ima eno rešitev, ko je diskriminanta $D=3^{2}-4 \\cdot 2(n-6)$ enaka $0$, kar nam da rešitev $n=\\frac{57}{8}$. Enačba tangente je $y=-x+\\frac{57}{8}$.\n\nZapis enačbe vzporednice grafa funkcije $g$: $y=-x+n$ ..... 1 točka\nZapis kvadratne enačbe $2x^{2}+3x+(n-6)=0$ ..... 1 točka\nZapis enačbe $3^{2}-4 \\cdot 2(n-6)=0$ ..... 1 točka\nIzračun $n=\\frac{57}{8}$ ..... 1 točka\nZapis enačbe tangente $y=-x+\\frac{57}{8}$ ..... 1 točka",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41882,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDani sta točki $A$ in $B$ ter krožnica $\\mathcal{K}$ s premerom $AB$. Na daljici $AB$ izberemo točko $T$ različno od $A$ in $B$. Pravokotnica na daljico $AB$ skozi točko $T$ naj seka krožnico $\\mathcal{K}$ v točkah $M$ in $N$. Označimo $|AT|=x$, $|TB|=y$ in $|TN|=z$. Izračunaj vrednost izraza\n$$\n\\frac{\\log_{y} z + \\log_{x} z}{\\log_{x} z \\log_{y} z}\n$$",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nPo Talesovem izreku o kotu v polkrogu je trikotnik $ABN$ pravokoten s pravim kotom pri $N$. Po višinskem izreku v pravokotnem trikotniku zato velja $z^{2} = x y$. Dan izraz zapišemo kot vsoto dveh ulomkov\n$$\n\\frac{\\log_{y} z + \\log_{x} z}{\\log_{x} z \\log_{y} z} = \\frac{1}{\\log_{x} z} + \\frac{1}{\\log_{y} z}\n$$\nin nato uporabimo formulo za zamenjavo osnove logaritma $\\log_{a} b = \\frac{\\log b}{\\log a}$, da dobimo\n$$\n\\frac{1}{\\log_{x} z} + \\frac{1}{\\log_{y} z} = \\frac{1}{\\frac{\\log z}{\\log x}} + \\frac{1}{\\frac{\\log z}{\\log y}} = \\frac{\\log x}{\\log z} + \\frac{\\log y}{\\log z} = \\log_{z} x + \\log_{z} y\n$$\nUpoštevamo formulo za vsoto logaritmov in zvezo $z^{2} = x y$, da dobimo\n$$\n\\log_{z} x + \\log_{z} y = \\log_{z}(x \\cdot y) = \\log_{z} z^{2} = 2\n$$\nTorej je\n$$\n\\frac{\\log_{y} z + \\log_{x} z}{\\log_{x} z \\log_{y} z} = 2\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41883,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoloči vse vrednosti, ki jih zavzame izraz $A = n \\cdot \\left\\lfloor \\frac{2017}{n} \\right\\rfloor$, ko $n$ preteče vsa naravna števila. Pri tem $\\lfloor x \\rfloor$ označuje največje celo število, ki ni večje od $x$.",
    "options": [],
    "answer": "0 and all integers from 1009 through 2017 inclusive",
    "solution": "Solution:\n\nOznačimo z $m < n$ ostanek števila $2017$ pri deljenju z $n$. Tedaj lahko število $2017$ zapišemo v obliki $2017 = k n + m$, kjer je $k$ nenegativno celo število. Tedaj je\n$$\nA = n \\cdot \\left\\lfloor \\frac{2017}{n} \\right\\rfloor = n k\n$$\nkar je zagotovo celo število. Če je $k = 0$, je $A = 0$. Če je $k = 1$, je $A = n$, hkrati pa je $n \\leq 2017 < 2 n$, torej je $1009 \\leq n \\leq 2017$. V tem primeru torej $A$ zavzame vsa naravna števila med vključno $1009$ in $2017$. Če pa je $k \\geq 2$, je $2 n \\leq 2017$ oziroma $m < n \\leq 1008$ in zato $A = n k = 2017 - m > 1009$ ter $A = 2017 - m \\leq 2017$. V tem primeru $A$ ne zavzame nobenih novih števil.\n\n\n2. način. Ker je $n$ naravno število, je $\\left\\lfloor \\frac{2017}{n} \\right\\rfloor$ nenegativno celo število, zato je tudi $A$ nenegativno celo število. Za vsako realno število $x$ velja $x - 1 < \\lfloor x \\rfloor \\leq x$, zato je\n$$\n2017 - n = n \\left( \\frac{2017}{n} - 1 \\right) < A \\leq n \\cdot \\frac{2017}{n} = 2017\n$$\nOd tod sledi, da za vsak $n \\leq 1008$ velja $1009 < A \\leq 2017$. Če je $1009 \\leq n \\leq 2017$, potem je $1 \\leq \\frac{2017}{n} < 2$, zato je $\\left\\lfloor \\frac{2017}{n} \\right\\rfloor = 1$ in $A = n$. S tem $A$ zavzame vsa naravna števila od vključno $1009$ do vključno $2017$. Če pa je $n \\geq 2018$, potem je $0 < \\frac{2017}{n} < 1$, zato je $\\left\\lfloor \\frac{2017}{n} \\right\\rfloor = 0$ in tudi $A = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41884,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $ABC$ tak ostrokotni trikotnik, da oglišči $A$ in $B$ ter središči trikotniku očrtane in včrtane krožnice ležijo na isti krožnici. Dokaži, da na tej krožnici leži tudi višinska točka trikotnika $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo z $O$, $I$ in $H$ zaporedoma središče očrtane krožnice, središče včrtane krožnice in višinsko točko trikotnika $ABC$. Kote trikotnika označimo kot običajno z $\\alpha$, $\\beta$ in $\\gamma$. Ker je $\\angle BAI = \\frac{\\alpha}{2}$ in $\\angle IBA = \\frac{\\beta}{2}$, je $\\angle AIB = 180^\\circ - \\frac{\\alpha}{2} - \\frac{\\beta}{2} = 90^\\circ + \\frac{\\gamma}{2}$. Po izreku o središčnem in obodnem kotu za očrtano krožnico velja $\\angle AOB = 2\\gamma$. Trikotnik $ABC$ je ostrokoten, zato točki $O$ in $I$ ležita znotraj trikotnika $ABC$. Iz koncikličnosti točk $A$, $B$, $O$ in $I$ zato sledi $\\angle AOB = \\angle AIB$ oziroma $2\\gamma = 90^\\circ + \\frac{\\gamma}{2}$. Od tod izračunamo $\\gamma = 60^\\circ$.\n\nKer je trikotnik $ABC$ ostrokoten, tudi točka $H$ leži znotraj trikotnika. Iz definicije višinske točke sledi $\\angle BAH = 90^\\circ - \\beta$ in $\\angle HBA = 90^\\circ - \\alpha$, torej je $\\angle AHB = 180^\\circ - (90^\\circ - \\beta) - (90^\\circ - \\alpha) = \\alpha + \\beta = 180^\\circ - \\gamma$. Ker je $\\gamma = 60^\\circ$, sledi $\\angle AHB = 120^\\circ = \\angle AIB = \\angle AOB$. Točke $A$, $H$, $I$, $O$ in $B$ torej res vse ležijo na isti krožnici.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41885,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nReši sistem enačb\n$$\n\\begin{aligned}\n\\sqrt{2 x-3 y} &= \\sqrt{x^{2}+4 y-1} \\quad \\text{in} \\\\\n\\sqrt{x-y+2}+2 &= x\n\\end{aligned}\n$$",
    "options": [],
    "answer": "(5, -2)",
    "solution": "Solution:\n\nPrvo enačbo kvadriramo in preoblikujemo v $x^{2}-2 x+7 y-1=0$. V drugi enačbi koren osamimo, kvadriramo in izrazimo $y=-x^{2}+5 x-2$. To vstavimo v zgornjo enačbo in dobimo $2 x^{2}-11 x+5=0$. Za $x_{1}=5$ dobimo $y_{1}=-2$, rešitev ustreza prvotni enačbi. Za $x_{2}=\\frac{1}{2}$ dobimo $y_{2}=\\frac{1}{4}$, rešitev ne ustreza prvotni enačbi.\n\nKvadriranje prve enačbe in preoblikovanje $\\mathbf{v}$ npr.: $x^{2}-2 x+7 y-1=0$ ..... 1 točka\nOsamitev korena v drugi enačbi ..... 1* točka\nKvadriranje in preoblikovanje $\\mathbf{v}$ $y=-x^{2}+5 x-2$ ali $x^{2}-5 x+y+2=0$ ..... 1 točka\nReševanje sistema nelinearnih enačb, npr. zamenjalni način ..... $1^{*}$ točka\nPreoblikovanje v $2 x^{2}-11 x+5=0$ ..... 1 točka\nRešitev $x_{1}=5$ in $y_{1}=-2$ ..... 1 točka\nRešitev $x_{2}=\\frac{1}{2}$ in $y_{2}=\\frac{1}{4}$ ..... 1 točka\nUgotovitev, da prva rešitev ustreza in druga ne ustreza ..... 1 točka",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41886,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDano je število $n=100\\ldots 001$, zapisano z 2017 ničlami in 2 enkama.\n\na. Ali je število $n$ deljivo z 11?\n\nb. Ali je število $n$ deljivo s 101?\n\nc. Ali je število $n$ deljivo s 1001?",
    "options": [],
    "answer": "a: no; b: yes; c: no",
    "solution": "Solution:\n\na.\nUporabimo pravilo za deljivost z 11. Naravno število $\\overline{a_{k} \\ldots a_{3} a_{2} a_{1}}$, kjer so $a_{i}$ števke, je deljivo z 11 natanko tedaj, ko je število $a_{1}-a_{2}+a_{3}-\\ldots+(-1)^{k+1} a_{k}$ deljivo z 11. Ker\n$$\n1-\\underbrace{0+0-0+\\ldots+0-0}_{2017 \\text{ ničel}}+1=2\n$$\nni deljivo z 11, tudi število $n$ ni deljivo z 11.\n\nPisno deljenje:\n```\n10000000...01: 11 = 90909\n100\n100\n1...\n```\nOpazimo, da se v vsaki naslednji vrstici vodilna števka 1 premakne za 2 mesti v desno. Ker ima število $n$ liho mnogo ničel, dobimo na zadnjem koraku račun\n$$\n\\begin{gathered}\n101: 11=9, \\\\\n2 \\text{ ost. }\n\\end{gathered}\n$$\nkar pomeni, da število $n$ ni deljivo z 11.\n\nb.\nUporabimo obrazec $x^{m}+y^{m}=(x+y)\\left(x^{m-1}-x^{m-2} y+\\ldots-x y^{m-2}+y^{m-1}\\right)$, kjer je $m$ liho število, da zapišemo\n$$\n\\begin{aligned}\nn & =10^{2018}+1=\\left(10^{2}\\right)^{1009}+1=\\left(10^{2}+1\\right)\\left(\\left(10^{2}\\right)^{1008}-\\left(10^{2}\\right)^{1007}+\\ldots-10^{2}+1\\right)= \\\\\n& =101 \\cdot\\left(\\left(10^{2}\\right)^{1008}-\\left(10^{2}\\right)^{1007}+\\ldots-10^{2}+1\\right)\n\\end{aligned}\n$$\nSledi, da je število $n$ deljivo s 101.\n\nPisno deljenje:\n```\n100000000...01:101=990099\n910\n1000\n910\n1.,\n```\nOpazimo, da se v lihih vrsticah vodilna števka 1 vsakič premakne za 4 mesta v desno. Ker ima število $n$ natanko $2017=504 \\cdot 4+1$ ničel, dobimo na zadnjem koraku račun $101: 101=1$. Ostanek je 0, torej je število $n$ deljivo s 101.\n\nc.\nŠtevilo $n$ ni deljivo s 1001, saj je $1001=7 \\cdot 11 \\cdot 13$ in po točki (a) število $n$ ni deljivo z 11.\n\nPisno deljenje:\n```\n1000000000000...01:1001=999000999\n9910\n9010\n10000\n9910\n9010\n1...\n```\nOpazimo, da se v vsaki tretji vrstici vodilna števka 1 vsakič premakne za 6 mest v desno. Ker ima število $n$ natanko $2017=336 \\cdot 6+1$ ničel, dobimo na zadnjem koraku 101, kar je ostanek deljenja. Število $n$ torej ni deljivo s 1001.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41887,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $n$ naravno število. Poišči vsa realna števila $x$, ki rešijo enačbo\n$$\n2^{n}(-x)^{n}+(-1)^{3 n+1} 2^{n+1} x^{n-1}(2 x+1)-(-2 x)^{n+1}=0\n$$",
    "options": [],
    "answer": "If n = 1: x = 2 or x = −1/2. If n ≥ 2: x = 0, x = 2, or x = −1/2.",
    "solution": "Solution:\n\nOpazimo, da je $(-1)^{3 n+1}=(-1)^{2 n}(-1)^{n+1}=(-1)^{n+1}$, zato lahko enačbo preoblikujemo do\n$$\n2^{n}(-1)^{n} x^{n}+(-1)^{n+1} 2^{n+1} x^{n-1}(2 x+1)-(-1)^{n+1} 2^{n+1} x^{n+1}=0\n$$\nNa levi strani enačbe izpostavimo skupni faktor, da dobimo\n$$\n(-1)^{n} 2^{n} x^{n-1}\\left(x-2(2 x+1)+2 x^{2}\\right)=0\n$$\nKer je $x-2(2 x+1)+2 x^{2}=x(2 x+1)-2(2 x+1)=(x-2)(2 x+1)$, sledi\n$$\n(-1)^{n} 2^{n} x^{n-1}(x-2)(2 x+1)=0\n$$\nČe je $n=1$, sta rešitvi enačbe $x=2$ in $x=-\\frac{1}{2}$, če pa je $n \\geq 2$, so rešitve enačbe $x=0$, $x=2$ in $x=-\\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41888,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nPoišči vsa praštevila $p, q, r$ in $s$, ki zadoščajo enačbama $p+q=r$ in $q+r=s^{2}$.",
    "options": [],
    "answer": "p=5, q=2, r=7, s=3",
    "solution": "Solution:\nIz prve enačbe vidimo, da je $r \\geq 4$, torej mora biti $r$ liho praštevilo. Če je tudi $q$ liho praštevilo, potem iz druge enačbe sledi, da je $s$ sodo praštevilo, torej $s=2$. Tedaj je $s^{2}=4$, hkrati pa je $q+r \\geq 6$, saj sta $q$ in $r$ lihi praštevili. Prišli smo do protislovja, saj enačba $q+r=s^{2}$ ni izpolnjena.\n\nOd tod sklepamo, da je $q$ sodo praštevilo, torej $q=2$. Iz prve enačbe izrazimo $r=p+2$ in to vstavimo v drugo enačbo, da dobimo $p+4=s^{2}$. To enačbo preoblikujemo v $p=(s-2)(s+2)$. Ker je $p$ praštevilo, od tod sledi $s-2=1$ in $s+2=p$, saj je $s-2<s+2$. Torej je $s=3$ in $p=5$. Iz $r=p+2$ dobimo še $r=7$.\n\nDobili smo eno samo rešitev $p=5$, $q=2$, $r=7$ in $s=3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41889,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nPoišči vsa realna števila $x$, ki rešijo enačbo\n$$\n\\log_{2}\\left(\\log_{2}(7x-6)\\right)+1=\\log_{2}\\left(\\log_{2}(3x-2)\\right)+\\log_{2} 3\n$$",
    "options": [],
    "answer": "2",
    "solution": "Solution:\nZ upoštevanjem zveze $1=\\log_{2} 2$ in formule za vsoto logaritmov $\\log_{2} a+\\log_{2} b=\\log_{2}(a \\cdot b)$ enačbo najprej preoblikujemo v\n$$\n\\log_{2}\\left(2 \\log_{2}(7x-6)\\right)=\\log_{2}\\left(3 \\log_{2}(3x-2)\\right)\n$$\nNato uporabimo še zvezo $a \\log_{2} b=\\log_{2} b^{a}$, da dobimo\n$$\n\\log_{2}\\left(\\log_{2}(7x-6)^{2}\\right)=\\log_{2}\\left(\\log_{2}(3x-2)^{3}\\right)\n$$\nDvakrat antilogatirmiramo, da pridemo do enačbe\n$$\n(7x-6)^{2}=(3x-2)^{3}\n$$\nOdpravimo oklepaje in enačbo preoblikujemo v\n$$\n27x^{3}-103x^{2}+120x-44=0\n$$\nUganemo, da je ena rešitev te enačbe $x_{1}=1$, in nato s Hornerjevim algoritmom razcepimo levo stran enačbe\n$$\n(x-1)\\left(27x^{2}-76x+44\\right)=0.\n$$\nS pomočjo formule za ničle kvadratnega polinoma izračunamo še ostali dve rešitvi $x_{2}=2$ in $x_{3}=\\frac{22}{27}$. Za vse vrednosti naredimo preizkus. Ugotovimo, da $x_{3}$ ni rešitev enačbe, saj je $7x_{3}-6=-\\frac{8}{27}<0$ in zato izraz $\\log_{2}\\left(7x_{3}-6\\right)$ ni definiran. Prav tako tudi $x_{1}$ ni rešitev enačbe, saj je $\\log_{2}\\left(7x_{1}-6\\right)=0$ in zato izraz $\\log_{2}\\left(\\log_{2}(7x-6)\\right)$ ni definiran. V primeru $x_{2}=2$ pa dobimo $\\log_{2}\\left(\\log_{2} 8\\right)+1=\\log_{2}\\left(\\log_{2} 4\\right)+\\log_{2} 3$, kar drži. Edina rešitev enačbe je torej $x=2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41890,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNajmanj kolikokrat moramo hkrati vreči dve pošteni igralni kocki, da bo verjetnost, da bomo vsaj enkrat na obeh kockah hkrati vrgli enako število pik, večja od $\\frac{1}{2}$ ?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nMoč algebre dogodkov pri metu dveh kock je $6 \\cdot 6 = 36$. Hkrati lahko na obeh kockah pade isto število pik na 6 načinov. Verjetnost, da na obeh kockah pade isto število pik, če kocki vržemo enkrat, je torej enaka $p = \\frac{6}{36} = \\frac{1}{6}$.\n\nDenimo, da kocki vržemo $n$-krat. Naj bo $A$ dogodek, da vsaj enkrat na obeh kockah pade enako število pik. Verjetnost nasprotnega dogodka $\\bar{A}$, da nikoli ne pade enako število pik, je enaka $P(\\bar{A}) = (1-p)^{n} = \\left(\\frac{5}{6}\\right)^{n}$. Verjetnost dogodka $A$ je zato enaka $P(A) = 1 - P(\\bar{A}) = 1 - \\left(\\frac{5}{6}\\right)^{n}$. Želimo, da je ta verjetnost večja od $\\frac{1}{2}$, zato mora veljati\n$$\n1 - \\left(\\frac{5}{6}\\right)^{n} > \\frac{1}{2}\n$$\nVrednost izraza $1 - \\left(\\frac{5}{6}\\right)^{n}$ je pri $n = 1, 2, 3, 4, \\ldots$ zaoredoma enaka $\\frac{1}{6}, \\frac{11}{36}, \\frac{91}{216}, \\frac{671}{1296}, \\ldots$ Prvič je vrednost večja od $\\frac{1}{2}$ pri $n = 4$, zato moramo kocki vreči vsaj 4-krat.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41891,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa točko $P$ znotraj trikotnika $ABC$ pravimo, da je izjemna, če velja naslednji pogoj:\nČe so $A'$, $B'$ in $C'$ zaporedoma presečišča premic $AP$, $BP$ in $CP$ s stranicami $BC$, $AC$ in $AB$, lahko daljice $AA'$, $BB'$ in $CC'$ vzporedno premaknemo tako, da oblikujemo trikotnik, katerega dolžine stranic so $|AA'|$, $|BB'|$ in $|CC'|$.\n\na. Dokaži, da je težišče trikotnika izjemna točka.\nb. Dokaži trditev \"Edina izjemna točka znotraj trikotnika je težišče trikotnika.\"",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOznačimo $\\vec{a}=\\overrightarrow{AB}$ in $\\vec{b}=\\overrightarrow{AC}$.\n\na.\nČe je $P$ težišče, tedaj je\n$$\n\\begin{aligned}\n& \\overrightarrow{AA'}=\\overrightarrow{AB}+\\frac{1}{2} \\overrightarrow{BC}=\\vec{a}+\\frac{1}{2}(-\\vec{a}+\\vec{b})=\\frac{1}{2} \\vec{a}+\\frac{1}{2} \\vec{b} \\\\\n& \\overrightarrow{BB'}=\\overrightarrow{BA}+\\frac{1}{2} \\overrightarrow{AC}=-\\vec{a}+\\frac{1}{2} \\vec{b} \\\\\n& \\overrightarrow{CC'}=\\overrightarrow{CA}+\\frac{1}{2} \\overrightarrow{AB}=-\\vec{b}+\\frac{1}{2} \\vec{a}\n\\end{aligned}\n$$\nSledi $\\overrightarrow{AA'}+\\overrightarrow{BB'}+\\overrightarrow{CC'}=0$, torej lahko z vzporednimi premiki daljic $AA'$, $BB'$ in $CC'$ sestavimo trikotnik, če postavimo daljice tako, da točki $A'$ in $B$ sovpadeta ter točki $B'$ in $C$ sovpadeta. Težišče trikotnika je torej izjemna točka.\n\nb.\nNaj bo $P$ izjemna točka. Ker sta vektorja $\\vec{a}$ in $\\vec{b}$ linearno neodvisna, obstajata realni števili $\\alpha$ in $\\beta$, da velja $\\overrightarrow{AP}=\\alpha \\vec{a}+\\beta \\vec{b}$. Ker $P$ leži v notranjosti trikotnika, je $\\alpha, \\beta>0$. Izrazimo vektorje $\\overrightarrow{AA'}$, $\\overrightarrow{BB'}$ in $\\overrightarrow{CC'}$ z vektorjema $\\vec{a}$ in $\\vec{b}$.\n\n$\\overrightarrow{AA'}=k(\\alpha \\vec{a}+\\beta \\vec{b})=\\vec{a}+\\ell(\\vec{b}-\\vec{a})$ za neki konstanti $k$ in $\\ell$, od koder sledi $k \\alpha=1-\\ell$ in $k \\beta=\\ell$. Iz teh dveh enačb sledi $k=\\frac{1}{\\alpha+\\beta}$, torej je $\\overrightarrow{AA'}=\\frac{1}{\\alpha+\\beta}(\\alpha \\vec{a}+\\beta \\vec{b})$.\n\n$\\overrightarrow{BB'}=k(\\alpha \\vec{a}+\\beta \\vec{b}-\\vec{a})=\\ell \\vec{b}-\\vec{a}$ za neki konstanti $k$ in $\\ell$, od koder sledi $k(\\alpha-1)=-1$ in $k \\beta=\\ell$. Torej je $k=\\frac{1}{1-\\alpha}$ in $\\overrightarrow{BB'}=\\frac{1}{1-\\alpha}(\\alpha \\vec{a}+\\beta \\vec{b}-\\vec{a})$.\n\n$\\overrightarrow{CC'}=k(\\alpha \\vec{a}+\\beta \\vec{b}-\\vec{b})=\\ell \\vec{a}-\\vec{b}$ za neki konstanti $k$ in $\\ell$, od koder sledi $k \\alpha=\\ell$ in $k(\\beta-1)=-1$. Torej je $k=\\frac{1}{1-\\beta}$ in $\\overrightarrow{CC'}=\\frac{1}{1-\\beta}(\\alpha \\vec{a}+\\beta \\vec{b}-\\vec{b})$.\n\nKer lahko z vzporednimi premiki iz daljic $AA'$, $BB'$ in $CC'$ sestavimo trikotnik, mora veljati $\\overrightarrow{AA'} \\pm \\overrightarrow{BB'} \\pm \\overrightarrow{CC'}=0$ za neko izbiro predznakov.\n\nDenimo najprej, da je $\\overrightarrow{AA'}+\\overrightarrow{BB'}+\\overrightarrow{CC'}=0$, torej\n$$\n\\frac{1}{\\alpha+\\beta}(\\alpha \\vec{a}+\\beta \\vec{b})+\\frac{1}{1-\\alpha}(\\alpha \\vec{a}+\\beta \\vec{b}-\\vec{a})+\\frac{1}{1-\\beta}(\\alpha \\vec{a}+\\beta \\vec{b}-\\vec{b})=0\n$$\nKer sta vektorja $\\vec{a}$ in $\\vec{b}$ linearno neodvisna, od tod sledi\n$$\n\\frac{\\alpha}{\\alpha+\\beta}-1+\\frac{\\alpha}{1-\\beta}=0 \\quad \\text{in} \\quad \\frac{\\beta}{\\alpha+\\beta}+\\frac{\\beta}{1-\\alpha}-1=0.\n$$\nV obeh enačbah odpravimo ulomke in ju poenostavimo, da dobimo\n$$\n\\alpha \\beta-\\beta+\\beta^2+\\alpha^2=0 \\quad \\text{in} \\quad \\alpha \\beta-\\alpha+\\beta^2+\\alpha^2=0\n$$\nOd tod sledi $\\alpha=\\beta$. Če to vstavimo v eno izmed enačb, dobimo $3 \\beta^2-\\beta=0$. Ker je $\\beta>0$, mora biti $\\beta=\\frac{1}{3}$ in $\\alpha=\\frac{1}{3}$.\n\nDenimo sedaj, da je $\\overrightarrow{AA'}+\\overrightarrow{BB'}-\\overrightarrow{CC'}=0$. Tedaj podobno kot zgoraj dobimo enačbi\n$$\n\\frac{\\alpha}{\\alpha+\\beta}-1-\\frac{\\alpha}{1-\\beta}=0 \\quad \\text{in} \\quad \\frac{\\beta}{\\alpha+\\beta}+\\frac{\\beta}{1-\\alpha}+1=0\n$$\nki ju preoblikujemo do\n$$\n\\beta^2-\\alpha^2-\\alpha \\beta-\\beta=0 \\quad \\text{in} \\quad \\beta^2-\\alpha^2-\\alpha \\beta+\\alpha+2\\beta=0\n$$\nOd tod sledi $\\alpha=-2\\beta$, kar pa je protislovje, saj sta $\\alpha$ in $\\beta$ pozitivni števili.\n\nČe je $\\overrightarrow{AA'}-\\overrightarrow{BB'}+\\overrightarrow{CC'}=0$, na podoben način dobimo protislovje $\\beta=-2\\alpha$.\n\nČe pa je $\\overrightarrow{AA'}-\\overrightarrow{BB'}-\\overrightarrow{CC'}=0$, dobimo enačbi\n$$\n\\frac{\\alpha}{\\alpha+\\beta}+1-\\frac{\\alpha}{1-\\beta}=0 \\quad \\text{in} \\quad \\frac{\\beta}{\\alpha+\\beta}-\\frac{\\beta}{1-\\alpha}+1=0\n$$\nki ju preoblikujemo do\n$$\n2\\alpha+\\beta-\\beta^2-\\alpha^2-3\\alpha\\beta=0 \\quad \\text{in} \\quad \\alpha+2\\beta-\\beta^2-\\alpha^2-3\\alpha\\beta=0\n$$\nOd tod sledi $\\alpha=\\beta$. Ko to vstavimo v prvi enačbo, dobimo $3\\beta-5\\beta^2=0$. Ker je $\\beta>0$, mora biti $\\beta=\\frac{3}{5}$ in $\\alpha=\\frac{3}{5}$. Toda v tem primeru $P$ leži zunaj trikotnika, saj je $\\overrightarrow{AP}=\\frac{3}{5}(\\vec{a}+\\vec{b})$, medtem ko za razpolovišče $M$ stranice $BC$ velja $\\overrightarrow{AM}=\\frac{1}{2}(\\vec{a}+\\vec{b})$.\n\nS tem smo pokazali, da znotraj trikotnika leži največ ena izjemna točka, iz točke (a) pa vemo, da je to težišče trikotnika.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41892,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPolona v pekarni redno kupuje žemlje in rogljičke. Med tednom so pri nakupu vsaj šestih rogljičkov le-ti $15 \\%$ cenejši, med vikendom pa na celoten nakup priznajo $10 \\%$ popust. V torek je Polona kupila 6 rogljičkov in 5 žemelj ter plačala $2,27 \\text{ €}$. V soboto pa je kupila 7 rogljičkov, 4 žemlje in vrečko, ki stane toliko, kot dva rogljička, ter plačala $2,52 \\text{ €}$. Kdaj se Poloni bolj splača kupiti 9 rogljičkov in 8 žemelj, med tednom ali med vikendom?",
    "options": [],
    "answer": "weekend",
    "solution": "Solution:\n\nNastavimo sistem enačb na podlagi podatkov za torkov nakup $6 r \\cdot 0{,}85 + 5 z = 2{,}27$ € ter za sobotni nakup $(7 r + 4 z + 2 r) \\cdot 0{,}9 = 2{,}52$. Pri tem smo pozorni na $10 \\%$ popust med vikendom, $15 \\%$ popust na rogljičke med tednom in vrednost vrečke. Sistem enačb lahko rešujemo z zamenjalnim načinom ali metodo nasprotnih koeficientov. Npr., prvo enačbo pomnožimo z $-4$, drugo pa s $5$, da dobimo sistem enačb $-20{,}4 r - 20 z = -9{,}08$ ter $45 r + 20 z = 14$. Enačbi seštejemo in dobimo linearno enačbo $24{,}6 r = 4{,}92$, od koder izračunamo, da je $r = 0{,}2$. Podatek vstavimo v eno od prvotnih enačb in dobimo na primer $9 \\cdot 0{,}2 + 4 z = 2{,}52$, od koder izračunamo $z = 0{,}25$.\n\nPolona bi med tednom plačala $9 \\cdot 0{,}2 \\cdot 0{,}85 + 8 \\cdot 0{,}25 = 3{,}53$ €, med vikendom pa $(9 \\cdot 0{,}2 + 8 \\cdot 0{,}25) \\cdot 0{,}9 = 3{,}42$ €, torej se ji nakup bolj splača med vikendom.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41893,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $\\sin \\alpha + \\sin \\beta = 1$ in $\\cos \\alpha + \\cos \\beta = -\\sqrt{3}$.\n\na. Izračunaj vrednost izraza $\\cos (\\alpha - \\beta)$.\n\nb. Poišči vse pare realnih števil $\\alpha$ in $\\beta$, ki ustrezajo danima enačbama.",
    "options": [],
    "answer": "cos(α − β) = 1; all solutions are α = 5π/6 + 2mπ and β = 5π/6 + 2nπ for integers m, n.",
    "solution": "Solution:\n\na. Adicijski izrek za kosinus nam da $\\cos (\\alpha - \\beta) = \\cos \\alpha \\cos \\beta + \\sin \\alpha \\sin \\beta$. Dani enačbi kvadriramo, da dobimo\n$$\n\\begin{array}{r}\n\\sin^2 \\alpha + 2 \\sin \\alpha \\sin \\beta + \\sin^2 \\beta = 1 \\\\\n\\cos^2 \\alpha + 2 \\cos \\alpha \\cos \\beta + \\cos^2 \\beta = 3\n\\end{array}\n$$\nDobljeni enačbi sedaj seštejemo in upoštevamo zvezo $\\sin^2 x + \\cos^2 x = 1$, da dobimo\n$$\n1 + 2(\\sin \\alpha \\sin \\beta + \\cos \\alpha \\cos \\beta) + 1 = 4\n$$\nod tod sledi $\\sin \\alpha \\sin \\beta + \\cos \\alpha \\cos \\beta = 1$ oziroma $\\cos (\\alpha - \\beta) = 1$.\n\nb. Ker je $\\cos (\\alpha - \\beta) = 1$, sledi $\\alpha - \\beta = 2k\\pi$ oziroma $\\alpha = \\beta + 2k\\pi$, $k \\in \\mathbb{Z}$. Torej je $\\cos \\alpha = \\cos (\\beta + 2k\\pi) = \\cos \\beta$ in podobno $\\sin \\alpha = \\sin \\beta$. Ko to upoštevamo v obeh danih enačbah, iz njiju izrazimo $\\sin \\beta = \\frac{1}{2}$ in $\\cos \\beta = -\\frac{\\sqrt{3}}{2}$. Od tod sklepamo, da je $\\beta = \\frac{5\\pi}{6} + 2n\\pi$, $n \\in \\mathbb{Z}$. Torej je $\\alpha = \\frac{5\\pi}{6} + 2n\\pi + 2k\\pi = \\frac{5\\pi}{6} + 2(n+k)\\pi$. Če označimo $k+n = m$, lahko zapišemo $\\alpha = \\frac{5\\pi}{6} + 2m\\pi$, $m \\in \\mathbb{Z}$. Danima enačbama torej ustrezajo vsi pari $\\alpha = \\frac{5\\pi}{6} + 2m\\pi$, $\\beta = \\frac{5\\pi}{6} + 2n\\pi$, kjer sta $m$ in $n$ poljubni celi števili.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41894,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoloči parametra $a$ in $b$ tako, da bosta premici podani z enačbama\n$$\n\\begin{aligned}\n(a+b) x - a y + a - 2 &= 0 \\\\\n(2b - a) x + (a - 4b) y - a &= 0\n\\end{aligned}\n$$\nidentični (sovpadali, se prekrivali).",
    "options": [],
    "answer": "a = 4, b = -1",
    "solution": "Solution:\n\nEnačbi premic zapišemo v eksplicitni obliki:\n$$\ny = \\frac{a+b}{a} x + \\frac{a-2}{a}\n$$\nin\n$$\ny = \\frac{a-2b}{a-4b} x + \\frac{a}{a-4b}.\n$$\nČe sta premici identični, imata zapisani v eksplicitni obliki enaka smerna koeficienta in enaka odseka na ordinatni osi. Dobimo enačbi\n$$\n\\frac{a+b}{a} = \\frac{a-2b}{a-4b}\n$$\nin\n$$\n\\frac{a-2}{a} = \\frac{a}{a-4b}.\n$$\nIz prve enačbe dobimo\n$$\nb \\cdot (a + 4b) = 0.\n$$\nRešitev $b = 0$ ne ustreza, ker se druga enačba preoblikuje v $a-2 = a$ in nima rešitve. Ostane še druga možnost $a + 4b = 0$, oziroma $4b = -a$. To uporabimo v drugi enačbi in dobimo\n$$\n\\frac{a-2}{a} = \\frac{a}{a+a} = \\frac{1}{2}\n$$\nin iz tega sledi $a = 4$ in $b = -1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41895,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse pare realnih števil $x$ in $y$, ki rešijo sistem enačb\n$$\n\\begin{aligned}\n\\frac{3}{x-4y} + \\frac{2}{x+y-5} &= 0 \\\\\n\\frac{2}{x^2-4y^2} + \\frac{1}{x^2+y^2-5} &= 0\n\\end{aligned}\n$$",
    "options": [],
    "answer": "x = -14, y = -17",
    "solution": "Solution:\n\nOdpravimo ulomke, da dobimo\n$$\n\\begin{aligned}\n3(x+y-5) + 2(x-4y) &= 0 \\\\\n2\\left(x^2 + y^2 - 5\\right) + \\left(x^2 - 4y^2\\right) &= 0\n\\end{aligned}\n$$\nin obe enačbi poenostavimo do\n$$\n\\begin{array}{r}\n5x - 5y - 15 = 0 \\\\\n3x^2 - 2y^2 - 10 = 0\n\\end{array}\n$$\nIz prve enačbe izrazimo $x = y + 3$. Ko slednje vstavimo v drugo enačbo, dobimo $y^2 + 18y + 17 = 0$. Levo stran enačbe razstavimo, da dobimo $(y + 1)(y + 17) = 0$. Rešitvi sta $y = -1$ in $y = -17$. V prvem primeru iz zveze $x = y + 3$ dobimo $x = 2$, v drugem pa $x = -14$. Z obema rešitvama naredimo preizkus in ugotovimo, da v primeru $x = 2$, $y = -1$ ulomka $\\frac{2}{x^2-4y^2}$ in $\\frac{1}{x^2+y^2-5}$ nista definirana. Edina rešitev sistema je torej par $x = -14$, $y = -17$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41896,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKot med nosilkama diagonal štirikotnika $ABCD$ je velik $60^{\\circ}$. Vsako oglišče štirikotnika prezrcalimo čez nosilko diagonale, ki poteka skozi njemu sosednji oglišči. Dokaži, da zrcalne slike oglišč ležijo na isti premici.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nZrcalne slike oglišč $A, B, C$ in $D$ pri zrcaljenju čez nosilke ustreznih diagonal označimo po vrsti z $A'$, $B'$, $C'$ in $D'$. Označimo presečišče nosilk diagonal s $P$.\n\nDenimo najprej, da $P$ leži v notranjosti štirikotnika. Zaradi simetrije lahko predpostavimo, da je $\\angle BPC = 60^{\\circ}$. Zaradi zrcaljenja potem velja $\\angle A'PD = \\angle DPA = 60^{\\circ}$, $\\angle APD' = \\angle DPA = 60^{\\circ}$, $\\angle C'PB = \\angle BPC = 60^{\\circ}$ in $\\angle CPB' = \\angle BPC = 60^{\\circ}$. Sledi\n$$\n\\angle A'PD' = \\angle A'PD + \\angle DPA + \\angle APD' = 180^{\\circ}\n$$\nin\n$$\n\\angle C'PB' = \\angle C'PB + \\angle BPC + \\angle CPB' = 180^{\\circ}.\n$$\nPrva enakost pove, da točke $A'$, $D'$ in $P$ ležijo na isti premici, druga pa, da točke $B'$, $C'$ in $P$ ležijo na isti premici. Zaradi zrcaljenja velja tudi $\\angle A'PA = \\angle C'PC = 120^{\\circ}$. Ker točke $A, C$ in $P$ ležijo na isti premici in točki $A'$ in $C'$ ležita na različnih bregovih premice $AC$, od tod sledi, da tudi točke $A'$, $C'$ in $P$ ležijo na isti premici. Od tod sledi, da vseh pet točk $A'$, $B'$, $C'$, $D'$ in $P$ leži na isti premici.\n\n2. način. Naj bo $P$ presečišče premic $AC$ in $BD$. S $p$ označimo premico skozi $P$, ki razpolavlja tisti kot med $AC$ in $BD$, ki meri $120^{\\circ}$. Premica $p$ zato seka obe nosilki diagonal pod kotom $60^{\\circ}$. Dokazati želimo, da točke $A'$, $B'$, $C'$ in $D'$ ležijo na $p$.\n\nKer se $AC$ in $BD$ sekata v točki $P$ pod kotom $60^{\\circ}$, mora tudi zrcalna slika premice $AC$ preko premice $BD$ sekati $BD$ v točki $P$ pod kotom $60^{\\circ}$. To je ravno premica $p$. Po drugi strani pa vemo, da točki $A'$ in $C'$ ležita na zrcalni sliki premice $AC$. Zato $A'$ in $C'$ ležita na $p$. Podobno mora zrcalna slika premice $BD$ preko premice $AC$ sekati $AC$ v točki $P$ pod kotom $60^{\\circ}$. To je ravno premica $p$. Ker točki $B'$ in $D'$ ležita na zrcalni sliki $BD$, ležita na premici $p$.\n\nDokazali smo, da točke $A'$, $B'$, $C'$ in $D'$ ležijo na premici $p$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41897,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDano je zaporedje $a_{1}, a_{2}, a_{3}, \\ldots$ z začetnima členoma $a_{1}=4$ in $a_{2}=16$, za katerega je $\\log _{2}\\left(\\log _{2} a_{1}\\right), \\log _{2}\\left(\\log _{2} a_{2}\\right), \\log _{2}\\left(\\log _{2} a_{3}\\right), \\ldots$ aritmetično zaporedje. Dokaži, da je\n$$\n\\log _{2}\\left(\\log _{2}\\left(4 a_{1} a_{2} \\ldots a_{n}\\right)\\right)=n+1\n$$\nza vsa naravna števila $n$.\n(20 točk)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nZaporedje $\\log _{2}\\left(\\log _{2}\\left(a_{n}\\right)\\right)$ je aritmetično z začetnima členoma $\\log _{2}\\left(\\log _{2} 4\\right)=\\log _{2} 2=1$ in $\\log _{2}\\left(\\log _{2} 16\\right)=\\log _{2} 4=2$, torej je $\\log _{2}\\left(\\log _{2} a_{n}\\right)=n$. Od tod sledi $\\log _{2} a_{n}=2^{n}$ in zato $a_{n}=2^{2^{n}}$. S pomočjo formule za vsoto geometrijskega zaporedja izračunamo\n$$\n4 a_{1} a_{2} \\ldots a_{n}=2^{2} 2^{2^{1}} 2^{2^{2}} 2^{2^{3}} \\ldots 2^{2^{n}}=2^{1+1+2^{1}+2^{2}+2^{3}+\\ldots+2^{n}}=2^{1+\\left(2^{n+1}-1\\right)}=2^{2^{n+1}}\n$$\nod koder sledi $\\log _{2}\\left(\\log _{2}\\left(4 a_{1} a_{2} \\ldots a_{n}\\right)\\right)=\\log _{2}\\left(\\log _{2}\\left(2^{2^{n+1}}\\right)\\right)=\\log _{2}\\left(2^{n+1}\\right)=n+1$.\n\n\n2. način. Kot v prvi rešitvi izpeljemo $\\log _{2}\\left(\\log _{2} a_{n}\\right)=n$ oziroma $\\log _{2} a_{n}=2^{n}$. Po formuli za logaritem produkta je\n$$\n\\begin{gathered}\n\\log _{2}\\left(\\log _{2}\\left(4 a_{1} a_{2} \\ldots a_{n}\\right)\\right)=\\log _{2}\\left(\\log _{2} 4+\\log _{2} a_{1}+\\log _{2} a_{2}+\\ldots+\\log _{2} a_{n}\\right)= \\\\\n=\\log _{2}\\left(2+2^{1}+2^{2}+\\ldots+2^{n}\\right)\n\\end{gathered}\n$$\nKer je vsota geometrijskega zaporedja $1+2^{1}+2^{2}+\\ldots+2^{n}$ enaka $2^{n+1}-1$, sledi $\\log _{2}\\left(\\log _{2}\\left(4 a_{1} a_{2} \\ldots a_{n}\\right)\\right)=\\log _{2}\\left(2^{n+1}\\right)=n+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41898,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse realne rešitve enačbe\n$$\n\\sqrt{x+4}+\\sqrt{2x+1}+\\sqrt{4x+2}=\\sqrt{3-x}\n$$",
    "options": [],
    "answer": "-1/2",
    "solution": "Solution:\n\nEnačbo preoblikujemo v\n$$\n\\sqrt{2x+1}+\\sqrt{x+4}=\\sqrt{3-x}-\\sqrt{4x+2}\n$$\njo kvadriramo\n$$\n(2x+1)+2\\sqrt{2x+1}\\sqrt{x+4}+(x+4)=(3-x)-2\\sqrt{3-x}\\sqrt{4x+2}+(4x+2)\n$$\nin poenostavimo, da dobimo\n$$\n\\sqrt{2x+1}\\sqrt{x+4}=-\\sqrt{3-x}\\sqrt{4x+2}\n$$\nKo enačbo še enkrat kvadriramo in poenostavimo, dobimo kvadratno enačbo\n$$\n6x^{2}-x-2=0\n$$\nPo formuli za kvadratno enačbo dobimo rešitvi $x_{1}=-\\frac{1}{2}$ in $x_{2}=\\frac{2}{3}$. Za obe rešitvi naredimo preizkus. Prva rešitev res ustreza enačbi, druga pa ne, saj dobimo $2\\sqrt{\\frac{14}{3}}+\\sqrt{\\frac{7}{3}}=\\sqrt{\\frac{7}{3}}$. Edina rešitev enačbe je torej $x=-\\frac{1}{2}$.\n\n\n2. način. Na enak način kot v prvi rešitvi dobimo\n$$\n\\sqrt{2x+1}\\sqrt{x+4}=-\\sqrt{3-x}\\sqrt{4x+2}\n$$\nVidimo, da je leva stran enačbe večja ali enaka 0, desna pa manjša ali enaka 0. Torej morata biti obe strani enaki 0, kar je možno le pri $x=-\\frac{1}{2}$.\n\n\n3. način. Opazimo, da je $4x+2=2(2x+1)$, zato lahko enačbo preoblikujemo v\n$$\n(1+\\sqrt{2})\\sqrt{2x+1}=\\sqrt{3-x}-\\sqrt{x+4}\n$$\nEnačbo kvadririramo in preoblikujemo, da dobimo\n$$\n(3+2\\sqrt{2})x+\\sqrt{2}-2=-\\sqrt{3-x}\\sqrt{x+4}\n$$\nŠe enkrat kvadriramo in preuredimo do\n$$\n(18+12\\sqrt{2})x^{2}+(-3-2\\sqrt{2})x-6-4\\sqrt{2}=0\n$$\nOpazimo, da enačbo lahko krajšamo s $3+2\\sqrt{2}$, da dobimo\n$$\n6x^{2}-x-2=0\n$$\nKot v prvi rešitvi ugotovimo, da ima ta enačba rešitvi $-\\frac{1}{2}$ in $\\frac{2}{3}$, od katerih pa le prva reši začetno enačbo.\n\n\n4. način. Enačbo preoblikujemo v\n$$\n\\sqrt{2x+1}+\\sqrt{4x+2}=\\sqrt{3-x}-\\sqrt{x+4}\n$$\nPo kvadriranju dobimo\n$$\n6x+3+2\\sqrt{2x+1}\\sqrt{4x+2}=7-2\\sqrt{3-x}\\sqrt{x+4}\n$$\nkar spet nekoliko preoblikujemo v\n$$\n\\sqrt{2x+1}\\sqrt{4x+2}+\\sqrt{3-x}\\sqrt{x+4}=2-3x\n$$\nEnačbo zopet kvadriramo in po preoblikovanju dobimo\n$$\n2\\sqrt{2x+1}\\sqrt{4x+2}\\sqrt{3-x}\\sqrt{x+4}=2x^{2}-19x-10\n$$\nŠe zadnjič kvadriramo in preoblikujemo do\n$$\n36x^{4}-12x^{3}-23x^{2}+4x+4=0\n$$\nLevo stran lahko razstavimo kot $(2x+1)^{2}(3x-2)^{2}$, torej ima ta enačba rešitvi $x=-\\frac{1}{2}$ in $x=\\frac{2}{3}$. Kot v prvi rešitvi preverimo, da le prva od teh dveh vrednosti reši začetno enačbo.\n\n\n5. način. Enačbo preoblikujemo v\n$$\n\\sqrt{x+4}+\\sqrt{4x+2}=\\sqrt{3-x}-\\sqrt{2x+1}\n$$\nter jo nato kvadriramo. Dobimo\n$$\n5x+6+2\\sqrt{x+4}\\sqrt{4x+2}=x+4-2\\sqrt{3-x}\\sqrt{2x+1}\n$$\nkar preoblikujemo v\n$$\n\\sqrt{x+4}\\sqrt{4x+2}+\\sqrt{3-x}\\sqrt{2x+1}=-1-2x\n$$\nEnačbo še enkrat kvadriramo in po preoblikovanju dobimo\n$$\n2\\sqrt{x+4}\\sqrt{4x+2}\\sqrt{3-x}\\sqrt{2x+1}=2x^{2}-19x-10\n$$\nOd tu dalje postopamo kot v četrti rešitvi, da pridemo do edine rešitve enačbe, ki je $x=-\\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41899,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $M$ točka na stranici $BC$ pravokotnega trikotnika $ABC$ s pravim kotom pri $A$. Razpolovišče daljice $AM$ označimo s $P$. Naj bosta $P_{B}$ in $P_{C}$ zrcalni sliki točke $P$ preko stranic $AC$ in $AB$, $B'$ in $C'$ pa zrcalni sliki točk $B$ in $C$ preko točke $P$. Denimo, da so točke $B'$, $C'$, $P_{B}$ in $P_{C}$ kolinearne. Dokaži, da je tedaj $M$ razpolovišče stranice $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nKer sta $P_{B}$ in $P_{C}$ zrcalni sliki točke $P$ preko stranic trikotnika, velja $\\Varangle PAC = \\Varangle CAP_{B}$ in $\\Varangle BAP = \\Varangle P_{C}AB$. Torej je\n$$\n\\Varangle P_{C}AP_{B} = \\Varangle P_{C}AB + \\Varangle BAP + \\Varangle PAC + \\Varangle CAP_{B} = 2(\\Varangle BAP + \\Varangle PAC) = 2 \\cdot 90^{\\circ} = 180^{\\circ},\n$$\nkar pomeni, da so točke $P_{B}$, $A$ in $P_{C}$ kolinearne. Sledi, da vseh 5 točk $B'$, $C'$, $P_{B}$, $P_{C}$ in $A$ leži na isti premici. Ker se daljici $BB'$ in $CC'$ razpolavljata v točki $P$, je $BCB'C'$ paralelogram, torej sta premici $BC$ in $B'C'$ vzporedni. Zato velja $\\Varangle MBA = \\Varangle C'AB = \\Varangle P_{C}AB = \\Varangle BAP = \\Varangle BAM$, kar pomeni, da je trikotnik $ABM$ enakokrak z vrhom pri $M$ in velja $|BM| = |AM|$. Podobno je tudi $\\Varangle ACM = \\Varangle CAB' = \\Varangle MAC$, torej je tudi trikotnik $CAM$ enakokrak in velja $|AM| = |CM|$. Sledi $|BM| = |CM|$, od koder sklepamo, da je $M$ razpolovišče stranice $BC$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41900,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoloči vrednosti parametrov $a$ in $b$ tako, da bo premica z enačbo $y=3 x+5$ tangenta na graf funkcije $f$, $f(x)=a x^{2}+b x$, v točki z absciso $-1$.",
    "options": [],
    "answer": "a = -5, b = -7",
    "solution": "Solution:\n\nIzračunamo ordinato dotikališča $y=3(-1)+5=2$.\n\nUpoštevamo koordinate dotikališča v $f(x)=a x^{2}+b x$ in dobimo enačbo $a-b=2$.\n\nOdvajamo funkcijo $f$ in dobimo $f'(x)=2 a x+b$.\n\nUpoštevamo dotikališče $D(-1,2)$ in smerni koeficient premice $k=3$ ter dobimo enačbo $-2 a+b=3$.\n\nRešimo sistem enačb in dobimo rešitev $a=-5$ in $b=-7$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41901,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nReši enačbo\n$$\n\\log^{2}\\left(\\sqrt{3^{x}}-1\\right)=1\n$$",
    "options": [],
    "answer": "x = log_3 121 or x = log_3(121/100)",
    "solution": "Solution:\nIz dane enačbe dobimo enačbi $\\log \\left(\\sqrt{3^{x}}-1\\right)=1$ in $\\log \\left(\\sqrt{3^{x}}-1\\right)=-1$.\n\nPo preoblikovanju prve enačbe dobimo $\\sqrt{3^{x}}-1=10$, ki jo preoblikujemo do enačbe $3^{x}=121$. Rešitev prve enačbe je $x=\\log_{3} 121 \\doteq 4,37$.\n\nPri drugi enačbi dobimo $\\sqrt{3^{x}}-1=10^{-1}$, kar preoblikujemo do $3^{x}=\\frac{121}{100}$. Rešitev druge enačbe je $x=\\log_{3} \\frac{121}{100} \\doteq 0,174$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41902,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $P$ poljubna točka na daljici $AB$ z dolžino $\\sqrt{27}\\ \\mathrm{cm}$. Nad daljico $AP$ z dolžino $x$ je konstruiran kvadrat $APDE$, nad daljico $BD$ pa enakostranični trikotnik $BCD$. Nariši skico.\n\na) Izrazi ploščino petkotnika $ABCDE$ z $x$.\n\nb) Določi $x$, da bo ploščina petkotnika $ABCDE$ najmanjša.",
    "options": [],
    "answer": "S(x) = ((1 + √3)/2) x^2 + ((3√3 − 9)/2) x + (27√3)/4; minimizing at x = (−9 + 6√3)/2 cm",
    "solution": "Solution:\n\nPloščina $S$ petkotnika $ABCDE$ je vsota ploščin kvadrata $APDE$ ($S_1$), pravokotnega trikotnika $PBD$ ($S_2$) in enakostraničnega trikotnika $BCD$ ($S_3$).\n\nPloščine vseh treh likov izrazimo z $x$:\n\nPloščina kvadrata je $S_1 = x^2$,\n\nploščina pravokotnega trikotnika je $S_2 = \\frac{x \\cdot (\\sqrt{27} - x)}{2}$.\n\nPloščina enakostraničnega trikotnika je $S_3 = \\frac{|BD|^2 \\sqrt{3}}{4}$.\n\nDolžino stranice $BD$ izračunamo z uporabo Pitagorovega izreka:\n\n$$\n|BD|^2 = y^2 = x^2 + (\\sqrt{27} - x)^2 = 2x^2 - 6\\sqrt{3}x + 27.\n$$\n\nTo upoštevamo v zgornji formuli in dobimo:\n\n$$\nS_3 = \\frac{2\\sqrt{3}x^2 - 18x + 27\\sqrt{3}}{4}.\n$$\n\nFormula za ploščino petkotnika $ABCDE$ izražena z $x$ je:\n\n$$\nS = S_1 + S_2 + S_3 = x^2 + \\frac{x \\cdot (\\sqrt{27} - x)}{2} + \\frac{2\\sqrt{3}x^2 - 18x + 27\\sqrt{3}}{4} = \\frac{1+\\sqrt{3}}{2}x^2 + \\frac{3\\sqrt{3} - 9}{2}x + \\frac{27\\sqrt{3}}{4}.\n$$\n\nUpoštevamo, da ima kvadratna funkcija minimum, ko je $x = p = -\\frac{b}{2a}$.\n\nIzračunamo:\n\n$$\np = -\\frac{\\frac{3\\sqrt{3} - 9}{2}}{1 + \\sqrt{3}} = \\frac{-3\\sqrt{3} + 9}{2(1 + \\sqrt{3})} = \\frac{-9 + 6\\sqrt{3}}{2}.\n$$\n\nTočka $p$ je oddaljena od točke $A$ za $\\frac{-9 + 6\\sqrt{3}}{2}\\ \\mathrm{cm}$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41903,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana je kvadratna enačba $a x^{2}+b x+a=0$ ($a, b \\neq 0$ in $a \\neq b$). Za njene koeficiente velja, da izrazi $1, \\frac{a+b}{a-b}, \\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$ tvorijo zaporedne člene aritmetičnega zaporedja. Zapiši zvezo med koeficientoma $a$ in $b$ in reši kvadratno enačbo.",
    "options": [],
    "answer": "b = -2a and the quadratic has the double root x = 1",
    "solution": "Solution:\n\nUpoštevamo definicijo aritmetičnega zaporedja in dobimo\n$$\n\\frac{a+b}{a-b} - 1 = \\frac{a^{2}+b^{2}}{a^{2}-b^{2}} - \\frac{a+b}{a-b}.\n$$\nUredimo enačbo:\n$$\n\\frac{a+3b}{a-b} = \\frac{a^{2}+b^{2}}{a^{2}-b^{2}}.\n$$\nPo odpravi ulomka dobimo enačbo:\n$$\nb^{2} + 2ab = b(b + 2a) = 0.\n$$\nČe upoštevamo, da je $b \\neq 0$, je rešitev $b = -2a$.\n\nPri upoštevanju $b = -2a$ v kvadratni enačbi, dobimo\n$$\na x^{2} - 2a x + a = 0.\n$$\nUpoštevamo, da je $a \\neq 0$ in dobimo enačbo\n$$\nx^{2} - 2x + 1 = 0.\n$$\nRešitev enačbe je $x = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41904,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIz 17 vžigalic sestavimo mrežo (glej sliko). Najmanj koliko vžigalic moramo odstraniti, da preostale vžigalice ne bodo tvorile nobenega kvadrata?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nVžigalice na sliki tvorijo 6 majhnih kvadratov s stranicami dolgimi 1 vžigalico ter 2 večja kvadrata s stranicami dolgimi 2 vžigalici. Osenčimo nekatere majhne kvadrate in oštevilčimo nekatere vžigalice, kot to prikazuje slika.\n\n![](attached_image_2.png)\n\nKer moramo iz vsakega osenčenega kvadrata odstraniti vsaj eno vžigalico, moramo skupaj odstraniti vsaj 3 vžigalice. Denimo, da zadostuje odstraniti 3 vžigalice. Tedaj mora biti vsaka odstranjena vžigalica stranica natanko 1 osenčenega in 1 neosenčenega majhnega kvadrata, hkrati pa moramo vsakemu majhnemu kvadratu odstraniti natanko 1 stranico. Vžigalic na robu mreže torej ne smemo odstraniti.\n\nČe odstranimo vžigalico 1, potem vžigalice 2 ne smemo odstraniti. Zato moramo v zgornjem desnem majhnem kvadratu odstraniti vžigalico 5, v spodnjem desnem majhnem kvadratu pa vžigalice 7 ne smemo odstraniti. To je protislovje, saj tedaj ne odstranimo nobene vžigalice z levega velikega kvadrata. Torej vžigalice 1 ne smemo odstraniti.\n\nZaradi simetrije lahko podobno pokažemo, da tudi vžigalice 6 ne smemo odstraniti. Toda tedaj ne odstranimo nobene vžigalice z desnega velikega kvadrata, kar je spet protislovje. S tem smo pokazali, da odstraniti 3 vžigalice ne zadostuje. Odstraniti moramo vsaj 4 vžigalice. Da to zadostuje, prikazuje naslednja slika.\n\n![](attached_image_2.png)\n\n![](attached_image_3.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41905,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. V domu za starejše občane so praznovali rojstni dan najstarejše oskrbovanke. Pripravili so 15 litrov napitka iz domačega hruškovega soka, razredčenega z vodo, tako da je bilo v napitku $20 \\%$ vode. Ker je bil še vedno presladek, so dolili še 5 litrov vode. Izračunaj delež naravnega soka v $\\%$ v dobljenem napitku.\n\nb. V tem domu imajo na voljo skupaj 141 sob. 70 sob je enoposteljnih, ostale so dvo in troposteljne sobe. Če so vse sobe popolnoma zasedene, je v domu 240 oskrbovancev. Koliko imajo dvoposteljnih in koliko triposteljnih sob?\n\n(10 točk)",
    "options": [],
    "answer": "a) 60%; b) 43 two-bed rooms and 28 three-bed rooms",
    "solution": "Solution:\n\na. V 15 litrih soka je $20 \\%$ od 15 litrov, torej 3 litri vode. Naravnega soka je torej 12 litrov v 20 litrih napitka, kar pomeni, da je delež naravnega soka $60 \\%$.\n\nb. Ugotovimo, da je dvoposteljnih in troposteljnih sob skupaj $141-70=71$. Zapišemo $D+T=71$. Skupaj je v dvoposteljnih in troposteljnih sobah $2D+3T=170$ oskrbovancev. Rešimo sistem enačb in dobimo $D=43$ in $T=28$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41906,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana je funkcija s predpisom $f(x)=\\frac{x^{2}-2x+1}{x^{2}+1}$.\n\na) Izračunaj ničle, pole, asimptoto in nariši graf funkcije $f$.\n\nb) Izračunaj presečišča grafa funkcije $f$ s premico $y=2$.\n\nc) Določi zalogo vrednosti funkcije $f$.",
    "options": [],
    "answer": "a) Zero at x = 1 (double). No real poles. Horizontal asymptote y = 1.\nb) Intersection with y = 2 at the point (−1, 2) (double contact).\nc) Range: [0, 2].",
    "solution": "Solution:\n\na.\nFunkcija ima dvojno ničlo v $x=1$. Pola nima. Vodoravna asimptota je $y=1$. Funkcija seka ordinatno os v točki $N(0,1)$. Narišemo graf funkcije $f$.\n\n![](attached_image_1.png)\n\nb.\nZapišemo enačbo: $\\frac{x^{2}-2x+1}{x^{2}+1}=2$. Po ureditvi dobimo enačbo: $x^{2}+2x+1=0$. Dvojna rešitev te enačbe je $x=-1$. Zapišemo presečišče $P(-1,2)$.\n\nc.\nZapišemo zalogo vrednosti funkcije $f$: $Z_{f}=[0,2]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41907,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoenostavi izraz:\n$$\n2\\left(x y^{-1}-1\\right)^{-p}\\left(x^{2} y^{-2}-1\\right)^{p}-\\left(\\frac{x+y}{y}\\right)^{p}\n$$\nZa $x=-2$, $y=-\\frac{1}{2}$ in $p=-3$ izračunaj vrednost izraza.",
    "options": [],
    "answer": "1/125",
    "solution": "Solution:\n\nUredimo prvi oklepaj $\\left(x y^{-1}-1\\right)^{-p}=\\left(\\frac{x}{y}-1\\right)^{-p}=\\left(\\frac{x-y}{y}\\right)^{-p}=\\left(\\frac{y}{x-y}\\right)^{p}$ in drugi oklepaj $\\left(x^{2} y^{-2}-1\\right)^{p}=\\left(\\frac{x^{2}-y^{2}}{y^{2}}\\right)^{p}$. Oklepaja pomnožimo med seboj, števec drugega oklepaja razstavimo in dobimo produkt $\\left(\\frac{y}{x-y}\\right)^{p} \\cdot\\left(\\frac{x^{2}-y^{2}}{y^{2}}\\right)^{p}=\\left(\\frac{y}{x-y}\\right)^{p} \\cdot\\left(\\frac{(x-y)(x+y)}{y^{2}}\\right)^{p}$. Ker je eksponent obeh oklepajev enak, damo v skupni oklepaj in okrajšamo $\\left(\\frac{y(x-y)(x+y)}{(x-y) y^{2}}\\right)^{p}=\\left(\\frac{x+y}{y}\\right)^{p}$. Dobimo izraz $2\\left(\\frac{x+y}{y}\\right)^{p}-\\left(\\frac{x+y}{y}\\right)^{p}$. Odštejemo in dobimo $\\left(\\frac{x+y}{y}\\right)^{p}$.\n\nVstavimo vrednosti za $x, y, p$ in dobimo rezultat $\\frac{1}{125}$.\n\nZapis potence $z$ negativnim eksponentom $x y^{-1}=\\frac{x}{y}$\nZapis v obliki ulomka $x y^{-1}-1=\\frac{x-y}{y}$\nZapis v obliki ulomka $x^{2} y^{-2}-1=\\frac{x^{2}-y^{2}}{y^{2}}$\nRazstavljanje $x^{2}-y^{2}=(x-y)(x+y)$\nKrajšanje $\\frac{y(x-y)(x+y)}{(x-y) y^{2}}=\\frac{x+y}{y}$\nPoenostavljen izraz $2\\left(\\frac{x+y}{y}\\right)^{p}-\\left(\\frac{x+y}{y}\\right)^{p}=\\left(\\frac{x+y}{y}\\right)^{p}$\nVstavljene vrednosti spremenljivk $x=-2$, $y=-\\frac{1}{2}$ in $p=-3$.\nIzračun $\\frac{-2-\\frac{1}{2}}{-\\frac{1}{2}}=5$\nRezultat $5^{-3}$ ali $\\frac{1}{125}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41908,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa realni števili $x$ in $y$, kjer $x \\neq 0$, $y \\notin \\{-2, 0, 2\\}$ in $x + y \\neq 0$, poenostavi izraz:\n$$\n\\frac{x y^{2018} + 2 x y^{2017}}{y^{2016} - 4 y^{2014}} \\cdot \\left(\\left(\\frac{x^{2}}{y^{3}} + x^{-1}\\right) : \\left(x y^{-2} - \\frac{1}{y} + x^{-1}\\right)\\right) : \\frac{(x-y)^{2} + 4 x y}{1 + \\frac{y}{x}} - \\frac{y^{2} + 2y}{y+2}\n$$",
    "options": [],
    "answer": "2y/(y - 2)",
    "solution": "Solution:\n\nŠtevec prvega ulomka $x y^{2018} + 2 x y^{2017}$ preoblikujemo v $x y^{2017}(y + 2)$, imenovalec $y^{2016} - 4 y^{2014}$ pa preoblikujemo v $y^{2014}(y - 2)(y + 2)$. Prvi ulomek okrajšamo in dobimo $\\frac{x y^{3}}{y - 2}$.\n\nIzraz $\\frac{x^{2}}{y^{3}} + x^{-1}$ preoblikujemo v $\\frac{x^{3} + y^{3}}{x y^{3}}$. Razstavimo vsoto kubov.\n\nIzraz $x y^{-2} - \\frac{1}{y} + x^{-1}$ preoblikujemo v $\\frac{x^{2} - x y + y^{2}}{x y^{2}}$.\n\nIzraz $\\frac{(x-y)^{2} + 4 x y}{1 + \\frac{y}{x}}$ preoblikujemo v $\\frac{x(x + y)^{2}}{x + y} = x(x + y)$.\n\nZadnji ulomek razstavimo in okrajšamo, dobimo $y$.\n\nPrvi člen poenostavimo in dobimo $\\frac{y^{2}}{y - 2}$. Od tega izraza odštejemo $y$ in dobimo $\\frac{2y}{y - 2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41909,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana je kvadratna funkcija $f$ s predpisom $f(x) = (m-1)x^{2} + m x + m$, kjer $m \\neq 1$.\n\na. Za $m=3$ izračunaj najmanjšo vrednost funkcije $f$.\n\nb. Poišči vsa realna števila $m$, da bo funkcija $f$ strogo negativna za vsak $x$.",
    "options": [],
    "answer": "a) 15/8; b) m < 0",
    "solution": "Solution:\n\na.\nNajmanjša vrednost kvadratne funkcije je pri $q = \\frac{-D}{4a}$.\nIzračunamo diskriminanto $D = -15$.\nIzračunamo najmanjšo vrednost $q = \\frac{15}{8}$.\n\nb.\nFunkcija $f$ bo negativna za vsak $x$, ko bosta izpolnjena pogoja $(m-1 < 0)$ in $D < 0$.\nDiskriminanta kvadratne funkcije $f$ je $D = m^{2} - 4(m-1)m = -3m^{2} + 4m$.\nUpoštevamo, da je $D < 0$ in dobimo kvadratno neenačbo $-3m^{2} + 4m < 0$.\nIzračunamo rešitvi kvadratne neenačbe $-3m^{2} + 4m = 0$ in dobimo $m_{1} = 0$, $m_{2} = \\frac{4}{3}$.\nRešitve kvadratne neenačbe so $m < 0$ ali $m > \\frac{4}{3}$.\nUpoštevamo pogoja $(m-1 < 0)$ in $D < 0$ ter dobimo rezultat $m < 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41910,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKolikšna je vrednost funkcije $f(x) = \\log_{3}\\left(\\frac{\\sin\\left(\\frac{x \\pi}{2}\\right)}{\\log_{\\frac{1}{2}} 2^{x}}\\right)$ za $x = -3$?\n\n(A) 0\n(B) -3\n(C) ne obstaja\n(D) 1\n(E) -1",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\n$f(-3) = \\log_{3}\\left(\\frac{\\sin\\left(\\frac{-3 \\pi}{2}\\right)}{\\log_{\\frac{1}{2}} 2^{-3}}\\right) = \\log_{3}\\left(\\frac{1}{3}\\right) = -1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41911,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa $3~\\mathrm{kg}$ pomaranč in $5~\\mathrm{kg}$ limon plačamo skupaj $8,40~€$. Za $5~\\mathrm{kg}$ pomaranč in $4~\\mathrm{kg}$ limon plačamo skupaj $8,80~€$. Koliko skupno plačamo za $2~\\mathrm{kg}$ pomaranč in $3~\\mathrm{kg}$ limon?\n\n(A) $7,20~€$\n(B) $5,20~€$\n(C) $3,60~€$\n(D) $5,60~€$\n(E) $4,80~€$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nZapišemo sistem enačb $3P + 5L = 8,40$ in $5P + 4L = 8,80$. Rešimo sistem enačb in dobimo $P = 0,80~€$, $L = 1,20~€$. Izračunamo $2P + 3L = 5,20~€$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41912,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKaj dobimo po racionalizaciji izraza $\\frac{4+2 \\sqrt{6}}{(1+\\sqrt{2})+\\sqrt{3}}$?\n(A) $1-\\sqrt{2}-\\sqrt{3}$\n(B) $1-\\sqrt{2}+\\sqrt{3}$\n(C) $1+\\sqrt{2}+\\sqrt{3}$\n(D) $-1+\\sqrt{2}+\\sqrt{3}$\n(E) $-1+\\sqrt{2}-\\sqrt{3}$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\nŠtevec in imenovalec pomnožimo z izrazom $(1+\\sqrt{2})-\\sqrt{3}$ in dobimo $\\frac{4+2 \\sqrt{6}}{(1+\\sqrt{2})+\\sqrt{3}} \\cdot \\frac{(1+\\sqrt{2})-\\sqrt{3}}{(1+\\sqrt{2})-\\sqrt{3}}$.\n\n$= \\frac{(4+2\\sqrt{6})((1+\\sqrt{2})-\\sqrt{3})}{((1+\\sqrt{2})+\\sqrt{3})((1+\\sqrt{2})-\\sqrt{3})}$\n\nImenovalec:\n$((1+\\sqrt{2})+\\sqrt{3})((1+\\sqrt{2})-\\sqrt{3}) = (1+\\sqrt{2})^2 - (\\sqrt{3})^2 = (1 + 2\\sqrt{2} + 2) - 3 = 3 + 2\\sqrt{2} - 3 = 2\\sqrt{2}$\n\nŠtevec:\n$(4+2\\sqrt{6})((1+\\sqrt{2})-\\sqrt{3}) = 4(1+\\sqrt{2}) - 4\\sqrt{3} + 2\\sqrt{6}(1+\\sqrt{2}) - 2\\sqrt{6}\\sqrt{3}$\n$= 4 + 4\\sqrt{2} - 4\\sqrt{3} + 2\\sqrt{6} + 2\\sqrt{12} - 2\\sqrt{18}$\n$= 4 + 4\\sqrt{2} - 4\\sqrt{3} + 2\\sqrt{6} + 4\\sqrt{3} - 6\\sqrt{2}$\n$= 4 + (4\\sqrt{2} - 6\\sqrt{2}) + (2\\sqrt{6}) + (-4\\sqrt{3} + 4\\sqrt{3})$\n$= 4 - 2\\sqrt{2} + 2\\sqrt{6}$\n\nTorej:\n$\\frac{4 - 2\\sqrt{2} + 2\\sqrt{6}}{2\\sqrt{2}} = \\frac{2 - \\sqrt{2} + \\sqrt{6}}{\\sqrt{2}}$\n\nDobljeni izraz racionaliziramo z $\\sqrt{2}$:\n$\\frac{2 - \\sqrt{2} + \\sqrt{6}}{\\sqrt{2}} = \\frac{(2 - \\sqrt{2} + \\sqrt{6})\\sqrt{2}}{2}$\n$= \\frac{2\\sqrt{2} - 2 + \\sqrt{12}}{2}$\n$= \\frac{2\\sqrt{2} - 2 + 2\\sqrt{3}}{2}$\n$= \\sqrt{2} - 1 + \\sqrt{3}$\n\nTorej je rezultat $-1 + \\sqrt{2} + \\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41913,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nČe bi plašč stožca razgrnili v ravnino, bi dobili četrtino kroga s polmerom $8~\\mathrm{cm}$. Koliko je visok stožec?\n(A) $10~\\mathrm{cm}$\n(B) $2 \\sqrt{15}~\\mathrm{cm}$\n(C) $\\sqrt{20}~\\mathrm{cm}$\n(D) $3~\\mathrm{cm}$\n(E) $16~\\mathrm{cm}$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\nUpoštevanje, da je polmer krožnega izseka enak dolžini stranice stožca $s=8~\\mathrm{cm}$. Dolžina krožnega loka razgrnjenega plašča stožca je enaka obsegu osnovne ploskve stožca $\\frac{1}{4} \\cdot 2 \\pi s = 2 \\pi r$. Izračun polmera stožca $r=\\frac{s}{4}=2~\\mathrm{cm}$. Izračun višine stožca $v^{2}=s^{2}-r^{2}=8^{2}-2^{2}$, $v=\\sqrt{60}~\\mathrm{cm}=2 \\sqrt{15}~\\mathrm{cm}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41914,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBratje Jure, Klemen, Luka, Miha in Nace so kupili čokolado. Ko so jo odvili, so ugotovili, da je zlomljena na 7 kosov (glej sliko), zato so si teh 7 kosov med sabo razdelili. Jure je pojedel največji kos čokolade. Klemen in Luka sta pojedla enako količino čokolade, toda Klemen je pojedel 3 kose, Luka pa le 1 kos. Miha je pojedel $\\frac{1}{7}$ celotne čokolade. Preostanek čokolade je pojedel Nace. Kateri kos čokolade\n\n![](attached_image_1.png)\n\nna sliki je pojedel Nace?\n(A) 1\n(B) 2\n(C) 3\n(D) 4\n(E) 5",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nTablica čokolade ima $4 \\cdot 7 = 28$ kvadratnih koščkov. Naj bo en tak košček velik 1 kvadratno enoto. Tedaj je 7 kosov čokolade označenih na sliki po vrsti velikih $\\frac{1 \\cdot 4}{2} = 2$, $\\frac{3 \\cdot 4}{2} = 6$, $\\frac{5 \\cdot 2}{2} = 5$, $\\frac{4 \\cdot 2}{2} = 4$, $\\frac{2 \\cdot 2}{2} = 2$, $2 \\cdot \\frac{3+5}{2} = 8$ in $\\frac{1 \\cdot 2}{2} = 1$ kvadratno enoto. Torej je Jure pojedel kos s številko 6, Miha pa kos s številko 4. Izmed preostalih kosov čokolade so kosi s številkami 1, 5 in 7 skupaj veliki $2 + 2 + 1 = 5$ kvadratnih enot, kar je enako kot kos s številko 3. Torej je Klemen pojedel kose s številkami 1, 5 in 7, Luka pa je pojedel kos s številko 3. Ostane le kos s številko 2, ki ga je pojedel Nace.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41915,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDaljica $AB$ je dolga $20~\\mathrm{cm}$, točka $O$ pa je njeno razpolovišče. Krožnica $\\mathcal{K}$ se od zunaj dotika krožnic s premeroma $AO$ in $BO$ ter od znotraj dotika krožnice s premerom $AB$ (glej sliko). Koliko centimetrov je polmer krožnice $\\mathcal{K}$?\n\n(A) $\\frac{5}{2}$\n(B) $\\frac{10}{3}$\n(C) $\\frac{5 \\sqrt{2}}{2}$\n(D) $\\sqrt{5}$\n(E) $2 \\sqrt{5}$\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nOznačimo razpolovišče daljice $AO$ s $P$, središče krožnice $\\mathcal{K}$ pa z $R$. Polmer krožnice $\\mathcal{K}$ označimo z $r$. Zaradi simetrije je premica $RO$ pravokotna na premico $AB$. Torej po Pitagorovem izreku velja $|PO|^{2} + |OR|^{2} = |PR|^{2}$. Merjeno v centimetrih je $|PO| = 5$, $|OR| = 10 - r$ in $|PR| = 5 + r$, zato je $5^{2} + (10 - r)^{2} = (5 + r)^{2}$. Enačbo poenostavimo, da dobimo $30r = 100$. Torej je $r = \\frac{10}{3}~\\mathrm{cm}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41916,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKoliko je takih trimestnih naravnih števil, pri katerih se poljubni dve števki razlikujeta za vsaj 4?\n\n(A) 10\n(B) 15\n(C) 18\n(D) 20\n(E) 21",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nObravnavajmo vse možnosti glede na prvo števko števila.\nČe je prva števka števila enaka $1$, tedaj moramo drugo in tretjo števko izbrati iz množice $\\{5,6,7,8,9\\}$. Imamo le dve možnosti, $59$ in $95$.\nPodobno, če je prva števka enaka $8$, morata biti preostali dve iz množice $\\{0,1,2,3,4\\}$. Spet imamo le dve možnosti, $04$ in $40$.\nČe je prva števka enaka $2$, morata biti preostali dve iz množice $\\{6,7,8,9\\}$, vendar se nobeni dve števili iz te množice ne razlikujeta za vsaj $4$. Podobno ugotovimo, da prva števka ne more biti enaka $7$ in tudi ne $3$ ali $6$.\nČe je prva števka enaka $4$, tedaj sta preostali dve lahko enaki $08$, $80$, $09$ ali $90$.\nČe je prva števka enaka $5$, tedaj sta preostali dve lahko enaki $09$, $90$, $19$ ali $91$.\nČe pa je prva števka enaka $9$, moramo preostali dve izbrati iz množice $\\{0,1,2,3,4,5\\}$, kar lahko storimo na $6$ načinov, tj. $04$, $40$, $05$, $50$, $15$, $51$.\nŠtevil z iskano lastnostjo je torej $4+8+6=18$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41917,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nV kateri točki graf funkcije $f(x)=2 \\log _{\\sqrt{2}}(\\sqrt{2} x-5)-4$ seka abscisno os?\n(A) $\\left(\\frac{7 \\sqrt{2}}{2}, 0\\right)$\n(B) $(7 \\sqrt{2}, 0)$\n(C) $(1-7 \\sqrt{2}, 0)$\n(D) $\\left(\\frac{\\sqrt{2}}{7}, 0\\right)$\n(E) $\\left(\\frac{\\sqrt{2}}{2}-7,0\\right)$",
    "options": [],
    "answer": "A",
    "solution": "Solution:\nPredpis funkcije $f$ enačimo z 0. Dobimo enačbo $2 \\log _{\\sqrt{2}}(\\sqrt{2} x-5)-4=0$. Rešitev enačbe je $x=\\frac{7 \\sqrt{2}}{2}$. Graf funkcije $f$ seka abscisno os v točki $\\left(\\frac{7 \\sqrt{2}}{2}, 0\\right)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41918,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKot $\\alpha$ je za $33^\\circ$ manjši od svojega komplementarnega kota. Koliko je velik komplementarni kot?\n(A) $29^\\circ 50^\\prime$\n(B) $61^\\circ 50^\\prime$\n(C) $61,5^\\circ$\n(D) $29,5^\\circ$\n(E) $61,30^\\circ$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\nZapis obeh enačb $\\alpha = \\beta - 33^\\circ$ in $\\alpha + \\beta = 90^\\circ$. Z reševanjem sistema dobimo, da je $\\beta = 61,5^\\circ$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 41919,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTonček je izpisal tretjo potenco izraza $2 x^{3} y - 3 x y^{2}$. Ugotovil je, da imata potenci z osnovama $x$ in $y$ v enem izmed členov enaka eksponenta. Kolikšen je koeficient v tem členu?\n\n(A) 36\n(B) -36\n(C) 18\n(D) -54\n(E) 54",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\n$\\left(2 x^{3} y - 3 x y^{2}\\right)^{3} = \\left(2 x^{3} y\\right)^{3} - 3\\left(2 x^{3} y\\right)^{2} \\cdot \\left(3 x y^{2}\\right) + 3\\left(2 x^{3} y\\right) \\cdot \\left(3 x y^{2}\\right)^{2} - \\left(3 x y^{2}\\right)^{3} = 8 x^{9} y^{3} - 36 x^{7} y^{4} + 54 x^{5} y^{5} - 27 x^{3} y^{6}$.\n\nTretji člen ima enaka eksponenta pri $x$ in $y$, iskana vrednost je $54$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41920,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nS števkami $1,2,3,4,5,6,7$ in $9$ sestavljamo soda šestmestna števila, katerih prva števka je praštevilo. Koliko je vseh takih šestmestnih števil?\n\n(A) 20160\n(B) 9440\n(C) 25552\n(D) 4320\n(E) 49152",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nŠestmestno število lahko sestavimo na $4 \\cdot 8 \\cdot 8 \\cdot 8 \\cdot 8 \\cdot 3 = 49152$ načinov.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41921,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa realno število $a$ velja $a^{2}-\\frac{1}{2} a=\\frac{1}{4}$. Koliko je vrednost izraza $a^{3}-\\frac{1}{2} a$?\n\n(A) $-\\frac{1}{4}$\n(B) $\\frac{1}{4}$\n(C) $\\frac{1}{2}$\n(D) 4\n(E) $\\frac{1}{8}$",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nS pomočjo dane enakosti izračunamo\n$$\n\\begin{aligned}\na^{3}-\\frac{1}{2} a & =\\left(a^{3}-\\frac{1}{2} a^{2}\\right)+\\left(\\frac{1}{2} a^{2}-\\frac{1}{4} a\\right)-\\frac{1}{4} a=a\\left(a^{2}-\\frac{1}{2} a\\right)+\\frac{1}{2}\\left(a^{2}-\\frac{1}{2} a\\right)-\\frac{1}{4} a= \\\\\n& =\\frac{1}{4} a+\\frac{1}{2} \\cdot \\frac{1}{4}-\\frac{1}{4} a=\\frac{1}{8}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41922,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nČe polinom $p(x)$ delimo s polinomom $x-18$, je ostanek pri deljenju enak 20. Če polinom $p(x)$ delimo s polinomom $x-20$, je ostanek pri deljenju enak 18. Koliko je ostanek pri deljenju, če polinom $p(x)$ delimo s polinomom $(x-20)(x-18)$ ?\n\n(A) 2018\n(B) $-x-2$\n(C) $x+2$\n(D) $-x+38$\n(E) $x+38$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nIz podatkov sklepamo, da je $p(x) = (x-18) r(x) + 20$ za nek polinom $r(x)$. Od tod sledi $p(18) = 20$. Podobno sklepamo, da je $p(20) = 18$. Naj bo $s(x)$ ostanek pri deljenju polinoma $p(x)$ z $(x-20)(x-18)$. Tedaj je $p(x) = (x-20)(x-18) t(x) + s(x)$ za nek polinom $t(x)$ in $s(x) = a x + b$ za neki realni števili $a$ in $b$.\n\nKer je $p(18) = 20$ in $p(20) = 18$, je tudi $s(20) = 18$ in $s(18) = 20$. Od tod sledi $20 a + b = 18$ in $18 a + b = 20$, od koder izračunamo $a = -1$ in $b = 38$. Torej je $s(x) = -x + 38$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41923,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKoliko je takih trimestnih naravnih števil, pri katerih se poljubni dve števki razlikujeta za vsaj 3?\n\n(A) 80\n(B) 88\n(C) 92\n(D) 100\n(E) 648",
    "options": [],
    "answer": "D",
    "solution": "Solution:\nObravnavajmo vse možnosti glede na prvo števko števila.\nČe je prva števka enaka $1$, potem moramo drugo in tretjo števko izbrati iz množice šestih zaporednih števil $\\{4,5,6,7,8,9\\}$, tako da se razlikujeta za vsaj $3$. To lahko storimo na $12$ načinov, tj. $47,48,49,58,59,69,74,84,85,94,95,96$. Podobno, če je prva števka enaka $8$, moramo drugo in tretjo števko izbrati iz množice šestih zaporednih števil $\\{0,1,2,3,4,5\\}$. To lahko spet storimo na $12$ načinov.\nČe je prva števka enaka $2$, moramo drugi dve izbrati iz množice petih zaporednih števil $\\{5,6,7,8,9\\}$, kar lahko storimo na $6$ načinov. Podobno imamo $6$ možnosti tudi, če je prva števka enaka $7$.\nČe je prva števka enaka $3$, moramo drugi dve izbrati iz množice $\\{0\\} \\cup \\{6,7,8,9\\}$, kar lahko storimo na $10$ načinov. Podobno, če je prva števka enaka $6$, moramo drugi dve izbrati iz množice $\\{9\\} \\cup \\{0,1,2,3\\}$, kar spet lahko storimo na $10$ načinov.\nČe je prva števka enaka $4$, moramo drugi dve izbrati iz množice $\\{0,1\\} \\cup \\{7,8,9\\}$, kar lahko storimo na $12$ načinov. Podobno imamo $12$ možnosti tudi, če je prva števka enaka $5$.\nČe pa je prva števka enaka $9$, potem moramo preostali dve števki izbrati iz množice $\\{0,1,2,3,4,5,6\\}$, kar lahko storimo na $20$ načinov.\nŠtevil z iskano lastnostjo je torej $24+12+20+24+20=100$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41924,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bosta $a$ in $b$ naravni števili, za kateri velja $2^{a}-2^{b}=240$. Koliko je vrednost izraza $a+b$ ?\n(A) 8\n(B) 11\n(C) 13\n(D) 16\n(E) Nič od naštetega.",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nOčitno mora biti $a > b$. Enačbo preoblikujemo do $2^{b}\\left(2^{a-b}-1\\right) = 2^{4} \\cdot 15$. Od tod sledi $b = 4$ in $2^{a-b} - 1 = 15$. Drugo enačbo preuredimo do $2^{a-b} = 16$ in sklepamo, da je $a-b = 4$. Od tod izračunamo še $a = 8$. Torej je $a + b = 12$ in pravilen odgovor je $(\\mathbf{E})$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41925,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nŠtiri pozitivna števila so v razmerju $1:2:3:4$. Vsota kvadratov najmanjših treh števil je za $1$ manjša od vsote največjih treh števil. Največ koliko nizov takih števil lahko najdemo?\n\n(A) $0$\n(B) $1$\n(C) $2$\n(D) $3$\n(E) $4$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nIz razmerja $a : b : c : d = 1 : 2 : 3 : 4$ sledi, da je $a = \\frac{1}{10} t$, $b = \\frac{2}{10} t$, $c = \\frac{3}{10} t$, $d = \\frac{4}{10} t$.\n\nIz enakosti $a^{2} + b^{2} + c^{2} + 1 = b + c + d$ dobimo $14 t^{2} - 90 t + 100 = 0$.\n\nRešitvi kvadratne enačbe sta $t_{1} = \\frac{10}{7}$, $t_{2} = 5$.\n\nObstajata dva niza.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41926,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKatera izmed navedenih funkcij je odvod funkcije $f(x)=\\frac{1-2x}{\\sqrt{2x}}$?\n(A) $f'(x)=\\frac{-\\sqrt{2x}(2x-1)}{4x^2}$\n(B) $f'(x)=\\frac{-\\sqrt{2x}(2x-1)}{8x^2}$\n(C) $f'(x)=\\frac{-\\sqrt{2x}(2x+1)}{4x^2}$\n(D) $f'(x)=\\frac{-\\sqrt{2x}(2x+1)}{8x^2}$\n(E) $f'(x)=\\frac{-\\sqrt{2x}(1-2x)}{4x^2}$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\nIzračunamo odvod funkcije $f$ po obrazcu $f'(x)=\\frac{-2\\sqrt{2x}-(1-2x)\\cdot\\frac{1}{2}(2x)^{-\\frac{1}{2}}\\cdot 2}{2x}$. Razširimo števec na skupni imenovalec, razrešimo dvojni ulomek in z racionaliziranjem ulomka dobimo $f'(x)=\\frac{-2x\\sqrt{2x}-\\sqrt{2x}}{4x^2}$. Izpostavimo skupni faktor v števcu in dobimo rezultat $f'(x)=\\frac{-\\sqrt{2x}(2x+1)}{4x^2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41927,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bosta $\\alpha$ in $\\beta$ ostra kota pravokotnega trikotnika. Kateri izmed navedenih izrazov je enakovreden izrazu\n$$\n\\frac{\\sin \\left(90^{\\circ}-\\alpha\\right)+\\cos \\beta}{4 \\sin \\beta \\cdot \\cot \\left(90^{\\circ}-\\alpha\\right)}\n$$\n?\n\n(A) $\\frac{\\sin \\beta+\\cos \\beta}{4 \\cos \\beta}$\n\n(B) $\\frac{\\sin \\beta+\\cos \\beta}{4 \\sin \\beta}$\n\n(C) $\\frac{1}{4 \\cos \\beta}$\n\n(D) $\\frac{\\sin \\beta-\\cos \\beta}{4 \\cos \\beta}$\n\n(E) $\\frac{\\sin \\beta+\\cos \\beta}{2 \\cos \\beta}$",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nUporabimo zveze $\\sin \\left(90^{\\circ}-\\alpha\\right)=\\cos \\beta$, $\\cot \\left(90^{\\circ}-\\alpha\\right)=\\frac{\\cos \\left(90^{\\circ}-\\alpha\\right)}{\\sin \\left(90^{\\circ}-\\alpha\\right)}=\\frac{\\cos \\beta}{\\sin \\beta}$. Prvotni izraz preoblikujemo v\n$$\n\\frac{(\\sin \\beta+\\cos \\beta) \\cdot \\sin \\beta}{4 \\sin \\beta \\cdot \\cos \\beta}\n$$\nokrajšamo in dobimo rezultat\n$$\n\\frac{\\sin \\beta+\\cos \\beta}{4 \\cos \\beta}\n$$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41928,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj za realni števili $x$ in $y$ velja $\\frac{x}{x+y}=101$. Kolikšna je vrednost izraza $\\frac{y-x}{y}$ ?\n\n(A) 1,02\n(B) 100\n(C) 201\n(D) 2,01\n(E) 1,01",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nEnakost pomnožimo z imenovalcem in izrazimo $x=-1,01 y$, to vstavimo v izraz $\\frac{y-x}{y}=$ $\\frac{y+1,01 y}{y}=\\frac{2,01 y}{y}=2,01$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41929,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV nekem trenutku kmalu po 4. uri urni in minutni kazalec na uri oklepata kot $119^{\\circ}$ (glej sliko). Koliko stopinj je velik kot, ki ga kazalca na uri oklepata natanko 1 uro in 20 minut po tem trenutku?\n(A) 30\n(B) 39\n(C) 41\n(D) 43\n(E) 45\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nKote bomo merili v smeri urinega kazalca. Veliki kazalec v eni uri opiše kot $360^{\\circ}$, v eni minuti pa kot $\\frac{360^{\\circ}}{60}=6^{\\circ}$. V 1 uri in 20 minutah se torej premakne za kot $20 \\cdot 6^{\\circ}=120^{\\circ}$. Mali kazalec v eni uri opiše kot $\\frac{360^{\\circ}}{12}=30^{\\circ}$, v eni minuti pa kot $\\frac{30^{\\circ}}{60}=0.5^{\\circ}$. V 1 uri in 20 minutah se torej premakne za kot $30^{\\circ}+20 \\cdot 0.5^{\\circ}=40^{\\circ}$. Kot med kazalcema po 1 uri in 20 minutah je zato enak $119^{\\circ}+40^{\\circ}-120^{\\circ}=39^{\\circ}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41930,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa stranicah $AB$, $BC$ in $CA$ trikotnika $ABC$ zaporedoma ležijo točke $D$, $F$ in $G$, tako da sta premici $FG$ in $AB$ vzporedni (glej sliko). Ploščini trikotnikov $GFC$ in $GFD$ sta zaporedoma enaki $4\\ \\mathrm{cm}^2$ in $2\\ \\mathrm{cm}^2$. Koliko kvadratnih centimetrov je ploščina trikotnika $ABC$?\n\n(A) 8\n(B) 9\n(C) 12\n(D) 16\n(E) Nemogoče je določiti.\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nTrikotnika $GFC$ in $GDF$ imata enako osnovnico $GF$, njuni ploščini pa sta v razmerju $2:1$. Torej morata biti tudi dolžini njunih višin na stranico $GF$ v razmerju $2:1$. Ker sta premici $FG$ in $AB$ vzporedni, je dolžina višine skozi $C$ trikotnika $ABC$ enaka vsoti dolžin prej omenjenih višin. Od tod sledi, da sta višini skozi $C$ trikotnikov $ABC$ in $GFC$ v razmerju $3:2$. Ker sta si ta dva trikotnika podobna, je tudi $|AB|:|GF|=3:2$. Ploščini trikotnikov $ABC$ in $GFC$ sta tako v razmerju $9:4$, zato je ploščina trikotnika $ABC$ enaka $9\\ \\mathrm{cm}^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41931,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nPoišči vsa realna števila $x$, ki rešijo enačbo\n$$\n\\sqrt[3]{x+1}+\\frac{6-6 \\sqrt[3]{x+1}}{\\sqrt{x+1}-\\sqrt[6]{x+1}}=1\n$$",
    "options": [],
    "answer": "63",
    "solution": "Solution:\nV enačbo vpeljemo novo spremenljivko $z=\\sqrt[6]{x+1}$, da dobimo $z^{2}+\\frac{6-6 z^{2}}{z^{3}-z}=1$, in ulomek na levi strani preoblikujemo $z^{2}+\\frac{-6\\left(z^{2}-1\\right)}{z\\left(z^{2}-1\\right)}=1$. Od tod sklepamo, da $z$ ne sme biti enak $0$, $1$ ali $-1$, saj sicer ulomek ne bi bil definiran, nato pa lahko ulomek okrajšamo, da dobimo enačbo $z^{2}-\\frac{6}{z}=1$. Enačbo pomnožimo z $z$ in poenostavimo do $z^{3}-z-6=0$. Izraz na levi strani enačbe razstavimo\n$$\nz^{3}-z-6=\\left(z^{3}-8\\right)-(z-2)=(z-2)\\left(z^{2}+2z+4\\right)-(z-2)=(z-2)\\left(z^{2}+2z+3\\right)\n$$\nKer je $z^{2}+2z+3=(z+1)^{2}+2>0$, je edina rešitev enačbe $z=2$. Sledi $\\sqrt[6]{x+1}=2$, od koder izračunamo $x=63$.\n\n\n2. način. Kot v prvi rešitvi vpeljemo novo spremenljivko $z=\\sqrt[6]{x+1}$, nato pa dobljeno enačbo $z^{2}+\\frac{6-6 z^{2}}{z^{3}-z}=1$ pomnožimo $z z^{3}-z$ in preoblikujemo do $z^{5}-2z^{3}-6z^{2}+z+6=0$.\nNekoliko bolj se moramo potruditi, da levo stran podobno kot v prvi rešitvi razstavimo\n$$\n\\begin{aligned}\nz^{5}-2z^{3}-6z^{2}+z+6 & =z\\left(z^{4}-2z^{2}+1\\right)-6\\left(z^{2}-1\\right)=z\\left(z^{2}-1\\right)^{2}-6\\left(z^{2}-1\\right)= \\\\\n& =\\left(z^{2}-1\\right)\\left(z^{3}-z-6\\right)=(z-1)(z+1)(z-2)\\left(z^{2}+2z+3\\right)\n\\end{aligned}\n$$\nRešitve enačbe so tako $z=1$, $z=-1$ in $z=2$, saj je spet $z^{2}+2z+3>0$. Prvi dve rešitvi odpadeta, ker v tem primeru ulomek v začetni enačbi za $z$ ni definiran, rešitev $z=2$ pa nam da $x=63$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41932,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nPoišči vsa naravna števila $n$, za katera je\n$$\n\\frac{100 n^{5}-n^{4}-50 n^{3}+2 n^{2}-290 n-2}{n^{2}-2}\n$$\ncelo število.\n(20 točk)",
    "options": [],
    "answer": "n = 1, 2, 3, 10",
    "solution": "Solution:\nČe polinom $100 n^{5}-n^{4}-50 n^{3}+2 n^{2}-290 n-2$ delimo s polinomom $n^{2}-2$, dobimo rezultat $100 n^{3}-n^{2}+150 n$ in ostanek $10 n-2$. Torej je\n$$\n\\frac{100 n^{5}-n^{4}-50 n^{3}+2 n^{2}-290 n-2}{n^{2}-2}=100 n^{3}-n^{2}+150 n+\\frac{10 n-2}{n^{2}-2}\n$$\nTo število bo celo natanko tedaj, ko bo $\\frac{10 n-2}{n^{2}-2}$ celo število. Opazimo, da je za $n>10$ števec tega ulomka manjši od imenovalca in oba sta pozitivna, zato je $0<\\frac{10 n-2}{n^{2}-2}<1$. V tem primeru $\\frac{10 n-2}{n^{2}-2}$ ni celo število. Torej mora biti $n \\leq 10$. Izračunamo vrednost izraza $\\frac{10 n-2}{n^{2}-2}$ za prvih 10 naravnih števil in opazimo, da dobimo celo število le, ko je $n=1,2,3,10$.\n\nIzračunan količnik $100 n^{3}-n^{2}+150 n$\n\nIzračunan ostanek $10 n-2$\n\nZapis $\\frac{100 n^{5}-n^{4}-50 n^{3}+2 n^{2}-290 n-2}{n^{2}-2}=100 n^{3}-n^{2}+150 n+\\frac{10 n-2}{n^{2}-2}$\n\nUgotovitev, da mora biti $\\frac{10 n-2}{n^{2}-2}$ celo število\n\nUgotovitev, da za $n>10$ ulomek $\\frac{10 n-2}{n^{2}-2}$ ni celo število\n\nIzračunana vrednost izraza $\\frac{10 n-2}{n^{2}-2}$ za prvih deset naravnih števil\n\nZapisana rešitev $n=1,2,3,10$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41933,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa praštevila $p$, $q$ in $r$, ki rešijo enačbo $r^{4} = p q + 4$.",
    "options": [],
    "answer": "p=7, q=11, r=3 and p=11, q=7, r=3",
    "solution": "Solution:\n\nEnac̆bo preoblikujemo v $p q = r^{4} - 4$ in desno stran razcepimo po formuli za razliko kvadratov, da dobimo\n$$\np q = (r^{2} - 2)(r^{2} + 2)\n$$\nObe števili $r^{2} - 2$ in $r^{2} + 2$ sta večji od $1$, zato imamo le dve možnosti. Bodisi je $r^{2} - 2 = p$ in $r^{2} + 2 = q$ ali pa je $r^{2} - 2 = q$ in $r^{2} + 2 = p$.\n\nDenimo, da velja prva možnost. Če je $r = 3$, tedaj sledi $p = 7$ in $q = 11$. Če pa je $r \\neq 3$, tedaj ima število $r^{2}$ pri deljenju s $3$ ostanek $1$, zato je število $q = r^{2} + 2$ deljivo s $3$. Ker je $q$ praštevilo, sledi $q = 3$ in $r = 1$, kar pa je protislovje, saj $1$ ni praštevilo. V tem primeru je torej edina rešitev $p = 7$, $q = 11$ in $r = 3$.\n\nDrugo možnost, $r^{2} - 2 = q$ in $r^{2} + 2 = p$, obravnavamo podobno, da dobimo še rešitev $p = 11$, $q = 7$ in $r = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41934,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNa krožnici s polmerom $r$ naključno izberemo dve točki. Kolikšna je verjetnost dogodka, da sta izbrani točki oddaljeni za več kot $\\sqrt{2} r$ in manj kot $\\sqrt{3} r$?",
    "options": [],
    "answer": "1/6",
    "solution": "Solution:\n![](attached_image_1.png)\nPrivzemimo oznake s slike, na kateri je prva izbrana točka na krožnici označena z $A$. Da bo razdalja med izbranima točkama večja od $\\sqrt{2} r$ in manjša od $\\sqrt{3} r$, mora druga izbrana točka $B$ ležati na enem od dveh odebeljenih lokov na sliki. Verjetnost tega dogodka je enaka razmerju med skupno dolžino obeh odebeljenih lokov in obsegom cele krožnice. Ker sta odebeljena loka očitno enako dolga, je to razmerje enako $\\frac{\\alpha r+\\alpha r}{2 \\pi r}=\\frac{\\alpha}{\\pi}$.\n\nDoločiti moramo še velikost kota $\\alpha$. Hitro opazimo, da je trikotnik $A P O$ polovica kvadrata, saj je enakokrak in ima osnovnico za faktor $\\sqrt{2}$ daljšo od obeh krakov. Od tod dobimo $\\angle A O P=\\frac{\\pi}{2}$. Podobno opazimo, da višina na osnovnico enakokrakega trikotnika $A Q O$ ta trikotnik razdeli na dva pravokotna trikotnika, ki sta enaka polovici enakostraničnega trikotnika, saj imata hipotenuzo dolžine $r$ in eno od katet dolžine $\\frac{\\sqrt{3}}{2} r$. Sledi, da je $\\angle A O Q=\\frac{2 \\pi}{3}$. Od tod izračunamo $\\alpha=\\angle A O Q-\\angle A O P=\\frac{\\pi}{6}$. Verjetnost iskanega dogodka je torej enaka $\\frac{\\alpha}{\\pi}=\\frac{1}{6}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41935,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa praštevila $p, q$ in $r$, za katera ima polinom $f(x)=x^{3}-p x^{2}+q x-r^{2}$ same racionalne ničle.",
    "options": [],
    "answer": "p = 11, q = 19, r = 3",
    "solution": "Solution:\n\nKer ima polinom $f$ celoštevilske koeficiente in vodilni koeficient enak 1, so vse njegove racionalne ničle v resnici celoštevilske. Če je $x$ negativno število, potem je tudi $f(x)=x^{3}-p x^{2}+q x-r^{2}$ negativno število, saj so vsi členi v tem primeru negativni. Torej polinom $f$ nima negativnih ničel. Ker tudi 0 ni ničla polinoma $f$, sledi, da so vse tri njegove ničle naravna števila. Označimo jih z $a, b$ in $c$.\n\nPo Vietovih formulah je $a+b+c=p$, $ab+ac+bc=q$ in $abc=r^{2}$. Ker je $r$ praštevilo, iz tretje enačbe sledi, da imamo le dve možnosti; bodisi sta dve ničli enaki $r$, tretja pa 1, ali pa sta dve ničli enaki 1, tretja pa $r^{2}$.\n\nV prvem primeru je $q=ab+ac+bc=r(r+2)$, kar pa ni mogoče, saj $r(r+2)$ ni praštevilo. Torej sta dve ničli enaki 1, tretja pa $r^{2}$. Od tod dobimo $p=a+b+c=r^{2}+2$. Če število $r$ ni deljivo s 3, potem je ostanek števila $r^{2}$ pri deljenju s 3 enak 1, zato je število $p=r^{2}+2$ deljivo s 3. To ni mogoče, saj je $p=r^{2}+2$ praštevilo, ki je očitno večje kot 3. Sledi, da je $r=3$, $p=r^{2}+2=11$ in $q=ab+ac+bc=2r^{2}+1=19$, ničle polinoma $f(x)=x^{3}-11 x^{2}+19 x-9$ pa so 1, 1 in 9.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41936,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV zavetišču imajo 15 psov in 20 mačk. Izmed njih naključno izberemo 8 živali. Izračunaj verjetnosti dogodkov na pet decimalnih mest natančno:\n\na) Izbrali smo 8 psov,\n\nb) Izbrali smo 5 psov in 3 mačke,\n\nc) Izbrali smo več psov kot mačk,\n\nd) Izbrali smo vsaj 6 mačk.",
    "options": [],
    "answer": "a) 0.00113; b) 0.09113; c) 0.11038; d) 0.23713",
    "solution": "Solution:\n\nIzračun števila vseh izidov:\n\nSkupno število živali je $15 + 20 = 35$. Izberemo $8$ živali:\n\n$$\nN = \\binom{35}{8}\n$$\n\na) Izbrali smo 8 psov:\n\nŠtevilo ugodnih izidov:\n$$\nN_a = \\binom{15}{8}\n$$\nVerjetnost:\n$$\nP_a = \\frac{\\binom{15}{8}}{\\binom{35}{8}} \\approx 0.00113\n$$\n\nb) Izbrali smo 5 psov in 3 mačke:\n\nŠtevilo ugodnih izidov:\n$$\nN_b = \\binom{15}{5} \\cdot \\binom{20}{3}\n$$\nVerjetnost:\n$$\nP_b = \\frac{\\binom{15}{5} \\cdot \\binom{20}{3}}{\\binom{35}{8}} \\approx 0.09113\n$$\n\nc) Izbrali smo več psov kot mačk:\n\nMožne kombinacije:\n- 8 psov, 0 mačk\n- 7 psov, 1 mačka\n- 6 psov, 2 mački\n- 5 psov, 3 mačke\n\nŠtevilo ugodnih izidov:\n$$\nN_c = \\binom{15}{8} \\cdot \\binom{20}{0} + \\binom{15}{7} \\cdot \\binom{20}{1} + \\binom{15}{6} \\cdot \\binom{20}{2} + \\binom{15}{5} \\cdot \\binom{20}{3}\n$$\nVerjetnost:\n$$\nP_c = \\frac{N_c}{\\binom{35}{8}} \\approx 0.11038\n$$\n\nd) Izbrali smo vsaj 6 mačk:\n\nMožne kombinacije:\n- 6 mačk, 2 psa\n- 7 mačk, 1 pes\n- 8 mačk, 0 psov\n\nŠtevilo ugodnih izidov:\n$$\nN_d = \\binom{20}{6} \\cdot \\binom{15}{2} + \\binom{20}{7} \\cdot \\binom{15}{1} + \\binom{20}{8} \\cdot \\binom{15}{0}\n$$\nVerjetnost:\n$$\nP_d = \\frac{N_d}{\\binom{35}{8}} \\approx 0.23713\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41937,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je izraz, v katerem je $x$ realno število in $x \\notin \\{-3, -1, 0, 1, 2, 3, 4\\}$.\n$$\n\\frac{x^{2}-4x}{5x-5} \\cdot \\left(\\frac{x^{3}+1}{x^{2}+x}-1\\right) \\cdot \\left(\\left(1-\\frac{3x-3}{x^{2}+x-6}\\right) \\cdot \\left(\\frac{6}{x+3}-\\frac{1}{x-2}\\right)^{-1}-1\\right)^{-1}\n$$\nPoenostavi ga.",
    "options": [],
    "answer": "x - 1",
    "solution": "Solution:\n\nRazstavljanje vsote kubov:\n\nIzračun $\\frac{x^{3}+1}{x^{2}+x}-1=\\frac{(x-1)^{2}}{x}$\n\nPreoblikovanje $1-\\frac{3x-3}{x^{2}+x-6} = \\frac{(x+1)(x-3)}{(x-2)(x+3)}$\n\nPreoblikovanje $\\frac{6}{x-3}-\\frac{1}{x-2} = \\frac{5(x-3)}{(x-2)(x+3)}$\n\nIzračun $\\left(1-\\frac{3x-3}{x^{2}+x-6}\\right) \\cdot \\left(\\frac{6}{x-3}-\\frac{1}{x-2}\\right)^{-1}=\\frac{x+1}{5}$\n\nIzračun $\\left(\\left(1-\\frac{3x-3}{x^{2}+x-6}\\right) \\cdot \\left(\\frac{6}{x-3}-\\frac{1}{x-2}\\right)^{-1}-1\\right)^{-1}=\\frac{5}{x-4}$\n\nRezultat $x-1$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41938,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKozarec valjaste oblike s polmerom $4~\\mathrm{cm}$ in višino $9~\\mathrm{cm}$ je do $\\frac{2}{9}$ višine napolnjen z vodo. Mark se je odločil, da bo vso vodo prelil v kozarec stožčaste oblike s polmerom $5~\\mathrm{cm}$ in višino $6~\\mathrm{cm}$ (glej sliko). Pri prelivanju je $5\\%$ vode polil. Koliko decilitrov vode je v stožčastem kozarcu in kako visoko sega voda (tj. koliko je $|MP|$)?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "0.955 dl; approximately 5.1 cm",
    "solution": "Solution:\n\nIzračun količine vode v valjastem kozarcu:\n\n$$\nV = \\frac{2}{9} \\pi \\cdot r^2 \\cdot v = \\frac{2}{9} \\pi \\cdot 4^2 \\cdot 9 = \\frac{2}{9} \\pi \\cdot 16 \\cdot 9 = 2 \\pi \\cdot 16 = 32\\pi \\approx 100,53~\\mathrm{cm}^3\n$$\n\nIzračun količine vode v stožčastem kozarcu po polivanju:\n\n$$\n0,95 \\cdot V = 0,95 \\cdot 100,53~\\mathrm{cm}^3 \\approx 95,5~\\mathrm{cm}^3\n$$\n\nOdgovor: $95,5~\\mathrm{cm}^3 = 0,955~\\mathrm{dl}$\n\nZapis zveze med polmerom gladine vode $r_1$ in višino vode $v_1$ v stožčastem kozarcu:\n\nKer je stožec podoben celotnemu stožcu, velja:\n$$\n\\frac{r_1}{v_1} = \\frac{5}{6} \\implies r_1 = \\frac{5}{6} v_1\n$$\n\nProstornina vode v stožcu:\n$$\nV_1 = \\frac{1}{3} \\pi r_1^2 v_1 = \\frac{1}{3} \\pi \\left(\\frac{5}{6} v_1\\right)^2 v_1 = \\frac{1}{3} \\pi \\cdot \\frac{25}{36} v_1^2 v_1 = \\frac{25}{108} \\pi v_1^3\n$$\n\nEnačimo s prostornino vode:\n$$\n\\frac{25}{108} \\pi v_1^3 = 95,5\n$$\n$$\nv_1^3 = \\frac{95,5 \\cdot 108}{25 \\pi} \\approx \\frac{10314}{78,54} \\approx 131,4\n$$\n$$\nv_1 \\approx \\sqrt[3]{131,4} \\approx 5,1~\\mathrm{cm}\n$$\n\nOdgovor: V stožčastem kozarcu je $0,955~\\mathrm{dl}$ vode, voda sega do višine približno $5,1~\\mathrm{cm}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41939,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV pravokotni trikotnik $ABC$ s pravim kotom pri $B$ včrtamo tri kvadrate, kot to prikazuje slika. Stranici manjših dveh kvadratov sta dolgi 3 oziroma 4 enote. Izračunaj dolžino stranice $AC$ trikotnika $ABC$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "245/12",
    "solution": "Solution:\n\nOznačimo oglišča kvadratov, kot prikazuje slika. Trikotniki $CKL$, $LJF$, $FHI$ in $IGA$ so podobni trikotniki, zato velja\n$$\n\\frac{|CK|}{3} = \\frac{3}{|JF|} = \\frac{|FH|}{4} = \\frac{4}{|AG|}\n$$\nOznačimo dolžino stranice kvadrata $BEFD$ z $x$. Potem iz enakosti (1) sledi\n$$\n3 \\cdot 4 = |FH||JF| = (x-4)(x-3)\n$$\nKo slednjo enakost poenostavimo in delimo z $x$, dobimo $x=7$. Torej je $|FH|=3$ in $|FJ|=4$. Iz enakosti (1) zato sledi $|CK|=\\frac{3^2}{|JF|}=\\frac{9}{4}$ in $|AG|=\\frac{4^2}{|FH|}=\\frac{16}{3}$. Od tod izračunamo\n$$\n|AB| = |AG| + |GD| + |DB| = \\frac{16}{3} + 4 + 7 = \\frac{49}{3}\n$$\nin\n$$\n|BC| = |BE| + |EK| + |KC| = 7 + 3 + \\frac{9}{4} = \\frac{49}{4}\n$$\nter naposled še\n$$\n|AC| = \\sqrt{|AB|^2 + |BC|^2} = \\sqrt{\\frac{49^2}{9} + \\frac{49^2}{16}} = \\sqrt{\\frac{49^2(16+9)}{9 \\cdot 16}} = \\frac{49 \\cdot 5}{3 \\cdot 4} = \\frac{245}{12}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41940,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana sta dva enako dolga vektorja $\\frac{1}{3} \\vec{a}+\\vec{b}$ in $\\frac{2}{3} \\vec{a}+\\vec{b}$, kjer sta $\\vec{a}$ in $\\vec{b}$ neničelna vektorja, za katera velja $|\\vec{a}|=\\sqrt{3}|\\vec{b}|$.\n\na) Izračunaj velikost kota med vektorjema $\\vec{a}$ in $\\vec{b}$.\n\nb) Izračunaj velikost kota med vektorjema $\\frac{1}{3} \\vec{a}+\\vec{b}$ in $\\frac{2}{3} \\vec{a}+\\vec{b}$.",
    "options": [],
    "answer": "Angle between a and b: 5π/6; angle between (1/3)a + b and (2/3)a + b: π/3",
    "solution": "Solution:\n\n(a) Vektorja $\\frac{1}{3} \\vec{a}+\\vec{b}$ in $\\frac{2}{3} \\vec{a}+\\vec{b}$ sta enako dolga, zato je $\\left|\\frac{1}{3} \\vec{a}+\\vec{b}\\right|^{2}=\\left|\\frac{2}{3} \\vec{a}+\\vec{b}\\right|^{2}$. Kot med vektorjema $\\vec{a}$ in $\\vec{b}$ označimo s $\\varphi$ in poračunamo\n$$\n\\left|\\frac{1}{3} \\vec{a}+\\vec{b}\\right|^{2}=\\left(\\frac{1}{3} \\vec{a}+\\vec{b}\\right) \\cdot\\left(\\frac{1}{3} \\vec{a}+\\vec{b}\\right)=\\frac{1}{9} \\vec{a} \\cdot \\vec{a}+\\frac{2}{3} \\vec{a} \\cdot \\vec{b}+\\vec{b} \\cdot \\vec{b}=\\frac{1}{9}|\\vec{a}|^{2}+\\frac{2}{3}|\\vec{a}||\\vec{b}| \\cos \\varphi+|\\vec{b}|^{2}\n$$\nPodobno dobimo še\n$$\n\\left|\\frac{2}{3} \\vec{a}+\\vec{b}\\right|^{2}=\\frac{4}{9}|\\vec{a}|^{2}+\\frac{4}{3}|\\vec{a}||\\vec{b}| \\cos \\varphi+|\\vec{b}|^{2}\n$$\nKer morata biti ta dva izraza enaka, sledi\n$$\n\\frac{1}{3}|\\vec{a}|^{2}+\\frac{2}{3}|\\vec{a}||\\vec{b}| \\cos \\varphi=0\n$$\nOd tod z upoštevanjem zveze $|\\vec{a}|=\\sqrt{3}|\\vec{b}|$ izrazimo $\\cos \\varphi=-\\frac{|\\vec{a}|}{2|\\vec{b}|}=-\\frac{\\sqrt{3}}{2}$, torej je $\\varphi=\\frac{5 \\pi}{6}$.\n\n\n(b) Kot med vektorjema $\\frac{1}{3} \\vec{a}+\\vec{b}$ in $\\frac{2}{3} \\vec{a}+\\vec{b}$ označimo z $\\alpha$. Tedaj je\n$$\n\\cos \\alpha=\\frac{\\left(\\frac{1}{3} \\vec{a}+\\vec{b}\\right) \\cdot\\left(\\frac{2}{3} \\vec{a}+\\vec{b}\\right)}{\\left|\\frac{1}{3} \\vec{a}+\\vec{b}\\right|\\left|\\frac{2}{3} \\vec{a}+\\vec{b}\\right|}=\\frac{\\frac{2}{9}|\\vec{a}|^{2}+\\vec{a} \\cdot \\vec{b}+|\\vec{b}|^{2}}{\\left|\\frac{2}{3} \\vec{a}+\\vec{b}\\right|^{2}}\n$$\nsaj je po predpostavki $\\left|\\frac{1}{3} \\vec{a}+\\vec{b}\\right|=\\left|\\frac{2}{3} \\vec{a}+\\vec{b}\\right|$. Z upoštevanjem enakosti (1), zveze $|\\vec{a}|=\\sqrt{3}|\\vec{b}|$ in rezultata $\\varphi=\\frac{5 \\pi}{6}$ iz točke (a) dobimo\n$$\n\\cos \\alpha=\\frac{\\frac{2}{9}|\\vec{a}|^{2}+|\\vec{a}||\\vec{b}| \\cos \\varphi+|\\vec{b}|^{2}}{\\frac{4}{9}|\\vec{a}|^{2}+\\frac{4}{3}|\\vec{a}||\\vec{b}| \\cos \\varphi+|\\vec{b}|^{2}}=\\frac{\\left(\\frac{2}{3}-\\frac{3}{2}+1\\right)|\\vec{b}|^{2}}{\\left(\\frac{4}{3}-2+1\\right)|\\vec{b}|^{2}}=\\frac{\\frac{1}{6}}{\\frac{1}{3}}=\\frac{1}{2}\n$$\nOd tod dobimo $\\alpha=\\frac{\\pi}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41941,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPokaži, da je izraz $2^{2 n+3}+3^{n+2} \\cdot 7^{n}$ deljiv s 17 za vsako naravno število $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNaj bo $n$ poljubno naravno število. Dani izraz preoblikujemo\n$$\n2^{2 n+3}+3^{n+2} \\cdot 7^{n}=2^{3} \\cdot 2^{2 n}+3^{2} \\cdot 3^{n} \\cdot 7^{n}=8 \\cdot 4^{n}+9 \\cdot 21^{n}=17 \\cdot 4^{n}+9 \\cdot\\left(21^{n}-4^{n}\\right)\n$$\nKer je prvi člen večkratnik števila 17, zadošča dokazati, da je $21^{n}-4^{n}$ deljivo s 17. Slednje sledi iz enakosti\n$$\n21^{n}-4^{n}=(21-4)\\left(21^{n-1}+21^{n-2} \\cdot 4+\\ldots+21 \\cdot 4^{n-2}+4^{n-1}\\right)\n$$\nTorej je $21^{n}-4^{n}$ deljivo s 17, s tem pa tudi celoten izraz.\n\n\nSolution 2:\n\nKot pri prvi rešitvi izraz preoblikujemo do $8 \\cdot 4^{n}+9 \\cdot 21^{n}$. Ker je ostanek števila 21 pri deljenju s 17 enak 4, ima število $8 \\cdot 4^{n}+9 \\cdot 21^{n}$ pri deljenju s 17 enak ostanek kot število $8 \\cdot 4^{n}+9 \\cdot 4^{n}=17 \\cdot 4^{n}$. Ker je slednje deljivo s 17, je tudi prvotno število deljivo s 17.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41942,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV trikotniku $ABC$ velja $\\angle BAC=\\frac{\\pi}{3}$ in $|AB|^{2}=|AC|^{2}+|AC| \\cdot |BC|$. Določi velikosti ostalih dveh notranjih kotov trikotnika $ABC$.",
    "options": [],
    "answer": "Angle B = 2π/9 and angle C = 4π/9",
    "solution": "Solution:\n\nKote trikotnika $ABC$ označimo kot običajno z $\\alpha, \\beta$ in $\\gamma$, torej je $\\alpha=\\frac{\\pi}{3}$. Naj bo $D$ taka točka na premici $AC$, da $C$ leži med $A$ in $D$ in velja $|CD|=|CB|$. Pogoj naloge lahko preoblikujemo v\n$$\n\\frac{|AB|}{|AC|}=\\frac{|AC|+|BC|}{|AB|}=\\frac{|AC|+|CD|}{|AB|}=\\frac{|AD|}{|AB|}\n$$\nTrikotnika $ABC$ in $ADB$ se torej ujemata v razmerju stranic in kotu $\\angle BAD$ med njima, zato sta si podobna. Sledi $\\angle ADB=\\beta$ in $\\angle DBA=\\gamma$. Ker je trikotnik $BCD$ enakokrak z vrhom v $C$, je $\\angle DBC=\\angle CDB=\\beta$. Torej je\n$$\n\\gamma=\\angle DBA=\\angle DBC+\\angle CBA=2\\beta\n$$\nOd tod sledi $\\pi=\\alpha+\\beta+\\gamma=\\frac{\\pi}{3}+3\\beta$, zato je $\\beta=\\frac{2\\pi}{9}$ in $\\gamma=\\frac{4\\pi}{9}$.\n\n\n2. način. Kote trikotnika $ABC$ označimo kot običajno z $\\alpha, \\beta$ in $\\gamma$. Po sinusnem izreku velja $\\frac{|BC|}{\\sin \\alpha}=\\frac{|AC|}{\\sin \\beta}=\\frac{|AB|}{\\sin \\gamma}$, od koder izrazimo $|BC|=\\frac{\\sin \\alpha}{\\sin \\gamma}|AB|$ in $|AC|=\\frac{\\sin \\beta}{\\sin \\gamma}|AB|$. Ko to vstavimo v enakost $|AB|^{2}=|AC|^{2}+|AC| \\cdot |BC|$, dobimo\n$$\n|AB|^{2}=\\frac{\\sin^{2} \\beta}{\\sin^{2} \\gamma} \\cdot |AB|^{2}+\\frac{\\sin \\alpha \\sin \\beta}{\\sin^{2} \\gamma} \\cdot |AB|^{2}\n$$\nEnakost okrajšamo z $|AB|^{2}$ in pomnožimo s $\\sin^{2} \\gamma$, da dobimo enakost $\\sin^{2} \\gamma=\\sin^{2} \\beta+\\sin \\alpha \\sin \\beta$, ki jo preuredimo v $\\sin^{2} \\gamma-\\sin^{2} \\beta=\\sin \\alpha \\sin \\beta$. S pomočjo faktorizacijskih formul in formul za dvojne kote levo stran preoblikujemo\n$$\n\\begin{aligned}\n\\sin^{2} \\gamma-\\sin^{2} \\beta &=(\\sin \\gamma+\\sin \\beta)(\\sin \\gamma-\\sin \\beta)=2 \\sin \\frac{\\gamma+\\beta}{2} \\cos \\frac{\\gamma-\\beta}{2} \\cdot 2 \\sin \\frac{\\gamma-\\beta}{2} \\cos \\frac{\\gamma+\\beta}{2}= \\\\\n&=\\sin(\\gamma+\\beta) \\sin(\\gamma-\\beta)=\\sin(\\pi-\\alpha) \\sin(\\gamma-\\beta)=\\sin \\alpha \\sin(\\gamma-\\beta)\n\\end{aligned}\n$$\nTorej velja $\\sin \\alpha \\sin \\beta=\\sin \\alpha \\sin(\\gamma-\\beta)$. Po predpostavki je $\\alpha=\\frac{\\pi}{3}$ oziroma $\\sin \\alpha=\\frac{\\sqrt{3}}{2} \\neq 0$, zato sledi $\\sin \\beta=\\sin(\\gamma-\\beta)$. Ker je $0<\\beta<\\pi$ in $-\\pi<\\gamma-\\beta<\\pi$, imamo le dve možnosti; bodisi je $\\gamma-\\beta=\\beta$ ali pa $\\gamma-\\beta=\\pi-\\beta$. V drugem primeru dobimo protislovje $\\gamma=\\pi$, torej je $\\gamma-\\beta=\\beta$, oziroma $\\gamma=2\\beta$. Hkrati je $\\gamma+\\beta=\\pi-\\alpha=\\frac{2\\pi}{3}$, torej je $3\\beta=\\frac{2\\pi}{3}$. Sledi $\\beta=\\frac{2\\pi}{9}$ in $\\gamma=\\frac{4\\pi}{9}$.\n\n\n3. način. Kote trikotnika $ABC$ označimo kot običajno z $\\alpha, \\beta$ in $\\gamma$. Po predpostavki je $|AB|^{2}=|AC|^{2}+|AC| \\cdot |BC|$, po kosinusnem izreku pa velja $|AB|^{2}=|AC|^{2}+|BC|^{2}-2|AC| \\cdot |BC| \\cos \\gamma$. Iz obeh enakosti sledi $|BC|^{2}=|AC| \\cdot |BC|(1+2 \\cos \\gamma)$ oziroma\n$$\n|BC|=(1+2 \\cos \\gamma)|AC|\n$$\nKo to vstavimo v enakost $|AB|^{2}=|AC|^{2}+|AC| \\cdot |BC|$, dobimo $|AB|^{2}=(2+2 \\cos \\gamma)|AC|^{2}$ oziroma\n$$\n|AB|=\\sqrt{2+2 \\cos \\gamma} \\cdot |AC|\n$$\nČe zapišemo še drugi kosinusni izrek $|BC|^{2}=|AB|^{2}+|AC|^{2}-2|AB| \\cdot |AC| \\cos \\alpha$, vanj vstavimo zgornji dve zvezi in upoštevamo predpostavko $\\alpha=\\frac{\\pi}{3}$, dobimo\n$$\n(1+2 \\cos \\gamma)^{2}|AC|^{2}=(2+2 \\cos \\gamma)|AC|^{2}+|AC|^{2}-\\sqrt{2+2 \\cos \\gamma} \\cdot |AC|^{2}\n$$\nEnakost okrajšamo z $|AC|^{2}$ in izrazimo člen s korenom\n$$\n\\begin{aligned}\n\\sqrt{2+2 \\cos \\gamma} &=3+2 \\cos \\gamma-(1+2 \\cos \\gamma)^{2}=2-2 \\cos \\gamma-4 \\cos^{2} \\gamma= \\\\\n&=(2+2 \\cos \\gamma)(1-2 \\cos \\gamma)\n\\end{aligned}\n$$\nEnakost sedaj kvadriramo, da dobimo $(2+2 \\cos \\gamma)=(2+2 \\cos \\gamma)^{2}(1-2 \\cos \\gamma)^{2}$. Ker $\\gamma \\neq \\pi$, je $2+2 \\cos \\gamma \\neq 0$, zato lahko enakost okrajšamo z $(2+2 \\cos \\gamma)$, in desno stran preuredimo\n$$\n\\begin{aligned}\n1 &=(2+2 \\cos \\gamma)(1-2 \\cos \\gamma)^{2}=2-6 \\cos \\gamma+8 \\cos^{3} \\gamma=2+2\\left(4 \\cos^{3} \\gamma-3 \\cos \\gamma\\right)= \\\\\n&=2+2 \\cos 3\\gamma\n\\end{aligned}\n$$\nSledi $\\cos 3\\gamma=-\\frac{1}{2}$. Ker je $\\alpha=\\frac{\\pi}{3}$, je $\\gamma \\leq \\frac{2\\pi}{3}$ oziroma $0<3\\gamma \\leq 2\\pi$. Zato imamo dve rešitvi, $3\\gamma=\\frac{2\\pi}{3}$ in $3\\gamma=\\frac{4\\pi}{3}$, od koder dobimo $\\gamma=\\frac{2\\pi}{9}$ in $\\gamma=\\frac{4\\pi}{9}$. Ker je $0<\\frac{2\\pi}{9}<\\frac{\\pi}{3}$, je $\\frac{1}{2}<\\cos \\frac{2\\pi}{9}<1$. Torej je pri vrednosti $\\gamma=\\frac{2\\pi}{9}$ desna stran enakosti (2) negativna, leva pa pozitivna, zato ta rešitev odpade. Rešitev $\\gamma=\\frac{4\\pi}{9}$ pa je res rešitev, saj je v tem primeru $\\gamma>\\frac{\\pi}{3}$ in zato $\\cos \\gamma<\\frac{1}{2}$, torej sta obe strani enakosti (2) pozitivni. Od tod izračunamo še $\\beta=\\pi-\\alpha-\\gamma=\\frac{2\\pi}{9}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41943,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $T$ težišče trikotnika $ABC$ in $D$ razpolovišče stranice $BC$. Premice $AT$, $BT$ in $CT$ naj drugič sekajo trikotniku $ABC$ očrtano krožnico zaporedoma v točkah $P$, $Q$ in $R$. Denimo, da je $|AD|=\\frac{\\sqrt{3}}{2}|AC|$. Dokaži, da je trikotnik $PQR$ enakokrak.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo dolžine stranic trikotnika $ABC$ in njegove kote kot običajno. Naj bo $E$ razpolovišče stranice $AC$. Potem je $|AE|=\\frac{b}{2}$, $|AC|=b$ in po predpostavki $|AD|=\\frac{\\sqrt{3}}{2}b$. Ker težišče deli težiščnico v razmerju $2:1$, lahko izračunamo še $|AT|=\\frac{2}{3}|AD|=\\frac{1}{\\sqrt{3}}b$. Torej je\n$$\n\\frac{|AE|}{|AT|}=\\frac{\\sqrt{3}}{2}=\\frac{|AD|}{|AC|}\n$$\nTo pa pomeni, da sta si trikotnika $ATE$ in $ACD$ podobna, saj imata skupen kot $\\angle DAC$. Torej je $\\angle ETA=\\angle ACD$, kar pomeni, da je štirikotnik $CETD$ tetiven. Sledi $\\angle ATR=\\angle DTC=\\angle DEC=\\alpha$, kjer smo upoštevali, da sta premici $AB$ in $DE$ vzporedni, saj sta $D$ in $E$ razpolovišči ustreznih stranic. Zaradi tetivnosti štirikotnika $ARBC$ velja $\\angle TRA=\\beta$, torej je $\\angle RAT=\\pi-\\angle ATR-\\angle TRA=\\pi-\\alpha-\\beta=\\gamma$. Iz tetivnosti zato sledi še $\\angle RQP=\\angle RAP=\\gamma$. Podobno je $\\angle RTB=\\angle CTE=\\angle CDE=\\beta$ in $\\angle BRT=\\alpha$, zato velja $\\angle TBR=\\gamma$ in $\\angle QPR=\\gamma$. Kota $\\angle QPR$ in $\\angle RQP$ sta tako oba enaka $\\gamma$, kar pomeni, da je trikotnik $PQR$ enakokrak z vrhom pri $R$.\n\n\n2. način. Kot v prvi rešitvi ugotovimo, da je $|AT|=\\frac{1}{\\sqrt{3}}b$, $\\frac{|AE|}{|AT|}=\\frac{|AD|}{|AC|}$ in $\\angle ETA=\\angle ACD$. Zdaj lahko izračunamo\n$$\n\\angle BTP=\\angle ETA=\\angle ACD=\\angle ACB=\\angle TPB\n$$\nkjer smo v zadnji enakosti upoštevali enakost obodnih kotov nad lokom $AB$. Sledi, da je trikotnik $BPT$ enakokrak in zato $|TB|=|PB|$.\nV trikotnikih $ATC$ in $RTP$ je $\\angle CTA=\\angle RTP$ in $\\angle ACT=\\angle TPR$, torej sta si trikotnika podobna. Zato je $|RP|=\\frac{|AC|\\cdot|RT|}{|AT|}=|RT|\\sqrt{3}$. Prav tako sta si trikotnika $BTC$ in $RTQ$ podobna, zato velja $|RQ|=\\frac{|BC|\\cdot|RT|}{|BT|}=\\frac{|BC|\\cdot|RT|}{|BP|}$, kjer smo upoštevali enakost $|BT|=|BP|$. Upoštevamo še podobnost trikotnikov $ADC$ in $BDP$ in dobimo $|BP|=\\frac{|AC|\\cdot|BD|}{|AD|}=\\frac{|BC|}{\\sqrt{3}}$. Sledi $|RQ|=|RT|\\sqrt{3}=|RP|$.\n\n\n3. način. Kot v drugem načinu ugotovimo, da je $|AT|=\\frac{1}{\\sqrt{3}}b$ in s pomočjo podobnosti trikotnikov $ATC$ in $RTP$ izračunamo $|RP|=|RT|\\sqrt{3}$, s pomočjo podobnosti trikotnikov $BTC$ in $RTQ$ pa $|RQ|=\\frac{|BC|\\cdot|RT|}{|BT|}$.\nS pomočjo dveh kosinusnih izrekov bomo pokazali, da je $|BT|=\\frac{|BC|}{\\sqrt{3}}$. To lahko naredimo na več načinov, eden od njih je naslednji: Po kosinusnem izreku v trikotniku $ACD$ je\n$$\n|BC|=2|CD|=\\sqrt{|AD|^2+|AC|^2-2|AC|\\cdot|AD|\\cos\\angle DAC}=\\sqrt{7-4\\sqrt{3}\\cos\\angle DAC}|AC|\n$$\nPo kosinusnem izreku v trikotniku $ATE$ in upoštevanju $|AT|=\\frac{|AC|}{\\sqrt{3}}$ pa je\n$$\n|BT|=2|ET|=2\\sqrt{|AT|^2+|AE|^2-2|AT|\\cdot|AE|\\cos\\angle DAC}=\\sqrt{\\frac{7}{3}-\\frac{4\\sqrt{3}}{3}\\cos\\angle DAC}|AC|.\n$$\nSledi $|BT|=\\frac{|BC|}{\\sqrt{3}}$ in iz enakosti $|RP|=|RT|\\sqrt{3}$ in $|RQ|=\\frac{|BC|\\cdot|RT|}{|BT|}$ sledi $|RP|=|RQ|$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41944,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $AB$ premer krožnice $\\mathcal{K}$, očrtane tetivnemu štirikotniku $ABCD$. Premici $AD$ in $BC$ se sekata v točki $E$, tangenti na krožnico $\\mathcal{K}$ v točkah $C$ in $D$ pa se sekata v točki $F$. Dokaži, da sta premici $EF$ in $AB$ pravokotni.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOznačimo $\\alpha=\\angle BAE$ in $\\beta=\\angle EBA$, in naj bo $O$ središče krožnice $\\mathcal{K}$. Potem je $\\angle AEB=\\pi-\\alpha-\\beta$. Zaradi tetivnosti štirikotnika $ABCD$ je $\\angle DCB=\\pi-\\alpha$ in $\\angle ADC=\\pi-\\beta$. Ker je $\\angle OCB=\\angle CBO=\\beta$ in $\\angle ADO=\\angle OAD=\\alpha$, sledi $\\angle ODC=\\angle DCO=\\pi-\\alpha-\\beta$.\n\nNaj bo $S$ središče trikotniku $CED$ očrtane krožnice. Po izreku o središčnem in obodnem kotu je $\\angle DSC=2\\angle DEC=2(\\pi-\\alpha-\\beta)$, zato je $\\angle CDS=\\angle SCD=\\frac{\\pi-\\angle DSC}{2}=\\alpha+\\beta-\\frac{\\pi}{2}$. Sledi $\\angle ODS=\\angle SCO=\\angle SCD+\\angle DCO=\\frac{\\pi}{2}$. Torej sta $CS$ in $DS$ tangenti na krožnico $\\mathcal{K}$, kar pomeni, da je $S=F$, oziroma $F$ je središče trikotniku $CED$ očrtane krožnice.\n\nNaj bo $H$ višinska točka trikotnika $ABE$. Po Talesovem izreku sta $AC$ in $BD$ višini tega trikotnika, torej je $H$ njuno presečišče. Prav tako po Talesovem izreku sledi, da je štirikotnik $CEDH$ tetiven in središče njemu očrtane krožnice leži na razpolovišču daljice $EH$. Pokazali smo že, da je točka $F$ središče trikotniku $CED$ očrtane krožnice, torej točke $E, F$ in $H$ ležijo na isti premici. Ker je $EH$ višina trikotnika $ABE$ na stranico $AB$, od tod sledi, da sta premici $EF$ in $AB$ pravokotni.\n\n\n2. način. Privzemimo enake oznake kot v prvi rešitvi. Ker sta $CF$ in $DF$ tangenti na krožnico $\\mathcal{K}$, je trikotnik $DCF$ enakokrak z vrhom pri $F$, hkrati pa velja $\\angle FCO=\\angle ODF=\\frac{\\pi}{2}$. Iz štirikotnika $OCFD$ zato dobimo $\\angle DFC=\\pi-\\angle COD$. Po izreku o središčnem in obodnem kotu je $\\angle COD=2\\angle CAD$, torej je $\\angle DFC=\\pi-2\\angle CAD$. Po Talesovem izreku velja $\\angle ACB=\\frac{\\pi}{2}$, zato je $\\angle DEC=\\frac{\\pi}{2}-\\angle CAD$. Od tod sledi $\\angle DFC=\\pi-2\\angle CAD=2\\angle DEC$. Ker točki $E$ in $F$ ležita na istem bregu premice $CD$ in je trikotnik $CFD$ enakokrak z vrhom pri $F$, po izreku o središčnem in obodnem kotu sledi, da je $F$ središče trikotniku $CED$ očrtane krožnice. Dokaz dokončamo podobno kot v prvi rešitvi.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41945,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Reši enačbi $9^{x} - 4 \\cdot 5^{2x-3} = 25^{x-1} + 2 \\cdot 3^{2x-3}$ in $\\sqrt{x+1+\\sqrt{x+1+\\sqrt{x+3}}} = \\sqrt{x+3}$.\n\nb) Zapiši splošno obliko polinoma z vodilnim koeficientom 2, katerega ničle so enake rešitvam enačb iz a).",
    "options": [],
    "answer": "a) First equation: x = 5/2; second equation: x = 1. b) p(x) = 2x^2 − 7x + 5.",
    "solution": "Solution:\nPreoblikovanje enačbe $3^{2x} - 2 \\cdot 3^{2x-3} = 5^{2x-2} + 4 \\cdot 5^{2x-3}$\n\nIzpostavitev skupnega faktorja\n\nUreditev do oblike $3^{2x-5} = 5^{2x-5}$\n\nRešitev enačbe $x = \\frac{5}{2}$\n\nReševanje korenske enačbe\n\nUreditev do oblike $x^{2} - 7x + 6 = 0$\n\nRešitev enačbe je $x = 1$, druga rešitev $x = 6$ ni ustrezna\n\nZapis polinoma v splošni obliki $p(x) = 2x^{2} - 7x + 5$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41946,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRačunsko določi vse možne pare naravnih števil $a$ in $b$, $a > b$, tako da bo razlika kvadratov teh dveh števil enaka $60$.",
    "options": [],
    "answer": "[(16, 14), (8, 2)]",
    "solution": "Solution:\n\nZapis $a^2 - b^2 = 60$\n\nRazstavljanje razlike kvadratov:\n$$a^2 - b^2 = (a - b)(a + b) = 60$$\n\nZapis vseh možnih razcepov števila $60$ na produkt dveh naravnih števil:\n$$1 \\cdot 60,\\ 2 \\cdot 30,\\ 3 \\cdot 20,\\ 4 \\cdot 15,\\ 5 \\cdot 12,\\ 6 \\cdot 10$$\n\nZapis vsaj enega sistema enačb:\n$$\n\\begin{cases}\n  a - b = m \\\\\n  a + b = n\n\\end{cases}\n$$\nkjer je $m \\cdot n = 60$ in $n > m > 0$.\n\nReševanje sistema enačb:\n$$\n\\begin{cases}\n  a = \\frac{m + n}{2} \\\\\n  b = \\frac{n - m}{2}\n\\end{cases}\n$$\n\nUgotovitev, da za razcepe $1 \\cdot 60$, $3 \\cdot 20$, $4 \\cdot 15$, $5 \\cdot 12$ rešitve niso naravna števila (ker vsaj eden izmed $a$ ali $b$ ni naravno število).\n\nRešitev $a = 16$ in $b = 14$ za razcep $60 = 2 \\cdot 30$:\n$$\n\\begin{cases}\n  a - b = 2 \\\\\n  a + b = 30\n\\end{cases}\n\\implies\n\\begin{cases}\n  a = 16 \\\\\n  b = 14\n\\end{cases}\n$$\n\nRešitev $a = 8$ in $b = 2$ za razcep $60 = 6 \\cdot 10$:\n$$\n\\begin{cases}\n  a - b = 6 \\\\\n  a + b = 10\n\\end{cases}\n\\implies\n\\begin{cases}\n  a = 8 \\\\\n  b = 2\n\\end{cases}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41947,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJure in Miha igrata igro z dvema posodama s kroglicami, v kateri poteze izvajata izmenično. Na začetku igre je v beli posodi $m$ kroglic, v črni pa $n$ kroglic. V vsaki potezi igralec bodisi odstrani eno kroglico iz ene od posod ali pa prestavi eno kroglico iz bele v črno posodo. Zmaga tisti igralec, ki odstrani zadnjo kroglico iz posod. V odvisnosti od $m$ in $n$ določi, kdo ima zmagovito strategijo, če je prvi na potezi Jure.",
    "options": [],
    "answer": "Miha wins if both starting counts are even; otherwise Jure wins.",
    "solution": "Solution:\n\nPokazali bomo, da če sta obe števili $m$ in $n$ sodi, ima zmagovito strategijo Miha, če pa je vsaj eno od števil $m$ in $n$ liho, ima zmagovito strategijo Jure.\n\nDenimo najprej, da sta $m$ in $n$ sodi. Tedaj lahko Miha vsakič ponovi Juretovo potezo, saj bo tako po vsaki njegovi potezi število kroglic v vsaki posodi sodo. Zadnjo kroglico bo v tem primeru odstranil Miha in tako zmagal igro.\n\nDenimo sedaj, da je vsaj eno od števil $m$ in $n$ liho. Tedaj lahko Jure v svoji prvi potezi poskrbi, da bo po njegovi potezi v obeh posodah sodo mnogo kroglic. Če je $m$ liho in $n$ sodo, potem Jure odstrani kroglico iz bele posode. Če je $m$ sodo in $n$ liho, Jure odstrani kroglico iz črne posode. Če pa sta $m$ in $n$ obe lihi, potem Jure prestavi kroglico iz bele v črno posodo. V vseh naslednjih potezah lahko nato Jure ponavlja Mihove poteze. S to strategijo bo Jure odstranil zadnjo kroglico in zmagal.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41948,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTaja in Lili igrata igro, pri kateri je na mizi postavljenih 10 kroglic, oštevilčenih z naravnimi števili od $1$ do $10$. V prvi potezi igre Taja izbere naravno število $n$, nato pa dekleti izmenjaje z mize jemljeta vsaka po eno kroglico, dokler kroglic ne zmanjka. Prvo kroglico z mize vzame Lili, zadnjo pa Taja. Zmaga tista, katere vsota števil na vseh njenih kroglicah je bližja številu $n$. Katero dekle ima zmagovito strategijo?",
    "options": [],
    "answer": "Lili",
    "solution": "Solution:\n\nPokazali bomo, da ima zmagovito strategijo Lili. Vsota števil vseh kroglic na mizi je enaka $\\frac{10 \\cdot 11}{2} = 55$. Torej bo na koncu vsota števil ene od deklet večja kot $\\frac{55}{2} = 27 \\frac{1}{2}$, vsota števil druge pa manjša od $27 \\frac{1}{2}$, hkrati pa bosta obe vsoti od števila $27 \\frac{1}{2}$ enako oddaljeni.\n\nČe Taja v prvi potezi izbere naravno število $n$, ki je večje od $27 \\frac{1}{2}$, potem mora Lili le poskrbeti, da bo njena vsota na koncu večja od Tajine, saj bo tedaj večja kot $27 \\frac{1}{2}$ in s tem bližja številu $n$. To lahko Lili stori tako, da vsakič z mize vzame kroglico z največjo številko.\n\nČe pa Taja v prvi potezi izbere naravno število manjše od $27 \\frac{1}{2}$, lahko na podoben način zmaga Lili, če z mize vsakič vzame kroglico z najmanjšo številko.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41949,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa celoštevilski mreži je označenih 16 točk (glej sliko). Največ koliko izmed teh točk lahko pobarvamo rdeče, tako da nobene tri rdeče točke ne bodo ležale na isti premici?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nKer izmed 4 točk, ki ležijo v isti vodoravni vrstici, lahko pobarvamo največ 2, lahko skupaj pobarvamo največ $4 \\cdot 2 = 8$ točk. Da 8 točk tudi res lahko pobarvamo, prikazuje spodnja slika.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41950,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Dani so prvi štirje členi neskončnega aritmetičnega zaporedja:\n$$\n-\\frac{3}{2}, \\frac{1}{2}, \\frac{5}{2}, \\frac{9}{2}\n$$\nIzračunaj, katero število je dvestoti člen tega zaporedja in kolikšna je vsota prvih dvestotih členov zaporedja.\n\nb) Med števili $\\frac{1}{2}$ in $\\frac{9}{2}$ vrinemo tri števila, tako da dobimo geometrijsko zaporedje. Določi vse možne vrednosti vrinjenih členov. Vrednosti naj bodo točne.",
    "options": [],
    "answer": "a) Two hundredth term: 793/2. Sum of first two hundred terms: 39500. b) Inserted triples: (sqrt(3)/2, 3/2, 3*sqrt(3)/2) and (-sqrt(3)/2, 3/2, -3*sqrt(3)/2).",
    "solution": "Solution:\n\na.\nIzračunamo diferenco aritmetičnega zaporedja:\n$$\nd = \\frac{1}{2} - \\left(-\\frac{3}{2}\\right) = 2.\n$$\nIzračunamo dvestoti člen zaporedja:\n$$\na_{200} = a_1 + (200-1) \\cdot d = -\\frac{3}{2} + 199 \\cdot 2 = -\\frac{3}{2} + 398 = \\frac{793}{2}.\n$$\nIzračunamo vsoto prvih dvestotih členov:\n$$\ns_{200} = \\frac{200}{2} (a_1 + a_{200}) = 100 \\left(-\\frac{3}{2} + \\frac{793}{2}\\right) = 100 \\cdot \\frac{790}{2} = 100 \\cdot 395 = 39500.\n$$\n\nb.\nMed števili $\\frac{1}{2}$ in $\\frac{9}{2}$ vrinemo tri števila $a_2, a_3, a_4$. Dobimo zaporedje:\n$$\n\\frac{1}{2}, a_2, a_3, a_4, \\frac{9}{2}.\n$$\nŠtevilo $\\frac{9}{2}$ je peti člen geometrijskega zaporedja, zato velja enakost:\n$$\na_1 \\cdot q^4 = \\frac{9}{2}.\n$$\nVstavimo $a_1$ v enakost in dobimo:\n$$\n\\frac{1}{2} \\cdot q^4 = \\frac{9}{2}.\n$$\nKo odpravimo ulomke, dobimo enačbo:\n$$\nq^4 = 9.\n$$\nEnačba ima dve rešitvi:\n$$\nq_1 = \\sqrt{3}, \\quad q_2 = -\\sqrt{3}.\n$$\nZapišemo obe rešitvi vrinjenih členov:\n$$\n\\frac{\\sqrt{3}}{2}, \\frac{3}{2}, \\frac{3\\sqrt{3}}{2} \\quad \\text{in} \\quad -\\frac{\\sqrt{3}}{2}, \\frac{3}{2}, -\\frac{3\\sqrt{3}}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41951,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoloči najmanjši skupni večkratnik in največji skupni delitelj izrazov:\n$$\n-3 x^{4}+24 x, \\quad x^{3}-6 x^{2}+12 x-8, \\quad a x^{2}-4 a x+4 a, \\quad 3 b x^{2}-12 b\n$$\n(10 točk)",
    "options": [],
    "answer": "gcd: x - 2; lcm: -3abx (x - 2)^3 (x + 2) (x^2 + 2x + 4)",
    "solution": "Solution:\n\nIzraze zapišemo v obliki produkta\n$$\n-3 x^{4}+24 x = -3 x (x^{3}-8) = -3 x (x-2)(x^{2}+2 x+4),\n$$\n$$\nx^{3}-6 x^{2}+12 x-8 = (x-2)^{3},\n$$\n$$\na x^{2}-4 a x+4 a = a(x^{2}-4 x+4) = a(x-2)^{2},\n$$\n$$\n3 b x^{2}-12 b = 3 b (x^{2}-4) = 3 b (x-2)(x+2).\n$$\n\nUgotovimo, da je največji skupni delitelj $D = x-2$ in najmanjši skupni večkratnik\n$$\nv = -3 a b x (x-2)^{3} (x+2) (x^{2}+2 x+4).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41952,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPodani sta premica $p$ z enačbo $-x+2y=1$ in premica $q$ z enačbo $-4x+3y=16$. Naj bo točka $A$ presečišče premice $p$ z osjo $x$, naj bo točka $B$ presečišče premice $q$ z osjo $x$, naj bo točka $C$ presečišče premic $p$ in $q$ ter naj bo točka $D$ pravokotna projekcija točke $C$ na os $x$.\n\na) Izračunaj in zapiši koordinate točk $A, B, C$ in $D$.\n\nb) Izračunaj velikost notranjega kota $\\beta$ v trikotniku $ABC$.\n\nc) Izračunaj ploščino trikotnika $ABC$.",
    "options": [],
    "answer": "a) A = (−1, 0), B = (−4, 0), C = (−29/5, −12/5), D = (−29/5, 0). b) β (at vertex B) ≈ 126.87° (since tan of the acute angle is 4/3, the interior angle is 180° − arctan(4/3)). c) Area S = 18/5 = 3.6.",
    "solution": "Solution:\n\na. Presečišče premice $p$ z osjo $x$ ima $y$ koordinato $0$. Tako dobimo $A(-1, 0)$ in $B(4, 0)$. Rešimo sistem enačb za $p$ in $q$, da dobimo koordinate točke $C\\left(-\\frac{29}{5}, -\\frac{12}{5}\\right)$. Točka $D$ ima koordinati $D\\left(-\\frac{29}{5}, 0\\right)$.\n\nb. Kot $\\beta$ iz trikotnika $ABC$ je hkrati zunanji kot pri oglišču $B$ v trikotniku $BDC$. Daljica $DC$ je dolga $\\frac{12}{5}$, daljica $BD$ pa ima dolžino $\\frac{29}{5} - 4 = \\frac{9}{5}$. Trikotnik $DBC$ je pravokoten, velja $\\tan \\beta' = \\frac{\\frac{12}{5}}{\\frac{9}{5}} = \\frac{4}{3}$ in dobimo $\\beta \\doteq 53,13^\\circ$ in $\\beta \\doteq 126,87^\\circ = 126^\\circ 52'$.\n\nc. Stranica $c$ v trikotniku $ABC$ ima dolžino $5$. Ploščino trikotnika $ABC$ pa lahko izračunamo na več različnih načinov:\n\n(a) Stranica $c$ v trikotniku $ABC$ ima dolžino $3$. Daljica $DC$ je pravokotna na os $x$ in s tem na nosilko stranice $c$ v trikotniku $ABC$. Daljica $DC$ je tako višina na stranico $c$ v trikotniku $ABC$, njena dolžina je $\\frac{12}{5}$. Ploščino trikotnika lahko izračunamo kot $S = \\frac{c \\cdot v_c}{2} = \\frac{3 \\cdot \\frac{12}{5}}{2} = \\frac{18}{5} = 3,6$.\n\n(b) Stranica $a$ v trikotniku $ABC$ je hkrati hipotenuza v pravokotnem trikotniku $BDC$, po Pitagorovem izreku dobimo njeno dolžino $3$. Isto dobimo z izračunom razdalje daljice $BC$. Ploščino trikotnika izračunamo kot $S = \\frac{c \\cdot a \\cdot \\sin \\beta}{2} = \\frac{3 \\cdot 3 \\cdot \\sin 126,87^\\circ}{2} = 3,6$.\n\n(c) Poleg $a = 3$ in $c = 3$ izračunamo še stranico $b$, torej dolžino daljice $AC$:\n$$\n|AC| = \\sqrt{\\left(\\frac{24}{5}\\right)^2 + \\left(\\frac{12}{5}\\right)^2} = \\sqrt{\\frac{144}{5}} = \\frac{12 \\sqrt{5}}{5} \\doteq 5,37.\n$$\nSedaj lahko z uporabo Heronovega obrazca izračunamo ploščino trikotnika.\n$$\ns = \\frac{a + b + c}{2} = \\frac{15 + 6 \\sqrt{5}}{5} \\doteq 5,68\n$$\nin\n$$\nS = \\sqrt{s \\cdot (s - a) \\cdot (s - b) \\cdot (s - c)} = \\sqrt{\\frac{15 + 6 \\sqrt{5}}{5} \\cdot \\frac{6 \\sqrt{5}}{5} \\cdot \\frac{6 \\sqrt{5}}{5} \\cdot \\frac{15 - 6 \\sqrt{5}}{5}} = \\frac{18}{5} = 3,6.\n$$\n\n(d) Ploščino trikotnika $ABC$ lahko izračunamo kot razliko ploščin trikotnika $ADC$ in trikotnika $BDC$:\n$$\nS = \\frac{\\frac{12}{5} \\cdot \\frac{24}{5}}{2} - \\frac{\\frac{12}{5} \\cdot \\frac{9}{5}}{2} = 3,6.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41953,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDani sta funkciji $f(x)=\\log_{3}(x+3)+1$ in $g(x)=\\log_{3}(3x)+1$.\n\na) Poišči absciso presečišča danih funkcij.\n\nb) Izračunaj ničlo, začetno vrednost in zapiši enačbo asimptote za obe funkciji. Funkciji nariši v isti koordinatni sistem.",
    "options": [],
    "answer": "Intersection abscissa: x = 3/2. For f: zero at x = −8/3, value at zero is 2, vertical asymptote x = −3. For g: zero at x = 1/9, value at zero does not exist, vertical asymptote x = 0.",
    "solution": "Solution:\n\na) Poiščemo absciso presečišča. Izenačimo funkciji $f(x)=g(x)$, ju vstavimo v enačbo in dobimo $\\log_{3}(x+3)+1=\\log_{3}(3x)+1$. Enačbo uredimo in dobimo linearno enačbo $x+3=3x$, katere rešitev je $x=\\frac{3}{2}$.\n\nb) Za funkcijo $f(x)=\\log_{3}(x+3)+1$ izračunamo začetno vrednost $f(0)=\\log_{3}(0+3)+1=2$. Izračunamo ničlo, tako da rešimo enačbo $0=\\log_{3}(x+3)+1$, ki jo preoblikujemo v $\\log_{3}(x+3)=-1$ in za tem s pomočjo definicije logaritma zapišemo enačbo $x+3=\\frac{1}{3}$. Dobimo rešitev $x=-2\\frac{2}{3}$, ki je ničla funkcije $f$. Zapišemo enačbo asimptote $x=-3$.\n\nZa funkcijo $g(x)=\\log_{3}(3x)+1$ ugotovimo, da začetna vrednost ne obstaja. Izračunamo ničlo, tako da rešimo enačbo $0=\\log_{3}(3x)+1$, jo preoblikujemo v $\\log_{3}(3x)=-1$ in za tem uporabimo definicijo logaritma ter dobimo enačbo $3x=\\frac{1}{3}$. Dobimo rešitev $x=\\frac{1}{9}$, ki je ničla funkcije $g$. Zapišemo enačbo asimptote $x=0$.\n\nNarišemo oba grafa.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41954,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je izraz\n$$\nX=\\left(a+a^{-1}\\right)^{-1}\\left(a^{2}+3 a+2\\right)\\left(a^{2}-3 a+2\\right)\\left(a^{2}-4\\right)^{-1}\n$$\na) Izraz $X$ poenostavi in zapiši v obliki produkta.\n(8 točk)\nb) Izračunaj vrednost izraza $X$ za $a=-\\frac{1}{3}$.",
    "options": [],
    "answer": "Simplified: X = a(a − 1)(a + 1)/(a^2 + 1). For a = −1/3: X = 4/15.",
    "solution": "Solution:\n\nPrvi faktor preoblikujemo v $\\left(a+a^{-1}\\right)^{-1}=\\frac{a}{a^{2}+1}$.\n\nV produktu $\\left(a^{2}+3 a+2\\right)\\left(a^{2}-3 a+2\\right)$ lahko vsak člen iz prvega oklepaja pomnožimo z vsakim členom iz drugega oklepaja. Hitrejša možnost pa je, da to preoblikujemo v produkt vsote in razlike istih členov in dobimo razliko kvadratov:\n$$\n\\left(a^{2}+3 a+2\\right)\\left(a^{2}-3 a+2\\right)=\\left(\\left(a^{2}+2\\right)+3 a\\right)\\left(\\left(a^{2}+2\\right)-3 a\\right)=\\left(a^{2}+2\\right)^{2}-(3 a)^{2}=a^{4}+4 a^{2}+4-9 a^{2}=a^{4}-5 a^{2}+4.\n$$\nTo s pomočjo Vietovega pravila razstavimo na $a^{4}-5 a^{2}+4=\\left(a^{2}-1\\right)\\left(a^{2}-4\\right)$.\n\nPreoblikujmo še zadnji faktor $\\left(a^{2}-4\\right)^{-1}=\\frac{1}{a^{2}-4}$.\n\nVse to zmnožimo in dobimo\n$$\nX=\\frac{a}{a^{2}+1} \\cdot \\left(a^{2}-1\\right)\\left(a^{2}-4\\right) \\cdot \\frac{1}{a^{2}-4}=\\frac{a}{a^{2}+1} \\cdot \\left(a^{2}-1\\right)=\\frac{a(a-1)(a+1)}{a^{2}+1}.\n$$\n\nV poenostavljen izraz $X$ vstavimo $a=-\\frac{1}{3}$ in dobimo\n$$\nX=\\frac{-\\frac{1}{3}\\left(-\\frac{1}{3}-1\\right)\\left(-\\frac{1}{3}+1\\right)}{\\left(-\\frac{1}{3}\\right)^{2}+1}=\\frac{-\\frac{1}{3}\\left(-\\frac{4}{3}\\right)\\left(\\frac{2}{3}\\right)}{\\frac{1}{9}+1}=\\frac{-\\frac{1}{3}\\cdot -\\frac{4}{3}\\cdot \\frac{2}{3}}{\\frac{10}{9}}=\\frac{-\\frac{1}{3}\\cdot -\\frac{8}{9}}{\\frac{10}{9}}=\\frac{\\frac{8}{27}}{\\frac{10}{9}}=\\frac{8}{27}\\cdot\\frac{9}{10}=\\frac{72}{270}=\\frac{4}{15}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41955,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIzračunaj vrednost izraza $\\left(-4 x y z^{-1}\\right)^{-2}$, če so $x$, $y$ in $z$ neznanke v sistemu\n$$\n\\begin{aligned}\n& \\frac{2}{x}-\\frac{1}{y}+\\frac{1}{z}=-3 \\\\\n& \\frac{1}{x}-\\frac{1}{y}-\\frac{1}{z}=3 \\\\\n& \\frac{3}{x}+\\frac{2}{y}+\\frac{2}{z}=4\n\\end{aligned}\n$$",
    "options": [],
    "answer": "9/64",
    "solution": "Solution:\n\nUvedemo nove neznanke $a=\\frac{1}{x}$, $b=\\frac{1}{y}$, $c=\\frac{1}{z}$ in dobimo sistem enačb $2a-b-c=-3$, $a-b-c=3$, $3a+2b+2c=4$, ki ga rešimo. Rešitev sistema je $a=2$, $b=3$, $c=-4$ in dobimo $x=\\frac{1}{2}$, $y=\\frac{1}{3}$, $z=-\\frac{1}{4}$.\n\nIzračunamo\n$$\n\\left(-4 x y z^{-1}\\right)^{-2} = \\left(-4 \\cdot \\frac{1}{2} \\cdot \\frac{1}{3} \\cdot \\left(-\\frac{1}{4}\\right)^{-1}\\right)^{-2} = \\left(\\frac{8}{3}\\right)^{-2} = \\frac{9}{64}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41956,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMaja je v dva enaka kozarca nalila limonin sok, tako da je bil prvi kozarec napolnjen do $\\frac{1}{3}$, drugi pa do $\\frac{2}{5}$. Nato je v oba kozarca dolivala vodo, dokler nista bila polna. Nazadnje je oba kozarca izpraznila v večjo skledo, pri čemer se tekočina ni prelila čez rob sklede. Kolikšen del tekočine v skledi je limonin sok?\n\n(A) $\\frac{1}{8}$\n(B) $\\frac{3}{16}$\n(C) $\\frac{11}{30}$\n(D) $\\frac{11}{19}$\n(E) $\\frac{11}{15}$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nOznačimo količino tekočine, ki jo drži en kozarec z $e$. Tedaj je v skledi $\\frac{1}{3} e + \\frac{2}{5} e = \\frac{11}{15} e$ limoninega soka od skupno $2e$ tekočine. Limonin sok torej predstavlja $\\frac{\\frac{11}{15} e}{2e} = \\frac{11}{30}$ tekočine v skledi.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41957,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $a=256$, $b$ pa zmnožek vseh pozitivnih deliteljev števila $a$. Katera od naslednjih enakosti je pravilna?\n\n(A) $b=a^{4}$\n(B) $b=a^{9}$\n(C) $b^{2}=a^{7}$\n(D) $b^{2}=a^{9}$\n(E) $b^{3}=a^{10}$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nKer je $a=2^{8}$, je $b=1 \\cdot 2 \\cdot 2^{2} \\cdot \\ldots \\cdot 2^{8}=2^{0+1+2+\\ldots+8}=2^{36}$. Torej velja $b^{2}=2^{72}=a^{9}$. Pravilen odgovor je (D).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41958,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTadej je na tablo zapisal ulomek, različen od $0$. Sara je števec Tadejevega ulomka povečala za $40\\%$, da je dobila nov ulomek. Za koliko odstotkov mora Katja zmanjšati imenovalec Sarinega novega ulomka, da bo dobila ulomek, katerega vrednost bo dvakrat tolikšna, kot je bila vrednost Tadejevega ulomka?\n\n(A) 25\n(B) 30\n(C) 40\n(D) 45\n(E) 50",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nČe je Tadejev ulomek enak $\\frac{a}{b}$, tedaj je Sarin ulomek $\\frac{1.4 \\cdot a}{b}$, Katjin ulomek pa $\\frac{2 \\cdot a}{b} = \\frac{1.4 \\cdot a}{0.7 \\cdot b}$. Katja mora imenovalec Sarinega novega ulomka zmanjšati za $30\\%$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41959,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKolikšen je prvi člen rekurzivno podanega zaporedja s formulo $a_{n}=2 a_{n-1}+1$, če je peti člen enak $7$?\n(A) $2$\n(B) $15$\n(C) $-\\frac{7}{2}$\n(D) $-\\frac{1}{2}$\n(E) $-5$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\nZapišemo $a_{5}=2 a_{4}+1$ in izračunamo $a_{4}=\\frac{a_{5}-1}{2}=\\frac{7-1}{2}=3$.\nPostopoma računamo vse člene do prvega člena:\n$a_{3}=\\frac{a_{4}-1}{2}=1$\n$a_{2}=\\frac{a_{3}-1}{2}=0$\n$a_{1}=\\frac{a_{2}-1}{2}=-\\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41960,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKolikšna je velikost središčnega kota, ki pripada krožnemu loku, ki ima enako dolžino kot polmer krožnice?\n(A) $\\frac{45^{\\circ}}{\\pi}$\n(B) $\\frac{90^{\\circ}}{\\pi}$\n(C) $\\frac{135^{\\circ}}{\\pi}$\n(D) $\\frac{180^{\\circ}}{\\pi}$\n(E) $\\frac{270^{\\circ}}{\\pi}$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\nDolžino krožnega loka izračunamo po formuli $l = \\frac{\\pi r \\alpha}{180^{\\circ}}$. Ker je dolžina krožnega loka enaka polmeru kroga, enačbo preuredimo v $r = \\frac{\\pi r \\alpha}{180^{\\circ}}$. Iz enačbe izrazimo $\\alpha$ in dobimo rezultat $\\frac{180^{\\circ}}{\\pi}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41961,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKateri od spodaj navedenih izrazov je enakovreden izrazu $(y-2)^{3}-y(y+5)(y-5)-29+6 y^{2}$?\n(A) $12 y^{2}+37 y-37$\n(B) $-13 y+37$\n(C) $37 y-21$\n(D) $37(y-1)$\n(E) $12 y^{2}-37 y-37$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n$(y-2)^{3}-y(y+5)(y-5)-29+6 y^{2}=y^{3}-6 y^{2}+12 y-8-y^{3}+25 y-29+6 y^{2}=37 y-37=37(y-1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41962,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV konveksnem štirikotniku so zunanji koti v razmerju $7: 4: 6: 1$. V kakšnem razmerju so notranji koti tega štirikotnika pri istih ogliččih?\n(A) $2: 5: 3: 8$\n(B) $8: 3: 5: 2$\n(C) $1: 6: 4: 7$\n(D) $7: 4: 6: 1$\n(E) $11: 14: 12: 17$",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nVemo, da je vsota zunanjih kotov v konveksnem večkotniku enaka $360^{\\circ}$. Zunanji koti torej merijo zapovrstjo $140^{\\circ}$, $80^{\\circ}$, $120^{\\circ}$ in $20^{\\circ}$. Notranji koti so zato v razmerju $40^{\\circ}: 100^{\\circ}: 60^{\\circ}: 160^{\\circ}$. Razmerje okrajšamo in dobimo $2: 5: 3: 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41963,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ poljubna soda funkcija in $g: \\mathbb{R} \\rightarrow \\mathbb{R}$ poljubna liha funkcija. Definirajmo funkcijo $\\varphi: \\mathbb{R} \\rightarrow \\mathbb{R}$ s predpisom $\\varphi(x)=f(g(x))+g(f(x))$ za vse $x \\in \\mathbb{R}$. Katera od naslednjih enakosti je zagotovo izpolnjena za vsa realna števila $x$ ?\n(A) $\\varphi(x)=f(f(x))$\n(B) $\\varphi(x)=g(g(x))$\n(C) $\\varphi(x)=\\varphi(-x)$\n(D) $\\varphi(-x)=-\\varphi(x)$\n(E) $\\varphi(x)=2 f(g(x))$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nKer je $f(x)$ soda funkcija in $g(x)$ liha funkcija, velja $f(-x)=f(x)$ in $g(-x)=-g(x)$ za vsa realna števila $x$. Torej je $\\varphi(-x)=f(g(-x))+g(f(-x))=f(-g(x))+g(f(x))=f(g(x))+g(f(x))=\\varphi(x)$. Pravilen odgovor je (C).\n\nPrimer $f(x)=x^{2}$ in $g(x)=2 x$ pokaže, da nobena od ostalih enakosti ni nujno izpolnjena, saj v tem primeru velja $\\varphi(x)=6 x^{2}$, $f(f(x))=x^{4}$, $g(g(x))=4 x$ in $2 f(g(x))=8 x^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41964,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDolžine diagonal mejnih ploskev kvadra so enake $|XY| = 8~\\mathrm{cm}$, $|YZ| = 9~\\mathrm{cm}$ in $|ZX| = \\sqrt{55}~\\mathrm{cm}$ (glej sliko). Koliko centimetrov je dolga telesna diagonala tega kvadra?\n\n(A) $\\sqrt{90}$\n\n(B) 10\n\n(C) $\\sqrt{120}$\n\n(D) $\\sqrt{200}$\n\n(E) 20\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nDolžino, širino in višino kvadra označimo z $a, b$ in $c$. Tedaj po Pitagorovem izreku velja $a^2 + b^2 = |XY|^2 = 64$, $b^2 + c^2 = |ZX|^2 = 55$ in $c^2 + a^2 = |YZ|^2 = 81$. Enačbe seštejemo, da dobimo $2\\left(a^2 + b^2 + c^2\\right) = 200$. Dolžina telesne diagonale kvadra je enaka $\\sqrt{a^2 + b^2 + c^2} = \\sqrt{\\frac{200}{2}} = 10~\\mathrm{cm}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41965,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKateri izraz je enakovreden izrazu $\\log _{a b}\\left(a^{-1} \\cdot \\sqrt{a^{3} b^{-1}}: \\sqrt[3]{b^{-1} a^{2}}\\right)$ ?\n(A) $a b$\n(B) $\\frac{a}{b}$\n(C) 1\n(D) $-\\frac{1}{6}$\n(E) $-\\frac{1}{2}$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\nIzraz v oklepaju preoblikujemo na skupni korenski eksponent $a^{-1} \\cdot \\sqrt{a^{3} b^{-1}}: \\sqrt[3]{b^{-1} a^{2}}=$ $\\sqrt[6]{a^{-6}} \\cdot \\sqrt[6]{a^{9} b^{-3}}: \\sqrt[6]{b^{-2} a^{4}}$. Izraz poenostavimo in damo pod skupni koren: $\\sqrt[6]{a^{-1} b^{-1}}=\\sqrt[6]{\\frac{1}{a b}}=$ $\\left(\\frac{1}{a b}\\right)^{\\frac{1}{6}}=a b^{-\\frac{1}{6}}$. Logaritem preuredimo v $\\log _{a b}\\left(a^{-1} \\cdot \\sqrt{a^{3} b^{-1}}: \\sqrt[3]{b^{-1} a^{2}}\\right)=\\log _{a b}(a b)^{-\\frac{1}{6}}=-\\frac{1}{6}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41966,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKateri interval je zaloga vrednosti funkcije $f(x)=\\cos x+1-\\pi$?\n\n(A) $[-1+\\pi, 2]$\n(B) $[0, \\pi+2]$\n(C) $[-\\pi, 2-\\pi]$\n(D) $[-\\pi, 1+\\pi]$\n(E) $[1-\\pi, 2-\\pi]$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nZaloga vrednosti funkcije kosinus je interval $[-1,1]$, zato sledi, da je zaloga vrednosti funkcije $f$ interval $[-\\pi, 2-\\pi]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41967,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV nekem večkotniku je vsota notranjih kotov enaka $4140^{\\circ}$. Kolikšno je število diagonal tega večkotnika?\n(A) 250\n(B) 275\n(C) 205\n(D) 725\n(E) 257",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nUporabimo obrazec za vsoto notranjih kotov v večkotniku $S_{n} = (n-2) \\cdot 180^{\\circ}$ in izračunamo število stranic (oglišč) iskanega večkotnika. Dobimo $n = 25$. Uporabimo še obrazec za število diagonal $d_{n} = \\frac{n(n-3)}{2}$ in dobimo $d_{25} = 275$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41968,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nMama je na vprašanje, koliko je star sin, odgovorila z uganko: Če od sedanje dvakratne sinove starosti odštejemo trikrat toliko let, kot jih je imel pred šestimi leti, dobimo, koliko je star danes. Koliko let je sin star danes?\n(A) 3 leta\n(B) 18 let\n(C) 9 let\n(D) 6 let\n(E) 27 let",
    "options": [],
    "answer": "C",
    "solution": "Solution:\nPredpostavimo, da je danes star $x$ let in zapišemo enačbo $2x - 3(x - 6) = x$. Odpravimo oklepaj, enačbo uredimo in dobimo rešitev $x = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41969,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDavid si je izmislil trimestno naravno število s tremi različnimi števkami. Vsako števko tega števila je nadomestil s črko, da je dobil besedo $E N A$, ki predstavlja njegovo število. Opazil je, da za njegovo število velja $c \\cdot E N A = 2331$, kjer je $c$ neko naravno število. Koliko je vrednost števila $c$?\n\n(A) 3\n(B) 7\n(C) 9\n(D) 21\n(E) 37",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nPozitivni delitelji števila $2331 = 3^{2} \\cdot 7 \\cdot 37$ so $1, 3, 7, 9, 21, 37, 63, 111, 259, 333, 777$ in $2331$. Med njimi je le $259$ trimestno naravno število s tremi različnimi števkami. Torej je $E N A = 259$ in zato $c = 2331 : 259 = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41970,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKlara ima 6 enakih kock. Na mejne ploskve vsake kocke je zapisala črke $B, A, B, I, C$ in $A$, na vsako mejno ploskev po črko. Nato je hkrati vrgla vseh 6 kock. Kolikšna je verjetnost, da lahko iz črk, ki so po metu na zgornih ploskvah kock, sestavi besedo $B A B I C A$ ?\n\n(A) $\\frac{6!}{(2!)^{2}}\\left(\\frac{1}{3}\\right)^{4}\\left(\\frac{1}{6}\\right)^{2}$\n(B) $\\frac{6!}{(2!)^{2}}\\left(\\frac{1}{6}\\right)^{6}$\n(C) $\\left(\\frac{1}{3}\\right)^{4}\\left(\\frac{1}{6}\\right)^{2}$\n(D) $\\left(\\frac{1}{6}\\right)^{6}$\n(E) 1",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nVerjetnosti, da na eni kocki pade črka $A, B, C$ oz. $I$ so zaporedoma enake $\\frac{1}{3}, \\frac{1}{3}, \\frac{1}{6}$ oz. $\\frac{1}{6}$. Mislimo si, da Klara kocke vrže tako, da so urejene v vrsto. Tedaj je verjetnost, da pade beseda $B A B I C A$ enaka $\\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot \\frac{1}{3}=\\left(\\frac{1}{3}\\right)^{4}\\left(\\frac{1}{6}\\right)^{2}$. Ugoden pa je katerikoli izid, kjer so črke v tej besedi premešane. Vseh takih izidov je skupaj $\\frac{6!}{(2!)^{2}}$, saj gre za permutacije s ponavljanjem na 6 elementih, pri čemer se črki $A$ in $B$ pojavita 2-krat. Verjetnost, da lahko Klara iz dobljenih črk sestavi besedo $B A B I C A$ je torej $\\frac{6!}{(2!)^{2}}\\left(\\frac{1}{3}\\right)^{4}\\left(\\frac{1}{6}\\right)^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41971,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKroglo prerežemo na dve polkrogli. Za koliko $\\%$ je vsota površin obeh polkrogel večja od površine krogle?\n\n(A) 25\n(B) 50\n(C) 75\n(D) 22,5\n(E) 40",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nPovršina krogle je enaka $P_{K} = 4 \\pi R^{2}$. Če kroglo prerežemo, dobimo še površini dveh glavnih krogov, ki sta enaki $2 \\pi R^{2}$. Nova površina je enaka $P_{PK} = 6 \\pi R^{2}$. Površina se torej poveča za $50\\%$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41972,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKateri od spodaj navedenih izrazov je enakovreden izrazu\n$$\n\\frac{16 a^{-2} b^{-1} c^{-4}}{2^{-1} a^{-3} b^{-2} c^{-3}} \\cdot \\left(a^{0} + (a b)^{0}\\right)^{-1}\n$$\n?\n(A) $64 a b c^{-1}$\n(B) $16 a b c^{-1}$\n(C) $\\frac{32}{a b c}$\n(D) $32 a b c^{-1}$\n(E) $16 a b c^{-7}$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nNajprej uredimo ulomek:\n$$\n\\frac{16 a^{-2} b^{-1} c^{-4}}{2^{-1} a^{-3} b^{-2} c^{-3}}\n$$\nŠtevec: $16 a^{-2} b^{-1} c^{-4}$\nImenovalec: $2^{-1} a^{-3} b^{-2} c^{-3}$\n\nDelimo potence z enakimi osnovami:\n$$\n= 16 \\cdot 2^{1} \\cdot a^{-2 - (-3)} \\cdot b^{-1 - (-2)} \\cdot c^{-4 - (-3)}\n$$\nKer $2^{-1}$ v imenovalcu je $2^{1}$ v števcu:\n$$\n= 16 \\cdot 2^{1} \\cdot a^{1} \\cdot b^{1} \\cdot c^{-1}\n$$\n$16 \\cdot 2 = 32$\n\nTorej:\n$$\n= 32 a b c^{-1}\n$$\n\nZdaj še drugi del izraza:\n$$\n\\left(a^{0} + (a b)^{0}\\right)^{-1}\n$$\n$a^{0} = 1$, $(a b)^{0} = 1$, torej $1 + 1 = 2$\n\nTorej:\n$$\n(2)^{-1} = \\frac{1}{2}\n$$\n\nCeloten izraz:\n$$\n32 a b c^{-1} \\cdot \\frac{1}{2} = 16 a b c^{-1}\n$$\n\nOdgovor je (B) $16 a b c^{-1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41973,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa sliki je graf polinoma $p$, ki seka os $y$ v $-4$, se dotika abscisne osi v $1$ in seka os $x$ v $4$. Kolikšna je vrednost izraza $p'(1) - 2 \\cdot p(0) - 4 \\cdot p(4)$?\n\n(A) 2\n(B) 4\n(C) 8\n(D) 0\n(E) -1\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nIz slike preberemo, da je $p'(1) = 0$, $p(0) = -4$ in $p(4) = 0$, torej je\n$$\np'(1) - 2 \\cdot p(0) - 4 \\cdot p(4) = 0 - 2 \\cdot (-4) - 4 \\cdot 0 = 0 + 8 - 0 = 8.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41974,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $b=\\sqrt[2019]{\\sqrt[2018]{\\ldots \\sqrt[4]{\\sqrt[3]{\\sqrt[2]{a}}}}},\\ a>0$. Kateri izraz je enakovreden $\\sqrt[2020]{\\sqrt[2019]{\\sqrt[2018]{\\ldots \\sqrt[4]{\\sqrt[3]{a^{404}}}}}}$ ?\n\n(A) $\\sqrt[2020]{b}$\n(B) $\\sqrt[5]{b}$\n(C) $\\sqrt[5]{b^{2}}$\n(D) $b^{404}$\n(E) $b^{202}$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nDrugi izraz je zelo podoben izrazu $b$, namesto $\\sqrt[2]{a}$ pa imamo $a^{404}$. Lahko pa $a^{404}$ pišemo kot $\\sqrt[2]{a^{808}}$. Vse skupaj pa lahko preoblikujemo v $\\sqrt[2020]{b^{808}}=\\sqrt[5]{b^{2}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41975,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nPoišči vsa naravna števila $n$, katerih kub je enak vsoti kvadratov treh ne nujno različnih deliteljev števila $n$.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\nNaj bodo $a, b$ in $c$ delitelji števila $n$, za katere je $n^{3}=a^{2}+b^{2}+c^{2}$. Tedaj velja $a, b, c \\leq n$, zato je $n^{3}=a^{2}+b^{2}+c^{2} \\leq 3 n^{2}$ oziroma $n \\leq 3$. Primer $n=1$ odpade, ker 1 ni vsota treh naravnih števil. Če je $n=2$, sta edina delitelja 1 in 2. Toda nobena od vsot $1^{2}+1^{2}+1^{2}$, $1^{2}+1^{2}+2^{2}$, $1^{2}+2^{2}+2^{2}$ in $2^{2}+2^{2}+2^{2}$ ni enaka $2^{3}$, zato tudi primer $n=2$ odpade. Edina rešitev je $n=3$, saj velja $3^{3}=3^{2}+3^{2}+3^{2}$.\n\n2. način. Naj bodo $a \\leq b \\leq c$ delitelji števila $n$, za katere je $n^{3}=a^{2}+b^{2}+c^{2}$. Ker $c^{2}$ deli $n^{3}$, mora deliti tudi $a^{2}+b^{2}$. Toda po naši predpostavki je $a^{2}+b^{2} \\leq 2 c^{2}$, zato imamo dve možnosti, $a^{2}+b^{2}=c^{2}$ ali $a^{2}+b^{2}=2 c^{2}$. V prvem primeru je $n^{3}=2 c^{2}$ oziroma $\\left(\\frac{n}{c}\\right)^{2} n=2$, kar pomeni, da $n$ deli 2. V drugem primeru je $n^{3}=3 c^{2}$ oziroma $\\left(\\frac{n}{c}\\right)^{2} n=3$, kar pomeni, da $n$ deli 3. Sledi, da je $n$ enak 1, 2 ali 3. Na enak način kot v prvi rešitvi pokažemo, da je rešitev le $n=3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41976,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKrožnici $\\mathcal{K}_1$ in $\\mathcal{K}_2$ s središčema $O_1$ in $O_2$ se od zunaj dotikata. Naj bo $p$ skupna tangenta krožnic $\\mathcal{K}_1$ in $\\mathcal{K}_2$, ki ne poteka skozi njuno dotikališče. Dokaži, da je $p$ tangenta tudi krožnice s premerom $O_1 O_2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNaj bo $S$ razpolovišče daljice $O_1 O_2$ ter $A$ dotikališče krožnic $\\mathcal{K}_1$ in $\\mathcal{K}_2$. Dotikališči premice $p$ s krožnicama $\\mathcal{K}_1$ in $\\mathcal{K}_2$ označimo zaporedoma z $B_1$ in $B_2$, presečišče premice $p$ s tangento na krožnici $\\mathcal{K}_1$ in $\\mathcal{K}_2$ skozi točko $A$ pa označimo z $D$. Trikotnika $O_1 A D$ in $O_1 B_1 D$ sta skladna, saj imata dve enako dolgi stranici in enak kot $\\frac{\\pi}{2}$ nasproti daljše stranice, zato velja $\\angle B_1 D O_1 = \\angle O_1 D A$. Podobno sklepamo, da velja tudi $\\angle A D O_2 = \\angle O_2 D B_2$, in od tod izpeljemo $\\angle O_1 D O_2 = \\angle O_1 D A + \\angle A D O_2 = \\frac{\\angle B_1 D A}{2} + \\frac{\\angle A D B_2}{2} = \\frac{\\pi}{2}$. Po Talesovem izreku torej točka $D$ leži na krožnici s premerom $O_1 O_2$, zato je dovolj dokazati, da je premica $S D$ pravokotna na premico $p$.\n\nKer točka $D$ leži na krožnici s premerom $O_1 O_2$, je $|S D| = |S O_1|$, torej je trikotnik $O_1 S D$ enakokrak z vrhom pri $S$. Sledi $\\angle S O_1 D = \\angle O_1 D S$ in zato je $\\angle B_1 D S = \\angle B_1 D O_1 + \\angle O_1 D S = \\frac{\\pi}{2} - \\angle D O_1 B_1 + \\angle S O_1 D = \\frac{\\pi}{2}$, kar je bilo treba dokazati.\n\n\n2. način. Primer, ko je premica $p$ vzporedna premici $O_1 O_2$ je očiten, zato lahko predpostavimo, da se omenjeni premici sekata in njuno presečišče označimo s $T$. Naj bo $S$ razpolovišče daljice $O_1 O_2$, $D'$ pa pravokotna projekcija točke $S$ na premico $p$. Dotikališči premice $p$ s krožnicama $\\mathcal{K}_1$ in $\\mathcal{K}_2$ označimo z $B_1$ in $B_2$, polmera krožnic $\\mathcal{K}_1$ in $\\mathcal{K}_2$ pa z $r_1$ in $r_2$. Iz podobnosti trikotnikov $T O_1 B_1$, $T S D'$ in $T O_2 B_2$ sledi\n$$\n\\frac{r_1}{|O_1 T|} = \\frac{|S D'|}{|S T|} = \\frac{r_2}{|O_2 T|}\n$$\ntorej je\n$$\n\\begin{gathered}\n|S D'| \\cdot |O_1 T| = r_1 \\cdot |S T| = r_1 \\cdot (|S O_1| + |O_1 T|) \\\\\nr_2 \\cdot |O_1 T| = r_1 \\cdot |O_2 T| = r_1 \\cdot (|O_2 O_1| + |O_1 T|) = r_1 \\cdot (2|S O_1| + |O_1 T|)\n\\end{gathered}\n$$\nPrvo enačbo pomnožimo z 2 in od nje odštejemo drugo enačbo, da dobimo\n$$\n(2|S D'| - r_2) \\cdot |O_1 T| = r_1 \\cdot |O_1 T|\n$$\nOd tod sledi $2|S D'| - r_2 = r_1$ oziroma $|S D'| = \\frac{r_1 + r_2}{2} = \\frac{|O_1 O_2|}{2} = |S O_1| = |S O_2|$. Torej točka $D'$ leži na krožnici s polmerom $O_1 O_2$ in premica $p$ je tangenta te krožnice.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41977,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nPoišči vsa realna števila $x$, ki rešijo enačbo\n$$\n6^{\\log_{10} x} + x^{\\log_{10} 6} = 72\n$$",
    "options": [],
    "answer": "100",
    "solution": "Solution:\nLogaritem je definiran le za pozitivna števila, zato mora biti $x > 0$. Število $x = 1$ očitno ni rešitev enačbe, zato lahko privzamemo, da je $x \\neq 1$. Za pozitivno realno število $x \\neq 1$ velja $\\log_{10} x = \\frac{1}{\\log_{x} 10}$, zato je $\\log_{10} 6 = \\frac{\\log_{x} 6}{\\log_{x} 10} = \\log_{10} x \\cdot \\log_{x} 6$. S pomočjo te zveze lahko dano enakost preoblikujemo do\n$$\n6^{\\log_{10} x} + \\left(x^{\\log_{x} 6}\\right)^{\\log_{10} x} = 72\n$$\nin nato upoštevamo enakost $x^{\\log_{x} 6} = 6$, da dobimo\n$$\n2 \\cdot 6^{\\log_{10} x} = 72\n$$\nOd tod sledi $6^{\\log_{10} x} = 36 = 6^{2}$, torej je $\\log_{10} x = 2$ oziroma $x = 100$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41978,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNajveč koliko je lahko največji skupni delitelj števil $a-2b+3$, $2a-3b-1$ in $3a+b-2$, če sta $a$ in $b$ naravni števili?",
    "options": [],
    "answer": "38",
    "solution": "Solution:\n\nNaj bo $d$ največji skupni delitelj danih števil. Tedaj $d$ deli tudi števila\n$$\n\\begin{aligned}\n& (2a-3b-1)-2(a-2b+3) = b-7 \\\\\n& (a-2b+3)+2(b-7) = a-11 \\\\\n& (3a+b-2)-(b-7) = 3a+5 \\\\\n& (3a+5)-3(a-11) = 38\n\\end{aligned}\n$$\nTorej je $d$ lahko največ $38$. Če izberemo $a=11$ in $b=7$, potem je $a-2b+3=0$, $2a-3b-1=0$ in $3a+b-2=38$, torej je največji skupni delitelj res lahko $38$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41979,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDokaži, da je za vsako naravno število $n$ število $7^{2018} + 9^{2020 n}$ deljivo s $5$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nŠtevilo $9^{2020 n} = 9^{2 \\cdot 1010 n} = 81^{1010 n}$ ima pri deljenju s $5$ ostanek $1$, saj je njegova zadnja števka v desetiškem zapisu enaka $1$, ne glede na to, koliko je vrednost naravnega števila $n$.\n\nŠtevilo $7^{2018} = 7^{4 \\cdot 504 + 2} = 2401^{504} \\cdot 49$ pa ima pri deljenju s $5$ ostanek $4$, saj je njegova zadnja števka v desetiškem zapisu enaka $9$.\n\nKer je vsota obeh ostankov $4 + 1 = 5$ deljiva s $5$, je tudi število $7^{2018} + 9^{2020 n}$ deljivo s $5$ za vsako naravno število $n$.\n\n\n2. način.\n\nZadnje števke potenc števila $9$ v desetiškem zapisu so enake: $9$ za $9^{2k+1}$ in $1$ za $9^{2k}$, kjer je $k$ poljubno nenegativno celo število. Zadnja števka števila $9^{2020 n}$ je torej $1$.\n\nPodobno so zadnje števke potenc števila $7$ v desetiškem zapisu enake: $7$ za $7^{4m+1}$, $9$ za $7^{4m+2}$, $3$ za $7^{4m+3}$ in $1$ za $7^{4m}$, kjer je $m$ poljubno nenegativno celo število. Ker je $2018 = 4 \\cdot 504 + 2$, je zadnja števka števila $7^{2018}$ enaka $9$.\n\nZa vsako naravno število $n$ je torej zadnja števka števila $7^{2018} + 9^{2020 n}$ v desetiškem zapisu enaka $0$, kar pomeni, da je to število deljivo s $5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41980,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRealni rešitvi enačbe $x^{4}-x^{3}-2 x-4=0$ sta prva dva člena padajočega aritmetičnega zaporedja s 40 členi.\na) Izračunaj prva dva člena in diferenco zaporedja.\nb) Izračunaj zadnji člen in zapiši splošni člen $a_{n}$ zaporedja.\nc) Izračunaj vsoto vseh členov zaporedja z lihimi indeksi.",
    "options": [],
    "answer": "a) a1 = 2, a2 = −1, d = −3; b) a40 = −115, an = 5 − 3n; c) sum of odd-indexed terms = −1100",
    "solution": "Solution:\n\na) Poiščemo realni rešitvi dane enačbe in zapišemo prva dva člena padajočega aritmetičnega zaporedja $a_{1}=2$ in $a_{2}=-1$. Diferenca zaporedja je $d=-3$.\n\nb) Izračunamo 40. člen zaporedja $a_{40}=a_{1}+39 d=2+39 \\cdot(-3)=-115$. Zapišemo splošni člen zaporedja $a_{n}=a_{1}+(n-1) d=2+(n-1) \\cdot(-3)=5-3 n$.\n\nc) Ugotovimo, da je diferenca členov zaporedja z lihimi indeksi $d=-6$ in da je v dani vrsti 20 členov. Uporabimo formulo za vsoto 20 členov aritmetičnega zaporedja in dobimo $s_{20}=\\frac{20 \\cdot(2 \\cdot 2+19 \\cdot(-6))}{2}=-1100$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41981,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa notranja kota $\\alpha$ in $\\beta$ trikotnika $ABC$ z obsegom 24 velja $\\cos \\alpha=\\frac{2}{3}$ in $\\cos \\beta=\\frac{2}{7}$. Izračunaj ploščino trikotnika $ABC$.",
    "options": [],
    "answer": "12 sqrt(5)",
    "solution": "Solution:\n\nOpazimo, da sta kota $\\alpha$ in $\\beta$ ostra, saj sta $\\cos \\alpha$ in $\\cos \\beta$ pozitivna. Označimo stranice trikotnika $ABC$ kot običajno z $a, b$ in $c$. Naj bo $C'$ nožišče višine $v$ skozi oglišče $C$. Dolžini daljic $AC'$ in $BC'$ označimo zaporedoma z $b_1$ in $a_1$. Tedaj velja\n$$\na_1 = a \\cos \\beta = \\frac{2}{7} a \\quad \\text{in} \\quad b_1 = b \\cos \\alpha = \\frac{2}{3} b\n$$\nPo sinusnem izreku velja $\\frac{a}{\\sin \\alpha} = \\frac{b}{\\sin \\beta}$. Ker sta kota $\\alpha$ in $\\beta$ ostra, sta $\\sin \\alpha$ in $\\sin \\beta$ pozitivna, zato je $\\sin \\alpha = \\sqrt{1-\\cos^2 \\alpha} = \\frac{\\sqrt{5}}{3}$ in $\\sin \\beta = \\sqrt{1-\\cos^2 \\beta} = \\frac{3 \\sqrt{5}}{7}$. Od tod sledi\n$$\nb = \\frac{\\sin \\beta}{\\sin \\alpha} a = \\frac{3 \\sqrt{5}}{7} \\cdot \\frac{3}{\\sqrt{5}} \\cdot a = \\frac{9}{7} a\n$$\nOd tod izrazimo še $b_1 = \\frac{2}{3} b = \\frac{6}{7} a$. Obseg trikotnika je torej\n$$\no = a + b + a_1 + b_1 = a + \\frac{9}{7} a + \\frac{2}{7} a + \\frac{6}{7} a = \\frac{24}{7} a\n$$\nin ker mora biti enak 24, je $a = 7$ ter posledično $b = 9$, $a_1 = 2$ in $b_1 = 6$. Višina $v$ trikotnika $ABC$ je zato enaka $v = a \\sin \\beta = 7 \\cdot \\frac{3 \\sqrt{5}}{7} = 3 \\sqrt{5}$, ploščina trikotnika pa je enaka\n$$\np = \\frac{(a_1 + b_1) v}{2} = \\frac{8 \\cdot 3 \\sqrt{5}}{2} = 12 \\sqrt{5}\n$$\n\n\nSolution 2:\n\nUporabimo oznake iz prve rešitve. Ker sta $\\cos \\alpha$ in $\\cos \\beta$ pozitivna, sta kota $\\alpha$ in $\\beta$ ostra, zato lahko na kotne funkcije gledamo kot na razmerje ustreznih stranic v pravokotnem trikotniku. Ker je $\\cos \\alpha = \\frac{2}{3}$, v trikotniku $AC'C$ velja $b_1 : b = 2 : 3$, zato smemo označiti $b_1 = 2t$ in $b = 3t$ za nek $t > 0$. Podobno smemo v trikotniku $C'BC$ označiti $a_1 = 2t'$ in $a = 7t'$ za nek $t' > 0$. Sedaj iz obeh pravokotnih trikotnikov po Pitagorovem izreku izračunamo višino\n$$\nv = \\sqrt{9t^2 - 4t^2} = \\sqrt{5} t \\quad \\text{in} \\quad v = \\sqrt{49 t'^2 - 4 t'^2} = 3 \\sqrt{5} t'\n$$\nSledi $\\sqrt{5} t = 3 \\sqrt{5} t'$, oziroma $t = 3 t'$. Obseg trikotnika je zato enak\n$$\no = a + b + a_1 + b_1 = 7 t' + 9 t' + 2 t' + 6 t' = 24 t'\n$$\nod koder sledi $t' = 1$ in $t = 3$. Stranica $c$ trikotnika je torej enaka $c = a_1 + b_1 = 2 + 6 = 8$, višina pa $v = 3 \\sqrt{5}$. Od tod izračunamo še ploščino trikotnika\n$$\np = \\frac{c \\cdot v}{2} = 12 \\sqrt{5}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41982,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNad stranicami trikotnika $ABC$ konstruiramo pozitivno orientirane rombe $BAA_{1}B_{2}$, $CBB_{1}C_{2}$ in $ACC_{1}A_{2}$. Dokaži, da lahko iz daljic $A_{1}A_{2}$, $B_{1}B_{2}$ in $C_{1}C_{2}$ z vzporednimi premiki sestavimo trikotnik, če te daljice niso vzporedne.\n\n(20 točk)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nVpeljimo vektorje $\\vec{x} = A A_{1}$, $\\vec{y} = B B_{1}$ in $\\vec{z} = C C_{1}$. Tedaj je $A A_{2} = \\vec{z}$, $B B_{2} = \\vec{x}$ in $C C_{2} = \\vec{y}$, saj sta nasprotni stranici vsakega romba vzporedni in enako dolgi. Sledi\n$$\n\\overrightarrow{A_{1}A_{2}} = \\vec{z} - \\vec{x}, \\quad \\overrightarrow{B_{1}B_{2}} = \\vec{x} - \\vec{y}, \\quad \\overrightarrow{C_{1}C_{2}} = \\vec{y} - \\vec{z}\n$$\nOpazimo, da je vsota vektorjev $\\overrightarrow{A_{1}A_{2}} + \\overrightarrow{B_{1}B_{2}} + \\overrightarrow{C_{1}C_{2}}$ enaka $\\overrightarrow{0}$. To pomeni, da če daljice $A_{1}A_{2}$, $B_{1}B_{2}$ in $C_{1}C_{2}$ vzporedno premaknemo tako, da krajišče $B_{1}$ postavimo na krajišče $A_{2}$ in nato krajišče $C_{1}$ postavimo na krajišče $B_{2}$, tedaj krajišči $C_{2}$ in $A_{1}$ sovpadeta. Tedaj premaknjene daljice $A_{1}A_{2}$, $B_{1}B_{2}$ in $C_{1}C_{2}$ tvorijo nov trikotnik, če niso vzporedne, oziroma, če na začetku niso bile vzporedne.\n\nOpomba. Trikotnik lahko sestavimo tudi tako, da krajišče $C_{1}$ postavimo na krajišče $A_{2}$ in nato krajišče $B_{1}$ postavimo na krajišče $C_{2}$. V tem primeru krajišči $B_{2}$ in $A_{1}$ sovpadeta.\n\nNarisana in označena skica (vključno z izbranimi vektorji $\\vec{x}$, $\\vec{y}$ in $\\vec{z}$ )\n\nZapis $A A_{2} = \\vec{z}$, $B B_{2} = \\vec{x}$ in $C C_{2} = \\vec{y}$\n\nZapis $\\overrightarrow{A_{1}A_{2}} = \\vec{z} - \\vec{x}$, $\\overrightarrow{B_{1}B_{2}} = \\vec{x} - \\vec{y}$, $\\overrightarrow{C_{1}C_{2}} = \\vec{y} - \\vec{z}$\n\nIzračun $\\overrightarrow{A_{1}A_{2}} + \\overrightarrow{B_{1}B_{2}} + \\overrightarrow{C_{1}C_{2}} = 0$\n\nUtemeljen sklep, da lahko iz daljic $A_{1}A_{2}$, $B_{1}B_{2}$ in $C_{1}C_{2}$ z ustreznimi vzporednimi premiki tvorimo nov trikotnik.\n\n\nSolution 2:\n\nNaj bo $A_{3}$ taka točka, da je $A_{2}A_{3}C_{2}C_{1}$ paralelogram. Tedaj so daljice $AC$, $A_{2}C_{1}$ in $A_{3}C_{2}$ enako dolge in vzporedne, saj sta nasprotni stranici vsakega paralelograma enako dolgi in vzporedni. Prav tako sta enako dolgi in vzporedni tudi daljici $BC$ in $B_{1}C_{2}$. Sledi, da sta trikotnika $ABC$ in $A_{3}B_{1}C_{2}$ skladna, saj imata dva para enako dolgih vzporednih stranic, in zato sta tudi daljici $AB$ in $A_{3}B_{1}$ enako dolgi in vzporedni. Posledično je tudi $A_{1}B_{2}B_{1}A_{3}$ paralelogram, torej sta tudi daljici $B_{1}B_{2}$ in $A_{3}A_{1}$ enako dolgi in vzporedni. Če torej daljice $A_{1}A_{2}$, $B_{1}B_{2}$ in $C_{1}C_{2}$ niso vzporedne, tedaj lahko iz njih z vzporednimi premiki sestavimo trikotnik $A_{1}A_{3}A_{2}$, tako da premaknemo $C_{1}C_{2}$ na $A_{2}A_{3}$ ter $B_{1}B_{2}$ na $A_{3}A_{1}$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41983,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZapiši enačbo stožnice, na kateri ležijo središča vseh tistih krožnic, ki se hkrati dotikajo krožnice $x^{2}+y^{2}=9$ in krožnice $x^{2}+(y-4)^{2}=1$, a ne vsebujejo točke $T(0,3)$. Enačbo stožnice zapiši v standardni obliki.\n(20 točk)",
    "options": [],
    "answer": "x^2/3 - (y - 2)^2 = -1",
    "solution": "Solution:\n\nOznačimo dani krožnici zaporedoma s $\\mathcal{K}_1$ in $\\mathcal{K}_2$. Krožnica $\\mathcal{K}_1$ ima središče v $S_1(0,0)$ in polmer $r_1=3$, krožnica $\\mathcal{K}_2$ pa ima središče v $S_2=(0,4)$ in polmer $r_2=1$. Krožnici se od zunaj dotikata v točki $T(0,3)$. Krožnice, ki se hkrati dotikajo krožnic $\\mathcal{K}_1$ in $\\mathcal{K}_2$ in vsebujejo točko $T(0,3)$, imajo središča na $y$-osi. Med preostalimi krožnicami $\\mathcal{K}$, ki se hkrati dotikajo obeh danih krožnic, so take, ki se obeh dotikajo od zunaj in take, ki se obeh dotikajo od znotraj, kot to prikazujeta sliki.\n\nObravnavajmo najprej primer, ko se krožnica $\\mathcal{K}$ dotika krožnic $\\mathcal{K}_1$ in $\\mathcal{K}_2$ od zunaj. Naj bo $S(x, y)$ središče krožnice $\\mathcal{K}$, kjer je $S \\neq T$. Tedaj je radij krožnice $\\mathcal{K}$ enak $r=|S S_1|-r_1=|S S_2|-r_2$. Ker je $|S S_1|=\\sqrt{x^{2}+y^{2}}$ in $|S S_2|=\\sqrt{x^{2}+(y-4)^{2}}$, sledi\n$$\n\\sqrt{x^{2}+y^{2}}=\\sqrt{x^{2}+(y-4)^{2}}+2\n$$\nEnačbo kvadriramo, da dobimo $x^{2}+y^{2}=x^{2}+(y-4)^{2}+4 \\sqrt{x^{2}+(y-4)^{2}}+4$. Slednje poenostavimo do $2y-5=\\sqrt{x^{2}+(y-4)^{2}}$ in ponovno kvadriramo, pri čemer mora veljati $y \\geq \\frac{5}{2}$. Dobimo $4y^{2}-20y+25=x^{2}+(y-4)^{2}$ oziroma $x^{2}-3y^{2}+12y=9$.\n\nObravnavajmo še primer, ko se krožnica $\\mathcal{K}$ dotika krožnic $\\mathcal{K}_1$ in $\\mathcal{K}_2$ od znotraj. Tedaj je njen polmer enak $r=|S S_1|+r_1=|S S_2|+r_2$, od koder sledi\n$$\n\\sqrt{x^{2}+y^{2}}=\\sqrt{x^{2}+(y-4)^{2}}-2\n$$\nPo kvadriranju in poenostavljanju dobimo $-2y+5=\\sqrt{x^{2}+(y-4)^{2}}$. Po ponovnem kvadriranju pridemo do enake enačbe kot v prvem primeru, le da mora tokrat veljati $y \\leq \\frac{5}{2}$.\n\nRešitev je torej stožnica z enačbo $x^{2}-3y^{2}+12y=9$. Standardna oblika enačbe je $\\frac{x^{2}}{3}-(y-2)^{2}=-1$ (kar predstavlja hiperbolo). Rešitev je enolično določena, saj je stožnica enolično določena že s 5 točkami. Opazimo lahko, da središča opazovanih krožnic pokrijejo celotno stožnico, razen točke $T(0,3)$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41984,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV trikotniku $ABC$ velja $|AC| \\neq |BC|$. Naj bo $S$ razpolovišče stranice $AB$, $D$ presečišče stranice $AB$ in simetrale kota $\\angle ACB$, ter $E$ zrcalna slika točke $D$ pri zrcaljenju čez točko $S$. Trikotniku $ABC$ očrtana krožnica seka premico $CD$ v točkah $C$ in $F$, premico $CS$ pa v točkah $C$ in $G$. Dokaži, da točke $E$, $G$, $F$ in $S$ ležijo na isti krožnici.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nZaradi simetrije lahko predpostavimo, da je $|AC| > |BC|$. Potem točka $D$ leži med $S$ in $B$, točka $E$ pa med $S$ in $A$. Hkrati točka $G$ leži na istem bregu premice $FS$ kot točka $E$. Simetrali stranice in nasprotnega kota v trikotniku se vedno sekata na trikotniku očrtani krožnici. Torej simetrala stranice $AB$ poteka skozi točko $F$, kar pomeni, da je premica $SF$ pravokotna na premico $AB$. Od tod sledi, da sta trikotnika $FSD$ in $FSE$ skladna, zato velja\n$$\n\\angle FES = \\angle SDF = \\angle ADF = \\angle ACD + \\angle DAC.\n$$\nPo izreku o obodnih kotih je\n$$\n\\angle FGS = \\angle FGC = \\angle FAC = \\angle FAB + \\angle BAC = \\angle FCB + \\angle BAC = \\angle ACF + \\angle BAC = \\angle ACD + \\angle DAC.\n$$\nOd tod sledi $\\angle FES = \\angle FGS$. Ker točki $E$ in $G$ ležita na istem bregu premice $FS$, po izreku o obodnih kotih sledi, da točke $E$, $G$, $F$ in $S$ ležijo na isti krožnici.\n\n\n2. način. Zaradi simetrije lahko predpostavimo, da je $|AC| > |BC|$. Potem točka $D$ leži med $S$ in $B$, točka $E$ pa med $S$ in $A$. Ker je $AGBC$ tetiven štirikotnik, je $\\angle BGC = \\angle BAC$, torej sta trikotnika $GBS$ in $ACS$ podobna, saj imata dva enaka kota. Podobno je $\\angle SGA = \\angle CBS$, zato sta tudi trikotnika $AGS$ in $CBS$ podobna. Iz prve podobnosti sledi $\\frac{|SG|}{|SA|} = \\frac{|BG|}{|AC|}$, iz druge pa $\\frac{|SG|}{|SB|} = \\frac{|AG|}{|BC|}$. Ker pa je $|SA| = |SB|$, iz obeh enakosti sledi $\\frac{|BG|}{|AC|} = \\frac{|AG|}{|BC|}$ oziroma $\\frac{|AC|}{|BC|} = \\frac{|BG|}{|AG|}$. Ker je premica $CD$ simetrala kota $\\angle ACB$, je $\\frac{|AD|}{|BD|} = \\frac{|AC|}{|BC|}$. Točki $D$ in $E$ sta simetrični glede na razpolovišče stranice $AB$, zato velja $|BE| = |AD|$ in $|AE| = |BD|$. Iz vsega tega izpeljemo\n$$\n\\frac{|BE|}{|AE|} = \\frac{|AD|}{|BD|} = \\frac{|AC|}{|BC|} = \\frac{|BG|}{|AG|}\n$$\nKer točka $E$ leži na daljici $AB$, od tod sledi, da je premica $GE$ simetrala kota $\\angle BGA$.\n\nOznačimo s $H$ drugo presečišče premice $GE$ in trikotniku $ABC$ očrtane krožnice. Ker je premica $GE$ simetrala kota $\\angle BGA$, sta krožna loka $\\widehat{BH}$ in $\\widehat{AH}$ enako dolga. Podobno je premica $CF$ simetrala kota $\\angle ACB$, zato sta tudi krožna loka $\\widehat{AF}$ in $\\widehat{BF}$ enako dolga. Od tod sledi, da je daljica $HF$ premer trikotniku $ABC$ očrtane krožnice, torej po Talesovem izreku velja $\\angle FGE = \\angle FGH = \\frac{\\pi}{2}$. Hkrati pa iz enakosti krožnih lokov sledi tudi, da premica $HF$ poteka skozi $S$ in je pravokotna na stranico $AB$, torej je $\\angle ESF = \\frac{\\pi}{2}$. Po Talesovem izreku torej točke $E$, $G$, $F$ in $S$ ležijo na isti krožnici.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41985,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je trikotnik $A B C$. Naj bosta $D$ in $E$ taki točki, ki ležita zaporedoma na poltrakih $C A$ in $C B$, a ne na stranicah trikotnika $A B C$, da velja $|A D|=|B E|=|A B|$. Presečišče premic $A E$ in $B D$ označimo z $G$, središče trikotniku $A B C$ včrtane krožnice pa z $I$. Dokaži, da premica GI poteka skozi razpolovišče stranice $A B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nKer je trikotnik $A B E$ enakokrak z vrhom pri $B$, velja\n$$\n\\Varangle E A B=\\frac{1}{2}(\\pi-\\Varangle A B E)=\\frac{1}{2} \\Varangle C B A .\n$$\nPremica $B I$ je simetrala kota $\\Varangle C B A$, zato je $\\frac{1}{2} \\Varangle C B A=\\Varangle I B A$. Od tod sledi $\\Varangle E A B=\\Varangle I B A$, torej sta premici $A E$ in $B I$ vzporedni.\n\nNa podoben način sklepamo, da je\n$$\n\\Varangle A B D=\\frac{1}{2}(\\pi-\\Varangle D A B)=\\frac{1}{2} \\Varangle B A C=\\Varangle B A I,\n$$\ntorej sta tudi premici $B D$ in $A I$ vzporedni.\n\nSledi da je $A G B I$ paralelogram, zato se njegovi diagonali razpolavljata. To pomeni, da premica GI poteka skozi razpolovišče stranice $A B$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41986,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse polinome stopnje $n \\geq 1$, ki imajo vse ničle racionalne, vsak izmed njihovih $n+1$ koeficientov pa je enak 1 ali -1.",
    "options": [],
    "answer": "±(x−1), ±(x+1), ±(x^3 + x^2 − x − 1), ±(x^3 − x^2 − x + 1)",
    "solution": "Solution:\n\nNaj bo $p(x) = a_n x^n + \\cdots + a_0$ iskani polinom. Če je $\\frac{k}{m}$ neka njegova racionalna ničla, tedaj $k$ deli $a_0$ in $m$ deli $a_n$. Ker pa sta $a_0$ in $a_n$ oba enaka 1 ali -1, sta tudi $k$ in $m$ enaka 1 ali -1. Torej so vse ničle polinoma $p(x)$ enake 1 ali -1. Hkrati je tudi vodilni koeficient polinoma $p(x)$ enak 1 ali -1, zato lahko pišemo $p(x) = \\pm(x-1)^m (x+1)^k$ za neki nenegativni celi števili $m$ in $k$ (ne obe enaki 0). Če je $m=0$, tedaj očitno le polinoma $\\pm(x+1)$ zadoščata pogojem naloge, in če je $k=0$, le polinoma $\\pm(x-1)$ zadoščata pogojem naloge. V nadaljevanju zato predpostavimo, da sta $k$ in $m$ naravni števili. Sledi\n$$\n\\pm p(x) = \\left(x^m - m x^{m-1} + \\cdots\\right)\\left(x^k + k x^{k-1} + \\cdots\\right) = x^{k+m} + (k-m)x^{k+m-1} + \\cdots\n$$\ntorej je $k-m = \\pm 1$.\nČe je $k = m+1$, potem je\n$$\n\\pm p(x) = \\left(x^2 - 1\\right)^m (x+1) = \\left(x^{2m} - m x^{2m-2} + \\cdots\\right)(x+1) = x^{2m+1} + x^{2m} - m x^{2m-1} + \\cdots\n$$\nod koder sledi $m=1$ in $p(x) = \\pm(x-1)(x+1)^2 = \\pm\\left(x^3 + x^2 - x - 1\\right)$.\nČe pa je $k = m-1$ oziroma $m = k+1$, pa podobno kot zgoraj sledi\n$$\n\\pm p(x) = \\left(x^2 - 1\\right)^k (x-1) = \\left(x^{2k} - k x^{2k-2} + \\cdots\\right)(x-1) = x^{2k+1} - x^{2k} - k x^{2k-1} + \\cdots\n$$\ntorej je $k=1$ in $p(x) = \\pm(x-1)^2(x+1) = \\pm\\left(x^3 - x^2 - x + 1\\right)$.\nVse rešitve so torej polinomi $\\pm(x-1)$, $\\pm(x+1)$, $\\pm\\left(x^3 + x^2 - x - 1\\right)$ in $\\pm\\left(x^3 - x^2 - x + 1\\right)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41987,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAli obstaja tako praštevilo $p$, da velja $p^{2}+p+1=n^{3}$ za neko naravno število $n$?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPokazali bomo, da takšno praštevilo ne obstaja. Enačbo preoblikujemo v $p^{2}+p=n^{3}-1$ in obe strani razstavimo, da dobimo $p(p+1)=(n-1)(n^{2}+n+1)$. Če $p$ deli $n-1$, tedaj $n^{2}+n+1$ deli $p+1$. Od tod sledi $n^{2}+n+1 \\leq p+1 \\leq n$, kar pa je protislovje. Torej mora $p$ deliti $n^{2}+n+1$ in zato $n-1$ deli $p+1$. To pomeni, da obstaja naravno število $k$, da velja $n^{2}+n+1=k p$ in $p+1=k(n-1)$. Iz druge enakosti izrazimo $p$ in ga vstavimo v prvo enakost, da dobimo $n^{2}+n+1=k(k(n-1)-1)=k^{2} n-k^{2}-k$. Slednje preoblikujemo v $n^{2}+(1-k^{2}) n+(k^{2}+k+1)=0$ in to pogledamo kot kvadratno enačbo v spremenljivki $n$. Ker mora imeti ta enačba rešitev, ki je naravno število, mora biti njena diskriminanta $D=k^{4}-6 k^{2}-4 k-3$ popoln kvadrat. Opazimo, da je $D=(k^{2}-3)^{2}-4 k-12< (k^{2}-3)^{2}$, saj je $4 k+12>0$. Prvi popoln kvadrat, ki je manjši od $(k^{2}-3)^{2}$, je $(k^{2}-4)^{2}$, zato mora veljati $D \\leq (k^{2}-4)^{2}$. Z upoštevanjem zveze $D=k^{4}-6 k^{2}-4 k-3$ to neenakost preoblikujemo v $2 k^{2}-4 k-19 \\leq 0$ oziroma $2 k(k-2) \\leq 19$. Opazimo, da za $k \\geq 5$ neenakost ni izpolnjena, torej so edine naravne rešitve $1,2,3$ in $4$. Toda diskriminanta $D$ je pri teh vrednostih števila $k$ zaporedoma enaka $-12,-19,12$ in $141$ in v nobenem primeru ni popoln kvadrat. Torej tako praštevilo $p$ ne obstaja.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41988,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n8 prijateljev, od tega 3 dekleta, se bo v zabaviščnem parku peljalo $z$ velikim razglednim kolesom. Vseh 8 prijateljev se bo naključno razporedilo v 4 proste kabine razglednega kolesa, v vsako kabino po 2 prijatelja. Kolikšna je verjetnost, da nobeni 2 dekleti ne bosta sedeli v isti kabini?\n(20 točk)",
    "options": [],
    "answer": "4/7",
    "solution": "Solution:\n\nNaj bo $A$ dogodek, da nobeni 2 dekleti ne sedita v isti kabini. Verjetnost dogodka $A$ izračunamo po formuli\n$$\nP(A)=\\frac{\\text{ugodne razporeditve}}{\\text{vse razporeditve}}\n$$\nVseh možnih razporeditev je $8!$, saj moramo 8 prijateljev na poljuben način razporediti na 8 sedišč, kar pomeni, da štejemo permutacije na 8 elementih.\n\nUgodne razporeditve preštejemo tako, da najprej izberemo 3 izmed 4 kabin, v katerih bodo sedela dekleta, in v vsaki od teh 3 kabin izberemo 1 od 2 sedišč, na katerem bo sedelo dekle. To lahko storimo na $\\binom{4}{3} \\cdot 2^{3}$ različnih načinov. Nato pa najprej razporedimo dekleta na njihova sedišča na $3!$ različnih načinov in nato še fante na njihova sedišča na $5!$ različnih načinov. Ugodnih razporeditev je torej $\\binom{4}{3} \\cdot 2^{3} \\cdot 3!\\cdot 5!$. Verjetnost, da nobeni 2 dekleti ne bosta sedeli v isti kabini je torej\n$$\nP(A)=\\frac{\\binom{4}{3} \\cdot 2^{3} \\cdot 3!\\cdot 5!}{8!}=\\frac{\\frac{4!}{3!\\cdot 1!} \\cdot 2^{3} \\cdot 3!}{8 \\cdot 7 \\cdot 6}=\\frac{4!\\cdot 2^{3}}{8 \\cdot 7 \\cdot 6}=\\frac{4 \\cdot 3 \\cdot 2 \\cdot 1 \\cdot 8}{8 \\cdot 7 \\cdot 6}=\\frac{4}{7}\n$$\n\n\n2. način. Rešitev lahko poiščemo tudi s pomočjo kombinacij, pri čemer upoštevamo, da ni pomembno, kako v posamezni kabini sedita prijatelja. Vse možnosti preštejemo tako, da najprej v prvo kabino postavimo 2 izmed 8 prijateljev, nato v drugo postavimo 2 izmed preostalih 6 prijateljev, zatem v tretjo postavimo 2 izmed preostalih 4 prijateljev in nazadnje še preostala 2 prijatelja postavimo v zadnjo kabino. To lahko storimo na $\\binom{8}{2} \\cdot\\binom{6}{2} \\cdot\\binom{4}{2}$ različnih načinov. Ugodne možnosti preštejemo tako, da najprej izberemo 3 izmed 4 kabin, kjer bodo sedela dekleta, kar lahko storimo na $\\binom{4}{3}$ različnih načinov. V prvo od teh kabin postavimo 1 od 3 deklet in 1 od 5 fantov na $3 \\cdot 5=15$ načinov, nato v drugo od teh kabin postavimo 1 od preostalih 2 deklet in 1 od preostalih 4 fantov na $2 \\cdot 4=8$ načinov in zatem v tretjo od teh kabin postavimo preostalo dekle in 1 od preostalih 3 fantov na $1 \\cdot 3=3$ načine. Preostala 2 fanta pa postavimo v preostalo kabino. Ugodnih možnosti je torej $\\binom{4}{3} \\cdot 15 \\cdot 8 \\cdot 3$, verjetnost opisanega dogodka $A$ pa je enaka\n$$\nP(A)=\\frac{\\binom{4}{3} \\cdot 15 \\cdot 8 \\cdot 3}{\\binom{8}{2} \\cdot\\binom{6}{2} \\cdot\\binom{4}{2}}=\\frac{4}{7}\n$$\n\n\n3. način. Namesto verjetnosti dogodka $A$, izračunamo verjetnost nasprotnega dogodka $A^{\\prime}$. Vse možnosti preštejemo na enak kot v 2. rešitvi. Ugodne možnosti za dogodek $A^{\\prime}$ preštejemo tako, da najprej izberemo 1 izmed 4 kabin v kateri bosta sedeli 2 dekleti in vanjo postavimo 2 izmed 3 deklet. Nato izmed preostalih 3 kabin izberemo 1, v kateri bosta sedela dekle in fant ter vanjo postavimo preostalo dekle in 1 od 5 fantov. V tretjo kabino postavimo 2 izmed preostalih 4 fantov, zadnja 2 fanta pa postavimo v zadnjo kabino. Verjetnost dogodka $A^{\\prime}$ je zato enaka\n$$\nP\\left(A^{\\prime}\\right)=\\frac{4 \\cdot\\binom{3}{2} \\cdot 3 \\cdot 5 \\cdot\\binom{4}{2}}{\\binom{8}{2} \\cdot\\binom{6}{2} \\cdot\\binom{4}{2}}=\\frac{3}{7}\n$$\nverjetnost dogodka $A$ pa je $P(A)=1-\\frac{3}{7}=\\frac{4}{7}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41989,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse pare realnih števil $x$ in $y$, ki rešijo sistem enačb\n$$\n\\begin{gathered}\n\\sqrt{x}-\\sqrt{y}=1 \\\\\n\\sqrt{8x+7}-\\sqrt{8y+7}=2\n\\end{gathered}\n$$",
    "options": [],
    "answer": "x = 9/4, y = 1/4",
    "solution": "Solution:\n\nEnačbi preoblikujemo v ekvivalentni enačbi $\\sqrt{x}=1+\\sqrt{y}$ in $\\sqrt{8x+7}=2+\\sqrt{8y+7}$ ter ju nato kvadriramo, da dobimo\n$$\n\\begin{aligned}\nx & = 1 + 2\\sqrt{y} + y \\, \\text{ in } \\\\\n8x + 7 & = 4 + 4\\sqrt{8y+7} + 8y + 7\n\\end{aligned}\n$$\nOd spodnje enačbe odštejemo 8-kratnik zgornje enačbe in dobljeno enačbo poenostavimo, da dobimo $1 + 4\\sqrt{y} = \\sqrt{8y+7}$. Enačbo kvadriramo in poenostavimo do $4\\sqrt{y} = 3 - 4y$. Ko enačbo še enkrat kvadriramo, dobimo kvadratno enačbo $16y^2 - 40y + 9 = 0$, katere levo stran lahko razstavimo kot $(4y-1)(4y-9) = 0$. Dobimo dve rešitvi, $y_1 = \\frac{1}{4}$ in $y_2 = \\frac{9}{4}$. Iz enačbe $x = 1 + 2\\sqrt{y} + y$ izračunamo še $x_1 = \\frac{9}{4}$ in $x_2 = \\frac{25}{4}$.\n\nPreverimo lahko, da par $\\left(x_1, y_1\\right)$ reši začetni sistem enačb, par $\\left(x_2, y_2\\right)$ pa ne, saj je\n$$\n\\sqrt{8 \\cdot \\frac{25}{4} + 7} - \\sqrt{8 \\cdot \\frac{9}{4} + 7} = \\sqrt{57} - 5\n$$\nkar ni celo število in zato ne more biti enako 2. Torej je edina rešitev naloge par $x = \\frac{9}{4}$ in $y = \\frac{1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41990,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse pare realnih števil $x$ in $y$, ki rešijo sistem enačb\n$$\n\\begin{gathered}\n\\frac{y^{3}+15 x^{2}}{y^{4}-x^{3}}=\\frac{y^{2}+15 x}{y^{3}-x^{2}} \\\\\n\\frac{1500 y^{3}+4 x^{2}}{9 y^{4}-4}=\\frac{1500 y^{2}+4 x}{9 y^{3}-4}\n\\end{gathered}\n$$",
    "options": [],
    "answer": "(0,1), (1,0), (-1/375,-1/375), (-5,1/3), (-15/2,1/2), (-5/2,1/6), (15,-1)",
    "solution": "Solution:\n\nV obeh enačbah odpravimo ulomke in ju poenostavimo, da dobimo\n$$\n\\begin{aligned}\n-x^{2} y^{3}+15 x^{2} y^{3} & =-x^{3} y^{2}+15 x y^{4} \\\\\n-6000 y^{3}+36 x^{2} y^{3}-16 x^{2} & =-6000 y^{2}+36 x y^{4}-16 x\n\\end{aligned}\n$$\nPrvo enačbo preoblikujemo v $x^{2} y^{2}(x-y)=-15 x y^{3}(x-y)$ in nato v\n$$\nx y^{2}(x-y)(x+15 y)=0\n$$\nŠtevili $x$ in $y$ torej zadoščata eni od naslednjih štirih možnosti:\n\n1. možnost: $x=0$. Ko to vstavimo v drugo od zgornjih dveh enačb, dobimo $y^{3}=y^{2}$ oziroma $y^{2}(y-1)=0$. Torej je $y=0$ ali $y=1$. Prva možnost odpade, saj pri $x=y=0$ nekateri ulomki v začetnih enačbah niso dobro definirani. Ostane torej le možnost $y=1$.\n\n2. možnost: $y=0$. Iz druge od zgornjih enačb tedaj dobimo $x^{2}=x$, torej je $x=0$ ali $x=1$. Možnost $x=0$ kot v prejšnjem primeru odpade, torej dobimo le rešitev $x=1$.\n\n3. možnost: $y=x$. To spet vstavimo v drugo enačbo in dobimo $-6000 x^{3}-16 x^{2}=-6000 x^{2}-16 x$, kar lahko preoblikujemo v $6000 x^{2}(1-x)=16 x(x-1)$, slednje pa v $16 x(x-1)(375 x+1)=0$. Možnost $x=0$ odpade, saj je v tem primeru tudi $y=0$ in nekateri ulomki v začetnih enačbah niso dobro definirani. Iz enakega razloga odpade tudi možnost $x=1$. Sledi $375 x+1=0$ oziroma $x=y=-\\frac{1}{375}$.\n\n4. možnost: $x=-15 y$. Ko to vstavimo v drugo enačbo in enačbo poenostavimo, dobimo\n$$\n8640 y^{5}-6000 y^{3}-2400 y^{2}-240 y=0\n$$\nKot v prejšnjem primeru možnost $y=0$ odpade, saj je tedaj tudi $x=0$. Zato lahko zadnjo enačbo delimo z $240 y$, da dobimo\n$$\n36 y^{4}-25 y^{2}+10 y-1=0\n$$\nOpazimo, da zadnji trije členi tvorijo popolni kvadrat, zato enačbo preoblikujemo v $36 y^{4}-(5 y-1)^{2}=0$, nato pa levo stran razstavimo po formuli za razliko kvadratov in dobimo\n$$\n\\left(6 y^{2}-5 y+1\\right)\\left(6 y^{2}+5 y-1\\right)=0\n$$\nOba faktorja na levi strani enačbe razstavimo in dobimo\n$$\n(3 y-1)(2 y-1)(6 y-1)(y+1)=0\n$$\nRešitve so torej $y=\\frac{1}{3}$ in $x=-5$, $y=\\frac{1}{2}$ in $x=-\\frac{15}{2}$, $y=\\frac{1}{6}$ in $x=-\\frac{5}{2}$ ter $y=-1$ in $x=15$. V vseh primerih so vsi ulomki v začetnih enačbah dobro definirani.\n\nVsi pari $(x, y)$, ki rešijo naloge so torej $(0,1),(1,0),\\left(-\\frac{1}{375},-\\frac{1}{375}\\right),\\left(-5, \\frac{1}{3}\\right),\\left(-\\frac{15}{2}, \\frac{1}{2}\\right),\\left(-\\frac{5}{2}, \\frac{1}{6}\\right)$ in $(15,-1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41991,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDano je naravno število $m$. Zaporedje $a_{n}$ je definirano s pogojema $a_{1}=m$ in $a_{n+1} = \\left[\\frac{a_{n}^{2} + 10 a_{n}}{2 a_{n} + 4}\\right]$ za vse $n \\in \\mathbb{N}$. Dokaži, da je zaporedje $a_{n}$ od nekega člena dalje konstantno.\n\nOpomba. Za poljubno realno število $x$ oznaka $[x]$ označuje največje celo število, ki ni večje od $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nČe je $a_{n} \\geq 0$ za nek $n$, tedaj je $\\frac{a_{n}^{2} + 10 a_{n}}{2 a_{n} + 4} \\geq 0$ in zato je tudi $a_{n+1} \\geq 0$. Ker je $a_{1} = m > 0$, so torej vsi členi zaporedja $a_{n}$ nenegativna cela števila.\n\nZa majhna naravna števila $m$ izračunamo nekaj členov zaporedja $a_{n}$:\n- pri $m=1$ dobimo $1, 1, 1, 1, \\ldots$\n- pri $m=2$ dobimo $2, 3, 3, 3, \\ldots$\n- pri $m=3$ dobimo $3, 3, 3, 3, \\ldots$\n- pri $m=4$ dobimo $4, 4, 4, 4, \\ldots$\n- pri $m=5$ dobimo $5, 5, 5, 5, \\ldots$\n- pri $m=6$ dobimo $6, 6, 6, 6, \\ldots$\n- pri $m=7$ dobimo $7, 6, 6, 6, \\ldots$\n- pri $m=8$ dobimo $8, 7, 6, 6, \\ldots$\n\nOpazimo, da se člen zaporedja na vsakem koraku zmanjša, razen v primeru, ko je enak 2. Zato obravnavamo dve primera:\n\n1. primer: Recimo, da je $a_{n} = 2$ za nek $n \\in \\mathbb{N}$. Potem je $a_{n+1} = 3$. Ker je $\\left[\\frac{3^{2} + 10 \\cdot 3}{2 \\cdot 3 + 4}\\right] = \\left[\\frac{39}{10}\\right] = 3$, sledi, da je $a_{k} = 3$ za vse $k \\geq n+1$. Naloga je v tem primeru dokazana.\n\n2. primer: Recimo, da je $a_{n} \\neq 2$ za vsak $n \\in \\mathbb{N}$. Dokažimo, da je zaporedje v tem primeru padajoče. Ker so vsi členi zaporedja $a_{n}$ cela števila, je pogoj $a_{n+1} \\leq a_{n}$ ekvivalenten pogoju $\\frac{a_{n}^{2} + 10 a_{n}}{2 a_{n} + 4} < a_{n} + 1$. Ta neenakost pa je ekvivalentna neenakosti $a_{n}^{2} - 4 a_{n} + 4 > 0$ oziroma $(a_{n} - 2)^{2} > 0$, kar očitno drži za vsak $n \\in \\mathbb{N}$, saj po predpostavki noben člen zaporedja ni enak 2. Zaporedje $a_{n}$ je torej padajoče, njegovi členi pa so nenegativna cela števila (različna od 2), zato mora biti zaporedje od nekega člena naprej konstantno.\n\n\n3. način. Če je $a_{n} \\geq 1$ za nek $n$, tedaj je $10 a_{n} \\geq 2 a_{n} + 4$, torej je tudi $a_{n+1} \\geq 1$. Ker je $a_{1} = m \\geq 1$, so torej vsi členi zaporedja $a_{n}$ naravna števila. Opazimo, da velja\n$$\n\\begin{gathered}\n{\\left[\\frac{1^{2} + 10 \\cdot 1}{2 \\cdot 1 + 4}\\right] = \\left[\\frac{11}{6}\\right] = 1, \\quad \\left[\\frac{2^{2} + 10 \\cdot 2}{2 \\cdot 2 + 4}\\right] = \\left[\\frac{24}{8}\\right] = 3, \\quad \\left[\\frac{3^{2} + 10 \\cdot 3}{2 \\cdot 3 + 4}\\right] = \\left[\\frac{39}{10}\\right] = 3,} \\\\\n{\\left[\\frac{4^{2} + 10 \\cdot 4}{2 \\cdot 4 + 4}\\right] = \\left[\\frac{56}{12}\\right] = 4, \\quad \\left[\\frac{5^{2} + 10 \\cdot 5}{2 \\cdot 5 + 4}\\right] = \\left[\\frac{75}{14}\\right] = 5, \\quad \\left[\\frac{6^{2} + 10 \\cdot 6}{2 \\cdot 6 + 4}\\right] = \\left[\\frac{96}{16}\\right] = 6} \\\\\n{\\left[\\frac{7^{2} + 10 \\cdot 7}{2 \\cdot 7 + 4}\\right] = \\left[\\frac{119}{18}\\right] = 6, \\quad \\left[\\frac{8^{2} + 10 \\cdot 8}{2 \\cdot 8 + 4}\\right] = \\left[\\frac{144}{20}\\right] = 7}\n\\end{gathered}\n$$\nIz prvih šestih enakosti sledi, da če je nek člen zaporedja manjši ali enak 6, potem je zaporedje od nekje dalje konstantno (in enako 1, 3, 4, 5 ali 6). Zadnji dve enakosti namigujeta, da če je nek člen zaporedja večji kot 6, potem bo naslednji člen strogo manjši. Pokažimo, da to res velja. Denimo torej, da je $a_{n} > 6$ za nek $n \\in \\mathbb{N}$. Pokazati želimo, da je tedaj $a_{n+1} < a_{n}$. Ker so vsi členi zaporedja naravna števila, je slednja neenakost ekvivalentna neenakosti $a_{n+1} \\leq a_{n} - 1$, ta pa neenakosti $\\frac{a_{n}^{2} + 10 a_{n}}{2 a_{n} + 4} < a_{n}$. Ko odpravimo ulomke, dobimo $a_{n}^{2} - 6 a_{n} > 0$ oziroma $a_{n}(a_{n} - 6) > 0$. Neenakost je izpolnjena, saj je $a_{n}$ naravno število in po predpostavki velja $a_{n} > 6$. Od tod sledi, da so za poljuben $m$ členi zaporedja prej ali slej manjši ali enaki 6, od prej pa že vemo, da je v tem primeru zaporedje od nekega člena naprej konstantno.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41992,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $K$ končna podmnožica celih števil s $k \\geq 3$ elementi. Pravimo, da sta števili $a, b \\in K$ povezani, če in samo če obstajajo števila $x_{1}, x_{2}, \\ldots, x_{k}$, za katera velja:\n- $\\{x_{1}, x_{2}, \\ldots, x_{k}\\}=K$,\n- $x_{1}=a, x_{k}=b$,\n- $|x_{i}-x_{i+1}|$ je liho število za vsak $i \\in\\{1,2, \\ldots, k-1\\}$.\nDokaži, da v množici $K$ obstajata dve števili, ki nista povezani.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTretji pogoj v nalogi pove, da sta števili $a$ in $b$ povezani, če lahko števila v množici $K$ razporedimo v vrsto, ki se začne z $a$ in konča z $b$, tako da sta vsaki dve sosednji števili v tej vrsti različne parnosti. Vrsto, v kateri sta vsaki dve sosednji števili različne parnosti, bomo imenovali alternirajoča vrsta.\n\nObravnavajmo različni možnosti glede na parnost števila $k$.\n\nDenimo, da je $k$ sodo število. Števili $a$ in $b$ sta povezani z alternirajočo vrsto natanko tedaj, ko sta različnih parnosti. Ker je $k \\geq 3$, lahko izmed števil v množici $K$ izberemo dve števili, ki sta iste parnosti in zato nista povezani.\n\nDenimo, da je $k$ liho število. Števili $a$ in $b$ sta povezani z alternirajočo vrsto natanko tedaj, ko sta iste parnosti. Ker je $k \\geq 3$, lahko izmed števil v množici $K$ izberemo dve števili, ki sta različne parnosti in zato nista povezani.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41993,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je trikotnik $A B C$. Naj bosta $D$ in $E$ taki točki, ki ležita zaporedoma na poltrakih $C A$ in $C B$, a ne na stranicah trikotnika $A B C$, da velja $|A D|=|B E|=|A B|$. Naj bo $F$ presečišče vzporednice $k A C$ skozi točko $E$ in vzporednice $k B C$ skozi točko $D$. Presečišče premic $A E$ in $B D$ označimo z $G$. Dokaži, da je premica $F G$ simetrala kota $\\Varangle E F D$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOznačimo s $P$ presečišče daljice $B C$ in simetrale kota $\\Varangle B A C$, z $Q$ pa presečišče daljice $A C$ in simetrale kota $\\Varangle C B A$. Presečišče premic $A P$ in $B Q$ je torej središče trikotniku $A B C$ včrtane krožnice, označimo ga z $I$.\n\nKer je trikotnik $B A D$ enakokrak z vrhom pri $A$, je $\\Varangle B D A = \\frac{1}{2}(\\pi - \\Varangle D A B) = \\frac{1}{2}(\\Varangle B A C) = \\Varangle P A C$. Torej sta premici $B D$ in $P A$ vzporedni in trikotnika $B C D$ in $P C A$ sta si podobna. Sledi\n$$\n\\frac{|C B|}{|C D|} = \\frac{|C P|}{|C A|} \\quad \\text{ oziroma } \\quad |C P| \\cdot |C D| = |C B| \\cdot |C A|.\n$$\nNa enak način pokažemo, da sta tudi premici $A E$ in $Q B$ vzporedni ter trikotnika $E C A$ in $B C Q$ podobna, zato velja še\n$$\n\\frac{|C E|}{|C A|} = \\frac{|C B|}{|C Q|} \\quad \\text{ oziroma } \\quad |C E| \\cdot |C Q| = |C B| \\cdot |C A|.\n$$\nIz zgornjih enakosti sledi\n$$\n|C P| \\cdot |C D| = |C E| \\cdot |C Q| \\quad \\text{ oziroma } \\quad \\frac{|C P|}{|C Q|} = \\frac{|C E|}{|C D|}\n$$\nTo pomeni, da sta trikotnika $P C Q$ in $E C D$ podobna. V posebnem sta premici $P Q$ in $E D$ vzporedni. Štirikotnik $F E C D$ je paralelogram, saj ima dva para vzporednih stranic, zato sta trikotnika $E C D$ in $D F E$ skladna. Posledično sta trikotnika $P C Q$ in $D F E$ podobna. Iz vzporednosti premic $B D$ in $P A$, $A E$ in $Q B$ ter $P Q$ in $E D$ sledi, da sta tudi trikotnika $Q I P$ in $E G D$ podobna. Torej sta celo štirikotnika $P C Q I$ in $D F E G$ podobna. Ker je diagonala $C I$ simetrala kota $\\Varangle Q C P$, je tudi diagonala $F G$ simetrala kota $\\Varangle E F D$.\n\n\n2. način. Uporabimo oznake iz prve rešitve. Podobnost trikotnikov $P C Q$ in $E C D$ lahko pokažemo tudi nekoliko drugače. Simetrala notranjega kota trikotnika razdeli nasprotno stranico v razmerju, ki je enako razmerju priležnih stranic. Tako velja\n$$\n\\frac{|B P|}{|C P|} = \\frac{|A B|}{|A C|} \\quad \\text{ oziroma } \\quad |B P| = \\frac{|A B| \\cdot |C P|}{|A C|}\n$$\nSlednje vstavimo v enakost $|B P| + |C P| = |B C|$ in izrazimo $|C P|$, da dobimo\n$$\n|C P| = \\frac{|C A| \\cdot |C B|}{|A B| + |A C|} = \\frac{|C A| \\cdot |C B|}{|C D|}\n$$\nkjer smo upoštevali še $|A B| = |A D|$. Na enak način izpeljemo še\n$$\n|C Q| = \\frac{|C A| \\cdot |C B|}{|A B| + |B C|} = \\frac{|C A| \\cdot |C B|}{|C E|}\n$$\nOd tod sledi $\\frac{|C P|}{|C Q|} = \\frac{|C E|}{|C D|}$, zato sta si trikotnika $P C Q$ in $E C D$ podobna in premici $P Q$ in $E D$ sta vzporedni. Ker sta tudi premici $E F$ in $A C$ ter premici $D F$ in $B C$ vzporedni, sta si podobna tudi trikotnika $P C Q$ in $D F E$.\nNa enak način kot v prvi rešitvi pokažemo, da sta premici $B D$ in $P A$ ter premici $A E$ in $Q B$ vzporedni. Skupaj z vzporednostjo premic $P Q$ in $E D$ to pomeni, da sta trikotnika $Q I P$ in $E G D$ podobna. Od tod sledi $\\frac{|D E|}{|D G|} = \\frac{|P Q|}{|P I|}$, iz podobnosti trikotnikov $P C Q$ in $E C D$ pa še $\\frac{|D F|}{|D E|} = \\frac{|P C|}{|P Q|}$. Zadnji dve enakosti zmnožimo, da dobimo\n$$\n\\frac{|D F|}{|D G|} = \\frac{|P C|}{|P I|}\n$$\nZaradi vzporednosti premic $D F$ in $P C$ ter premic $D G$ in $P I$, pa sledi še $\\Varangle F D G = \\Varangle C P I$. Trikotnika $F D G$ in $C P I$ se tako ujemata v enem kotu in razmerju stranic ob tem kotu, zato sta si podobna. Na enak način pokažemo, da sta si podobna tudi trikotnika $F E G$ in $C Q I$. Torej je $\\Varangle E F G = \\Varangle Q C I = \\Varangle I C P = \\Varangle G F D$.\n\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 41994,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je izraz $Z=5 a^{-x}\\left(1-a^{-x}\\right)^{-1}-3 a^{-x}\\left(1+a^{-x}\\right)^{-1}-2 a^{x}\\left(a^{2 x}-1\\right)^{-1}$, kjer je $a^{x} \\neq 0,1,-1$.\n\na) Poenostavi izraz $Z$.\n\nb) Izračunaj vrednost izraza $Z$ za $a=9^{b+c} \\cdot 3^{2 b+c}: 27^{\\frac{4}{3} b+c+\\frac{1}{3}}$ in $x=1$.",
    "options": [],
    "answer": "Z = 8 / (a^(2x) − 1); for a = 9^(b+c) · 3^(2b+c) : 27^((4/3)b + c + 1/3) and x = 1, Z = −9",
    "solution": "Solution:\n\nPoenostavimo vsak člen posebej. Prvi člen preoblikujemo v\n$$\n5 a^{-x}\\left(1-a^{-x}\\right)^{-1}=\\frac{5}{a^{x}}\\left(\\frac{a^{x}-1}{a^{x}}\\right)^{-1}=\\frac{5}{a^{x}-1}\n$$\nPodobno drugi člen preoblikujemo v\n$$\n3 a^{-x}\\left(1+a^{-x}\\right)^{-1}=\\frac{3}{a^{x}+1}\n$$\nImenovalec tretjega člena razstavimo na produkt vsote in razlike\n$$\n2 a^{x}\\left(a^{2 x}-1\\right)^{-1}=\\frac{2 a^{x}}{a^{2 x}-1}=\\frac{2 a^{x}}{\\left(a^{x}-1\\right)\\left(a^{x}+1\\right)}\n$$\nOpazimo, da je to hkrati skupni imenovalec za prva dva člena. Prva dva člena razširimo, izraz poenostavimo in dobimo rezultat\n$$\n\\frac{5}{a^{x}-1}-\\frac{3}{a^{x}+1}-\\frac{2 a^{x}}{\\left(a^{x}-1\\right)\\left(a^{x}+1\\right)}=\\frac{8}{\\left(a^{x}-1\\right)\\left(a^{x}+1\\right)}=\\frac{8}{a^{2 x}-1}\n$$\nIzračunamo vrednost izraza\n$$\na=9^{b+c} \\cdot 3^{2 b+c}: 27^{\\frac{4}{3} b+c+\\frac{1}{3}}=3^{2 b+2 c} \\cdot 3^{2 b+c}: 3^{3\\left(\\frac{4}{3} b+c+\\frac{1}{3}\\right)}=\\frac{1}{3} .\n$$\nIzračunamo vrednost izraza $Z$ za dobljeni vrednosti $a$ in $x: \\frac{8}{\\left(\\frac{1}{3}\\right)^{2}-1}=-9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41995,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDani sta realni funkciji $f(x)=x+1$ in $g(x)=x^{2}+3$.\na) Funkcija $h$ je podana s predpisom $h(x)=\\frac{f(x)+g(x)}{g(x)}$. Izračunaj stacionarne točke funkcije $h$. Zapiši enačbo vodoravne asimptote grafa funkcije $h$ in izračunaj presečišče grafa z vodoravno asimptoto.\nb) Izračunaj, za katere $a \\in \\mathbb{R}$ bo imela funkcija $j(x)=g(x)+a f(x)-a$ vsaj eno realno ničlo.",
    "options": [],
    "answer": "a) Stationary points: x = 1 and x = −3; horizontal asymptote: y = 1; intersection with the asymptote: (−1, 1). b) The function has a real zero for a ≤ −2√3 or a ≥ 2√3.",
    "solution": "Solution:\n\na) Zapišemo predpis funkcije $h(x)=\\frac{x^{2}+x+4}{x^{2}+3}$. Izračunamo odvod funkcije $h'(x)=\\frac{-x^{2}-2 x+3}{\\left(x^{2}+3\\right)^{2}}$. Ničle odvoda funkcije $h$ so rešitve enačbe $-x^{2}-2 x+3=0$. Rešitvi enačbe sta $x_{1}=1$, $x_{2}=-3$, to sta stacionarni točki funkcije $h$.\n\nEnačba vodoravne asimptote grafa funkcije $h$ je $y=1$. Abscisa presečišča grafa funkcije $h$ z vodoravno asimptoto je ničla ostanka pri deljenju števca z imenovalcem racionalne funkcije $h$. Ničla ostanka $r(x)=x+1$ je $x=-1$. Presečišče je točka $P(-1,1)$.\n\nb) Zapišemo predpis funkcije $j(x)=x^{2}+a x+3$. Funkcija $j$ bo imela vsaj eno realno ničlo, ko bo veljalo $D \\geq 0$, torej je $a^{2}-12 \\geq 0$. Rešitve neenačbe so $-2 \\sqrt{3} \\geq a$ ali $a \\geq 2 \\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41996,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je izraz $A = |x + 3| + |x - 5|$.\n\na) Izračunaj vrednost izraza za $x = \\sqrt{2}$.\n\nb) Poišči vse možne vrednosti $x$, za katere velja $A = 4$.",
    "options": [],
    "answer": "a) 8; b) no real solutions",
    "solution": "Solution:\n\na)\nZa $x$ vstavimo $\\sqrt{2}$ in izračunamo vrednost izraza:\n$$\nA = |\\sqrt{2} + 3| + |\\sqrt{2} - 5| = (\\sqrt{2} + 3) + (5 - \\sqrt{2}) = 8.\n$$\n\nb)\nZa $x < -3$ ima enačba:\n$$\nA = |x + 3| + |x - 5| = -(x + 3) - (x - 5) = -x - 3 - x + 5 = -2x + 2.\n$$\nPostavimo $A = 4$:\n$$\n-2x + 2 = 4 \\implies -2x = 2 \\implies x = -1.\n$$\nKer $x < -3$, rešitev $x = -1$ ne ustreza pogoju.\n\nZa $-3 \\leq x < 5$ je:\n$$\nA = (x + 3) + (5 - x) = x + 3 + 5 - x = 8.\n$$\nKer $A = 8 \\neq 4$, v tem območju ni rešitve.\n\nZa $x \\geq 5$ je:\n$$\nA = (x + 3) + (x - 5) = x + 3 + x - 5 = 2x - 2.\n$$\nPostavimo $A = 4$:\n$$\n2x - 2 = 4 \\implies 2x = 6 \\implies x = 3.\n$$\nKer $x \\geq 5$, rešitev $x = 3$ ne ustreza pogoju.\n\nTorej enačba nima rešitev.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41997,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa državno tekmovanje v računanju se lahko uvrsti največ 30 tekmovalcev. Na letošnjem državnem tekmovanju so tekmovalci reševali 4 naloge, pri čemer je $\\frac{1}{3}$ tekmovalcev rešila natanko 3 naloge, $\\frac{1}{4}$ tekmovalcev je rešila natanko 2 nalogi, $\\frac{1}{6}$ tekmovalcev je rešila natanko 1 nalogo, $\\frac{1}{8}$ tekmovalcev pa ni rešila nobene naloge. Koliko tekmovalcev je rešilo vse 4 naloge?\n\n(A) 1\n(B) 2\n(C) 3\n(D) 4\n(E) 5",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nNajmanjši skupni večkratnik števil $3, 4, 6$ in $8$ je $24$. Število tekmovalcev na tekmovanju mora biti torej deljivo s $24$. Ker pa so vsi večkratniki števila $24$, razen števila $24$, večji od $30$, je bilo na tekmovanju $24$ tekmovalcev, od katerih so vse $4$ naloge rešili\n$$\n\\left(1-\\frac{1}{3}-\\frac{1}{4}-\\frac{1}{6}-\\frac{1}{8}\\right) \\cdot 24 = \\frac{3}{24} \\cdot 24 = 3\n$$\ntekmovalci.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41998,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVsi liki na slikah so omejeni s polkrožnimi loki, pri čemer je največji polkrožni lok pri vseh likih enak. Kateri izmed likov z najmanjšim obsegom ima največjo ploščino?\n(A)\n![](attached_image_1.png)\n(B)\n![](attached_image_2.png)\n(C)\n![](attached_image_3.png)\n(D)\n![](attached_image_4.png)\n(E)\n![](attached_image_5.png)",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nSkupna dolžina polkrožnih lokov nad daljico dolžine $d$ je enaka $\\frac{\\pi d}{2}$ neglede na to, koliko je teh polkrožnih lokov in kako veliki so (glej primer na sliki).\n![](attached_image_6.png)\nTorej je obseg vseh 5 likov enak, saj je enak $2 \\cdot \\frac{\\pi d}{2}$, kjer je $d$ premer največjega krožnega loka. Največjo ploščino izmed vseh likov ima lik (C), saj je ta edini, ki ima ploščino večjo od ploščine polkroga omejenega z največjim krožnim lokom.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 41999,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV katerih točkah na krivulji, podani $z$ enačbo $f(x)=x^{3}-2 x^{2}+3$, tangenta $z$ abscisno osjo oklepa kot $135^{\\circ}$?\n\n(A) $T_{1}(1,-4)$ in $T_{2}(-2,0)$.\n(B) $T_{1}(1,-4)$ in $T_{2}(2,0)$.\n(C) $T_{1}(1,0)$ in $T_{2}(-1,4)$.\n(D) $T_{1}(1,2)$ in $T_{2}\\left(\\frac{1}{3}, \\frac{28}{9}\\right)$\n(E) $T_{1}(1,2)$ in $T_{2}\\left(\\frac{1}{3}, \\frac{76}{27}\\right)$.",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nVrednost odvoda funkcije $f$ v iskanih točkah mora biti enaka tangensu naklonskega kota tangente $v$ teh točkah $f'(x)=\\tan 135^{\\circ}=-1$. Rešimo enačbo $3 x^{2}-4 x=-1$. Enačbo uredimo in izračunamo abscisi iskanih točk $x_{1}=1$ in $x_{2}=\\frac{1}{3}$. Izračunamo funkcijski vrednosti $f\\left(x_{1}\\right)=$ $f(1)=2$ in $f\\left(x_{2}\\right)=f\\left(\\frac{1}{3}\\right)=\\frac{76}{27}$. Zapišemo točki $T_{1}(1,2)$ in $T_{2}\\left(\\frac{1}{3}, \\frac{76}{27}\\right)$. Pravilen je odgovor E.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42000,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj za števili $x$ in $y$ velja zveza $(x+2y)^2 - 3y(y-1) = (2x+y)^2 - 3x(x+1)$. Kolikšna je njuna vsota $x+y$?\n\n(A) $-100$\n(B) $-1$\n(C) $0$\n(D) $1$\n(E) se ne da enolično določiti",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nPo kvadriranju in odpravi oklepajev dobimo $x^2 + 4xy + 4y^2 - 3y^2 + 3y = 4x^2 + 4xy + y^2 - 3x^2 - 3x$. Enačba se preoblikuje v enačbo $3y = -3x$, oziroma $x + y = 0$. Pravilen je odgovor C.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42001,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPravilna, pokončna, štiristrana prizma ima površino $128~\\mathrm{m}^2$ in višino $6~\\mathrm{m}$. Koliko meri njena telesna diagonala?\n\n(A) $D \\doteq 8,23~\\mathrm{m}$\n(B) $D \\doteq 8,24~\\mathrm{m}$\n(C) $D \\doteq 7,22~\\mathrm{m}$\n(D) $D \\doteq 8,25~\\mathrm{m}$\n(E) $D \\doteq 7,21~\\mathrm{m}$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nS pomočjo obrazca za izračun površine pravilne štiristrane prizme $P=2 a^{2}+4 a v$, dobimo kvadratno enačbo $128=2 a^{2}+24 a$. Rešitvi kvadratne enačbe sta $a=4$ in $a=-16$. Ugotovimo, da je pravilna rešitev $a=4$. Osnovni rob prizme meri $a=4~\\mathrm{m}$. Z obrazcem $D=\\sqrt{2 a^{2}+v^{2}}$ izračunamo dolžino telesne diagonale $D \\doteq 8,25~\\mathrm{m}$. Pravilen je odgovor $\\mathrm{D}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42002,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKolikšno vrednost mora imeti število $a \\in \\mathbb{R}$, $a \\neq 4$ in $a \\neq 2$, da se bosta premici z enačbama $a x-(a-2) y-2=0$ in $(a-1) x+(4-a) y+2=0$ sekali na ordinatni osi?\n\n(A) -5\n(B) 5\n(C) -1\n(D) -3\n(E) 3",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nČe se premici sekata na ordinatni osi, je $x$ koordinata presečišča enaka $0$. V obe enačbi premice vstavimo za $x=0$ ter ju poenostavimo do npr. $-a y+2 y-2=0$ in $4 y-a y+2=0$. Dobimo sistem dveh enačb z dvema neznankama, ki ga rešimo na katerikoli način. Rešitvi sta $y=-2$ in $a=3$. Pravilen odgovor je E.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42003,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nAna in Meta sta se hkrati odpeljali iz vasi Zabukovje v vas Zahrastje, Ana s kolesom in Meta z avtom. Ana je vozila s konstantno hitrostjo $30~\\mathrm{km}/\\mathrm{h}$, Meta pa s konstantno hitrostjo $70~\\mathrm{km}/\\mathrm{h}$. Ko je Meta prišla v Zahrastje, je bila tam $1~\\mathrm{h}$, nato pa se je z enako hitrostjo $70~\\mathrm{km}/\\mathrm{h}$ odpeljala nazaj v Zabukovje. Na poti nazaj je srečala Ano $105~\\mathrm{km}$ od Zahrastja. Koliko kilometrov je razdalja med vasema Zabukovje in Zahrastje?\n(A) 262,5\n(B) 300\n(C) 315\n(D) 345\n(E) 375",
    "options": [],
    "answer": "C",
    "solution": "Solution:\nOznačimo z $x$ razdaljo, ki jo je prevozila Ana do srečanja z Meto. Potem je Meta do srečanja z Ano prevozila $x + 2 \\cdot 105~\\mathrm{km}$. Če čas, ki je pretekel do srečanja, zapišemo z Aninega in Metinega stališča v urah, dobimo\n$$\n\\frac{x}{30} = \\frac{x + 2 \\cdot 105}{70} + 1\n$$\nod koder sledi, da je $x = 210$ km. Razdalja med vasema Zabukovje in Zahrastje je $210 + 105 = 315~\\mathrm{km}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42004,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLucijana je iz množice števil $\\{1,2,3,4,5,6,7,8,9\\}$ izbrala 3 različna števila. Z njimi je zapisala največje možno trimestno število in najmanjše možno trimestno število. Dobljeni števili je seštela in dobila število 545. Koliko je vsota 3 števil, ki jih je izbrala Lucijana?\n\n(A) 6\n(B) 7\n(C) 9\n(D) 11\n(E) 13",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nNaj bodo $a, b$ in $c$ števila, ki jih je izbrala Lucijana, pri čemer je $a > b > c$. Največje možno število, ki ga z njimi lahko zapiše, je tedaj $\\overline{abc}$, najmanjše možno število pa $\\overline{cba}$. Za njuno vsoto velja, da je $100a + 10b + c + 100c + 10b + a = 101(a + c) + 20b = 545$. Če primerjamo enice na obeh straneh enačbe, vidimo, da je $a + c = 5$ ali $a + c = 15$, če pa primerjamo še stotice na obeh straneh enačbe, vidimo, da je $a + c = 5$. Torej je $b = 2$, $a = 4$ in $c = 1$ ter $a + b + c = 7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42005,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVsota števk petmestnega števila je $44$. Koliko je produkt števk tega petmestnega števila?\n\n(A) $2^{3} \\cdot 3^{8}$\n(B) $2^{3} \\cdot 9^{3}$\n(C) $8 \\cdot 4^{9}$\n(D) $8 \\cdot 3^{4}$\n(E) Nič od predhodno naštetega.",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nVsota $5$ enomestnih števil je enaka $44$ samo, če je eno od teh števil enako $8$ in so ostala štiri števila enaka $9$, torej $8 + 9 + 9 + 9 + 9 = 44$. Produkt je zato enak $8 \\cdot 9^{4} = 2^{3} \\cdot 3^{8}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42006,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPodana imamo tri števila $A=26^{351}$, $B=5^{702}$ in $C=3^{1053}$. Števila uredi po velikosti. Kateri spodnji zapis je pravilen?\n\n(A) $A<B<C$\n(B) $A<C<B$\n(C) $B<C<A$\n(D) $B<A<C$\n(E) $C<A<B$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\n$B=5^{702}=(5^{2})^{351}=25^{351}$, $C=(3^{3})^{351}=27^{351}$. Ker je $25^{351}<26^{351}<27^{351}$, dobimo $B<A<C$. Pravilen je odgovor $\\mathrm{D}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42007,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa neko celo število $x$ je končno zaporedje $\\sqrt{x}+2, 3 \\sqrt{x+1}, 2 \\sqrt{x}+4$ geometrijsko. Kolikšen je količnik tega zaporedja?\n\n(A) 2\n(B) $\\sqrt{3}$\n(C) $\\sqrt{2}$\n(D) 3\n(E) $3 \\sqrt{2}$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nUpoštevamo zvezo med zaporednimi členi geometrijskega zaporedja. Rešimo iracionalno enačbo in dobimo rešitvi $x_{1}=1$ in $x_{2}=\\frac{1}{49}$. Edina celoštevilska rešitev je $1$. Nato izračunamo člene zaporedja $a_{1}=3$, $a_{2}=3 \\sqrt{2}$, $a_{3}=6$. Izračunamo količnik $q=\\sqrt{2}$. Pravilen je odgovor C.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42008,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa katere vrednosti parametra $a \\in \\mathbb{R}$ je funkcija $f(x)=\\log \\left(x^{2}+(a+4) x+9\\right)$ definirana na množici vseh realnih števil?\n\n(A) $a<2$\n(B) $a>-10$\n(C) $-2<a<10$\n(D) $-10<a<2$\n(E) $a=-4$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nUgotovimo, da bo definicijsko območje dane funkcije $f$ množica vseh realnih števil ob pogoju $x^{2}+(a+4) x+9>0$. To pa bo izpolnjeno, če bo $D<0$. Poenostavimo levo stran neenačbe:\n$$\nD=b^{2}-4 a c=(a+4)^{2}-4 \\cdot 1 \\cdot 9=a^{2}+8 a-20=(a+10)(a-2)\n$$\nRešimo kvadratno neenačbo $(a+10)(a-2)<0$. Rešitvi kvadratne enačbe $(a+10)(a-2)=0$ sta $a_{1}=-10$ in $a_{2}=2$. Rešitev kvadratne neenačbe pa usterza intervalu $-10<a<2$. Pravilen je odgovor D.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42009,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKvadrat s stranico dolžine $2$ je razdeljen na $4$ trikotnike (glej sliko). Vsi $3$ osenčeni trikotniki imajo enako ploščino. Koliko je ploščina belega trikotnika?\n\n(A) $\\frac{1+\\sqrt{5}}{2}$\n(B) $\\frac{8}{5}$\n(C) $2$\n(D) $3 \\sqrt{5}-5$\n(E) $6-2 \\sqrt{5}$\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nOznačimo z $A, B, C$ in $D$ oglišča kvadrata in dodatno z $E$ in $F$ oglišči belega trikotnika (glej sliko).\n\n![](attached_image_2.png)\n\nOznačimo $x=|AE|$. Ker imata pravokotna trikotnika $AED$ in $FCD$ eno od stranic enako stranici kvadrata in imata enaki ploščini, je $|CF|=|AE|=x$. Če upoštevamo, da imata tudi trikotnika $AED$ in $EBF$ enaki ploščini, dobimo\n$$\n\\frac{2 \\cdot x}{2} = \\frac{(2-x) \\cdot (2-x)}{2}.\n$$\nEnačbo preoblikujemo v $x^2 - 6x + 4 = 0$ in jo rešimo kot kvadratno enačbo, da dobimo $x = 3 \\pm \\sqrt{5}$. Ker mora biti $x < 2$, je prava rešitev $x = 3 - \\sqrt{5}$.\n\nPloščina belega trikotnika je zato enaka\n$$\np = 2^2 - 3 \\cdot \\frac{2 \\cdot (3 - \\sqrt{5})}{2} = 3 \\sqrt{5} - 5\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42010,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKvadrat s stranico dolžine $2$ je razdeljen na $4$ trikotnike, od teh sta $2$ trikotnika enakokraka (glej sliko). Ploščina enega od enakokrakih trikotnikov je dvakrat tolikšna, kot je ploščina drugega enakokrakega trikotnika. Koliko je ploščina osenčenega trikotnika?\n\n![](attached_image_1.png)\n\n(A) $\\frac{2}{3}$\n\n(B) $\\frac{\\sqrt{2}}{2}$\n\n(C) $\\frac{8}{13}$\n\n(D) $2-\\sqrt{2}$\n\n(E) $\\frac{3-\\sqrt{2}}{2}$",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nNarišemo diagonalo kvadrata in označimo z $A, B, C, D, E$ in $F$ nekatere točke na kvadratu (glej sliko).\n\n![](attached_image_2.png)\n\nOznačimo z $x=|B F|$. Ker je ploščina trikotnika $A C E$ dvakrat tolikšna kot ploščina trikotnika $A B C$, je $|E F|=2|B F|=2 x$. Trikotnik $B F C$ je pravokoten in enakokrak, zato je $|B F|=|F C|=|F A|=x$. Torej je ploščina trikotnika $A B C$ enaka $\\frac{2 x \\cdot x}{2}=x^{2}$, ploščina trikotnika $A C E$ pa je enaka $\\frac{2 x \\cdot 2 x}{2}=2 x^{2}$. Ker je $3 x=|B E|=2 \\sqrt{2}$, je $x=\\frac{2 \\sqrt{2}}{3}$, ploščina osenčenega trikotnika pa je enaka\n\n$$\np=\\frac{1}{2}\\left(2^{2}-x^{2}-2 x^{2}\\right)=\\frac{1}{2}\\left(4-\\frac{8}{9}-\\frac{16}{9}\\right)=\\frac{2}{3}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42011,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDiagonali $AC$ in $BD$ trapeza $ABCD$ se sekata v točki $E$ in razdelita trapez na 4 trikotnike s ploščinami $25~\\mathrm{cm}^2$, $36~\\mathrm{cm}^2$, $X~\\mathrm{cm}^2$ in $X~\\mathrm{cm}^2$ (glej sliko). Kolikšna je vrednost $X$?\n(A) 25\n(B) 30\n(C) 32\n(D) 36\n(E) 61\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nKer imata trikotnika $AED$ in $ABE$ enaki višini iz točke $A$, velja $X : 36 = |DE| : |EB|$. Podobno imata trikotnika $CDE$ in $CEB$ enaki višini iz točke $C$, zato velja $25 : X = |DE| : |EB|$. Sledi $X : 36 = 25 : X$, od koder izračunamo $X = \\sqrt{25 \\cdot 36} = 30$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42012,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKateri izraz je ekvivalenten izrazu\n$$\n\\frac{\\sin^{2} x-\\cos^{2} x}{\\sin^{-1} x+\\cos^{-1} x} \\cdot \\frac{\\tan x}{\\cos x-\\sin x}\n$$\n?\n(A) $-\\cos x$\n(B) $-\\sin x$\n(C) $-\\sin^{2} x$\n(D) $-\\cos^{2} x$\n(E) $\\cos x$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\nŠtevec prvega ulomka razstavimo kot razliko kvadratov $\\sin^{2} x-\\cos^{2} x=(\\sin x-\\cos x)(\\sin x+\\cos x)$, imenovalec pa zapišemo kot vsoto obratnih vrednosti $\\sin^{-1} x+\\cos^{-1} x=\\frac{1}{\\sin x}+\\frac{1}{\\cos x}$ in dobljena ulomka seštejemo. Števec drugega ulomka zapišemo $\\tan x=\\frac{\\sin x}{\\cos x}$, v imenovalcu drugega ulomka pa izpostavimo $\\cos x-\\sin x=-(\\sin x-\\cos x)$. Sledi krajšanje ulomkov in dobimo rezultat $-\\sin^{2} x$. Pravilen je odgovor C.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42013,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV pravokotnem trikotniku $ABC$ s pravim kotom pri $C$, višina iz oglišča $C$ seka stranico $AB$ v točki $D$. Kolikšna je velikost kota $\\angle BAC$, če velja $|CD| = \\sqrt{12}\\ \\mathrm{cm}$ in $|AD| = 4\\ \\mathrm{cm}$?\n\n(A) $\\alpha \\doteq 49,06^\\circ$\n(B) $\\alpha \\doteq 40,89^\\circ$\n(C) $\\alpha \\doteq 40,54^\\circ$\n(D) $\\alpha \\doteq 49,11^\\circ$\n(E) $\\alpha \\doteq 40,45^\\circ$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nUporabimo kotne funkcije v pravokotnem trikotniku $\\tan \\alpha = \\frac{v_c}{b_1} = \\frac{\\sqrt{12}}{4}$ in izračunamo kot $\\alpha \\doteq 40,89^\\circ$. Pravilen je odgovor B.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42014,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $x \\in \\mathbb{R}$ in $x>0$, $x \\neq 4$, $x \\neq 9$. Kateri izraz je ekvivalenten izrazu $\\left((2+\\sqrt{x})^{-1}+1\\right)^{-1}$?\n\n(A) $\\frac{x+\\sqrt{x}-6}{x-9}$\n\n(B) $\\frac{x-\\sqrt{x}-6}{x-9}$\n\n(C) $\\frac{x-\\sqrt{x}+6}{x-9}$\n\n(D) $\\frac{x-\\sqrt{x}-6}{x-4}$\n\n(E) $\\frac{x-\\sqrt{x}+6}{x-4}$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\n$\\left((2+\\sqrt{x})^{-1}+1\\right)^{-1}$ preoblikujemo v $\\left(\\frac{1}{2+\\sqrt{x}}+1\\right)^{-1}=\\left(\\frac{3+\\sqrt{x}}{2+\\sqrt{x}}\\right)^{-1}=\\frac{2+\\sqrt{x}}{3+\\sqrt{x}}$, $x \\neq 9$, $x \\neq 4$. Izraz še racionaliziramo in dobimo $\\frac{x-\\sqrt{x}-6}{x-9}$. Pravilen je odgovor B.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42015,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDolžina akvarija je $50~\\mathrm{cm}$, širina $20~\\mathrm{cm}$ in višina $25~\\mathrm{cm}$. Koliko $\\mathrm{cm}$ od zgornjega roba akvarija bo nivo vode, če vanj vlijemo $19$ litrov vode?\n\n(A) $19~\\mathrm{cm}$\n(B) $1{,}9~\\mathrm{cm}$\n(C) $10{,}9~\\mathrm{cm}$\n(D) $6~\\mathrm{cm}$\n(E) $0{,}6~\\mathrm{cm}$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nUgotovimo, da je prostornina vode $19~\\mathrm{dm}^3$. Zapišemo podatke v obrazec za prostornino vode ($x$ je višina vode v akvariju) in dobimo enačbo $V = 5 \\cdot 2 \\cdot x = 19$. Izračunamo $x = 1{,}9~\\mathrm{dm} = 19~\\mathrm{cm}$. Nivo vode bo torej $6~\\mathrm{cm}$ nižje od zgornjega roba akvarija. Pravilen je odgovor D.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42016,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDokaži, da ne obstajata naravni števili $a$ in $b$, za kateri velja $\\sqrt{a}+\\sqrt{b}=\\sqrt{2021}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDenimo, da taki naravni števili obstajata. Tedaj iz dane enakosti očitno sledi $a, b<2021$. Enakost preoblikujemo v $\\sqrt{a}=\\sqrt{2021}-\\sqrt{b}$ in jo kvadriramo, da dobimo $a=2021-2 \\sqrt{2021 b}+b$. Ker sta $a$ in $b$ naravni števili, je $2 \\sqrt{2021 b}$ celo število in zato je $\\sqrt{2021 b}$ racionalno število. Spomnimo se, da je za naravno število $n$ število $\\sqrt{n}$ racionalno natanko takrat, ko je število $n$ popoln kvadrat. Od tod sledi, da je $2021 b$ popoln kvadrat. Ker pa je $2021=43 \\cdot 47$ in sta $43$ in $47$ praštevili, mora biti $b=43 \\cdot 47 \\cdot k^{2}=2021 k^{2}$ za neko naravno število $k$. Sledi $b \\geq 2021$, kar pa je protislovje. Taki naravni števili $a$ in $b$ torej ne obstajata.\n\n2. način. Ker sta $a$ in $b$ naravni števili, iz dane enakosti sledi $a, b<2021$. Enakost kvadriramo, da dobimo $a+2 \\sqrt{a b}+b=2021$. Nato jo preoblikujemo v $2 \\sqrt{a b}=2021-a-b$ in ponovno kvadriramo, da dobimo $4 a b=(2021-a-b)^{2}$. Leva stran enakosti je deljiva s $4$, torej mora biti tudi desna stran deljiva s $4$. To pomeni, da je $2021-a-b$ sodo število in zato je $\\sqrt{a b}=\\frac{2021-a-b}{2}$ naravno število. Začetno enakost sedaj pomnožimo s $\\sqrt{b}$, da dobimo $\\sqrt{a b}+b=\\sqrt{2021 b}$. Ker je po pravkar dokazanem leva stran enakosti naravno število, mora biti tudi $\\sqrt{2021 b}$ naravno število in zato je $2021 b$ popoln kvadrat. Od tod na enak način kot v prvi rešitvi pridemo do protislovja.\n\n3. način. Ker sta $a$ in $b$ naravni števili, je $a, b<2021$. Prvotno enačbo kvadriramo in dobimo $a+2 \\sqrt{a b}+b=2021$. To enačbo preoblikujemo v $2 \\sqrt{a b}=2021-a-b$ in jo ponovno kvadriramo. Dobljeno enačbo preoblikujemo v $(a-b)^{2}=2021(2 a+2 b-2021)=43 \\cdot 47 \\cdot(2 a+2 b-2021)$. Ker sta $43$ in $47$ praštevili sledi, da $43 \\cdot 47 \\mid a-b$.\nČe je $a-b=0$, je $a=b$ in sledi $4 a-2021=0$, kar pa ni mogoče, saj je $4 a$ sodo število, $2021$ pa liho.\nSledi $a-b \\neq 0$. Ampak potem iz $2021 \\mid a-b$ sledi, da je $|a-b| \\geq 2021$. Ker pa sta $a, b \\in \\mathbb{N}$ sledi, da je vsaj eno izmed števil $a$ in $b$ večje od $2021$. To pa je v protislovju s sklepom, da sta $a, b<2021$.\nTorej ne obstajata naravni števili $a$ in $b$, ki bi rešili prvotno enačbo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42017,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSestre Neža, Meta in Ajda so med počitniškim delom v mesecu juliju zaslužile denar v razmerju $3:4:5$. V mesecu avgustu je Neža zaslužila $40\\%$ več kot v mesecu juliju, Meta petino manj kot v juliju, Ajda pa za $200$ manj kot v juliju. Skupaj so sestre v avgustu zaslužile $2280$. Kakšen je bil zaslužek vsake od sester v mesecu juliju?",
    "options": [],
    "answer": "Neža 600, Meta 800, Ajda 1000",
    "solution": "Solution:\n\n1. Zapišemo zaslužek Neže v juliju $3x$, zaslužek Mete v juliju $4x$, zaslužek Ajde v juliju $5x$.\n\nZapis zaslužka Neže v avgustu je $3x \\cdot 1.4 = 4.2x$.\n\nZapis zaslužka Mete v avgustu je $4x \\cdot 0.8 = 3.2x$.\n\nZapis zaslužka Ajde v avgustu je $5x - 200$.\n\nIzračunamo skupen zaslužek vseh treh sester v avgustu:\n$$\n4.2x + 3.2x + 5x - 200 = 2280\n$$\nPreoblikujemo:\n$$\n12.4x = 2480\n$$\nDobimo $x = 200$.\n\nV juliju je Neža zaslužila $3 \\cdot 200 = 600$, Meta $4 \\cdot 200 = 800$ in Ajda $5 \\cdot 200 = 1000$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42018,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZaporedje $\\{a_{n}\\}_{n \\in \\mathbb{N}}$ je podano s prvim členom $a_{1}=3$ in rekurzivno zvezo $(3-a_{n+1})(6+a_{n})=18$ za vse $n \\geq 1$. Dokaži, da za vsako naravno število $n$ velja\n$$\n\\sum_{k=1}^{n} \\frac{1}{a_{k}}=\\frac{2^{n+1}}{3}-\\frac{n+2}{3}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nZa vsak $n \\geq 1$ iz rekurzivne zveze izrazimo $a_{n+1}=3-\\frac{18}{6+a_{n}}=\\frac{3 a_{n}}{6+a_{n}}$. Od tod sledi\n$$\n\\frac{1}{a_{n+1}}=\\frac{a_{n}+6}{3 a_{n}}=\\frac{1}{3}+\\frac{2}{a_{n}}\n$$\nTo zvezo uporabimo večkrat zapored, da izpeljemo\n$$\n\\begin{aligned}\n\\frac{1}{a_{n+1}} & =\\frac{1}{3}+\\frac{2}{a_{n}}=\\frac{1}{3}+2\\left(\\frac{1}{3}+\\frac{2}{a_{n-1}}\\right)=\\frac{1}{3}+\\frac{2}{3}+4\\left(\\frac{1}{3}+\\frac{2}{a_{n-2}}\\right)=\\ldots \\\\\n& =\\frac{1}{3}+\\frac{2}{3}+\\frac{4}{3}+\\frac{8}{3}+\\ldots+2^{n-1}\\left(\\frac{1}{3}+\\frac{2}{a_{1}}\\right)=\\frac{1}{3}+\\frac{2}{3}+\\frac{4}{3}+\\frac{8}{3}+\\ldots+\\frac{2^{n-1}}{3}+\\frac{2^{n}}{a_{1}}\n\\end{aligned}\n$$\nZ upoštevanjem vrednosti $a_{1}=3$ ter formule za vsoto členov geometrijskega zaporedja dobimo\n$$\n\\frac{1}{a_{n+1}}=\\frac{1}{3}\\left(1+2+2^{2}+2^{3}+\\ldots+2^{n}\\right)=\\frac{1}{3}\\left(2^{n+1}-1\\right)\n$$\nod koder sledi $\\frac{1}{a_{k}}=\\frac{1}{3}\\left(2^{k}-1\\right)$ za vse $k \\geq 2$, ta formula pa očitno velja tudi za $k=1$. Torej je\n$$\n\\sum_{k=1}^{n} \\frac{1}{a_{k}}=\\sum_{k=1}^{n} \\frac{1}{3}\\left(2^{k}-1\\right)=\\frac{1}{3} \\sum_{k=1}^{n} 2^{k}-\\frac{1}{3} \\sum_{k=1}^{n} 1=\\frac{1}{3} \\cdot 2 \\cdot \\frac{2^{n}-1}{2-1}-\\frac{1}{3} \\cdot n=\\frac{2^{n+1}}{3}-\\frac{n+2}{3}\n$$\npri čemer smo ponovno uporabili formulo za vsoto členov geometrijskega zaporedja.\n\n\n2. način. Trditev bomo pokazali z matematično indukcijo. Če je $n=1$, je leva stran enakosti enaka $\\sum_{k=1}^{1} \\frac{1}{a_{k}}=\\frac{1}{a_{1}}=\\frac{1}{3}$, desna stran pa $\\frac{4}{3}-\\frac{1}{3}-\\frac{2}{3}=\\frac{1}{3}$. Ker sta obe vrednoti enaki, je baza indukcije izpolnjena. Denimo sedaj, da dana enakost velja za neko naravno $n$. Iz rekurzivne zveze zaporedja izrazimo $\\frac{1}{a_{n+1}}=\\frac{1}{3}+\\frac{2}{a_{n}}$. S pomočjo te zveze izpeljemo in vrednosti $a_{1}=3$ izpeljemo\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n+1} \\frac{1}{a_{k}} & =\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\frac{1}{a_{3}}+\\ldots+\\frac{1}{a_{n+1}}=\\frac{1}{3}+\\left(\\frac{1}{3}+\\frac{2}{a_{1}}\\right)+\\left(\\frac{1}{3}+\\frac{2}{a_{2}}\\right)+\\ldots+\\left(\\frac{1}{3}+\\frac{2}{a_{n}}\\right)= \\\\\n& =(n+1) \\cdot \\frac{1}{3}+2\\left(\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\frac{1}{a_{3}}+\\ldots+\\frac{1}{a_{n}}\\right)=\\frac{n+1}{3}+2 \\sum_{k=1}^{n} \\frac{1}{a_{k}}\n\\end{aligned}\n$$\nPo indukcijski predpostavki je torej\n$$\n\\sum_{k=1}^{n+1} \\frac{1}{a_{k}}=\\frac{n+1}{3}+2\\left(\\frac{2^{n+1}}{3}-\\frac{n+2}{3}\\right)=\\frac{2^{n+2}}{3}-\\frac{n+3}{3}\n$$\nTorej dana enakost velja tudi za število $n+1$. S tem smo dokazali indukcijksi korak. Po principu matematične indukcije, je enakost izpolnjena za vsa naravna števila $n$.\n\n\n3. način. Iz rekurzivne zveze izrazimo $a_{n+1}=\\frac{3 a_{n}}{6+a_{n}}$ za vse $n \\geq 1$. Izračunajmo prvih nekaj členov zaporedja:\n$$\n\\begin{aligned}\na_{1} & =3 \\\\\na_{2} & =\\frac{3 \\cdot 3}{3+6}=1 \\\\\na_{3} & =\\frac{3 \\cdot 1}{1+6}=\\frac{3}{7}=\\frac{3}{8-1} \\\\\na_{4} & =\\frac{3 \\cdot \\frac{3}{7}}{\\frac{3}{7}+6}=\\frac{9}{45}=\\frac{3}{15}=\\frac{3}{16-1}\n\\end{aligned}\n$$\nUganemo, da je $a_{n}=\\frac{3}{2^{n}-1}$ in to dokažemo z indukcijo. Za $n=1$ je $a_{1}=3$ in $\\frac{3}{2^{1}-1}=3$, torej zveza velja. Predpostavimo, da je $a_{n}=\\frac{3}{2^{n}-1}$ in izračunajmo\n$$\na_{n+1}=\\frac{3 a_{n}}{6+a_{n}}=\\frac{3 \\frac{3}{2^{n}-1}}{6+\\frac{3}{2^{n}-1}}=\\frac{\\frac{3}{2^{n}-1}}{\\frac{2\\left(2^{n}-1\\right)+1}{2^{n}-1}}=\\frac{3}{2^{n+1}-1}\n$$\nVidimo, da zveza velja tudi za $n+1$, če velja za $n$, in s tem je indukcija zaključena. Če $a_{k}=\\frac{3}{2^{k}-1}$ vstavimo v iskano vsoto, dobimo\n$$\n\\sum_{k=1}^{n} \\frac{1}{a_{k}}=\\sum_{k=1}^{n} \\frac{1}{3}\\left(2^{k}-1\\right)=\\frac{1}{3} \\sum_{k=1}^{n} 2^{k}-\\frac{1}{3} \\sum_{k=1}^{n} 1=\\frac{1}{3} \\cdot 2 \\cdot \\frac{2^{n}-1}{2-1}-\\frac{1}{3} \\cdot n=\\frac{2^{n+1}}{3}-\\frac{n+2}{3}\n$$\npri čemer smo uporabili formulo za vsoto členov geometrijskega zaporedja.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42019,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa cela števila $a$, za katera je tudi $\\log_{2}\\left(a^{2}-4a-1\\right)$ celo število.",
    "options": [],
    "answer": "a = -1 and a = 5",
    "solution": "Solution:\n\nOznačimo $\\log_{2}\\left(a^{2}-4a-1\\right)=n$, kjer je $n$ celo število. Potem je $a^{2}-4a-1=2^{n}$ oziroma $a^{2}-4a-(1+2^{n})=0$. To je kvadratna enačba za $a$, ki ima rešitvi\n$$\na_{1}=2+\\sqrt{5+2^{n}}, \\quad a_{2}=2-\\sqrt{5+2^{n}}.\n$$\nEnačba ima celoštevilsko rešitev le v primeru, ko je $\\sqrt{5+2^{n}}$ celo število. Naj bo $\\sqrt{5+2^{n}}=k$ oziroma $5+2^{n}=k^{2}$ in ločimo tri primere glede na predznak števila $n$.\n\nČe je $n<0$, enakost preuredimo v $2^{n}=k^{2}-5$ in opazimo, da je na desni strani celo število, na levi pa ne. Zato v tem primeru nimamo rešitev.\n\nČe je $n=0$, sledi $k^{2}=6$, kar pa ni mogoče, saj $6$ ni kvadrat celega števila.\n\nČe je $n>0$, je število $5+2^{n}$ liho, zato mora biti tudi $k$ liho število. Pišimo $k=2m-1$, kjer je $m$ celo število. Enačbo $5+2^{n}=(2m-1)^{2}$ lahko preuredimo do\n$$\nm(m-1)=2^{n-2}+1.\n$$\nOpazimo, da je $m(m-1)$ sodo celo število, zato mora biti tako tudi število $2^{n-2}+1$. Pri $n=1$ število $2^{n-2}+1$ ni celo, pri $n>2$ pa ni sodo. Ostane nam torej le možnost $n=2$. V tem primeru je $a_{1}=5$ in $a_{2}=-1$. Edini rešitvi naloge sta torej $a=-1$ in $a=5$. V obeh primerih je $\\log_{2}\\left(a^{2}-4a-1\\right)=2$.\n\n\n2. način.\n\nOznačimo $\\log_{2}\\left(a^{2}-4a-1\\right)=n$, kjer je $n$ celo število. Potem je $a^{2}-4a-1=2^{n}$. Ker je leva stran enakosti celo število, mora biti tudi na desna stran celo število, torej je $n \\geq 0$. Če je $n=0$, sledi $a^{2}-4a-2=0$. Ta enačba nima celoštevilskih rešitev, saj njena diskriminanta $D=16+8=24$ ni popoln kvadrat. Torej je $n \\geq 1$, kar pomeni, da je $2^{n}$ sodo število. Iz enakosti $a^{2}-4a-1=2^{n}$ zato sledi, da mora biti $a$ liho število. Pišimo $a=2k+1$, kjer je $k$ celo število. Če to vstavimo v zadnjo enakost in enakost poenostavimo, dobimo\n$$\nk^{2}-k-1=2^{n-2}.\n$$\nKer je $k^{2}-k=k(k-1)$ sodo število, je leva stran enakosti liho število, zato mora biti tudi desna stran liho število. To pomeni, da je $n=2$ in zato $k^{2}-k-2=0$. Slednja enačba ima rešitvi $k=2$ in $k=-1$. V prvem primeru je $a=5$, v drugem pa $a=-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42020,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNad stranico $CD$ kvadrata $ABCD$ z zunanje stani narišemo pravokotni trikotnik $DCE$ s pravim kotom pri $E$. Dokaži, da simetrala kota $\\angle DEC$ razdeli kvadrat $ABCD$ na dva ploščinsko enaka dela.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOznačimo z $G$ presečišče diagonal kvadrata $ABCD$. Ker je $\\angle DEC=90^\\circ=\\angle CGD$, po Talesovem izreku točke $G, C, E$ in $D$ ležijo na isti krožnici $\\mathcal{K}$. Točka $G$ zaradi simetrije razpolavlja lok $\\overparen{DC}$ krožnice $\\mathcal{K}$. Ker tudi simetrala obodnega kota $\\angle DEC$ razpolavlja lok $\\overparen{DC}$, točka $G$ leži na simetrali kota $\\angle DEC$. Od tod sledi, da sta trikotnika $AHG$ in $CFG$ skladna, ker imata enake notranje kote in enako dolžino stranice $|AG|=|CG|$ nasproti enakega kota. Torej je $|AH|=|CF|$.\n\nPloščina levega dela kvadrata, ki ga odreže simetrala kota $\\angle DEC$, je zato po formuli za ploščino pravokotnega trapeza enaka\n$$\np_{AHFD}=\\frac{|AH|+|DF|}{2} \\cdot|AD|=\\frac{|CF|+|DF|}{2} \\cdot|AD|=\\frac{|CD||AD|}{2}\n$$\nkar je ravno polovica ploščine celotnega kvadrata $ABCD$. Torej simetrala kota $\\angle DEC$ razdeli kvadrat $ABCD$ na dva ploščinsko enaka dela.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42021,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV pokončno prizmo, katere osnovna ploskev je paralelogram z enim notranjim kotom $\\alpha$, je včrtana sfera z radijem $r$, ki se dotika vseh mejnih ploskev prizme. Izrazi prostornino prizme s pomočjo $r$ in $\\alpha$.",
    "options": [],
    "answer": "8 r^3 / sin α",
    "solution": "![](attached_image_1.png)\n\nKer se sfera dotika spodnje in zgornje ploskve prizme, je višina prizme enaka premeru sfere, torej $h=2r$.\n\nNaj bo $\\Sigma$ ravnina, ki vsebuje središče sfere in je vzporedna osnovni ploskvi prizme. Označimo s $\\mathcal{P}$ presečišče prizme in ravnine $\\Sigma$ ter s $\\mathcal{K}$ presečišče sfere in ravnine $\\Sigma$, kot je prikazano na levi sliki. Tedaj je $\\mathcal{P}$ paralelogram, skladen z osnovno ploskvijo prizme z notranjim kotom $\\alpha$, in $\\mathcal{K}$ krožnica z radijem $r$.\n\nDotikališča sfere s stranskimi ploskvami prizme ležijo vsa na krožnici $\\mathcal{K}$ in hkrati tudi na robu paralelograma $\\mathcal{P}$. To pomeni, da je krožnica $\\mathcal{K}$ včrtana paralelogramu $\\mathcal{P}$. Obe višini paralelograma $\\mathcal{P}$ sta torej enaki $v=2r$, zato je $\\mathcal{P}$ romb.\n\nPrivzemimo najprej, da je $\\alpha$ manjši od notranjih kotov romba $\\mathcal{P}$ in označimo stranico romba z $a$, kot je prikazano na desni sliki. Tedaj iz pravokotnega trikotnika dobimo $\\sin \\alpha = \\frac{2r}{a}$ oziroma $a = \\frac{2r}{\\sin \\alpha}$, ploščina romba $\\mathcal{P}$ pa je enaka $p = a \\cdot v = \\frac{2r}{\\sin \\alpha} \\cdot 2r = \\frac{4r^2}{\\sin \\alpha}$.\n\nProstornina prizme je zato enaka $V = p \\cdot h = \\frac{4r^2}{\\sin \\alpha} \\cdot 2r = \\frac{8r^3}{\\sin \\alpha}$.\n\nKer pa je drugi notranji kot paralelograma enak $\\beta = \\pi - \\alpha$ in velja $\\sin \\beta = \\sin (\\pi - \\alpha) = \\sin \\alpha$, je ta formula za prostornino prizme pravilna tudi, če je $\\alpha$ večji izmed notranjih kotov osnovne ploskve prizme.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42022,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bosta $a$ in $b$ taki realni števili, da ima polinom $p(x)=x^{2}+a x+b$ dve realni ničli, polinom $p\\left(q(x)\\right)$, kjer je $q(x)=x^{2}+2 x+7$, pa nima realnih ničel. Dokaži, da je $p(8)>4$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n1. Naj bosta $x_{1}$ in $x_{2}$ realni ničli polinoma $p(x)$. Tedaj je $p(x)=\\left(x-x_{1}\\right)\\left(x-x_{2}\\right)$ in\n$$\np(q(x))=\\left(q(x)-x_{1}\\right)\\left(q(x)-x_{2}\\right)=\\left(x^{2}+2 x+\\left(7-x_{1}\\right)\\right)\\left(x^{2}+2 x+\\left(7-x_{2}\\right)\\right)\n$$\nPolinom $p(q(x))$ nima realnih ničel, zato tudi polinoma $x^{2}+2 x+\\left(7-x_{1}\\right)$ in $x^{2}+2 x+\\left(7-x_{2}\\right)$ nimata realnih ničel. Ker sta to polinoma z realnimi koeficienti, morata biti torej njuni diskriminanti negativni, se pravi $D_{1}=4-4\\left(7-x_{1}\\right)<0$ in $D_{2}=4-4\\left(7-x_{2}\\right)<0$. Od tod sledi $x_{1}<6$ ter $x_{2}<6$ in zato je $p(8)=\\left(8-x_{1}\\right)\\left(8-x_{2}\\right)>2 \\cdot 2=4$.\n\n\n2. način. Naj bosta $x_{1}$ in $x_{2}$ realni ničli polinoma $p(x)$, torej $p(x)=\\left(x-x_{1}\\right)\\left(x-x_{2}\\right)$. Opazimo, da je $q(x)=(x+1)^{2}+6$, torej lahko polinom $q(x)$ zavzame vsa realna števila, ki so večja ali enaka 6. Ker pa polinom $p(q(x))$ nima realnih ničel, polinom $q(x)$ ne sme zadeti števil $x_{1}$ in $x_{2}$, torej mora veljati $x_{1}, x_{2}<6$. Od tod sledi $p(8)=\\left(8-x_{1}\\right)\\left(8-x_{2}\\right)>2 \\cdot 2=4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42023,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nPoišči vsa realna števila $x$, za katera velja $\\left(x^{2}-7 x+11\\right)^{x^{2}-13 x+42}=1$.",
    "options": [],
    "answer": "2, 3, 4, 5, 6, 7",
    "solution": "Solution:\nVrednost potence $a^{b}$ je enaka $1$ le v primeru, ko je $a=1$, $a=-1$ in $b$ sodo celo število ali $b=0$ in $a \\neq 0$.\n\nČe je $x^{2}-7 x+11=1$, sledi $x^{2}-7 x+10=0$ oziroma $(x-2)(x-5)=0$. Od tod dobimo rešitvi $x=2$ in $x=5$.\n\nČe je $x^{2}-7 x+11=-1$ oziroma $x^{2}-7 x+12=0$, sledi $(x-3)(x-4)=0$ in zato $x=3$ ali $x=4$. V obeh primerih je $x^{2}-13 x+42$ sodo število, zato sta tudi to rešitvi.\n\nČe pa je $x^{2}-13 x+42=0$ oziroma $(x-6)(x-7)=0$, sledi $x=6$ ali $x=7$ in v obeh primerih je $x^{2}-7 x+11 \\neq 0$.\n\nRešitev so torej števila $2, 3, 4, 5, 6$ in $7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42024,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je paralelogram $ABCD$ z dolžinama stranic $|AB|=7~\\mathrm{cm}$ in $|BC|=5~\\mathrm{cm}$. Višina na stranico $AB$ meri $4~\\mathrm{cm}$ in kot $\\alpha=\\angle BAD$ je oster.\n\na) Z ravnilom in šestilom konstruiraj dani paralelogram in trikotniku $BCD$ očrtaj krožnico. Središče trikotniku očrtane krožnice označi s $S$.\n\nb) Velikost večjega od obeh kotov (udrtega kota) $\\angle BSD$ zaokrožimo na $254^\\circ$ in uporabimo v nadaljnjih izračunih. Izračunaj koliko merijo notranji koti paralelograma.\n\nc) V paralelogramu izračunaj velikost topega kota med simetralo stranice $AD$ in višino na stranico $AB$.",
    "options": [],
    "answer": "Interior angles of the parallelogram: 53° and 127°. Obtuse angle between the perpendicular bisector of AD and the altitude to AB: 127°.",
    "solution": "Solution:\n\na)\nS šestilom in ravnilom konstruiramo paralelogram. Trikotniku $BCD$ očrtamo krog in središče kroga označimo s $S$.\n\nb)\nUdrti kot $\\angle BSD$ meri $254^\\circ$, vbočeni kot $\\angle BSD$ meri $106^\\circ$, saj skupaj tvorita polni kot. Vbočeni kot $\\angle BSD$ je središčni kot, kot $\\angle BCD$ paralelograma pa je obodni kot nad istim lokom, torej kot $\\angle BCD$ meri $53^\\circ$. Od tod sledi, da kot $\\angle BAD$ meri $53^\\circ$. Ker je v paralelogramu vsota kotov ob osnovnici $180^\\circ$, je velikost kota $\\angle ABC=127^\\circ$.\n\nc)\nSimetrala stranice $AD$, višina na stranico $AB$ ter stranici $AB$ in $BC$ oblikujejo štirikotnik, z dvema pravima kotoma, kotom $\\angle BAD$ in iskanim kotom. S pomočjo vsote notranjih kotov štirikotnika izračunamo iskani kot, ki meri $127^\\circ$.\n\n![](attached_image_1.png)\n\nIzračun kota $\\angle ABC=127^\\circ$ ali $\\angle ADC=127^\\circ$ ..... $1^*$ točka\nNarisana simetrala na stranico $AD$ in ugotovitev, da je pravokotna na stranico $AD$ ..... 1 točka\nNarisana višina na stranico $AB$ in ugotovitev, da je pravokotna na stranico $AB$ ..... 1 točka\nIzračun ostrega kota med simetralo in višino $127^\\circ$ ..... 1 točka",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42025,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Vsota prvih osmih členov aritmetičnega zaporedja je 124, prvi člen pa je enak 5. Izračunaj prve štiri člene aritmetičnega zaporedja.\n\nb) Vsota prvih sedmih členov nekega aritmetičnega zaporedja je enaka 105. Prvi, tretji in sedmi člen danega aritmetičnega zaporedja so zaporedni trije členi nekega geometrijsko zaporedje. Izračunaj prve štiri člene aritmetičnega zaporedja.",
    "options": [],
    "answer": "a) 5, 8, 11, 14; b) 15, 15, 15, 15 or 6, 9, 12, 15",
    "solution": "Solution:\n\na)\nZapišimo obrazec za vsoto prvih $n$ členov aritmetičnega zaporedja:\n$$S_n = \\frac{n}{2}\\left(2 a_1 + (n-1) d\\right).$$\nVstavimo podatke za vsoto prvih $8$ členov in dobimo:\n$$S_8 = \\frac{8}{2}(2 \\cdot 5 + 7 d) = 124.$$ \nPoenostavimo:\n$$4(10 + 7d) = 124$$\n$$40 + 28d = 124$$\n$$28d = 84$$\n$$d = 3.$$ \nPo obrazcu za splošni člen aritmetičnega zaporedja $a_n = a_1 + (n-1)d$ izračunamo prve štiri člene:\n$$a_1 = 5$$\n$$a_2 = 5 + 3 = 8$$\n$$a_3 = 5 + 2 \\cdot 3 = 11$$\n$$a_4 = 5 + 3 \\cdot 3 = 14$$\nTorej so prvi štirje členi: $5, 8, 11, 14$.\n\nb)\nZapišimo obrazec za vsoto prvih $n$ členov aritmetičnega zaporedja. Za $n=7$ dobimo:\n$$S_7 = \\frac{7}{2}\\left(2 a_1 + 6 d\\right) = 105.$$ \nPoenostavimo:\n$$7(a_1 + 3d) = 105$$\n$$a_1 + 3d = 15$$\n\nUpoštevamo podatek, da prvi, tretji in sedmi člen aritmetičnega zaporedja sami zase tvorijo geometrijsko zaporedje. Torej:\n$$q = \\frac{a_3}{a_1} = \\frac{a_7}{a_3}$$\nali\n$$\\frac{a_1 + 2d}{a_1} = \\frac{a_1 + 6d}{a_1 + 2d}$$\nKrižno množimo:\n$$(a_1 + 2d)^2 = a_1(a_1 + 6d)$$\n$$a_1^2 + 4a_1 d + 4d^2 = a_1^2 + 6a_1 d$$\n$$4a_1 d + 4d^2 = 6a_1 d$$\n$$4d^2 - 2a_1 d = 0$$\n$$2d(2d - a_1) = 0$$\nTorej $d = 0$ ali $a_1 = 2d$.\n\nČe $d = 0$, je zaporedje konstantno:\n$$a_1 + 3 \\cdot 0 = 15 \\implies a_1 = 15$$\nTorej so prvi štirje členi: $15, 15, 15, 15$.\n\nČe $a_1 = 2d$, vstavimo v $a_1 + 3d = 15$:\n$$2d + 3d = 15$$\n$$5d = 15$$\n$$d = 3$$\n$$a_1 = 2 \\cdot 3 = 6$$\nTorej so prvi štirje členi: $6, 9, 12, 15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42026,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nReši enačbo: $4x^{4} + 2x^{3} - 4x^{2} - x = -1$. Izračunaj razliko kvadratov vsote racionalnih rešitev in vsote iracionalnih rešitev enačbe.",
    "options": [],
    "answer": "1/4",
    "solution": "Solution:\n\nEnačbo lahko rešimo s pomočjo uporabe Hornerjevega algoritma in dobimo rešitve $x_{1} = -1$, $x_{2} = \\frac{1}{2}$, $x_{3} = \\frac{\\sqrt{2}}{2}$, $x_{4} = -\\frac{\\sqrt{2}}{2}$. Ugotovimo, da sta prvi dve rešitvi racionalni, drugi dve pa iracionalni.\n\nIzračunamo kvadrat vsote prvih dveh rešitev:\n\n$(-1 + \\frac{1}{2})^{2} = (-\\frac{1}{2})^{2} = \\frac{1}{4}$\n\nIzračunamo kvadrat vsote drugih dveh rešitev:\n\n$\\left(\\frac{\\sqrt{2}}{2} + \\left(-\\frac{\\sqrt{2}}{2}\\right)\\right)^{2} = (0)^{2} = 0$\n\nNazadnje izračunamo še razliko kvadratov vsote racionalnih rešitev in vsote iracionalnih rešitev enačbe:\n\n$\\frac{1}{4} - 0 = \\frac{1}{4}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42027,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNataša je zlepila kvadrat in enakostranični trikotnik v petkotnik z idejo, da lahko z njimi oblikuje neskončen vzorec, ki ravnine ne pokrije v celoti. Ali Natašin vzorec pokrije več kot $75\\%$ površine ravnine?\n![](attached_image_1.png)\n![](attached_image_2.png)",
    "options": [],
    "answer": "Yes, the pattern covers more than 75 percent of the plane.",
    "solution": "Solution:\nOznačimo z $a$ dolžino stranice Natašinega petkotnika. Potem je njegova ploščina enaka\n$$\na^2\\left(1+\\frac{\\sqrt{3}}{4}\\right).\n$$\nNajvečji notranji kot petkotnika je enak $90^{\\circ}+60^{\\circ}=150^{\\circ}$. Luknja v Natašinem vzorcu je torej romb z manjšim notranjim kotom enakim $360^{\\circ}-2 \\cdot 150^{\\circ}=60^{\\circ}$. Ta romb je torej sestavljen iz dveh enakokrakih trikotnikov s stranico dolžine $a$, zato je njegova ploščina enaka\n$$\n2 \\cdot a^2 \\frac{\\sqrt{3}}{4}=a^2 \\frac{\\sqrt{3}}{2}.\n$$\nOpazimo, da lahko ravnino prekrijemo z osemkotnikom\n![](attached_image_3.png)\nprikazuje spodnja slika. Ploščina tega osemkotnika je enaka\n$$\n2 \\cdot a^2\\left(1+\\frac{\\sqrt{3}}{4}\\right)+a^2 \\frac{\\sqrt{3}}{2}=a^2(2+\\sqrt{3}).\n$$\nDelež ravnine, ki ga pokriva Natašin vzorec je zato enak deležu osemkotnika, ki ga prekrivata petkotnika, tj.\n$$\n\\frac{2 a^2\\left(1+\\frac{\\sqrt{3}}{4}\\right)}{a^2(2+\\sqrt{3})}=\\frac{2+\\frac{\\sqrt{3}}{2}}{2+\\sqrt{3}}=\\frac{4+\\sqrt{3}}{4+2 \\sqrt{3}}.\n$$\nNeenakost $\\frac{4+\\sqrt{3}}{4+2 \\sqrt{3}} > \\frac{3}{4}$ je ekvivalentna neenakosti $4(4+\\sqrt{3}) > 3(4+2 \\sqrt{3})$, ki jo lahko preuredimo v $4 > 2 \\sqrt{3}$. Slednja neenakost je izpolnjena, saj je $\\sqrt{3}<2$. Natašin vzorec torej pokrije več kot $75\\%$ površine ravnine.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42028,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je enakokrak trikotnik $ABC$ z vrhom pri $C$, v katerem je $\\angle ACB < 90^\\circ$. Naj bo $X$ od $C$ različna točka na stranici $AC$ in $Y$ od $C$ različna točka na stranici $BC$. Naj bo $D$ taka točka, da je premica $DX$ vzporedna premici $AB$, premica $AC$ pa je notranja simetrala kota $\\angle BAD$. Podobno naj bo $E$ taka točka, da je premica $EY$ vzporedna premici $AB$, premica $BC$ pa je notranja simetrala kota $\\angle EBA$. Denimo, da obstaja taka točka $T$ na stranici $AB$, da velja $|XT| = |XD|$ in $|YT| = |YE|$. Izrazi vrednost izraza $|AX| + |BY|$ z dolžinami stranic trikotnika $ABC$.",
    "options": [],
    "answer": "AB^2/BC",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOznačimo $\\angle BAC = \\angle CBA = \\alpha$. Ker je premica $AC$ simetrala kota $\\angle BAD$, je $\\angle XAD = \\alpha$. Zaradi vzporednosti premic $XD$ in $AB$ pa je tudi $\\angle DXA = \\alpha$. Trikotnik $AXD$ je torej enakokrak z vrhom pri $D$, zato velja $|XD| = |AD| = |TX|$. Ker sta premici $AB$ in $DX$ vzporedni, od tod sledi, da je štirikotnik $ATXD$ bodisi romb bodisi enakokrak trapez. Podobno sklepamo, da je tudi štirikotnik $TBEY$ bodisi romb bodisi enakokrak trapez. Pokažimo, da sta ta dva štirikotnika oba romba.\n\nPo predpostavki je kot $\\angle ACB = 180^\\circ - 2\\alpha < 90^\\circ$, zato je $\\alpha > 45^\\circ$. Če je štirikotnik $ATXD$ romb, je $\\angle XTA = 180^\\circ - \\angle TAD = 180^\\circ - 2\\alpha$, če pa je enakokrak trapez, je $\\angle XTA = \\angle TAD = 2\\alpha$. Podobno je tudi $\\angle BTY = 180^\\circ - 2\\alpha$, če je $TBEY$ romb, in $\\angle BTY = 2\\alpha$, če je $TBEY$ enakokrak trapez. Ker sta točki $X$ in $Y$ različni od $C$ in ležita na ustreznih stranicah trikotnika, je $\\angle BTY + \\angle XTA < 180^\\circ$. Ker pa kombinacije $2\\alpha + 2\\alpha = 4\\alpha > 4 \\cdot 45^\\circ = 180^\\circ$, $\\left(180^\\circ - 2\\alpha\\right) + 2\\alpha = 180^\\circ$ in $2\\alpha + \\left(180^\\circ - 2\\alpha\\right) = 180^\\circ$ ne ustrezajo temu pogoju, je po zgornjem možna le kombinacija $\\angle BTY = \\angle XTA = 180^\\circ - 2\\alpha$, torej sta štirikotnika $ATXD$ in $TBEY$ romba.\n\nOd tod sledi, da sta trikotnika $ATX$ in $YTB$ enakokraka z vrhom pri $T$ in zato podobna trikotniku $BCA$, saj se z njim ujemata v kotu ob osnovnici. Torej je\n$$\n\\frac{|AX|}{|AB|} = \\frac{|AT|}{|BC|} \\quad \\text{in} \\quad \\frac{|BY|}{|AB|} = \\frac{|BT|}{|AC|} = \\frac{|BT|}{|BC|}\n$$\nS pomočjo teh enakosti izrazimo\n$$\n|AX| + |BY| = \\frac{|AB| \\cdot |AT|}{|BC|} + \\frac{|AB| \\cdot |BT|}{|BC|} = \\frac{|AB| \\cdot (|AT| + |BT|)}{|BC|} = \\frac{|AB|^2}{|BC|}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42029,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa cela števila $n$, ki jih lahko zapišemo v obliki $n=\\frac{m+2021}{2021-m}$, kjer je $m$ celo število.",
    "options": [],
    "answer": "-4043, -2022, -95, -87, -48, -44, -3, -2, 0, 1, 42, 46, 85, 93, 2020, 4041",
    "solution": "Solution:\n\nEnakost pomnožimo z $2021-m$, da dobimo $2021 n - m n = m + 2021$. Nato jo preuredimo do $2021(n-1) = m(n+1)$ in izrazimo $m = \\frac{2021(n-1)}{n+1} = 2021 - \\frac{2 \\cdot 2021}{n+1}$. Torej lahko v predpisani obliki zapišemo vsa tista cela števila $n$, za katera je $\\frac{2 \\cdot 2021}{n+1}$ celo število. To pomeni, da mora biti $n+1$ delitelj števila $2 \\cdot 2021 = 2 \\cdot 43 \\cdot 47$. Delitelji tega števila so $\\pm 1, \\pm 2, \\pm 43, \\pm 47, \\pm 86, \\pm 94, \\pm 2021$ in $\\pm 4042$. Rešitev naloge so torej cela števila\n$$\n-4043, -2022, -95, -87, -48, -44, -3, -2, 0, 1, 42, 46, 85, 93, 2020, 4041\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42030,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $ABC$ ostrokotni trikotnik. Krožnica s središčem v $A$, ki se dotika stranice $BC$, seka stranico $AB$ v točki $B_{1}$ in stranico $CA$ v točki $C_{2}$. Krožnica s središčem v $B$, ki se dotika stranice $CA$, seka stranico $BC$ v točki $C_{1}$ in stranico $AB$ v točki $A_{2}$. Krožnica s središčem v $C$, ki se dotika stranice $AB$, seka stranico $CA$ v točki $A_{1}$ in stranico $BC$ v točki $B_{2}$. Dokaži, da je trikotnik, ki ga določajo premice $A_{1}A_{2}$, $B_{1}B_{2}$ in $C_{1}C_{2}$, podoben trikotniku $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDokažimo, da sta premici $AB$ in $C_{1}C_{2}$ vzporedni. Naj bodo $D, E$ in $F$ zaporedoma nožišča višin iz $A, B$ in $C$ trikotnika $ABC$. Kot med tangento in tetivo je enak obodnemu kotu nad tetivo, ta pa je enak polovici središčenega kota nad tetivo. Ker je stranica $AC$ tangenta na krožnico $s$ središčem v $B$, zato velja $\\angle C_{1}EC_{2} = \\frac{1}{2} \\angle C_{1}BE = \\frac{1}{2} \\angle CBE$. Podobno zaradi tangentnosti velja tudi $\\angle C_{1}DC_{2} = \\frac{1}{2} \\angle DAC_{2} = \\frac{1}{2} \\angle DAC$. Ker se trikotnika $ADC$ in $BEC$ ujemata v dveh kotih, sta podobna in se ujemata tudi v tretjem kotu. Zato je $\\angle CBE = \\angle DAC$ in iz zgoraj dokazanega sledi še $\\angle C_{1}EC_{2} = \\angle C_{1}DC_{2}$. Od tod sklepamo, da so točke $C_{1}, C_{2}, D$ in $E$ konciklične. Ker velja $\\angle AEB = 90^{\\circ} = \\angle ADB$, so tudi točke $A, B, D$ in $E$ konciklične. Iz obeh koncikličnosti sledi\n$$\n\\angle BAC = \\angle BAE = 180^{\\circ} - \\angle EDB = \\angle C_{1}DE = 180^{\\circ} - \\angle EC_{2}C_{1} = \\angle C_{1}C_{2}C\n$$\ntorej sta premici $AB$ in $C_{1}C_{2}$ vzporedni. Na podoben način dokažemo vzporednost premic $BC$ in $A_{1}A_{2}$ ter premic $CA$ in $B_{1}B_{2}$. Ker imata trikotnik, ki ga določajo premice $A_{1}A_{2}$, $B_{1}B_{2}$ in $C_{1}C_{2}$ ter trikotnik $ABC$ paroma vzporedne stranice, sta torej podobna.\n\n\n2. način. Dokažimo vzporednost $AB$ in $C_{1}C_{2}$ še na drugačen način. Iz navodil naloge razberemo, da velja $|AC_{2}| = |AD|$ ter $|BC_{1}| = |BE|$. Ker sta trikotnika $ADC$ in $BEC$ podobna, od tod sledi\n$$\n\\frac{|AC_{2}|}{|BC_{1}|} = \\frac{|AD|}{|BE|} = \\frac{|AC|}{|BC|}\n$$\nkar pomeni, da sta premici $AB$ in $C_{1}C_{2}$ vzporedni. Na podoben način dokažemo še ostali dve vzporednosti in dokaz zaključimo kot v prvi rešitvi.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42031,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je polinom $p(x)=x^{6}+x^{5}+\\ldots+x+1$. Dokaži, da polinom $p(x)$ deli polinom $p\\left(x^{9}\\right)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPri dokazu bomo nekajkrat uporabiti razcep\n$$\na^{n}-1=(a-1)\\left(a^{n-1}+a^{n-2}+\\ldots+a+1\\right), \\quad n \\in \\mathbb{N}\n$$\nNajprej opazimo, da je $(x-1) p(x)=x^{7}-1$. Torej je\n$$\n\\begin{aligned}\n\\left(x^{9}-1\\right) p\\left(x^{9}\\right) & =\\left(x^{9}\\right)^{7}-1=x^{7 \\cdot 9}-1=\\left(x^{7}\\right)^{9}-1= \\\\\n& =\\left(x^{7}-1\\right)\\left(x^{7 \\cdot 8}+x^{7 \\cdot 7}+\\ldots+x^{7}+1\\right)= \\\\\n& =(x-1) p(x)\\left(x^{7 \\cdot 8}+x^{7 \\cdot 7}+\\ldots+x^{7}+1\\right)\n\\end{aligned}\n$$\nKer pa јe $x^{9}-1=(x-1)\\left(x^{8}+x^{7}+\\ldots+x^{2}+x+1\\right)=(x-1)\\left(x^{2} p(x)+x+1\\right)$, sledi\n$$\n(x-1)\\left(x^{2} p(x)+x+1\\right) p\\left(x^{9}\\right)=(x-1) p(x)\\left(x^{7 \\cdot 8}+x^{7 \\cdot 7}+\\ldots+x^{7}+1\\right)\n$$\nkar nam po preoblikovanju da\n$$\n(x+1) p\\left(x^{9}\\right)=p(x)\\left(x^{7 \\cdot 8}+x^{7 \\cdot 7}+\\ldots+x^{7}+1-x^{2} p\\left(x^{9}\\right)\\right)\n$$\nPolinoma $x+1$ in $p(x)=(x+1)\\left(x^{5}+x^{3}+x\\right)+1$ sta tuja, zato mora $p(x)$ deliti $p\\left(x^{9}\\right)$.\n\n\n2. način. Nalogo lahko rešimo tudi z neposrednim deljenjem polinoma $p\\left(x^{9}\\right)$ s polinomom $p(x)$, kar pa zahteva precej računanja. Kvocient je enak\n$$\n\\begin{aligned}\n& x^{48}-x^{47}+x^{41}-x^{40}+x^{39}-x^{38}+x^{34}-x^{33}+x^{32}-x^{31}+x^{30}-x^{29}+x^{27}-x^{26}+x^{25} \\\\\n& -x^{24}+x^{23}-x^{22}+x^{21}-x^{19}+x^{18}-x^{17}+x^{16}-x^{15}+x^{14}-x^{10}+x^{9}-x^{8}+x^{7}-x+1\n\\end{aligned}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42032,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa tabelo velikosti $2 \\times n$, kjer je $n$ naravno število, naključno na različni polji postavimo dva žetona. Kolikšna je verjetnost, da se da preostanek tabele prekriti z dominami velikosti $2 \\times 1$? Domine se ne smejo prekrivati in ne smejo segati čez rob tabele.",
    "options": [],
    "answer": "n/(2n-1)",
    "solution": "Solution:\n\nTabelo pobarvamo kot šahovnico. Tedaj je v tabeli enako število belih in črnih polj. Ker vsaka domina prekrije 1 belo in 1 črno polje, vse domine skupaj prekrijejo enako število belih in črnih polj. Da lahko preostanek tabele prekrijemo z dominami velikosti $2 \\times 1$, morata biti torej žetona postavljena na polji različnih barv. Pokažimo, da lahko v tem primeru preostanek tabele res prekrijemo z dominami velikosti $2 \\times 1$.\n\nNa tabelo narišimo sklenjeno krivuljo, ki poteka skozi vsa polja tabele, kot je prikazano na levi sliki na konkretnem primeru tabele velikosti $2 \\times 10$ pri konkretni postavitvi žetonov.\n\n![](attached_image_1.png)\n\nIz tabele sedaj odstranimo polji, na katerih sta žetona, kot je prikazano na desni sliki. Ker sta bila žetona na poljih različne barve, narisana krivulja razpade na dva dela, ki oba potekata skozi sodo mnogo polj tabele. Torej lahko oba dela prekrijemo z dominami velikosti $2 \\times 1$, tako da začnemo na enem koncu krivulje in ji pri prekrivanju sledimo, na vsakem koraku pa z domino prekrijemo naslednji dve polji na krivulji (to lahko vedno storimo).\n\nVerjetnost iz naloge lahko sedaj izračunamo kot\n$$\nP=\\frac{\\text{ugodne postavitve žetonov}}{\\text{vse možne postavitve žetonov}}\n$$\nkjer so ugodne postavitve žetonov tiste, pri katerih sta žetona postavljana na polji različne barve. Vseh možnih postavitev žetonov je $2n \\cdot (2n-1)$, saj lahko prvi žeton postavimo na poljubno polje tabele (teh je $2n$), za postavitev drugega žetona pa nam ostane $2n-1$ polj tabele. Ugodnih postavitev žetonov je $2n \\cdot n$, saj lahko prvi žeton prav tako postavimo na poljubno polje tabele, za postavitev drugega žetona pa imamo na voljo le $n$ polj tabele, saj mora biti ta postavljen na polje druge barve kot prvi žeton. Iskana verjetnost je torej enaka\n$$\nP=\\frac{2n^2}{2n(2n-1)}=\\frac{n}{2n-1}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42033,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vsa realna števila $x$, $y$ in $z$, ki rešijo sistem enačb\n$$\n\\frac{3 x y}{x-y}=2, \\quad \\frac{2 y z}{y+2 z}=3, \\quad \\frac{x z}{z-4 x}=3\n$$",
    "options": [],
    "answer": "x = -1, y = 2, z = -3",
    "solution": "Solution:\n\nV enačbah najprej odpravimo ulomke in dobimo\n$$\n\\begin{aligned}\n3 x y & =2 x-2 y \\\\\n2 y z & =3 y+6 z \\\\\nx z & =3 z-12 x\n\\end{aligned}\n$$\nPrvo enačbo preoblikujemo v $(3 y-2) x=-2 y$. Če je $3 y-2=0$ oziroma $y=\\frac{2}{3}$, tedaj sledi $y=0$, kar pa je protislovje. Torej enačbo lahko delimo z $3 y-2$ in izrazimo $x=\\frac{-2 y}{3 y-2}$. Drugo enačbo preoblikujemo v $(2 y-6) z=3 y$ in s podobnim sklepom izrazimo $z=\\frac{3 y}{2 y-6}$. Oboje sedaj vstavimo v zadnjo enačbo, da dobimo\n$$\n\\frac{-2 y}{3 y-2} \\cdot \\frac{3 y}{2 y-6}=\\frac{9 y}{2 y-6}+\\frac{24 y}{3 y-2}\n$$\nEnačbo pomnožimo z $(3 y-2)(2 y-6)$, da dobimo\n$$\n-6 y^{2}=9 y(3 y-2)+24 y(2 y-6)\n$$\nkar lahko poenostavimo do $81 y^{2}-162 y=0$. Levo stran razstavimo in dobimo $81 y(y-2)=0$. Če je $y=0$, iz zgornjih zvez sledi še $x=0$ in $z=0$. Toda to ni rešitev sistema enačb, saj ulomki v enačbah v tem primeru niso definirani. Zato mora biti $y=2$ in posledično $x=\\frac{-4}{4}=-1$ ter $z=\\frac{6}{-2}=-3$. Edina rešitev sistema je torej $x=-1$, $y=2$ in $z=-3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42034,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDani sta enačbi $(y+3)^2 - x(x-4) = 7 + (y-x)(y+x)$ in $x + \\frac{y-1}{2} = \\frac{x}{2} - y$.\n\na) Reši sistem enačb in rešitev zapiši kot točko $A$ v koordinatnem sistemu.\n\nb) Zapiši enačbo premice, ki poteka skozi točko $A$ in ima isto presečišče z osjo $x$ kot premica $3x + 2y - 15 = 0$ v eksplicitni, implicitni in odsekovni obliki.",
    "options": [],
    "answer": "Point A: (-2, 1). Line through A with the same x-intercept as 3x + 2y - 15 = 0:\n- Explicit: y = -1/7 x + 5/7\n- Implicit: x + 7y - 5 = 0\n- Intercept: x/5 + y/(5/7) = 1",
    "solution": "Solution:\n\na.\nPoenostavimo prvo enačbo do zapisa npr. $4x + 6y + 2 = 0$. Poenostavimo drugo enačbo do zapisa npr. $x + 3y - 1 = 0$. Rešimo sistem enačb s katerokoli metodo. Rešitev sistema enačb zapišemo kot točko $A(-2, 1)$.\n\nb.\nIzračunamo presečišče premice $3x + 2y - 15 = 0$ z $x$ osjo $(5, 0)$. Izračunamo smerni koeficient $k = \\frac{y_2 - y_1}{x_2 - x_1} = \\frac{0 - 1}{5 + 2} = \\frac{-1}{7}$. Zapišemo enačbo premice npr. $y = kx + n$ in izračunamo začetno vrednost $n = \\frac{5}{7}$. Zapišemo vse tri oblike enačbe premice. Eksplicitna oblika $y = \\frac{-1}{7}x + \\frac{5}{7}$, implicitna oblika $x + 7y - 5 = 0$ in odsekovna oblika enačbe premice $\\frac{x}{5} + \\frac{y}{\\frac{5}{7}} = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42035,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIzračunaj vrednost izraza $0 . \\overline{27} \\cdot\\left(x^{2}+2 y x+y^{2}\\right):\\left(3 y^{2}-3 x\\right)$, če velja $x-y=5$ in $\\frac{7 x-7 y}{10}-\\left(\\frac{2}{2 x+y}\\right)^{-1}=5^{0}$.",
    "options": [],
    "answer": "-5/11",
    "solution": "Solution:\nŠtevilo $0 . \\overline{27}$ zapišemo v obliki okrajšanega ulomka $\\frac{3}{11}$. Enačbo $\\frac{7 x-7 y}{10}-\\left(\\frac{2}{2 x+y}\\right)^{-1}=5^{0}$ preoblikujemo v $\\frac{7 \\cdot(x-y)}{10}-\\frac{2 x+y}{2}=1$. Upoštevamo 1. enačbo $x-y=5$ in dobimo $\\frac{7 \\cdot 5}{10}-\\frac{2 x+y}{2}=1$. Ta enačba se preoblikuje v $2 x+y=5$, tako da dobimo sistem dveh linearnih enačb z dvema neznankama $x-y=5$ in $2 x+y=5$. V primeru, da pri poenostavljanju druge enačbe ne bi uporabili $x-y=5$, bi se druga enačba poenostavila v $3 x+12 y=-10$. Rešitev sistema je $x=\\frac{10}{3}$ in $y=-\\frac{5}{3}$. Izračunamo še vrednost izraza $0 . \\overline{27} \\cdot\\left(x^{2}+2 y x+y^{2}\\right):\\left(3 y^{2}-3 x\\right)=\\frac{3}{11} \\cdot \\frac{25}{9}:\\left(\\frac{25}{3}-10\\right)=\\frac{25}{33} \\cdot\\left(-\\frac{3}{5}\\right)=-\\frac{5}{11}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42036,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIzračunaj $x+y$, če je $\\left(x+\\sqrt{1+x^{2}}\\right)\\left(y+\\sqrt{1+y^{2}}\\right)=1$.",
    "options": [],
    "answer": "0",
    "solution": "Solution:\nDano enakost pomnožimo z $\\left(x-\\sqrt{1+x^{2}}\\right)$, da dobimo $\\left(x^{2}-\\left(1+x^{2}\\right)\\right)\\left(y+\\sqrt{1+y^{2}}\\right) = x-\\sqrt{1+x^{2}}$ oziroma $-y-\\sqrt{1+y^{2}} = x-\\sqrt{1+x^{2}}$. Slednje lahko preoblikujemo v $\\sqrt{1+x^{2}}-\\sqrt{1+y^{2}}=x+y$.\n\nČe dano enakost pomnožimo z $\\left(y-\\sqrt{1+y^{2}}\\right)$, dobimo $-x-\\sqrt{1+x^{2}} = y-\\sqrt{1+y^{2}}$ oziroma $\\sqrt{1+x^{2}}-\\sqrt{1+y^{2}}=-(x+y)$. Iz obeh enačb sledi $x+y=-(x+y)$ oziroma $x+y=0$.\n\nNa enak način kot v prvi rešitvi izpeljemo $x+y=\\sqrt{1+x^{2}}-\\sqrt{1+y^{2}}$. Enačbo kvadriramo, da dobimo $x^{2}+2 x y+y^{2}=\\left(1+x^{2}\\right)-2 \\sqrt{1+x^{2}} \\sqrt{1+y^{2}}+\\left(1+y^{2}\\right)$, ter jo nato poenostavimo do $\\sqrt{1+x^{2}} \\sqrt{1+y^{2}}=1-x y$. Po ponovnem kvadriranju dobimo $\\left(1+x^{2}\\right)(1+y^{2})=1-2 x y+x^{2} y^{2}$, kar lahko zopet poenostavimo do $x^{2}+2 x y+y^{2}=0$. Dobljena enačba je ekvivalentna enačbi $(x+y)^{2}=0$, od koder sledi $x+y=0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42037,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $x$ takšno realno število, za katerega velja $\\cos \\left(60^\\circ-x\\right) \\neq 0$ in $\\sin \\left(120^\\circ-x\\right) \\neq 0$. Brez uporabe žepnega računala izračunaj natančno vrednost izraza: \n$$\n\\frac{\\sqrt{3}+4 \\sin x \\cos x}{\\cos \\left(60^\\circ-x\\right) \\cdot \\sin \\left(120^\\circ-x\\right)}\n$$",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nNajprej dvakrat uporabimo adicijska izreka in izraza kar se da poenostavimo:\n\n$\\cos \\left(60^\\circ-x\\right) = \\cos 60^\\circ \\cdot \\cos x + \\sin 60^\\circ \\cdot \\sin x = \\frac{1}{2} \\cdot \\cos x + \\frac{\\sqrt{3}}{2} \\cdot \\sin x$\n\nin\n\n$\\sin \\left(120^\\circ-x\\right) = \\sin 120^\\circ \\cdot \\cos x - \\sin x \\cdot \\cos 120^\\circ = \\frac{1}{2} \\cdot \\sin x + \\frac{\\sqrt{3}}{2} \\cdot \\cos x$\n\nProdukt dobljenih poenostavljenih izrazov je potem\n\n$$\n\\left(\\frac{1}{2} \\cos x + \\frac{\\sqrt{3}}{2} \\sin x\\right)\\left(\\frac{1}{2} \\sin x + \\frac{\\sqrt{3}}{2} \\cos x\\right)\n$$\n\nIzračunamo produkt:\n\n$$\n= \\frac{1}{2} \\cos x \\cdot \\frac{1}{2} \\sin x + \\frac{1}{2} \\cos x \\cdot \\frac{\\sqrt{3}}{2} \\cos x + \\frac{\\sqrt{3}}{2} \\sin x \\cdot \\frac{1}{2} \\sin x + \\frac{\\sqrt{3}}{2} \\sin x \\cdot \\frac{\\sqrt{3}}{2} \\cos x\n$$\n\n$$\n= \\frac{1}{4} \\sin x \\cos x + \\frac{\\sqrt{3}}{4} \\cos^2 x + \\frac{\\sqrt{3}}{4} \\sin^2 x + \\frac{3}{4} \\sin x \\cos x\n$$\n\n$$\n= \\left(\\frac{1}{4} + \\frac{3}{4}\\right) \\sin x \\cos x + \\frac{\\sqrt{3}}{4}(\\sin^2 x + \\cos^2 x)\n$$\n\n$$\n= \\sin x \\cos x + \\frac{\\sqrt{3}}{4} \\cdot 1\n$$\n\n$$\n= \\sin x \\cos x + \\frac{\\sqrt{3}}{4}\n$$\n\nTorej je produkt:\n\n$$\n\\cos \\left(60^\\circ-x\\right) \\cdot \\sin \\left(120^\\circ-x\\right) = \\sin x \\cos x + \\frac{\\sqrt{3}}{4}\n$$\n\nZdaj izraz v nalogi postane:\n\n$$\n\\frac{\\sqrt{3} + 4 \\sin x \\cos x}{\\cos \\left(60^\\circ-x\\right) \\cdot \\sin \\left(120^\\circ-x\\right)} = \\frac{\\sqrt{3} + 4 \\sin x \\cos x}{\\sin x \\cos x + \\frac{\\sqrt{3}}{4}}\n$$\n\nŠtevec lahko zapišemo kot $4\\left(\\sin x \\cos x + \\frac{\\sqrt{3}}{4}\\right)$, torej:\n\n$$\n= \\frac{4\\left(\\sin x \\cos x + \\frac{\\sqrt{3}}{4}\\right)}{\\sin x \\cos x + \\frac{\\sqrt{3}}{4}} = 4\n$$\n\nTorej je vrednost izraza $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42038,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV nekem domu za ostarele so pred enim tednom ugotovili prve okužene s covid-19. Od takrat se je njihovo število popeterilo, torej jih je sedaj petkrat toliko kot pred enim tednom. 21 od teh okuženih so zjutraj odpeljali v bolnišnico, tako je med oskrbovanci, ki so ostali v domu, $9 . \\overline{09}\\ \\%$ okuženih. Zvečer bodo v bolnišnico odpeljali še 9 okuženih, tako da bo med oskrbovanci, ki bodo ostali v domu, le $5\\ \\%$ okuženih. Koliko je bilo prvotno število okuženih pred enim tednom in koliko je bilo takrat vseh oskrbovancev v domu?",
    "options": [],
    "answer": "8 infected originally; 230 total residents originally",
    "solution": "Solution:\n\nNaj bo prvotno $x$ okuženih in v domu skupno $y$ oskrbovancev. Od vseh okuženih $5x$ so jih 21 odpeljali, okuženih je ostalo $5x-21$. Oskrbovancev je ostalo $y-21$. Delež okuženih je $\\frac{5x-21}{y-21}=9 . \\overline{09}\\ \\%$. Če periodično število $9 . \\overline{09}$ spremenimo v ulomek, dobimo $\\frac{100}{11}$ in $9 . \\overline{09}\\ \\%=\\frac{1}{11}$. Zato delež okuženih lahko pišemo kot $\\frac{5x-21}{y-21}=\\frac{1}{11}$, kar lahko preoblikujemo v $y=55x-210$.\n\nČe odpeljejo še 9 okuženih, jih bodo skupno odpeljali 30 in dobimo delež okuženih $\\frac{5x-30}{y-30}=5\\ \\%$. Upoštevamo, da je $5\\ \\%=\\frac{1}{20}$ in delež preoblikujemo v enačbo $y=100x-570$.\n\nZ enačenjem obeh enačb dobimo $x=8$ in $y=230$, kar pomeni, da je bilo vseh oskrbovancev 230, od tega 8 okuženih.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42039,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV eksplicitni, implicitni in odsekovni obliki zapiši enačbo premice s pozitivnim smernim koeficientom, ki s koordinatnima osema oblikuje pravokotni trikotnik s ploščino $9$ kvadratnih enot in abscisno os seka pri $x = -3$.",
    "options": [],
    "answer": "Intercept form: x/(-3) + y/6 = 1; Implicit form: -2x + y - 6 = 0; Explicit form: y = 2x + 6",
    "solution": "Solution:\n\nPremica s koordinatnima osema oblikuje pravokotni trikotnik, katerega ploščino izračunamo kot polovični produkt med katetama $p = \\frac{k_1 \\cdot k_2}{2}$. Iz naloge je razvidno, da je dolžina ene katete $3$, dolžino druge katete pa izračunamo s pomočjo ploščine in dobimo rezultat $6$. Ker je smerni koeficient premice pozitiven, premica seka koordinatni osi v točkah $(-3, 0)$ in $(0, 6)$, s pomočjo teh dveh točk lahko zapišemo vse tri oblike enačbe premice.\n\nOdsekovna oblika:\n$$\n\\frac{x}{-3} + \\frac{y}{6} = 1\n$$\n\nImplicitna oblika:\n$$\n-2x + y - 6 = 0\n$$\n\nEksplicitna oblika:\n$$\ny = 2x + 6\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42040,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVeronika ima list karirastega papirja z $78 \\times 78$ kvadratki. List želi razrezati na manjše kose, od katerih bo vsak imel bodisi 14 bodisi 15 kvadratkov, pri čemer z vsakim rezom prereže enega od kosov papirja na dva dela vzdolž ene od črt na papirju. Najmanj kolikokrat mora Veronika prerezati papir?",
    "options": [],
    "answer": "405",
    "solution": "Solution:\n\nZ vsakim rezom se število kosov papirja poveča za 1. Na koncu bo torej število kosov papirja za 1 večje od števila rezov, ki jih je Veronika izvedla. Da bo izvedla čim manj rezov, mora imeti na koncu čim manj kosov papirja. Denimo, da ima na koncu $k$ kosov s 14 kvadratki in $n$ kosov s 15 kvadratki, torej skupaj $n+k$ kosov papirja. Tedaj mora veljati $14 k+15 n=78^{2}$. Enakost zapišemo v obliki $15(n+k)=78^{2}+k$ in izrazimo $n+k=\\frac{78^{2}+k}{15}$. Od tod sledi, da mora biti število $78^{2}+k$ deljivo s $15$. Da bo $n+k$ čim manjše, mora biti $k$ čim manjši. Ker ima število $78^{2}=6084$ pri deljenju s $15$ ostanek $9$, mora biti $k \\geq 6$. Ko je $k=6$, je $n+k=406$, torej mora Veronika papir prerezati vsaj $405$-krat.\n\nPreverimo še, da lahko Veronika s $405$ rezi papir res razreže na kose, ki imajo bodisi $14$ bodisi $15$ kvadratkov. Veronika najprej od lista s $5$ rezi odreže $5$ trakov velikosti $15 \\times 78$, vsakega od teh trakov pa s $77$ rezi razreže na kose velikosti $15 \\times 1$. Ostane ji še kos velikosti $3 \\times 78$. Z $10$ rezi od tega kosa odreže $10$ kosov velikosti $3 \\times 5$, da ji ostane kos velikosti $3 \\times 28$. Z $1$ rezom ta kos prereže na $2$ kosa velikosti $3 \\times 14$, nazadnje pa vsakega od teh kosov z $2$ rezoma razreže na kose velikosti $1 \\times 14$. Tako imajo vsi dobljeni kosi $14$ ali $15$ kvadratov, za kar je bilo potrebnih $5+5 \\cdot 77+10+1+2 \\cdot 2=405$ rezov.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42041,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je izraz $\\frac{\\sqrt{x+1}+\\sqrt{x-1}}{\\sqrt{x+1}-\\sqrt{x-1}}$.\n\na) Racionaliziraj imenovalec in izraz poenostavi.\n\nb) Izračunaj vrednost izraza za $x=\\frac{5}{4}$.",
    "options": [],
    "answer": "x + sqrt(x^2 - 1); at x = 5/4, the value is 2",
    "solution": "Solution:\n\na.\nDani izraz $\\frac{\\sqrt{x+1}+\\sqrt{x-1}}{\\sqrt{x+1}-\\sqrt{x-1}}$ v števcu in imenovalcu pomnožimo z $\\sqrt{x+1}+\\sqrt{x-1}$ ter poenostavimo:\n\n$$\n\\frac{\\sqrt{x+1}+\\sqrt{x-1}}{\\sqrt{x+1}-\\sqrt{x-1}} \\cdot \\frac{\\sqrt{x+1}+\\sqrt{x-1}}{\\sqrt{x+1}+\\sqrt{x-1}} = \\frac{(\\sqrt{x+1}+\\sqrt{x-1})^2}{(\\sqrt{x+1})^2-(\\sqrt{x-1})^2}\n$$\n\n$$\n= \\frac{x+1+2 \\sqrt{x+1} \\sqrt{x-1}+x-1}{x+1-(x-1)} = \\frac{2x+2 \\sqrt{(x+1)(x-1)}}{2} = x+\\sqrt{x^2-1}\n$$\n\nb.\nVstavimo $x=\\frac{5}{4}$ v racionaliziran izraz:\n\n$$\n\\frac{5}{4}+\\sqrt{\\left(\\frac{5}{4}\\right)^2-1} = \\frac{5}{4}+\\sqrt{\\frac{25}{16}-\\frac{16}{16}} = \\frac{5}{4}+\\sqrt{\\frac{9}{16}} = \\frac{5}{4}+\\frac{3}{4} = \\frac{8}{4} = 2\n$$\n\nLahko pa vstavimo $x=\\frac{5}{4}$ v dani izraz:\n\n$$\n\\frac{\\sqrt{\\frac{5}{4}+1}+\\sqrt{\\frac{5}{4}-1}}{\\sqrt{\\frac{5}{4}+1}-\\sqrt{\\frac{5}{4}-1}} = \\frac{\\sqrt{\\frac{9}{4}}+\\sqrt{\\frac{1}{4}}}{\\sqrt{\\frac{9}{4}}-\\sqrt{\\frac{1}{4}}} = \\frac{\\frac{3}{2}+\\frac{1}{2}}{\\frac{3}{2}-\\frac{1}{2}} = \\frac{\\frac{4}{2}}{\\frac{2}{2}} = 2\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42042,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKolikšna je absolutna vrednost razlike rešitev enačbe $\\left(1-\\left(1+x^{-2}\\right)^{-1}\\right)^{-1}=3,25$?\n\n(A) -3\n(B) -1\n(C) 1\n(D) 3\n(E) 0",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nEnačbo prevedemo v obliko $x^{-2}=\\frac{4}{9}$. Rešitvi sta $\\frac{3}{2}=1 \\frac{1}{2}$ in $-\\frac{3}{2}=-1 \\frac{1}{2}$. Števili se razlikujeta za 3. Pravilen je odgovor $D$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42043,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDiagonali kvadratov $ABCD$ in $BEFG$ sta zaporedoma dolgi $8~\\mathrm{cm}$ in $11~\\mathrm{cm}$. Točka $P$ je presečišče diagonal kvadrata $BEFG$ (glej sliko). Koliko kvadratnih centimetrov je ploščina trikotnika $APF$?\n\n(A) 8\n(B) 9\n(C) 10\n(D) 11\n(E) 12\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nStranici kvadratov sta dolgi $|AB| = \\frac{8}{\\sqrt{2}}~\\mathrm{cm}$ in $|BE| = \\frac{11}{\\sqrt{2}}~\\mathrm{cm}$. Ker je točka $P$ razpolovišče diagonale $BF$, je ploščina trikotnika $APF$ enaka ploščini trikotnika $ABP$, ta pa je enaka\n$$\n\\frac{|AB| \\cdot \\frac{|BG|}{2}}{2} = \\frac{\\frac{8}{\\sqrt{2}} \\cdot \\frac{11}{2\\sqrt{2}}}{2} = \\frac{88}{8} = 11~\\mathrm{cm}^2\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42044,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKaj je rešitev enačbe $\\sqrt[5]{9^{x-3}}-\\frac{1}{\\sqrt{3}}=0$?\n(A) $x=-\\frac{7}{4}$\n(B) $x=-\\frac{4}{7}$\n(C) $x=\\frac{4}{7}$\n(D) $x=\\frac{7}{4}$\n(E) $x=-\\frac{9}{4}$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\nEnačbo preoblikujemo v obliko $9^{\\frac{x-3}{5}}=\\frac{1}{\\sqrt{3}}$ in nadalje v obliko $3^{\\frac{2x-6}{5}}=3^{-\\frac{1}{2}}$. Ko rešimo to eksponentno enačbo, dobimo rešitev $\\frac{7}{4}$. Pravilen je odgovor $D$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42045,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKateri izraz je ekvivalenten danemu izrazu\n$$\n\\frac{\\sqrt[3]{4 a^{2} \\cdot \\sqrt[3]{a b} \\cdot (a+b)^{0}}}{\\left(8 b^{-2} \\sqrt{a}\\right)^{-\\frac{2}{3}}}\n$$\nkjer $a, b, a+b \\neq 0$?\n\n(A) $2^{\\frac{8}{3}} a^{\\frac{10}{9}} b^{-\\frac{11}{9}}$\n\n(B) $2^{\\frac{2}{3}} a^{\\frac{5}{6}} b^{\\frac{4}{9}}$\n\n(C) $2^{-3} a^{-\\frac{5}{3}} b^{\\frac{7}{9}}$\n\n(D) $2^{-\\frac{7}{6}} a^{-\\frac{10}{3}} b^{\\frac{5}{9}}$\n\n(E) $2^{\\frac{7}{3}} a^{-2} b^{\\frac{1}{6}}$",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nIzraz zapišemo s potencami in dobimo\n$$\n\\frac{4^{\\frac{1}{3}} \\frac{2}{3} a^{\\frac{1}{9}} b^{\\frac{1}{9}} \\cdot 1}{8^{-\\frac{2}{3}} b^{\\frac{4}{3}} a^{-\\frac{1}{3}}}\n$$\nŠtevili $4$ in $8$ zapišemo kot potenco števila $2$ in izraz zapišemo brez ulomka ter dobimo\n$$\n2^{\\frac{2}{3}} \\cdot 2^{2} \\cdot a^{\\frac{2}{3}} \\cdot a^{\\frac{1}{9}} \\cdot a^{\\frac{1}{3}} \\cdot b^{\\frac{1}{9}} \\cdot b^{-\\frac{4}{3}}\n$$\nKo združimo potence, dobimo rezultat\n$$\n2^{\\frac{8}{3}} a^{\\frac{10}{9}} b^{-\\frac{11}{9}}\n$$\nPravilen je odgovor $A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42046,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKoliko je $\\frac{13}{5} \\sin x$, če je $\\frac{13}{12} \\cos x = -1$?\n(A) 1\n(B) $\\frac{5}{13}$\n(C) $-\\frac{5}{13}$\n(D) -1\n(E) $\\pm 1$",
    "options": [],
    "answer": "E",
    "solution": "Solution:\nIzrazimo $\\cos x = -\\frac{12}{13}$ in uporabimo zvezo $\\sin^2 x + \\cos^2 x = 1$. Izračunamo $\\sin^2 x$:\n$$\n\\sin^2 x = 1 - \\cos^2 x = 1 - \\left(-\\frac{12}{13}\\right)^2 = \\frac{25}{169}.\n$$\nKer je $\\sin x = \\pm \\frac{5}{13}$, je vrednost izraza\n$$\n\\frac{13}{5} \\sin x = \\frac{13}{5} \\cdot \\left( \\pm \\frac{5}{13} \\right) = \\pm 1.\n$$\nPravilen je odgovor $E$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42047,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNajveč koliko izmed vsot $x+y$, $x+z$, $x+w$, $y+z$, $y+w$ in $z+w$ je lahko lihih, če so $x$, $y$, $z$ in $w$ naravna števila?\n\n(A) 2\n(B) 3\n(C) 4\n(D) 5\n(E) 6",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nOpazimo, da so dane vsote ravno vse možne vsote po 2 izmed števil $x$, $y$, $z$ in $w$. Če nobeno od števil $x$, $y$, $z$ in $w$ ni liho, tudi nobena vsota ni liha. Če je liho natanko 1 izmed števil, so lihe natanko 3 vsote, če sta lihi 2 števili, so lihe 4 vsote, če so liha 3 števila, so lihe 3 vsote, če pa so liha vsa 4 števila, ni nobena vsota liha. Torej so lihe največ 4 vsote.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42048,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVsota dveh naravnih števil je enaka trikratniku njune razlike, njun zmnožek pa je enak štirikratniku njune vsote. Koliko je vsota teh dveh naravnih števil?\n\n(A) 9\n(B) 10\n(C) 12\n(D) 15\n(E) 18",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nOznačimo ti dve naravni števili z $m$ in $n$. Tedaj je $m+n=3(m-n)$ in $m n=4(m+n)$. Iz prve enakosti dobimo $4 n=2 m$ oziroma $m=2 n$. Ko slednje vstavimo v drugo enakost, dobimo $2 n^{2}=12 n$, od koder sledi $n=6$. Torej je $m=12$ in $m+n=18$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42049,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKolikšen je natančen volumen vrtenine, ki jo dobimo, če pravokotnik s stranicama $a=2$ cm in $b=3$ cm zavrtimo okoli simetrale krajše stranice?\n(A) $2\\pi$ cm$^3$\n(B) $4\\pi$ cm$^3$\n(C) $12\\pi$ cm$^3$\n(D) $\\pi$ cm$^3$\n(E) $3\\pi$ cm$^3$",
    "options": [],
    "answer": "E",
    "solution": "Solution:\nČe pravokotnik s stranicama $a=2$ cm in $b=3$ cm zavrtimo okoli simetrale krajše stranice, dobimo vrtenino v obliki valja s polmerom $r=1$ cm in višino $v=3$ cm. Volumen valja izračunamo po obrazcu $V=\\pi r^{2} v=3\\pi$ cm$^3$. Pravilen je odgovor $E$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 42050,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNaj bosta $p$ in $q$ različni praštevili. Za koliko različnih vrednosti $a$ sta obe rešitvi kvadratne enačbe $x^{2}+a x+p q=0$ celoštevilski?\n(A) 4\n(B) 3\n(C) 2\n(D) 1\n(E) 0",
    "options": [],
    "answer": "A",
    "solution": "Solution:\nKer je koeficient pri $x^{2}$ enak 1, je po Vietovih pravilih produkt $x_{1} x_{2}$ obeh rešitev enačbe enak $p q$. Ker morata biti obe rešitvi celoštevilski in sta $p$ in $q$ praštevili, imamo za množico rešitev $\\{x_{1}, x_{2}\\}$ le možnosti $\\{1, p q\\},\\{-1,-p q\\},\\{p, q\\}$ ali $\\{-p,-q\\}$. Po Vietovih pravilih je koeficient $a$ enak $-\\left(x_{1}+x_{2}\\right)$, zato ima lahko 4 različne vrednosti, to so $1+p q,-1-p q, p+q$ in $-p-q$. Vse 4 vrednosti so med sabo različne.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42051,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nŽelimo splesti $20~\\mathrm{m}$ dolg navijaški šal. V koliko dneh ga bomo dokončali, če prvi dan spletemo $18~\\mathrm{cm}$, nato pa vsak naslednji dan za $4~\\mathrm{cm}$ več kot predhodni dan?\n\n(A) v 27 dneh\n(B) v 18 dneh\n(C) v 36 dneh\n(D) v 28 dneh\n(E) v 497 dneh",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nUgotovimo, da gre za aritmetično zaporedje s prvim členom $18$ in diferenco $4$ ter vsoto prvih $n$ členov $2000$. Zapišemo njegov splošni člen $a_{n} = 18 + (n-1) 4$ in formulo za vsoto prvih $n$ členov aritmetičnega zaporedja $S_{n} = \\frac{n}{2} (a_{1} + a_{n}) = 2000$. Dobimo enačbo oblike $n^{2} + 8n - 1000 = 0$. Rešitvi enačbe sta $n_{1} \\doteq 27,87$ in $n_{2} \\doteq -35,87$. Ugotovimo, da ustreza samo pozitivna rešitev, torej da bi $20~\\mathrm{m}$ šala spletli v $28$ dneh. Pravilen je odgovor $D$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42052,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV trapezu $ABCD$ je dolžina kraka $AD$ enaka $13~\\mathrm{cm}$ in dolžina kraka $BC$ enaka $9~\\mathrm{cm}$. Kot $\\angle BAD$ meri $37^\\circ$. Kolikšna je velikost kota $\\angle BCD$, zaokrožena na dve decimalni mesti, če je kot $\\angle CBA$ ostri?\n\n(A) $37^\\circ$\n(B) $143^\\circ$\n(C) $48,29^\\circ$\n(D) $119,62^\\circ$\n(E) $71,59^\\circ$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nIzračunamo višino trapeza $v = d \\cdot \\sin \\alpha \\doteq 7,82~\\mathrm{cm}$. Potem iz $\\sin \\beta = \\frac{v}{b}$ dobimo $\\beta_1 \\doteq 60,38^\\circ$ in $\\beta_2 = 180^\\circ - \\beta_1 \\doteq 119,62^\\circ$. Pravilen je odgovor $D$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42053,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKristina je narisala 2 kvadrata, katerih dolžine stranice $v$ centimetrih so naravna števila, in osenčila del večjega kvadrata, ki leži zunaj manjšega kvadrata (glej sliko). Ploščina osenčenega območja je enaka $43~\\mathrm{cm}^2$. Koliko kvadratnih centimetrov je vsota ploščin obeh Kristininih kvadratov?\n![](attached_image_1.png)\n(A) 882\n(B) 925\n(C) 968\n(D) 1685\n(E) 2022",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nOznačimo dolžino stranice večjega kvadrata v centimetrih z $a$, manjšega pa z $b$. Tedaj je $43 = a^2 - b^2 = (a + b)(a - b)$. Ker pa sta $a + b$ in $a - b$ naravni števili in je $43$ praštevilo, sledi $a + b = 43$ in $a - b = 1$. Torej je $a = 22$ in $b = 21$. Vsota ploščin obeh Kristininih kvadratov je enaka $a^2 + b^2 = 484 + 441 = 925~\\mathrm{cm}^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42054,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV posodi je $60 \\%$ rdečih in $40 \\%$ modrih bonbonov. $30 \\%$ rdečih in $15 \\%$ modrih bonbonov je čokoladnih. Koliko procentov bonbonov ni čokoladnih?\n\n(A) $24 \\%$\n(B) $85 \\%$\n(C) $90 \\%$\n(D) $76 \\%$\n(E) $45 \\%$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nRdečih čokoladnih je $0,6 \\cdot 0,3 = 0,18$, kar je $18 \\%$. Modrih čokoladnih je $0,4 \\cdot 0,15 = 0,06$, kar je $6 \\%$. Vseh čokoladnih je vsota, torej $24 \\%$ in tistih, ki niso čokoladni $76 \\%$. Pravilen je odgovor $D$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42055,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDrugi največji delitelj nekega naravnega števila $n$ je 2022. Kateri je tretji največji delitelj tega naravnega števila $n$?\n(A) 337\n(B) 674\n(C) 1011\n(D) 1348\n(E) 2021",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nDrugi največji delitelj naravnega števila $n$ je enak $\\frac{n}{p}$, kjer je $p$ najmanjše praštevilo, ki deli $n$. Torej je $n = 2022 p = 2 \\cdot 3 \\cdot 337 \\cdot p$. Od tod sledi, da je $n$ deljiv z $2$, torej je $p = 2$. Tretji največji delitelj števila $n$ je zato enak $2 \\cdot 337 \\cdot 2 = 1348$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42056,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZa padajočo linearno funkcijo $f(x)=k \\cdot x+n$ velja $f(x+y)=f(x)+f(y)-3$ in $f(k)=2 f(1)$, za vsak $x, y \\in \\mathbb{R}$. Koliko je vrednost $f(-1)$ ?\n\n(A) 4\n(B) 0\n(C) 2\n(D) -2\n(E) -4",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\n$f(x+y)=k(x+y)+n=k x+k y+n$ in $f(x)+f(y)-3=k x+n+k y+n-3$. Iz tega dobimo $n=3$.\n\nIz $f(k)=2 f(1)$ dobimo enačbo $k \\cdot k+3=2 k+6$. Preoblikujemo jo v $k^{2}-2 k-3=0$.\n\nRešitev $3$ ne ustreza. Za rešitev $-1$ dobimo $f(x)=-x+3$.\n\nZato je $f(-1)=4$.\n\nPravilen je odgovor $A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42057,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je izraz $\\frac{x^{n-1}}{x^{n}-2 x^{n-1}}-\\frac{x^{n}}{x^{n+1}-4 x^{n-1}}$. Kateri izraz je ekvivalenten izrazu za $x \\neq 0$?\n\n(A) $\\frac{1}{(x-2)}$\n\n(B) $\\frac{2}{(x-2)(x+2)}$\n\n(C) $\\frac{1}{(x+2)}$\n\n(D) $\\frac{2 x}{(x-2)(x+2)}$\n\n(E) $\\frac{1-x}{(x-2)(x+2)}$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nV imenovalcih ulomkov izpostavimo skupni faktor ter krajšamo, kar se da\n\n$\\frac{x^{n-1}}{x^{n}-2 x^{n-1}}-\\frac{x^{n}}{x^{n+1}-4 x^{n-1}} = \\frac{x^{n-1}}{x^{n-1}(x-2)}-\\frac{x^{n}}{x^{n-1}\\left(x^{2}-4\\right)} = \\frac{1}{(x-2)}-\\frac{1}{x^{-1}\\left(x^{2}-4\\right)}$.\n\nSeštejemo ulomka\n\n$\\frac{1}{(x-2)}-\\frac{x}{\\left(x^{2}-4\\right)} = \\frac{x+2}{(x-2)(x+2)}-\\frac{x}{(x-2)(x+2)} = \\frac{2}{(x-2)(x+2)}$.\n\nPravilen je odgovor $B$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42058,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ funkcija, za katero velja\n$$\nf(x)= \\begin{cases}x ; & x \\geq 2 \\\\ f(4-x) ; & 0 \\leq x<2 \\\\ f(x+2) ; & x<0\\end{cases}\n$$\nKoliko je $f(-5)$ ?\n(A) -3\n(B) -1\n(C) 1\n(D) 3\n(E) 5",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\n$$\nf(-5)=f(-5+2)=f(-3)=f(-3+2)=f(-1)=f(-1+2)=f(1)=f(4-1)=f(3)=3\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42059,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKatera tangenta na parabolo $z$ enačbo $y = x^{2} + x + 9$ je vzporedna premici $z$ enačbo $-4x + 2y - 5 = 0$?\n\n(A) $4x - 8y + 37 = 0$\n(B) $4x - 8y - 37 = 0$\n(C) $8x - 4y - 35 = 0$\n(D) $-4x - 8y + 37 = 0$\n(E) $8x - 4y + 35 = 0$",
    "options": [],
    "answer": "E",
    "solution": "Solution:\nZapišemo enačbo $x^{2} + x + 9 = 2x + n$ in jo preoblikujemo do oblike $x^{2} - x + 9 - n = 0$. Potem upoštevamo pogoj, da ima kvadratna enačba eno dvojno realno rešitev, če je vrednost diskriminante kvadratne enačbe enaka $0$. V ta pogoj vstavimo vrednosti parametrov in dobimo enačbo $1 - 4(9 - n) = 0$. Rešitev te enačbe je $n = \\frac{35}{4}$ in tako je enačba tangente $8x - 4y + 35 = 0$. Pravilen je odgovor $E$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42060,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHkrati vržemo 3 poštene igralne kocke različnih barv. V koliko primerih lahko dobimo vsoto pik 10?\n\n(A) 30\n(B) 27\n(C) 10\n(D) 6\n(E) 36",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nVsoto $10$ pik dobimo v šestih možnih izidih, in sicer, v izidih $\\{1,3,6\\}$, $\\{1,4,5\\}$ in $\\{2,3,5\\}$ vsakič $3! = 6$, v izidih $\\{2,2,6\\}$, $\\{3,3,4\\}$ in $\\{4,4,2\\}$ pa vsakič $\\frac{3!}{2!} = 3$. Če seštejemo vse možnosti $6 \\cdot 3 + 3 \\cdot 3$ dobimo $27$ možnosti. Pravilen je odgovor $B$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42061,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse pare naravnih števil $k$ in $n$, za katere ima ulomek $\\frac{4^{k-1}}{n}$ v decimalnem zapisu obliko $0, \\bar{n}$, kjer $\\bar{n}$ označuje periodo. Na primer, če je $n=720$, tedaj je $0, \\bar{n}=0,720720720 \\ldots$",
    "options": [],
    "answer": "k=1, n=3; k=2, n=6",
    "solution": "Solution:\n\nIz navodil naloge sledi $\\frac{4^{k-1}}{n}=0, \\bar{n}$. Označimo z $m$ število števk števila $n$. Če enačbo pomnožimo z $10^{m}$, dobimo $\\frac{4^{n}-1 \\cdot 10^{m}}{n}=n, \\bar{n}$. Prvo enačbo odštejemo od druge, da dobimo $\\frac{4^{k-1} \\cdot 10^{m}}{n}-\\frac{4^{k-1}}{n}=n$, kar lahko poenostavimo do\n$$\n4^{k-1} \\cdot\\left(10^{m}-1\\right)=n^{2}\n$$\nKer je $10^{m}-1=\\underbrace{99 \\ldots 99}_{m}=9 \\cdot \\underbrace{11 \\ldots 11}_{m}$, sledi\n$$\n4^{k-1} \\cdot 9 \\cdot \\underbrace{11 \\ldots 11}_{m}=n^{2}\n$$\nDesna stran enačbe je popoln kvadrat, hkrati pa je tudi število $4^{k-1} \\cdot 9=\\left(2^{k-1} \\cdot 3\\right)^{2}$ popoln kvadrat. Od tod sklepamo, da mora biti tudi število $\\underbrace{11 \\ldots 11}_{m}$ popoln kvadrat. Če je $m \\geq 2$ je ostanek števila $\\underbrace{11 \\ldots 11}_{m}$ pri deljenju s 4 enak ostanku števila 11 pri deljenju s 4, torej 3. Toda popoln kvadrat ima pri deljenju s 4 lahko ostanek le 0 ali 1. Od tod sledi, da mora biti $m=1$, torej je $n$ cifra. Dobimo enačbo $4^{k-1} \\cdot 9=n^{2}$. Ker pa je $n$ cifra, je $n^{2} \\leq 81$, zato mora biti $4^{k-1} \\leq 9$ oziroma $k \\leq 2$. Rešitvi sta torej $k=1$ in $n=3$ ter $k=2$ in $n=6$. V prvem primeru imamo $\\frac{1}{3}=0, \\overline{3}$, v drugem primeru pa $\\frac{4}{6}=0, \\overline{6}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42062,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNaj bo $p$ praštevilo in $n \\leq p^{p}$ naravno število. Dokaži, da vsaj eno od števil $n+p^{p}$ in $n \\cdot p^{p}$ ni popoln kvadrat.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDenimo, da sta obe števili popolna kvadrata, torej $n+p^{p}=a^{2}$ in $n \\cdot p^{p}=b^{2}$, kjer sta $a$ in $b$ naravni števili.\n\nČe je $p=2$, tedaj je $n \\leq 2^{2}=4$. Toda nobeno od števil $1+2^{2}$, $2+2^{2}$, $3+2^{2}$ in $4+2^{2}$ ni popoln kvadrat. Torej mora biti $p$ liho praštevilo.\n\nPišimo $p=2k+1$, kjer je $k$ naravno število. Sledi $n \\cdot p^{2k+1}=b^{2}$. Torej mora biti $b^{2}$ deljiv s $p^{2k+1}$ in zato je $b$ deljiv s $p^{k+1}$. Pišimo $b=p^{k+1} d$, kjer je $d$ naravno število. Tedaj je $n \\cdot p^{2k+1}=p^{2k+2} d^{2}$ oziroma $n=p d^{2}$.\n\nKer je $n \\leq p^{p}=p^{2k+1}$, sledi $d \\leq p^{k}$. Pišimo $d=p^{m} u$ in $a=p^{r} v$, kjer sta $m$ in $r$ nenegativni celi števili, $u$ in $v$ pa naravni števili tuji $p$. Tedaj je $m \\leq k$ in $n=p d^{2}=p^{2m+1} u^{2}$, zato iz enačbe $n+p^{p}=a^{2}$ sledi $p^{2m+1} u^{2}+p^{2k+1}=p^{2r} v^{2}$ oziroma $p^{2m+1}\\left(u^{2}+p^{2k-2m}\\right)=p^{2r} v^{2}$.\n\nČe je $m<k$, tedaj $p$ deli $p^{2k-2m}$, a ne deli $u^{2}$, zato je največja potenca praštevila $p$, ki deli levo stran enakosti, $p^{2m+1}$, največja potenca praštevila $p$, ki deli desno stran, pa $p^{2r}$. Prišli smo do protislovja, saj je prva potenca liha, druga pa soda.\n\nTorej ostane le možnost $k=m$. Toda v tem primeru je $n=p^{2k+1} u^{2}=p^{p} u^{2}$, zato iz $n \\leq p^{p}$ sledi $u^{2}=1$ in $n=p^{p}$. V tem primeru pa število $n+p^{p}=2 p^{p}$ ni popoln kvadrat, saj $p \\neq 2$.\n\nPrišli smo do protislovja, od koder sklepamo, da vsaj eno od števil $n+p^{p}$ in $n \\cdot p^{p}$ ni popoln kvadrat.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42063,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoišči vse pare naravnih števil $a$ in $b$, za katere velja\n$$\nv = a b - 2 a - 4 b\n$$\nkjer je $v$ najmanjši skupni večkratnik števil $a$ in $b$.",
    "options": [],
    "answer": "(7, 7), (6, 12), (10, 5), (16, 4)",
    "solution": "Solution:\n\nNaj bo $d$ največji skupni delitelj števil $a$ in $b$ in naj bosta $m$ in $n$ taki naravni števili, da je $a = d m$ in $b = d n$. Tedaj sta števili $m$ in $n$ tuji in velja $v = d m n$. Z upoštevanjem teh zvez iz dane enačbe sledi\n$$\nd m n = d^2 m n - 2 d m - 4 d n,\n$$\nod koder po krajšanju z $d$ dobimo\n$$\nm n = d m n - 2 m - 4 n\n$$\nLeva stran enačbe je deljiva z $m$, zato mora biti tudi desna, kar pomeni, da $m$ deli $4 n$. Števili $m$ in $n$ sta tuji, zato $m$ deli $4$, torej je $m \\in \\{1, 2, 4\\}$. Podobno sklepamo, da $n$ deli $2 m$ in zato $n$ deli $2$, torej je $n \\in \\{1, 2\\}$. Ker sta števili $m$ in $n$ tuji, ne moreta biti hkrati sodi, zato imamo za pare $(m, n)$ naslednje štiri možnosti: $(1, 1), (1, 2), (2, 1), (4, 1)$. Iz zgornje enačbe izrazimo\n$$\nd = \\frac{m n + 4 n + 2 m}{m n} = 1 + \\frac{4}{m} + \\frac{2}{n}\n$$\nin za vse štiri možnosti po vrsti izračunamo $d = 7, 6, 5, 4$. Z upoštevanjem enakosti $a = d m$ in $b = d n$, dobimo štiri rešitve za pare $(a, b)$, to so $(7, 7), (6, 12), (10, 5)$ in $(16, 4)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42064,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV kvadratu $ABCD$ so včrtane krožnica $\\mathcal{K}$, polkrožnica $\\mathcal{P}$ in četrtina krožnice $\\mathcal{Q}$. Četrtina krožnice $\\mathcal{Q}$ ima središče v oglišču $A$ in vsebuje točki $B$ in $D$. Polkrožnica $\\mathcal{P}$ ima središče v razpolovišču stranice $AD$ in vsebuje točki $A$ in $D$. Krožnica $\\mathcal{K}$ se dotika polkrožnice $\\mathcal{P}$ v točki $E$, četrtine krožnice $\\mathcal{Q}$ v točki $F$ in stranice $AB$ kvadrata $ABCD$ v točki $G$.\n\na) Izrazi polmer krožnice $\\mathcal{K}$ z dolžino stranice kvadrata $ABCD$.\n\nb) Dokaži, da je premica $EF$ vzporedna stranici $AB$.",
    "options": [],
    "answer": "The radius equals one quarter of the side length, and the line through the tangency points is parallel to the side.",
    "solution": "Solution:\n\n(a) Označimo dolžino stranice kvadrata $ABCD$ z $a$, polmer krožnice $\\mathcal{K}$ pa z $r$. Naj bo $S$ središče polkrožnice $\\mathcal{P}$, $T$ središče krožnice $\\mathcal{K}$ in $H$ pravokotna projekcija točke $T$ na stranico $AD$. Dolžino daljice $AG$ označimo z $x$. Ker je $|AT| = |AF| - |FT| = a - r$, po Pitagorovem izreku za trikotnik $AGT$ velja $x^2 + r^2 = (a - r)^2$, kar lahko poenostavimo do $x^2 = a^2 - 2ar$. Ker je $AGTH$ pravokotnik, je $|HT| = x$ in $|SH| = |SA| - |AH| = \\frac{a}{2} - r$. Hkrati je $|ST| = |SE| + |ET| = \\frac{a}{2} + r$, zato po Pitagorovem izreku za trikotnik $THS$ velja $x^2 + \\left(\\frac{a}{2} - r\\right)^2 = \\left(\\frac{a}{2} + r\\right)^2$, kar lahko poenostavimo do $x^2 = 2ar$. Iz obeh enakosti sledi $a^2 - 2ar = 2ar$ oziroma $r = \\frac{a}{4}$.\n\n(b) Iz rezultata pri točki (a) sledi $|AH| = \\frac{a}{4}$ in zato tudi $|HS| = |SA| - |AH| = \\frac{a}{2} - \\frac{a}{4} = \\frac{a}{4}$, torej je trikotnik $ATS$ enakokrak z vrhom pri $T$. Hkrati je tudi trikotnik $FTE$ enakokrak z vrhom pri $T$. Od tod izpeljemo\n$$\n\\angle EFT = \\frac{\\pi - \\angle FTE}{2} = \\frac{\\angle ETA}{2} = \\angle HTA = \\angle GAT\n$$\ntorej je premica $EF$ vzporedna stranici $AB$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42065,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrapez $ABCD$ je včrtan krožnici $\\mathcal{K}$. Nosilki stranic $AD$ in $BC$ se sekata v točki $M$, tangenti na krožnico $\\mathcal{K}$ v točkah $B$ in $D$ pa se sekata v točki $N$. Dokaži, da sta daljici $MN$ in $AB$ vzporedni.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNaj bo $S$ središče krožnice $\\mathcal{K}$ in $T$ presečišče premice $MS$ s stranico $AB$. Ker je trapez $ABCD$ tetiven, je enakokrak s krakoma $AD$ in $BC$. Označimo $\\alpha=\\angle BAD=\\angle CBA$. Zaradi simetrije lahko predpostavimo, da je $|AB|>|CD|$. Trikotnik $BMA$ je enakokrak z vrhom pri $M$, premica $MS$ pa je zaradi simetrije njegova višina, torej je pravokotna na stranico $AB$. Kot $\\angle BSD$ je središčni kot nad lokom $\\widehat{BD}$ krožnice $\\mathcal{K}$, kot $\\angle BAD=\\alpha$ pa obodni kot nad istim lokom, zato je $\\angle BSD=2\\alpha$. Ker sta trikotnika $SND$ in $SNB$ skladna, sledi $\\angle NSD=\\angle BSN=\\alpha$. Torej sta trikotnika $ATM$ in $SND$ podobna, saj se ujemata v dveh kotih (pravem kotu in kotu $\\alpha$), zato je $\\angle AMT=\\angle SND$. Po izreku o obodnem kotu sledi, da so točke $D, S, N$ in $M$ konciklične, torej je $\\angle SMN=\\angle SDN=\\frac{\\pi}{2}$. S tem smo dokazali, da je premica $MS$ pravokotna na daljici $MN$ in $AB$, zato sta ti dve daljici vzporedni.\n\n\n2. način. Naj bo $S$ središče krožnice $\\mathcal{K}$ in $T$ presečišče premice $MS$ s stranico $AB$. Podobno kot v prvi rešitvi sklepamo, da je trapez $ABCD$ enakokrak in označimo $\\alpha=\\angle BAD=\\angle CBA$. Zaradi simetrije lahko zopet predpostavimo, da je $|AB|>|CD|$ oziroma $\\alpha<\\frac{\\pi}{2}$. Tedaj je $\\angle DMB=\\angle AMB=\\pi-\\angle MBA-\\angle BAM=\\pi-2\\alpha$. Kot $\\angle BSD$ je središčni kot nad lokom $\\widehat{BD}$ krožnice $\\mathcal{K}$, kot $\\angle BAD=\\alpha$ pa obodni kot nad istim lokom, zato je $\\angle BSD=2\\alpha$. Štirikotnik $SBN D$ je po Talesovem izreku tetiven, zato je $\\angle DNB=\\pi-\\angle BSD=\\pi-2\\alpha$. S tem smo pokazali, da je $\\angle DNB=\\angle DMB$, torej je tudi štirikotnik $DBNM$ tetiven. Od tod sledi\n$$\n\\angle AMN=\\angle DMN=\\pi-\\angle NBD=\\pi-\\left(\\frac{\\pi}{2}-\\angle DBS\\right)=\\frac{\\pi}{2}+\\angle DBS=\\frac{\\pi}{2}+\\angle DMS\n$$\nZaradi simetrije je premica $MS$ oziroma $MT$ višina enakokrakega trikotnika $BMA$, zato je $\\angle MTA=\\frac{\\pi}{2}$ in $\\angle DMS=\\angle DMT=\\frac{\\pi}{2}-\\angle TAM=\\frac{\\pi}{2}-\\alpha$. Iz zgornje enakosti zato sledi\n$$\n\\angle AMN=\\frac{\\pi}{2}+\\angle DMS=\\frac{\\pi}{2}+\\left(\\frac{\\pi}{2}-\\alpha\\right)=\\pi-\\alpha=\\angle ADC\n$$\ntorej sta daljici $MN$ in $AB$ vzporedni.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42066,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKvocient dolžin katete $a$ in hipotenuze $c$ v pravokotnem trikotniku je $3:4$.\na) Pod katerim kotom se sekata simetrali ostrih kotov?\nb) Pod katerim kotom seka simetrala kota $\\alpha$ nasprotno kateto?",
    "options": [],
    "answer": "a) 45°; b) 114° 18′",
    "solution": "Solution:\n\na) Kot med simetralama ostrih kotov je enak $\\varphi = 180^\\circ - \\frac{\\alpha}{2} - \\frac{\\beta}{2}$. Ker v pravokotnem trikotniku velja, da je $\\alpha = 90^\\circ - \\beta$, je kot $\\varphi$ enak\n$$\n\\varphi = 180^\\circ - \\frac{90^\\circ - \\beta}{2} - \\frac{\\beta}{2} = 180^\\circ - 45^\\circ = 135^\\circ.\n$$\nIskani kot je suplementaren kotu $\\varphi$ in je enak $45^\\circ$.\n\nb) Kot med simetralo kota $\\alpha$ in stranico $a$ je enak $\\varphi = 180^\\circ - \\frac{\\alpha}{2} - \\beta$. Ker v pravokotnem trikotniku velja, da je $\\beta = 90^\\circ - \\alpha$, je kot $\\varphi$ enak\n$$\n\\varphi = 180^\\circ - \\frac{\\alpha}{2} - (90^\\circ - \\alpha) = 90^\\circ + \\frac{\\alpha}{2}.\n$$\nKer je velikost kota $\\alpha$ enak $\\alpha = \\sin^{-1}\\left(\\frac{3}{4}\\right) = 48^\\circ 35'$, je velikost iskanega kota enaka $\\varphi = 114^\\circ 18'$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42067,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana je tabela velikosti $1 \\times n$, kjer je $n>10$ naravno število. Polja tabele so po vrsti od leve proti desni oštevilčena z naravnimi števili od $1$ do $n$. Polje številka $10$ je črno in na njem je postavljen žeton, vsa ostala polja tabele so bela. Dva igralca izmenjaje igrata naslednjo igro. Igralec, ki je prvi na potezi, premakne žeton na poljubno belo polje tabele in to polje pobarva črno. V vsaki naslednji potezi igralec, ki je na potezi, premakne žeton na eno od belih polj tabele, pri čemer pa mora z žetonom preskočiti vsaj eno črno polje tabele. Polje, na katerega postavi žeton, nato pobarva črno. Igralec, ki prvi ne more izvesti poteze, izgubi igro. V odvisnosti od števila $n$ določi, kateri igralec ima zmagovito strategijo, tisti, ki je prvi na potezi, ali tisti, ki je drugi na potezi.",
    "options": [],
    "answer": "The second player has a winning strategy if and only if the row length is nineteen; otherwise, the first player has a winning strategy.",
    "solution": "Solution:\n\nPokažimo, da ima v primeru $n=19$ zmagovito strategijo drugi igralec, za vse ostale $n$ pa prvi igralec.\n\nNaj bo $n=19$. Tedaj je na vsaki strani polja $10$ natanko $9$ polj tabele. Če prvi igralec v prvi potezi prestavi žeton na polje manjše od $10$, tedaj lahko drugi igralec odigra strategijo $11,12,13, \\ldots, 19$, pri kateri prvega igralca prisili, da vedno premakne žeton na polje manjše od $10$, sam pa premakne žeton na polje večje od $10$. Prepričajmo se, da res lahko tako odigra. Ko je na vrsti prvi igralec in je žeton na polju $k>10$, so vsa polja med $10$ in $k$ črna, vsa polja večja od $k$ pa bela. Prvi igralec mora zato žeton premakniti v levo in pri tem preskočiti polje $10$. Ko je na vrsti drugi igralec, lahko izvede svojo potezo, saj preskoči črno polje $10$. Ker ima prvi igralec pri tem na voljo največ $9$ potez, drugi pa natanko $9$ potez, bo prvemu igralcu prej zmanjkalo ustreznih belih polj kot drugemu in bo zato izgubil. Podobno sklepamo, da če prvi igralec v prvi potezi prestavi žeton na polje večje od $10$, tedaj lahko drugi igralec odigra zmagovito strategijo $9,8,7, \\ldots, 1$.\n\nNaj bo $n \\neq 19$. Tedaj je na eni strani polja $10$ več polj tabele kot na drugi strani. Če je več polj na levi strani polja $10$, lahko prvi igralec odigra zmagovito strategijo $9,8,7, \\ldots$, saj bo tedaj podobno kot zgoraj drugemu igralcu prej zmanjkalo ustreznih belih polj. Če pa je več polj na desni strani polja $10$, lahko prvi igralec odigra zmagovito strategijo $11,12,13, \\ldots$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42068,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTone, Luka in Tine so zbirali star papir in dobili denarno nagrado. Prvotno naj bi bila nagrada razdeljena v razmerju $7:6:5$. Kasneje so dogovor spremenili in razdelili nagrado v razmerju $6:5:4$. Obe razmerji sta zapisani v istem vrstem redu, kot so navedena imena.\n\na. Katera delitev je za Tineta ugodnejša? Utemelji odgovor.\n\nb. Tone je dobil pri drugi delitvi 216 evrov več kot Tine. Koliko evrov je dobil vsak izmed njih?",
    "options": [],
    "answer": "a) The first division is better for Tine. b) Tone 648 euros, Luka 540 euros, Tine 432 euros.",
    "solution": "Solution:\n\na.\nTine bi po prvotnem dogovoru dobil $\\frac{5}{7+6+5} = \\frac{5}{18} = \\frac{25}{90}$ celotne nagrade, po spremembi pa dobi $\\frac{4}{6+5+4} = \\frac{4}{15} = \\frac{24}{90}$ nagrade. Zato je za Tineta ugodnejša prva delitev.\n\nb.\nVemo, da so razdelili nagrado v razmerju $6:5:4$. Tako je Tonetov delež enak $6x$, Lukov $5x$, Tinetov pa $4x$. Zapišemo zvezo med Tonetovim in Tinetovim deležem $6x - 216 = 4x$ in dobimo $x = 108$. Tone je torej dobil $6x = 6 \\cdot 108 = 648$ evrov, Luka $5x = 5 \\cdot 108 = 540$ evrov, Tine pa $4x = 4 \\cdot 108 = 432$ evrov.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42069,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Izračunaj presečišči in velikost kota med krivuljama $y = -x^{-2}$ in $y = -\\frac{x^{2}}{2} + \\frac{1}{2}$.\n\nb) Dana je funkcija $f(x) = \\frac{x}{\\ln x}$. Določi definicijsko območje funkcije $f$ in intervale, na katerih funkcija $f$ narašča in pada.",
    "options": [],
    "answer": "a) Intersections: (√2, −1/2) and (−√2, −1/2). The angle between the curves at each intersection is 90 degrees.\nb) Domain: (0, 1) ∪ (1, ∞). Increasing on (e, ∞). Decreasing on (0, 1) and (1, e).",
    "solution": "Solution:\na)\nEnačimo krivulji: $-x^{-2} = -\\frac{x^{2}}{2} + \\frac{1}{2}$, nato množimo s skupnim imenovalcem in uredimo do oblike $x^{4} - x^{2} - 2 = 0$. Zapišemo v obliki produkta $(x^{2} - 2)(x^{2} + 1) = 0$, nato razstavimo še prvi oklepaj na $(x - \\sqrt{2})(x + \\sqrt{2})(x^{2} + 1) = 0$. Dobimo rešitvi $x_{1} = \\sqrt{2}$ in $x_{2} = -\\sqrt{2}$. Izračunamo ustrezni ordinati in zapišemo presečišči $P_{1}(\\sqrt{2}, -\\frac{1}{2})$ in $P_{2}(-\\sqrt{2}, -\\frac{1}{2})$.\n\nOdvajamo prvo krivuljo $y' = 2x^{-3}$ in zapišemo njen smerni koeficient $k_{1} = \\frac{\\sqrt{2}}{2}$. Odvajamo drugo krivuljo $y' = -x$ in zapišemo njen smerni koeficient $k_{2} = -\\sqrt{2}$. Vstavimo podatke v obrazec za izračun kota med krivuljama in dobimo $\\tan \\alpha = \\left| \\frac{k_{2} - k_{1}}{1 + k_{1} \\cdot k_{2}} \\right| = \\left| \\frac{-\\sqrt{2} - \\frac{\\sqrt{2}}{2}}{1 + \\left(\\frac{\\sqrt{2}}{2}\\right)(-\\sqrt{2})} \\right| = \\infty$. Kot med krivuljama je $\\alpha = 90^{\\circ}$.\n\nb)\nFunkcija ni definirana v polu pri $x = 1$. Zapišemo definicijsko območje funkcije $D_{f} = (0, 1) \\cup (1, \\infty)$. Funkcijo odvajamo in dobimo odvod $f'(x) = \\frac{\\ln x - 1}{(\\ln x)^{2}}$. Izračunamo ničlo odvoda $\\ln x = 1$ in dobimo rešitev $x = e$. Pol funkcije je pri $x = 1$. Določimo interval naraščanja $(e, \\infty)$ in intervala padanja $(0, 1) \\cup (1, e)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42070,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nKoliko je vrednost izraza $2^{0^{2^{3}}}+0^{2^{3^{2}}}+2^{3^{2^{0}}}+3^{2^{0^{2}}}$?\n(A) 3\n(B) 4\n(C) 7\n(D) 12\n(E) Večja od 100.",
    "options": [],
    "answer": "D",
    "solution": "Solution:\nUpoštevamo, da je $0^{m}=0$ in $n^{0}=1$ za katerikoli naravni števili $m$ in $n$, in dobimo\n$$\n2^{0^{2^{3}}}+0^{2^{3^{2}}}+2^{3^{2^{0}}}+3^{2^{0^{2}}}=2^{0}+0+2^{3}+3=1+0+8+3=12\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42071,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n2023-mesto naravno število $n$ ima vse števke enake 1. Koliko je vsota števk naravnega števila $n \\cdot 1111$?\n\n(A) 8080\n(B) 8083\n(C) 8086\n(D) 8092\n(E) 8101",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nČe pisno množimo $n \\cdot 1111$, najprej v računu 4-krat podpišemo število $n$ in nato seštevamo števke po stolpcih. Pri izračunu vsot števk nikoli ne pride do prenosa števk. Torej je vsota števk produkta $n \\cdot 1111$ enaka 4-kratniku vsote števk števila $n$, to je $4 \\cdot 2023 = 8092$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42072,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDelitelj naravnega števila $n$ je pravi delitelj, če je različen od 1 in $n$. Za koliko naravnih števil $n$ je število 17 največji pravi delitelj?\n\n(A) 0\n(B) 1\n(C) 3\n(D) 5\n(E) 7",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nKer je 17 pravi delitelj števila $n$, je $n = 17k$ za neko naravno število $k \\geq 2$. Ker je tudi $k$ pravi delitelj števila $n$, mora veljati $k \\leq 17$. Hkrati mora biti $k$ praštevilo, saj če bi bilo $k = ab$ za neki naravni števili $a, b > 1$, tedaj bi bil $17a$ pravi delitelj števila $n$, ki je večji od 17. Torej mora biti $k \\in \\{2, 3, 5, 7, 11, 13, 17\\}$ in vse te možnosti so dobre. Takih naravnih števil $n$ je 7.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42073,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDano je dvomestno naravno število $n$. Če seštejemo naslednja tri zaporedna naravna števila, ki sledijo številu $n$, dobimo dvomestno število, ki ima enaki števki kot število $n$. Koliko pozitivnih deliteljev ima število $n$ ?\n(A) 3\n(B) 4\n(C) 5\n(D) 6\n(E) 8",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nOznačimo $n=\\overline{ab}=10a+b$. Tedaj je $(n+1)+(n+2)+(n+3)=\\overline{ba}=10b+a$. Od tod sledi $10b+a=3n+6=30a+3b+6$ oziroma $29a=7b-6$. Ker pa sta $a$ in $b$ števki, je $29a \\leq 7 \\cdot 9-6=57$, od koder sledi $a=1$. Torej je $7b=35$ oziroma $b=5$ in zato $n=15$. Število $n$ ima 4 pozitivne delitelje, to so $1,3,5$ in $15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42074,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa matematičnem tekmovanju so tekmovalci reševali 3 naloge. Prvo nalogo je pravilno rešilo $90 \\%$ tekmovalcev, drugo $80 \\%$ in tretjo $70 \\%$ tekmovalcev. Najmanj koliko odstotkov tekmovalcev je pravilno rešilo vse tri naloge?\n(A) 30\n(B) 35\n(C) 40\n(D) 50\n(E) 70",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nDenimo, da je na tekmovanju sodelovalo $n$ tekmovalcev. Če sta $A$ in $B$ dve množici tekmovalcev, potem je $n \\geq |A \\cup B| = |A| + |B| - |A \\cap B|$ oziroma $|A \\cap B| \\geq |A| + |B| - n$. Ker je prvo nalogo pravilno rešilo $\\frac{9}{10} n$ tekmovalcev, drugo nalogo pa $\\frac{8}{10} n$ tekmovalcev, je po zgornji oceni prvi dve nalogi pravilno rešilo vsaj $\\frac{9}{10} n + \\frac{8}{10} n - n = \\frac{7}{10} n$ tekmovalcev, vse tri naloge pa je pravilno rešilo vsaj $\\frac{7}{10} n + \\frac{7}{10} n - n = \\frac{4}{10} n$ tekmovalcev, kar je $40 \\%$ tekmovalcev.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42075,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKatero je najmanjše naravno število $n$, za katero interval $\\left[\\frac{n+8}{2}, \\frac{2 n+14}{3}\\right]$ vsebuje vsaj 4 naravna števila?\n\n(A) 16\n(B) 17\n(C) 18\n(D) 19\n(E) Nobeno izmed naštetih.",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nDolžina intervala je enaka $\\frac{2 n+14}{3}-\\frac{n+8}{2}=\\frac{(4 n+28)-(3 n+24)}{6}=\\frac{n+4}{6}$. Ker mora interval vsebovati vsaj 4 naravna števila, mora biti njegova dolžina vsaj 3 (saj sta lahko obe krajišči naravni števili). Torej mora biti $\\frac{n+4}{6} \\geq 3$ oziroma $n \\geq 14$. Če je $n=14$, je interval enak [11,14] in vsebuje 4 naravna števila. Pravilen odgovor je torej (E).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42076,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV trikotniku $ABC$ je $\\angle ACB = 32^\\circ$. Na nosilki stranice $AB$ ležita točki $D$ in $E$, za kateri velja $|AD| = |AC|$ in $|BE| = |BC|$ (glej sliko). Koliko stopinj je velikost kota $\\angle DCE$?\n\n(A) 90\n(B) 96\n(C) 100\n(D) 106\n(E) 116\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nOznačimo $\\angle BAC = \\alpha$. Tedaj je $\\angle CBA = 180^\\circ - 32^\\circ - \\alpha = 148^\\circ - \\alpha$. Torej je $\\angle CAD = 180^\\circ - \\angle BAC = 180^\\circ - \\alpha$ in $\\angle EBC = 180^\\circ - \\angle CBA = \\alpha + 32^\\circ$. Trikotnika $CAD$ in $EBC$ sta enakokraka z vrhoma pri $A$ in $B$, zato je $\\angle DCA = \\frac{180^\\circ - \\angle CAD}{2} = \\frac{\\alpha}{2}$ in $\\angle BCE = \\frac{180^\\circ - \\angle EBC}{2} = 74^\\circ - \\frac{\\alpha}{2}$. Od tod izračunamo $\\angle DCE = \\angle DCA + \\angle ACB + \\angle BCE = 32^\\circ + 74^\\circ = 106^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42077,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je pravokotni trikotnik $ABC$ s pravim kotom pri $A$. Na stranicah $AB$, $BC$ in $CA$ zaporedoma ležijo točke $D$, $E$ in $F$, tako da velja $|BD| = |BE|$ in $|CF| = |CE|$ (glej sliko). Koliko stopinj je velikost kota $\\angle FED$?\n\n(A) 30\n(B) 37,5\n(C) 45\n(D) 52,5\n![](attached_image_1.png)\n(E) Nemogoče je določiti, saj je odgovor odvisen od velikosti kota $\\angle CBA$.",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nOznačimo kote trikotnika $ABC$ kot običajno z $\\alpha$, $\\beta$ in $\\gamma$. Tedaj je $\\alpha = 90^\\circ$ in zato je $\\beta + \\gamma = 90^\\circ$.\n\nKer sta trikotnika $DBE$ in $ECF$ enakokraka z vroma pri $B$ in $C$, je\n$$\n\\angle DEB = \\frac{180^\\circ - \\beta}{2} = 90^\\circ - \\frac{\\beta}{2}\n$$\nin\n$$\n\\angle CEF = \\frac{180^\\circ - \\gamma}{2} = 90^\\circ - \\frac{\\gamma}{2}.\n$$\nTorej je\n$$\n\\angle FED = 180^\\circ - \\angle CEF - \\angle DEB = \\frac{\\beta + \\gamma}{2} = 45^\\circ.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42078,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZlatar ima dve zlitini. Prva zlitina vsebuje $90\\%$ zlata, druga pa $54\\%$ zlata. Zlatar je zmešal $320~\\mathrm{g}$ prve zlitine in $160~\\mathrm{g}$ druge zlitine, da je dobil novo zlitino. Koliko odstotkov zlata vsebuje nova zlitina?\n(A) 33\n(B) 48\n(C) 65\n(D) 72\n(E) 78",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nV $320~\\mathrm{g}$ prve zlitine je $320 \\cdot \\frac{90}{100} = 32 \\cdot 9 = 288~\\mathrm{g}$ zlata, v $160~\\mathrm{g}$ druge zlitine pa $160 \\cdot \\frac{54}{100} = 16 \\cdot \\frac{27}{5}~\\mathrm{g}$ zlata. Delež zlata v novi zlitini je torej enak\n$$\n\\frac{32 \\cdot 9 + 16 \\cdot \\frac{27}{5}}{320 + 160} = \\frac{16 \\cdot \\left(18 + \\frac{27}{5}\\right)}{16 \\cdot (20 + 10)} = \\frac{\\frac{117}{5}}{30} = \\frac{117}{150} = \\frac{39}{50} = \\frac{78}{100}\n$$\nkar znaša $78\\%$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42079,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV kraju Zmajski Vrh so se prebivalci odločili, da bodo uporabljali nov način merjenja dnevnega časa. Vsak dan so s poldnevom in polnočjo razdelili na dve enaki polovici. Namesto, da bi vsako polovico razdelili na 12 ur s po 60 minutami, so jo razdelili na 10 zmajskih ur s po 100 zmajskimi minutami. Župan Zmajskega Vrha ima uro, ki je bila izdelana za merjenje časa v zmajskih urah in zmajskih minutah. Opoldan županova ura kaže čas 10.00, običajna ura pa kaže čas 12.00. Kakšen čas kaže običajna ura v trenutku, ko županova ura kaže čas 8.25 ?\n\n(A) 7.54\n(B) 8.15\n(C) 8.25\n(D) 9.15\n(E) 9.54",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nČas $8.25$ v zmajskih urah in zmajskih minutah ustreza $8 \\frac{25}{100} = 8 \\frac{1}{4} = \\frac{33}{4}$ zmajskih ur. Ker $10$ zmajskih ur ustreza $12$ običajnim uram, $1$ zmajska ura ustreza $\\frac{12}{10}$ običajne ure. Torej $\\frac{33}{4}$ zmajskih ur ustreza $\\frac{33}{4} \\cdot \\frac{12}{10} = \\frac{99}{10} = 9 \\frac{9}{10} = 9 \\frac{54}{60}$ običajnih ur. Torej bi običajna ura kazala $9.54$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42080,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV pravilni 8-kotnik je včrtan kvadrat z oglišči v ogliščih pravilnega 8-kotnika (glej sliko). Ploščina kvadrata je enaka $2~\\mathrm{cm}^2$. Koliko kvadratnih centimetrov je ploščina pravilnega 8-kotnika?\n\n(A) $1+\\sqrt{2}$\n(B) $\\frac{5}{2}$\n(C) $2 \\sqrt{2}$\n(D) 3\n(E) $3 \\sqrt{2}-1$\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nKvadrat s ploščino $2~\\mathrm{cm}^2$ ima stranico dolgo $\\sqrt{2}~\\mathrm{cm}$ in diagonalo dolgo $2~\\mathrm{cm}$. Če povežemo oglišča 8-kotnika z njegovim središčem, dobimo torej 8 enakokrakih trikotnikov s kraki dolžine $r=1~\\mathrm{cm}$ in notranjim kotom med kraki enakim $\\alpha=45^\\circ$. Ploščina enega takega trikotnika je $\\frac{1}{2} r^2 \\sin \\alpha = \\frac{1}{2} \\cdot \\frac{\\sqrt{2}}{2} = \\frac{\\sqrt{2}}{4}~\\mathrm{cm}^2$, ploščina 8-kotnika pa $8 \\cdot \\frac{\\sqrt{2}}{4} = 2 \\sqrt{2}~\\mathrm{cm}^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42081,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrapez $ABCD$ ima osnovnici dolgi $|AB| = 120\\ \\mathrm{mm}$ in $|CD| = 30\\ \\mathrm{mm}$ in prava kot pri ogliščih $B$ in $C$. Diagonali trapeza se sekata v točki $E$. Koliko milimetrov je točka $E$ oddaljena od stranice $BC$?\n\n(A) 21\n(B) 22\n(C) 23\n(D) 24\n(E) 25",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nNaj bo $F$ pravokotna projekcija točke $E$ na stranico $BC$. Trikotnika $ABE$ in $CDE$ sta podobna. Razmerje med njunima stranicama nasproti oglišča $E$ je enako $|AB| : |CD| = 120 : 30 = 4 : 1$, zato je tudi razmerje med njunima višinama na ti dve stranici enako $|FB| : |FC| = 4 : 1$. Tudi trikotnika $ABC$ in $EFC$ sta podobna, razmerje njunih višin skozi oglišče $C$ pa je enako $|CB| : |CF| = 5 : 1$. Torej je tudi razmerje med njunima stranicama nasproti oglišča $C$ enako $|AB| : |EF| = 5 : 1$ in zato je $|EF| = \\frac{|AB|}{5} = 24\\ \\mathrm{mm}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42082,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nV koordinatni sistem nariši množico točk $(x, y)$, ki ustrezajo pogoju $(|x| \\leq 2) \\wedge (-2 \\leq y \\leq 4) \\wedge (y \\leq x)$. Izračunaj ploščino in obseg lika, ki nastane.",
    "options": [],
    "answer": "Area = 8; Perimeter = 8 + 4√2",
    "solution": "Solution:\n\n1. V koordinatni sistem narišemo množico točk, ki ustreza pogoju $(|x| \\leq 2) \\wedge (-2 \\leq y \\leq 4) \\wedge (y \\leq x)$. Dobljena množica točk oblikuje pravokotni trikotnik s katetama $k_{1} = k_{2} = 4$. Izračunamo ploščino trikotnika $S = \\frac{k_{1} \\cdot k_{2}}{2} = 8$. S Pitagorovim izrekom izračunamo dolžino hipotenuze nastalega trikotnika $h = \\sqrt{32} = 4 \\sqrt{2}$ in njegov obseg $o = 8 + 4 \\sqrt{2}$.\n\n![](attached_image_1.png)\n\nNarisana množica točk $|x| \\leq 2$ ..... 1 točka\nNarisana množica točk $-2 \\leq y \\leq 4$ ..... 1 točka\nNarisana množica točk $y \\leq x$ ..... 1 točka\nOznačen presek množic ..... 1 točka\nZapis ali uporaba dolžin obeh katet $k_{1} = k_{2} = 4$ ..... 1 točka\nZapis ali uporaba obrazca za izračun ploščine pravokotnega trikotnika $S = \\frac{k_{1} \\cdot k_{2}}{2}$ ..... 1 točka\nIzračun ploščine trikotnika $S = 8$ ..... 1 točka\nZapis ali uporaba Pitagorovega izreka $h^{2} = k_{1}^{2} + k_{2}^{2}$ ..... 1 točka\nIzračun dolžine hipotenuze $h = \\sqrt{32} = 4 \\sqrt{2}$ ..... 1 točka\nIzračun obsega $o = 8 + 4 \\sqrt{2}$ ..... 1 točka",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 42083,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIzračunaj koordinate presečišč grafov funkcije $f(x)=x^{4}-2 x^{3}-7 x+2$ in funkcije $g(x)=3 x^{3}-8 x^{2}-1$. Zapiši smerni koeficient premice skozi ti dve presečišči. Izračunaj tangens manjšega od kotov med to premico in premico z enačbo $3 x+2 y-11=0$.",
    "options": [],
    "answer": "Intersections: (1, -6) and (3, 8); slope: 7; tangent of the smaller angle: 17/19.",
    "solution": "Solution:\n\n1. Presečišče grafov funkcij določimo tako, da enačimo funkcijska predpisa $f(x)=g(x)$. Dobimo enačbo višje stopnje $x^{4}-2 x^{3}-7 x+2=3 x^{3}-8 x^{2}-1$. Enačbo preoblikujemo in dobimo $x^{4}-5 x^{3}+8 x^{2}-7 x+3=0$.\n\nUgotovimo, da je $x_{1}=1$ rešitev te enačbe, naredimo Hornerjev algoritem in dobimo razcep $(x^{3}-4 x^{2}+4 x-3)(x-1)=0$. Prvi faktor je enak nič za $x_{2}=3$. Še enkrat naredimo Hornerjev algoritem in dobimo razcep $(x^{2}-x+1)(x-3)(x-1)=0$. Kvadratni faktor ima negativno diskriminanto in zato drugih realnih rešitev ni.\n\nZa $x_{1}=1$ in $x_{2}=3$ izračunamo še ordinati presečišč. $y_{1}=g(x_{1})=-6$ in $y_{2}=g(x_{2})=8$. Iskani presečišči sta $P_{1}(1,-6)$ in $P_{2}(3,8)$.\n\nIzračunamo smerni koeficient premice skozi ti dve presečišči $k=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=7$. Smerni koeficient prve premice je $k_{1}=7$.\n\nPremico $3 x+2 y-11=0$ zapišemo v eksplicitni obliki $y=-\\frac{3}{2} x+\\frac{11}{2}$. Smerni koeficient druge premice je torej $k_{2}=-\\frac{3}{2}$.\n\nUporabimo obrazec za tangens vmesnega kota $\\tan \\varphi=\\left|\\frac{k_{2}-k_{1}}{1+k_{1} \\cdot k_{2}}\\right|=\\frac{17}{19}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42084,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoloči najmanjšo možno vrednost izraza $\\left|25^{m}-36^{n}\\right|$, če sta $m$ in $n$ naravni števili.",
    "options": [],
    "answer": "11",
    "solution": "Solution:\n\nUpoštevanjem $25=5^{2}$ in $36=6^{2}$ lahko izraz razstavimo kot\n$$\n\\left|25^{m}-36^{n}\\right|=\\left|5^{2 m}-6^{2 n}\\right|=\\left|\\left(5^{m}\\right)^{2}-\\left(6^{n}\\right)^{2}\\right|=\\left|\\left(5^{m}+6^{n}\\right)\\left(5^{m}-6^{n}\\right)\\right|=\\left(5^{m}+6^{n}\\right)\\left|\\left(5^{m}-6^{n}\\right)\\right|\n$$\nKer sta $m$ in $n$ naravni števili, je $5^{m}+6^{n} \\geq 5+6=11$. Število $5^{m}-6^{n}$ je celo in očitno ne more biti enako 0, torej je $\\left|5^{m}-6^{n}\\right| \\geq 1$. Sledi $\\left|25^{m}-36^{n}\\right| \\geq 11 \\cdot 1=11$. Vrednost 11 je dosežena pri $m=n=1$.\n\n2. način. Zadnji števki izrazov $25^{m}$ in $36^{n}$ sta 5 in 6. Če je $36^{n}>25^{m}$, je zadnja števka izraza $\\left|25^{m}-36^{n}\\right|=36^{n}-25^{m}$ enaka 1. Če bi bilo $36^{n}-25^{m}=1$, bi imeli $36^{n}-1=25^{m}$ in $\\left(6^{n}-1\\right)\\left(6^{n}+1\\right)=25^{m}$. Toda število $6^{n}+1$ ni večkratnik števila 5, saj je njegova zadnja števka enaka 7. Torej je $36^{n}-25^{m} \\geq 11$ in vrednost 11 je dosežena pri $m=n=1$. Če pa je $25^{m}>36^{n}$, je zadnja števka izraza $\\left|25^{m}-36^{n}\\right|=25^{m}-36^{n}$ enaka 9. Če bi bilo $25^{m}-36^{n}=9$, bi veljalo $25^{m}=36^{n}+9$, kar pa ni mogoče, saj je desna stran deljiva s 3, leva pa ne. Torej je v tem primeru $25^{m}-36^{n} \\geq 19$. Najmanjša vrednost izraza $\\left|25^{m}-36^{n}\\right|$ je zato enaka 11.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42085,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoloči natančno zgornjo mejo zaporedja s splošnim členom $a_{n} = n \\left(\\frac{20}{23}\\right)^{n}$ za vse $n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "7*(20/23)^7",
    "solution": "Solution:\n\nPoglejmo, kdaj zaporedje narašča. Neenakost $a_{n+1} > a_{n}$ se z upoštevanjem formule za splošni člen zaporedja glasi\n$$(n+1)\\left(\\frac{20}{23}\\right)^{n+1} > n\\left(\\frac{20}{23}\\right)^{n}$$\nin je ekvivalentna neenakosti\n$$(n+1) \\cdot \\frac{20}{23} > n.$$\nNeenakost poenostavimo do\n$$20n + 20 > 23n,$$\nod koder sledi\n$$n < \\frac{20}{3} = 6\\frac{2}{3}.$$\nNeenakost $a_{n+1} > a_{n}$ je zato izpolnjena za $n \\in \\{1,2,3,4,5,6\\}$. To pomeni, da zaporedje $a_{n}$ narašča do člena $a_{7}$, nato pa začne padati. Njegova natančna zgornja meja je torej enaka vrednosti\n$$a_{7} = 7\\left(\\frac{20}{23}\\right)^{7}.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42086,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nPoišči vsa realna števila $a \\neq -1$, za katera je razmerje med rešitvama kvadratne enačbe $(2a+2)x^{2} + (2a+3)x + 1 = 0$ enako $1 : 3$.",
    "options": [],
    "answer": "a = 1/2 or a = -5/6",
    "solution": "Solution:\nOpazimo, da lahko levo stran enačbe razstavimo in dobimo\n$$\n((2a+2)x+1)(x+1)=0\n$$\nRešitvi sta torej $x_{1} = -1$ in $x_{2} = -\\frac{1}{2a+2}$. Torej mora biti bodisi $x_{2} = -3$ bodisi $x_{2} = -\\frac{1}{3}$. V prvem primeru sledi $2a+2 = \\frac{1}{3}$, od koder dobimo $a = -\\frac{5}{6}$. V drugem primeru je $2a+2 = 3$ oziroma $a = \\frac{1}{2}$.\n\n2. način. Označimo rešitvi enačbe $x_{1} = t$ in $x_{2} = 3t$. Po Vietovih pravilih sledi\n$$\n\\begin{gathered}\n4t = x_{1} + x_{2} = -\\frac{2a+3}{2a+2} \\\\\n3t^{2} = x_{1} x_{2} = \\frac{1}{2a+2}\n\\end{gathered}\n$$\nIz prve enačbe izrazimo $t = -\\frac{2a+3}{4(2a+2)}$ in vstavimo v drugo enačbo, da dobimo\n$$\n\\frac{1}{2a+2} = \\frac{3(2a+3)^{2}}{16(2a+2)^{2}}\n$$\nNato se znebimo ulomkov $16(2a+2) = 3(2a+3)^{2}$ in enačbo poenostavimo do $12a^{2} + 4a - 5 = 0$. Rešitvi te kvadratne enačbe sta $a = \\frac{1}{2}$ in $a = -\\frac{5}{6}$.\n\n3. način.\nRešitvi kvadratne enačbe sta $x_{1,2} = \\frac{-2a-3 \\pm \\sqrt{(2a+3)^{2} - 8(a+1)}}{4(a+1)}$. Diskriminanta je popoln kvadrat $(2a+3)^{2} - 8(a+1) = (2a+1)^{2}$, zato se ničli poenostavita v $x_{1} = -\\frac{1}{2(a+1)}$, $x_{2} = -1$. Razmerje ničel je tedaj $x_{1}/x_{2} = 3^{\\pm 1}$, kar vodi do para rešitev $a = -5/6$ in $a = 1/2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42087,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDana je enačba $6 z^{10}+a_{9} z^{9}+a_{8} z^{8}+a_{7} z^{7}+\\ldots+a_{3} z^{3}+a_{2} z^{2}+a_{1} z+9216=0$, kjer so $a_{1}, a_{2}, \\ldots, a_{9}$ realna števila in $z$ neznanka. Poznamo 5 rešitev dane enačbe, in sicer $1+i, 2+2 i, 3+3 i, 4+4 i$ ter $\\frac{1}{2}$, kjer je $i$ imaginarna enota. Določi vrednost koeficienta $a_{9}$.",
    "options": [],
    "answer": "-125",
    "solution": "Solution:\nKer ima polinom na levi strani enačbe realne koeficiente, njegove kompleksne ničle nastopanjo v kompleksnih parih. Poleg danih rešitev enačbe so zato rešitve enačbe tudi $1-i, 2-2 i$, $3-3 i$ in $4-4 i$. Poznamo torej 9 ničel polinoma. Ker je polinom stopnje 10, ima 10 ničel štetih z večkratnostjo. Označimo njegovo zadnjo ničlo z $z_{0}$. Dana enačba je torej ekvivalentna enačbi\n$$\n\\begin{aligned}\n& 6\\left(z-\\frac{1}{2}\\right)(z-(1+i))(z-(1-i))(z-(2+2 i))(z-(2-2 i)) \\\\\n& \\quad \\cdot(z-(3+3 i))(z-(3-3 i))(z-(4+4 i))(z-(4-4 i))\\left(z-z_{0}\\right)=0\n\\end{aligned}\n$$\nČe levo stran enačbe zmnožimo, kot konstantni člen dobimo produkt\n$$\n6 \\cdot \\frac{1}{2} \\cdot(1+i)(1-i)(2+2 i)(2-2 i)(3+3 i)(3-3 i)(4+4 i)(4-4 i) z_{0}=3 \\cdot 2 \\cdot 8 \\cdot 18 \\cdot 32 \\cdot z_{0}\n$$\npri čemer smo kompleksna števila zmnožili po konjugiranih parih. Ker pa mora biti konstantni člen v enačbi enak 9216 , sledi $z_{0}=\\frac{9216}{3 \\cdot 2 \\cdot 8 \\cdot 18 \\cdot 32}=\\frac{1}{3}$. Da določimo $a_{9}$, moramo razmisliti, kaj dobimo pri potenci $z^{9}$, ko zmnožimo levo stran zgornje enačbe. Potenco $z^{9}$ dobimo, ko pri 9 faktorjih vzamemo $z$, pri enem pa konstanto, torej bo\n$$\n\\begin{aligned}\na_{9} & =-6\\left(\\frac{1}{2}+(1+i)+(1-i)+(2+2 i)+(2-2 i)+(3+3 i)+(3-3 i)+(4+4 i)+(4-4 i)+z_{0}\\right) \\\\\n& =-6\\left(\\frac{1}{2}+20+\\frac{1}{3}\\right)=-125\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42088,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoenostavi izraz v množici realnih števil: $$\\left(\\frac{1}{x^{4}-27 x}-\\frac{1}{x^{4}-3 x^{3}}\\right): \\frac{x^{n+2}-9 x^{n}}{x^{n+5}+3 x^{n+4}+9 x^{n+3}} ;\\ x \\neq 0,3,-3.$$",
    "options": [],
    "answer": "-3/(x-3)^2",
    "solution": "Solution:\n\nNajprej razstavimo prvi imenovalec $x^{4}-27 x = x(x^{3}-27) = x(x-3)(x^{2}+3 x+9)$ in drugi imenovalec $x^{4}-3 x^{3} = x^{3}(x-3)$ ter razširimo oba ulomka na najmanjši skupni imenovalec in dobimo\n$$\n\\frac{-3 x-9}{x^{3}(x-3)(x^{2}+3 x+9)}.\n$$\nZa tem razstavimo števec tretjega ulomka $x^{n+2}-9 x^{n} = x^{n}(x^{2}-9) = x^{n}(x-3)(x+3)$ in imenovalec tretjega ulomka $x^{n+5}+3 x^{n+4}+9 x^{n+3} = x^{n+3}(x^{2}+3 x+9)$. Ko ulomke okrajšamo kolikor je mogoče dobimo rezultat\n$$\n-\\frac{3}{(x-3)^{2}}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42089,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je štirikotnik $A B C D$ s pravima kotoma pri ogliščih $A$ in $C$. Pravokotni projekciji oglišč $D$ in $B$ na daljico $A C$ zaporedoma označimo z $E$ in $F$. Koliko je dolžina daljice $B F$, če je $|A E|=3, |D E|=5$ in $|C E|=7$?",
    "options": [],
    "answer": "21/5",
    "solution": "Solution:\n\nKer sta kota štirokotnika $A B C D$ pri ogliščih $A$ in $C$ prava, točki $A$ in $C$ po Talesovem izreku ležita na krožnici s premerom $B D$. Torej je štirikotnik $A B C D$ tetiven. Po Pitagorovem izreku je $|A D|=\\sqrt{|A E|^{2}+|D E|^{2}}=\\sqrt{9+25}=\\sqrt{34}$ in $|C D|=\\sqrt{|C E|^{2}+|D E|^{2}}=\\sqrt{49+25}=\\sqrt{74}$. Ker je štirikotnik $A B C D$ tetiven, je $\\angle C B D=\\angle C A D$. Torej sta trikotnika $B C D$ in $A E D$ podobna, saj imata dva para skladnih kotov. Od tod sledi $\\frac{|B D|}{|C D|}=\\frac{|A D|}{|E D|}$ in zato $|B D|=\\frac{|C D| \\cdot|A D|}{|E D|}=\\frac{\\sqrt{74} \\cdot \\sqrt{34}}{5}$. Po Pitagorovem izreku sledi\n$$\n|A B|=\\sqrt{|B D|^{2}-|A D|^{2}}=\\sqrt{\\frac{74 \\cdot 34}{25}-34}=\\sqrt{\\frac{74 \\cdot 34-25 \\cdot 34}{25}}=\\sqrt{\\frac{49 \\cdot 34}{25}}=\\frac{7}{5} \\sqrt{34}\n$$\nin\n$$\n|B C|=\\sqrt{|B D|^{2}-|C D|^{2}}=\\sqrt{\\frac{74 \\cdot 34}{25}-74}=\\sqrt{\\frac{74 \\cdot 34-25 \\cdot 74}{25}}=\\sqrt{\\frac{9 \\cdot 74}{25}}=\\frac{3}{5} \\sqrt{74}\n$$\nTudi trikotnika $A B F$ in $D B C$ sta podobna, saj zaradi tetivnosti štirikotnika $A B C D$ velja $\\angle B A C=\\angle B D C$. Torej je $\\frac{|B F|}{|A B|}=\\frac{|B C|}{|D B|}$, od koder sledi\n$$\n|B F|=\\frac{|A B| \\cdot|B C|}{|D B|}=\\frac{\\frac{7}{5} \\sqrt{34} \\cdot \\frac{3}{5} \\sqrt{74}}{\\frac{\\sqrt{74} \\cdot \\sqrt{34}}{5}}=\\frac{21}{5}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42090,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa kocki $A B C D E F G H$ označimo središče stranske ploskve $E F G H$ s $P$, središče stranskega roba $A B$ pa z $R$. Ravnina $\\Sigma$, ki vsebuje točke $C, P$ in $R$, seka robova $E F$ in $G H$ zaporedoma v točkah $X$ in $Y$.\n\na) Dokaži, da je $|E X|:|X F|=|G Y|:|Y H|=1: 3$.\n\nb) Ravnina $\\Sigma$ razdeli kocko na dva dela.\nDoloči razmerje med prostorninama teh dveh delov.",
    "options": [],
    "answer": "EX:XF = GY:YH = 1:3; volume ratio = 5:3",
    "solution": "Solution:\n\na)\nDaljici $R C$ in $K G$ sta vzporedni, ker je $R C G K$ pravokotnik. Tudi daljici $R C$ in $X Y$ sta vzporedni, saj sta ploskvi $A B C D$ in $E F G H$ vzporedni, daljici pa sta presek teh dveh ploskev z ravnino $\\Sigma$. Torej sta daljici $K G$ in $X Y$ vzporedni, zato velja $|K X|=|G Y|$. Ker pa daljica $X Y$ poteka skozi središče $P$ ploskve $E F G H$, je hkrati tudi $|G Y|=|E X|$. Torej je $|K X|=|E X|$ in zato je $X$ je razpolovišče daljice $E K$. Ker je $R$ središče daljice $A B$, je $K$ središče daljice $E F$. To pomeni, da je $|E X|=\\frac{1}{4}|E F|$, od koder sledi $|E X|:|X F|=1: 3$. Zaradi simetrije je tudi $|G Y|:|Y H|=1: 3$.\n\nb)\nNaj bo dolžina roba konce $A B C D E F G H$ enaka $a$. Ker je $P$ središče ploskve $E F G H$, je štiristrana prizma $A L M D E X Y H$ polovica kocke, zato je njena prostornina enaka $V_{1}=\\frac{a^{3}}{2}$. Ker je $R$ središče roba $A B$, je tristrana prizma $R B C K F G$ četrtina kocke, zato je njena prostornina enaka $V_{2}=\\frac{a^{3}}{4}$. Torej je tudi štiristrana prizma $L R C M X K G Y$ četrtina kocke. Ravnina $\\Sigma$ razdeli to prizmo na dva prostorninsko enaka dela s prostorninama $V_{3}=\\frac{a^{3}}{8}$. Od tod sledi, da ravnina $\\Sigma$ razdeli kocko na dva dela z razmerjem prostornin\n$$\n\\left(V_{1}+V_{3}\\right):\\left(V_{3}+V_{2}\\right)=\\left(\\frac{a^{3}}{2}+\\frac{a^{3}}{8}\\right):\\left(\\frac{a^{3}}{8}+\\frac{a^{3}}{4}\\right)=\\frac{5}{8}: \\frac{3}{8}=5: 3\n$$\n\nOpomba. Razmerje prostornin obeh delov kocke lahko določimo tudi tako, da določimo razmerje njunih povprečnih višin glede na osnovni ploskvi $A D H E$ in $B C G F$, saj sta ti dve osnovni ploskvi skladni, stranski robovi pa so pravokotni naju. Povprečna višina levega dela je enaka\n$$\n\\frac{|A R|+|D C|+|H Y|+|E X|}{4}=\\frac{\\frac{1}{2} a+a+\\frac{3}{4} a+\\frac{1}{4} a}{4}=\\frac{5}{8} a\n$$\npovprečna višina desnega dela pa\n$$\n\\frac{|B R|+|C C|+|G Y|+|F X|}{4}=\\frac{\\frac{1}{2} a+0+\\frac{1}{4} a+\\frac{3}{4} a}{4}=\\frac{3}{8} a\n$$\nIskano razmerje je torej $\\frac{5}{8} a: \\frac{3}{8} a=5: 3$.\n\n\n2. način. Nad kocko $A B C D E F G H$ dorišimo skladno kocko $E F G H A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime}$. Razpolovišče stranice $C^{\\prime} D^{\\prime}$ označimo z $R^{\\prime}$, dolžino stranice kocke pa z $a$.\n![](attached_image_1.png)\n\na) Ker je $P$ središče ploskve $E F G H$, so točke $C, P$ in $A^{\\prime}$ kolinearne ter točke $R, P$ in $R^{\\prime}$ kolinearne. Torej ravnina $\\Sigma$ vsebuje tudi točki $A^{\\prime}$ in $R^{\\prime}$. Iz podobnosti trikotnikov $A^{\\prime} E X$ in $A^{\\prime} A R$ sledi $|E X|=\\frac{1}{2}|A R|=\\frac{1}{4}|A B|$, torej je $|E X|:|X F|=1: 3$ in podobno je tudi $|G Y|:|Y H|=1: 3$.\n\nb) Opazimo, da je prostornina telesa $A R C D E X Y H$ enaka prostornini telesa $X F G Y A^{\\prime} B^{\\prime} C^{\\prime} R^{\\prime}$, saj sta telesi zrcalno simetrični glede na točko $P$. Podobno je prostornina telesa $R B C X F G Y$ enaka prostornini telesa $E X Y H A^{\\prime} R^{\\prime} D^{\\prime}$. Hkrati opazimo, da če telesi $X F G Y A^{\\prime} B^{\\prime} C^{\\prime} R^{\\prime}$ in RBCXFGY zamenjamo, tako da ju zadrsamo vzdolž ravnine $\\Sigma$, potem dobimo dva kvadra, ki imata vodoravni stranici dolžin $\\frac{5}{4} a$ in $\\frac{3}{4} a$, vse ostale stranice pa dolžine $a$. Razmerje prostornin teh dveh kvadrov je torej $5: 3$ in je po zgornjem enako iskanemu razmerju prostornin obeh delov kocke.\n\n\nPrvi del naloge lahko rešimo z uporabo vektorjev. Naj bodo $\\vec{a}=\\overrightarrow{A B}, \\vec{b}=\\overrightarrow{A D}$ ter $\\vec{c}=\\overrightarrow{A E}$. Naj bo $\\overrightarrow{E X}=\\alpha \\cdot \\overrightarrow{E F}$. Izračunamo $\\overrightarrow{R C}=\\frac{1}{2} \\vec{a}+\\vec{b}$ ter\n$$\n\\overrightarrow{R P}=\\overrightarrow{R A}+\\overrightarrow{A E}+\\overrightarrow{E P}=\\overrightarrow{R A}+\\overrightarrow{A E}+\\frac{1}{2} \\overrightarrow{E G}=\\vec{c}+\\frac{1}{2} \\vec{b}\n$$\nSedaj izračunamo vektor $\\overrightarrow{R X}$ na dva načina. Velja\n$$\n\\overrightarrow{R X}=\\overrightarrow{R A}+\\overrightarrow{A E}+\\overrightarrow{E X}=\\left(\\alpha-\\frac{1}{2}\\right) \\vec{a}+\\vec{c}\n$$\nhkrati pa so vektorji $\\overrightarrow{R X}, \\overrightarrow{R P}$ in $\\overrightarrow{R C}$ koplanarni, zato\n$$\n\\overrightarrow{R P}=\\beta \\cdot \\overrightarrow{R C}+\\gamma \\cdot \\overrightarrow{R P}=\\frac{1}{2} \\beta \\vec{a}+\\left(\\beta+\\frac{1}{2} \\gamma\\right) \\vec{b}+\\gamma \\vec{c}\n$$\nPo primerjavi po komponentah dobimo $\\alpha=\\frac{1}{4}$, kar pomeni, da sta iskani razmerji res $1: 3$.\nMožno je tudi opaziti, da je $\\overrightarrow{X Y}=\\overrightarrow{R C}$ in dobiti isti rezultat s pomočjo zapisa\n$$\n\\overrightarrow{E X}=\\overrightarrow{E H}+\\overrightarrow{H Y}+\\overrightarrow{Y X}=\\left(\\frac{1}{2}-\\alpha\\right) \\vec{a}\n$$\nhkrati pa vemo, da je $\\overrightarrow{E X}=\\alpha \\cdot \\vec{a}$, od koder sledi $\\alpha=\\frac{1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42091,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDana je krožnica $\\mathcal{K}$ s premerom $AB$. Točki $C$ in $D$ ležita na krožnici $\\mathcal{K}$ na istem bregu premice $AB$, tako da je premica $BD$ simetrala kota $\\angle CBA$. Tetivi $AC$ in $BD$ se sekata v točki $E$. Koliko je dolžina daljice $DE$, če je $|AE|=169$ in $|CE|=119$?",
    "options": [],
    "answer": "65",
    "solution": "Solution:\n\nKer je premica $BE$ simetrala kot $\\angle CBA$, je po izreku o simetrali kota trikotnika $\\frac{|AB|}{|BC|}=\\frac{|AE|}{|CE|}=\\frac{169}{119}$ oziroma $|AB|=\\frac{169}{119}|BC|$. Po Talesovem izreku o kotu v polkrogu je $\\angle ACB=\\angle ADB=90^{\\circ}$, zato po Pitagorovem izreku velja $|AB|^{2}=|AC|^{2}+|BC|^{2}$ oziroma $\\frac{169^{2}}{119^{2}}|BC|^{2}=(169+119)^{2}+|BC|^{2}$. Od tod izrazimo $|BC|^{2}=\\frac{119^{2}(169+119)^{2}}{169^{2}-119^{2}}=\\frac{119^{2}(169+119)}{169-119}=\\frac{119^{2} \\cdot 288}{50}=\\frac{119^{2} \\cdot 144}{25}$.\n\nPo Pitagorovem izreku v trikotniku $BCE$ je $|BE|=\\sqrt{|CE|^{2}+|BC|^{2}}=\\sqrt{119^{2}+\\frac{119^{2} \\cdot 144}{25}}=119 \\sqrt{\\frac{169}{25}}=119 \\cdot \\frac{13}{5}$.\n\nKer sta trikotnika $ADE$ in $BCE$ podobna, je $\\frac{|DE|}{|AE|}=\\frac{|CE|}{|BE|}$, od koder izračunamo $|DE|=\\frac{|AE| \\cdot|CE|}{|BE|}=\\frac{169 \\cdot 119}{119 \\cdot \\frac{13}{5}}=5 \\cdot 13=65$.\n\nOpomba. Enakost $\\frac{|AB|}{|BC|}=\\frac{|AE|}{|CE|}$ iz prvega dela rešitve lahko dokažemo tudi s pomočjo sinusnega izreka. Označimo $\\angle CBA=\\beta$ in $\\angle BEC=\\varphi$. Po sinusnem izreku v trikotniku $BCE$ je $\\frac{|CE|}{|BC|}=\\frac{\\sin \\frac{\\beta}{2}}{\\sin \\varphi}$, po sinusnem izreku v trikotniku $BEA$ pa velja $\\frac{|AE|}{|AB|}=\\frac{\\sin \\frac{\\beta}{2}}{\\sin (\\pi-\\varphi)}$. Ker je $\\sin (\\pi-\\varphi)=\\sin \\varphi$, sledi $\\frac{|AE|}{|AB|}=\\frac{|CE|}{|BC|}$ oziroma $\\frac{|AB|}{|BC|}=\\frac{|AE|}{|CE|}$.\nSolution:\n\nNaj bo $F$ pravokotna projekcija točke $E$ na daljico $AB$. Ker je premica $BE$ simetrala kota $\\angle CBA$, sta trikotnika $BCE$ in $BFE$ skladna, saj se ujemata v dveh kotih in stranici nasproti večjemu kotu. Torej velja $|BF|=|BC|$ in $|EF|=|CE|$. Po Talesovem izreku o kotu v polkrogu je $\\angle ACB=\\angle ADB=90^{\\circ}$, zato po Pitagorovem izreku sledi\n$$\n|BC|^{2}+|AC|^{2}=|AB|^{2}=(|AF|+|BF|)^{2}=(|AF|+|BC|)^{2}=|AF|^{2}+2 \\cdot|AF| \\cdot|BC|+|BC|^{2}\n$$\nOd tod izrazimo $|BC|=\\frac{|AC|^{2}-|AF|^{2}}{2|AF|}$. Po Pitagorovem izreku za trikotnik $AFE$ sledi $|AF|^{2}=|AE|^{2}-|EF|^{2}=|AE|^{2}-|CE|^{2}=169^{2}-119^{2}$, zato je\n$$\n\\begin{aligned}\n|BC| & =\\frac{(169+119)^{2}-\\left(169^{2}-119^{2}\\right)}{2 \\sqrt{169^{2}-119^{2}}}=\\frac{2 \\cdot 119 \\cdot(169+119)}{2 \\sqrt{(169+119)(169-119)}} \\\\\n& =119 \\cdot \\sqrt{\\frac{169+119}{169-119}}=119 \\cdot \\sqrt{\\frac{288}{50}}=119 \\cdot \\sqrt{\\frac{144}{25}}=119 \\cdot \\frac{12}{5}\n\\end{aligned}\n$$\nPo Pitagorovem izreku za trikotnik $BCE$ sledi $|BE|=\\sqrt{|BC|^{2}+|CE|^{2}}=\\sqrt{119^{2} \\cdot \\frac{144}{25}+119^{2}}=119 \\cdot \\frac{13}{5}$. Ker sta trikotnika $ADE$ in $BCE$ podobna, je $\\frac{|DE|}{|AE|}=\\frac{|CE|}{|BE|}$, od koder izračunamo $|DE|=\\frac{|AE| \\cdot|CE|}{|BE|}=\\frac{169 \\cdot 119}{119 \\cdot \\frac{13}{5}}=5 \\cdot 13=65$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42092,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKrog $K$ s polmerom $R$ razdelimo na tri krožne izseke, tako da je vsota ploščin manjših dveh izsekov enaka ploščini največjega izseka, razlika ploščin manjših dveh izsekov pa je enaka tretjini ploščine največjega izseka. V vsakega izmed izsekov včrtamo največji možen krog in te kroge označimo s $K_{1}, K_{2}$ in $K_{3}$. S parametrom $R$ izrazi ploščino območja $K \\backslash\\left(K_{1} \\cup K_{2} \\cup K_{3}\\right)$.",
    "options": [],
    "answer": "π R^2 (12√3 − 733/36)",
    "solution": "Solution:\n\nOznačimo središčne kote izsekov z $\\alpha_{1}, \\alpha_{2}$ in $\\alpha_{3}$, kjer je $\\alpha_{1}<\\alpha_{2}<\\alpha_{3}$. Ker je ploščina vsakega krožnega izseka premosorazmerna z njegovim s središčnim kotom, iz podatkov sledi $\\alpha_{1}+\\alpha_{2}+\\alpha_{3}=360^{\\circ}, \\alpha_{1}+\\alpha_{2}=\\alpha_{3}$ in $\\alpha_{2}-\\alpha_{1}=\\frac{1}{3} \\alpha_{3}$. Rešitev tega sistema enačb je $\\alpha_{1}=60^{\\circ}, \\alpha_{2}=120^{\\circ}$ in $\\alpha_{3}=180^{\\circ}$. Privzamemo lahko, da je krog $K_{i}$ včrtan v krožni izsek s središčnim kotom $\\alpha_{i}$. Tedaj je polmer največjega kroga $K_{3}$ enak $\\frac{R}{2}$, saj je včrtan v polovico kroga $K$. Polmera krogov $K_{1}$ in $K_{2}$ označimo z $r_{1}$ in $r_{2}$, njuni središči s $S_{1}$ in $S_{2}$, njuni dotikališči z robno krožnico kroga $K$ pa z $D_{1}$ in $D_{2}$. Pravokotni projekciji točk $S_{1}$ in $S_{2}$ na premer kroga $K$, ki ga določa največji izsek, označimo s $P_{1}$ in $P_{2}$, središče kroga $K$ pa s $S$.\n\n![](attached_image_1.png)\n\nKer je $\\angle P_{1} S S_{1}=\\frac{\\alpha_{1}}{2}=30^{\\circ}$, je trikotnik $S P_{1} S_{1}$ polovica enakostraničnega trikotnika, torej je $\\left|S S_{1}\\right|=2 r_{1}$. Od tod izpeljemo $R=\\left|S S_{1}\\right|+\\left|S_{1} D_{1}\\right|=2 r_{1}+r_{1}=3 r_{1}$ in zato je $r_{1}=\\frac{R}{3}$. Ker je $\\angle S S_{2} P_{2}=\\frac{\\alpha_{2}}{2}=60^{\\circ}$, je tudi trikotnik $S S_{2} P_{2}$ polovica enakostraničnega trikotnika, torej velja $r_{2}=\\frac{\\left|S S_{2}\\right| \\sqrt{3}}{2}$ oziroma $\\left|S S_{2}\\right|=\\frac{2 r_{2}}{\\sqrt{3}}$. Sledi $R=\\left|S S_{2}\\right|+\\left|S_{2} D_{2}\\right|=\\frac{2 r_{2}}{\\sqrt{3}}+r_{2}=\\left(\\frac{2}{\\sqrt{3}}+1\\right) r_{2}=\\frac{2+\\sqrt{3}}{\\sqrt{3}} r_{2}$, od koder izrazimo $r_{2}=\\frac{\\sqrt{3}}{2+\\sqrt{3}} R=\\sqrt{3}(2-\\sqrt{3}) R=(2 \\sqrt{3}-3) R$. Ploščina območja $K \\backslash\\left(K_{1} \\cup K_{2} \\cup K_{3}\\right)$ je zato enaka\n\n$$\np=\\pi R^{2}-\\pi\\left(\\frac{R}{2}\\right)^{2}-\\pi\\left(\\frac{R}{3}\\right)^{2}-\\pi(2 \\sqrt{3}-3)^{2} R^{2}=\\pi\\left(12 \\sqrt{3}-20 \\frac{13}{36}\\right) R^{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42093,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa 1. nivoju računalniške igre Zakladnica se igralec sprehaja po podzemni zakladnici, ki je sestavljana iz 13 osemkotnih in 12 kvadratnih sob. Edini vhod in izhod iz zakladnice je v osrednji sobi, v preostalih 24 sobah pa je po en zlat cekin (glej sliko). Med vsakima dvema sobama je prehod, ki se zapre takoj, ko igralec stopi skozenj. Ko igralec izstopi iz sobe, v kateri je pobral zlat cekin, se v tej sobi pojavi nov zlat cekin. Če igralec ostane ujet v zakladnici, izgubi vse zbrane cekine, igra pa se ponastavi. Največ koliko zlatih cekinov lahko igralec prinese iz zakladnice?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "44",
    "solution": "Solution:\n\nPreštejmo, koliko vrat ima posamezna soba s cekinom. Opazimo, da je v zakladnici 12 sob s po 3 vrati, 4 sobe s po 4 vrati, 4 sobe s po 5 vrati in 4 sobe s po 8 vrati (srednje sobe ne štejemo, ker nima cekina). Skozi sobo s 3 vrati lahko gre igralec le enkrat, saj bi sicer ostal ujet v njej. S tem pobere 1 cekin. Skozi sobo s 4 vrati lahko gre dvakrat in pobere 2 cekina. Skozi sobo s 5 vrati lahko gre prav tako dvakrat in pobere 2 cekina. Skozi sobo z 8 vrati pa lahko gre štirikrat in pobere 4 cekine. Skupaj lahko torej igralec pobere največ\n$$\nn = 12 \\cdot 1 + 4 \\cdot 2 + 4 \\cdot 2 + 4 \\cdot 4 = 44\n$$\ncekinov. Preveriti moramo še, da res obstaja pot, ki se začne in konča v srednji sobi in gre 44-krat skozi sobo s cekinom. Primer take poti je prikazan na sliki.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42094,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDan je karirast list papirja velikosti $7 \\times 7$ kvadratkov in enako velika plastična karirasta šablona, na kateri so nekateri kvadratki zeleni, ostali pa prosojni. Če šablono postavimo na list papirja tako, da se stranice šablone ujemajo s stranicami papirja, se karirast vzorec na šabloni ujema s karirastim vzorcem na papirju. Če izberemo katerikoli kvadratek papirja in ga obarvamo rdeče, potem lahko šablono postavimo na papir tako, da se njene stranice ujemajo s stranicami papirja in je rdeč kvadratek papirja prekrit z enim od zelenih kvadratkov šablone (šablono lahko pred tem poljubno zavrtimo in po potrebi obrnemo na hrbtno stran). Najmanj koliko kvadratkov šablone je zelenih?",
    "options": [],
    "answer": "10",
    "solution": "Solution:\n\nŠablono lahko na papir postavimo na 8 različnih načinov; na 4 načine, če je obrnjena na prednjo stran, in na 4 načine, če je obrnjena na zadnjo stran. Med različnimi postavitvami prehajamo z rotacijo šablone za $90^\\circ$ okrog njenega središča in z obračanjem šablone na drugo stran preko ene od njenih 4 simetral (kar ustreza zrcaljenju preko te simetrale). Torej lahko z istim zelenim kvadratkom šablone prekrijemo tiste kvadratke papirja, do katerih pridemo s temi operacijami. Označimo kvadratke papirja s števili, kot je prikazano na sliki.\n\n| 1 | 2 | 3 | 4 | 3 | 2 | 1 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 2 | 5 | 6 | 7 | 6 | 5 | 2 |\n| 3 | 6 | 8 | 9 | 8 | 6 | 3 |\n| 4 | 7 | 9 | 10 | 9 | 7 | 4 |\n| 3 | 6 | 8 | 9 | 8 | 6 | 3 |\n| 2 | 5 | 6 | 7 | 6 | 5 | 2 |\n| 1 | 2 | 3 | 4 | 3 | 2 | 1 |\n\nVse množice kvadratkov z istim številom so simetrične glede na vse 4 simetrale papirja in ostanejo nespremenjene, če jih zarotiramo za $90^\\circ$ okrog središča papirja. Torej lahko en zelen kvadratek šablone prekrije le eno množico kvadratkov z istim številom. Na šabloni mora biti torej pobarvanih vsaj 10 kvadratkov. To hkrati zadostuje, saj vsako množico kvadratkov z istim številom lahko prekrijemo z enim samim zelenim kvadratkom na šabloni (z uporabo zgoraj omenjenih operacij). Torej če je na šabloni pobarvanih 10 kvadratkov, od katerih vsak prekrije kvadratek z drugim številom, kot je na primer prikazano na sliki, potem lahko s tako šablono z zelenim kvadratkom prekrijemo katerikoli kvadratek papirja.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42095,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, $S(n)$ denotes the sum of its digits and $U(n)$ its unit digit. Determine all positive integers $n$ with the property that\n$$\nn = S(n) + U(n)^2.\n$$",
    "options": [],
    "answer": "13, 46, 99",
    "solution": "Write $n$ as $a_0 + 10a_1 + 100a_2 + \\dots$, where $a_0, a_1, a_2, \\dots$ are the digits of $n$. Then the stated equation is equivalent to\n$$\na_0 + 10a_1 + 100a_2 + \\dots = a_0 + a_1 + a_2 + \\dots + a_0^2\n$$\nor\n$$\n9a_1 + 99a_2 + 999a_3 + \\dots = a_0^2.\n$$\nThe right hand side is at most $9^2 = 81 < 99$. Therefore, $a_2, a_3, \\dots$ have to be 0 (otherwise, the left hand side would be strictly greater than the right hand side). Hence we obtain\n$$\n9a_1 = a_0^2.\n$$\nIt follows that $a_0$ must be divisible by 3 (since $9a_1$ is), which leaves us with the possibilities $a_0 = 3$ ($a_1 = 1$), $a_0 = 6$ ($a_1 = 4$) and $a_0 = 9$ ($a_1 = 9$). So there are three solutions: 13, 46 and 99.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42096,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a triangle $ABC$ with $BC = 3$. Choose a point $D$ on $BC$ such that $BD = 2$. Find the value of\n$$\nAB^2 + 2AC^2 - 3AD^2.\n$$",
    "options": [],
    "answer": "6",
    "solution": "Drop the altitude from $A$ to $BC$, and let $F$ be its foot. Furthermore, suppose that $BF = x$ (if $F$ lies on the extension of $BC$ beyond $B$, assign a negative sign to $x$) and $AF = y$. Then, by the Pythagorean theorem,\n$$\nAB^2 = BF^2 + AF^2 = x^2 + y^2,\n$$\n$$\nAC^2 = CF^2 + AF^2 = (3-x)^2 + y^2,\n$$\n$$\nAD^2 = DF^2 + AF^2 = (2-x)^2 + y^2.\n$$\nIt follows that\n$$\n\\begin{aligned}\nAB^2 + 2AC^2 - 3AD^2 &= x^2 + y^2 + 2(3-x)^2 + 2y^2 - 3(2-x)^2 - 3y^2 \\\\\n&= x^2 + y^2 + 18 - 12x + 2x^2 + 2y^2 - 12 + 12x - 3x^2 - 3y^2 \\\\\n&= 6,\n\\end{aligned}\n$$\nregardless of the values of $x$ and $y$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42097,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ such that $5^n - 1$ can be written as a product of an even number of consecutive integers.",
    "options": [],
    "answer": "2",
    "solution": "$5^n - 1$ cannot be a product of more than five consecutive integers: since one of the factors would have to be divisible by $5$, this would have to be the case for the product as well, but $5^n - 1$ is not divisible by $5$.\n\nNow we have to consider two cases:\n\nCase 1: If $5^n - 1$ is a product of two consecutive integers, say $m$ and $m + 1$, then\n$$\n5^n = m(m+1) + 1 = m^2 + m + 1.\n$$\nConsider this equation modulo $5$: clearly, the left hand side is divisible by $5$. On the other hand, the right hand side is never divisible by $5$:\n\n| $m \\bmod 5$ | 0 | 1 | 2 | 3 | 4 |\n|-------------|---|---|---|---|---|\n| $m^2 + m + 1 \\bmod 5$ | 1 | 3 | 2 | 3 | 1 |\n\nIt follows that there is no solution in this case.\n\nCase 2: If $5^n - 1$ is a product of four consecutive integers, say $m, \\dots, m+3$, then\n$$\n\\begin{aligned}\n5^n &= m(m+1)(m+2)(m+3) + 1 = (m^2 + 3m)(m^2 + 3m + 2) + 1 \\\\\n&= (m^2 + 3m + 1)^2 - 1 + 1 \\\\\n&= (m^2 + 3m + 1)^2.\n\\end{aligned}\n$$\nTherefore $n$ must be even, and $m^2 + 3m + 1$ must be a power of $5$:\n$$\nm^2 + 3m + 1 = 5^k,\n$$\nwhere $k = \\frac{n}{2} > 0$. Solving for $m$ yields\n$$\nm = \\frac{-3 \\pm \\sqrt{5 + 4 \\cdot 5^k}}{2} = \\frac{-3 \\pm \\sqrt{5(1 + 4 \\cdot 5^{k-1})}}{2}.\n$$\nIf now $k > 1$, then $5(1 + 4 \\cdot 5^{k-1})$ is divisible by $5$, but not by $25$ ($4 \\cdot 5^{k-1}$ is a multiple of $5$ in this case, so $4 \\cdot 5^{k-1} + 1$ is not), and so it cannot be a square. This shows that there cannot be any solutions for $k > 1$. On the other hand, $k = 1$ and $m = 1$ satisfy the equation, which leads us to the only solution\n$$\n5^2 - 1 = 1 \\cdot 2 \\cdot 3 \\cdot 4.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42098,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $n$ positive real numbers satisfying $x_1 \\ge x_2 \\ge \\dots \\ge x_n \\ge 0$ and $x_1^2 + x_2^2 + \\dots + x_n^2 = 1$, prove that\n$$\n\\frac{x_1}{\\sqrt{1}} + \\frac{x_2}{\\sqrt{2}} + \\dots + \\frac{x_n}{\\sqrt{n}} \\ge 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that for any $k \\le n$, we have\n$$\nkx_k^2 \\le x_1^2 + x_2^2 + \\dots + x_k^2 \\le 1\n$$\nby the given conditions. This implies that $x_k \\le \\frac{1}{\\sqrt{k}}$ and thus $x_k^2 \\le \\frac{x_k}{\\sqrt{k}}$. We conclude that\n$$\n\\frac{x_1}{\\sqrt{1}} + \\frac{x_2}{\\sqrt{2}} + \\cdots + \\frac{x_n}{\\sqrt{n}} \\ge x_1^2 + x_2^2 + \\cdots + x_n^2 = 1,\n$$\nwhich proves the statement.\nWe prove the inequality by induction on $n$: for $n = 1$, we have $x_1 = 1$ by the given conditions, and there is nothing to prove. Now assume that the inequality holds for $n-1$, and consider real numbers $x_1, x_2, \\dots, x_n$ that satisfy the given conditions. As in the first solution, we find that $x_n \\le \\frac{1}{\\sqrt{n}}$. Now set\n$$\ny_k = \\frac{x_k}{\\sqrt{1 - x_n^2}}.\n$$\nThen $y_1 \\ge y_2 \\ge \\dots \\ge y_{n-1}$ and\n$$\ny_1^2 + y_2^2 + \\dots + y_{n-1}^2 = \\frac{1}{1-x_n^2} (x_1^2 + x_2^2 + \\dots + x_{n-1}^2) = \\frac{1}{1-x_n^2} (1-x_n^2) = 1,\n$$\nso that we can apply the induction hypothesis to $y_1, y_2, \\dots, y_{n-1}$. We conclude that\n$$\n\\begin{aligned}\n\\frac{x_1}{\\sqrt{1}} + \\frac{x_2}{\\sqrt{2}} + \\dots + \\frac{x_{n-1}}{\\sqrt{n-1}} + \\frac{x_n}{\\sqrt{n}} &= \\sqrt{1-x_n^2} \\left( \\frac{y_1}{\\sqrt{1}} + \\frac{y_2}{\\sqrt{2}} + \\dots + \\frac{y_{n-1}}{\\sqrt{n-1}} \\right) + \\frac{x_n}{\\sqrt{n}} \\\\ &\\ge \\sqrt{1-x_n^2} + \\frac{x_n}{\\sqrt{n}}. \\end{aligned}\n$$\nIt remains to show that the last expression is $\\ge 1$ to complete the induction. However, this is equivalent to\n$$\n\\sqrt{1 - x_n^2} \\ge 1 - \\frac{x_n}{\\sqrt{n}}\n$$\nor\n$$\n1 - x_n^2 \\ge 1 - \\frac{2x_n}{\\sqrt{n}} + \\frac{x_n^2}{n},\n$$\nwhich finally simplifies to\n$$\nx_n \\left( \\frac{2}{\\sqrt{n}} - \\left( 1 + \\frac{1}{n} \\right) x_n \\right) \\ge 0.\n$$\nSince we know that $x_n \\le \\frac{1}{\\sqrt{n}}$, we have\n$$\n\\frac{2}{\\sqrt{n}} - \\left(1 + \\frac{1}{n}\\right) x_n \\ge \\frac{2}{\\sqrt{n}} - 2x_n \\ge 0,\n$$\nso that both factors on the left hand side are nonnegative. This completes our proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42099,
    "subject": "Mathematics (Multi-modal)",
    "question": "Write either $1$ or $-1$ in each of the cells of a $(2n) \\times (2n)$-table, in such a way that there are exactly $2n^2$ entries of each kind. Let the minimum of the absolute values of all row sums and all column sums be $M$. Determine the largest possible value of $M$.",
    "options": [],
    "answer": "Maximum M is n if n is even, and n − 1 if n is odd.",
    "solution": "Split the table into four smaller tables of size $n \\times n$. The upper left quarter is now filled with $1$s, the lower right quarter with $-1$s, and each of the remaining two quarters in a checkerboard pattern (if $n$ is odd, fill them in such a way that one of the quarters contains more $1$s than $-1$s, and the other more $-1$s than $1$s). If $n$ is even, then each of the rows and columns contains either $n/2$ $1$s and $3n/2$ $-1$s, or vice versa, so that $M = n$. If $n$ is odd, then each of the rows and columns contains either $(n - 1)/2$ $1$s ($-1$s) and $(3n + 1)/2$ $-1$s ($1$s), or $(n + 1)/2$ $1$s ($-1$s) and $(3n - 1)/2$ $-1$s ($1$s); it follows that $M = n - 1$ in this case.\n\nNow we show that $M$ cannot be larger. If there is a row or column that contains as many $1$s as $-1$s, then $M = 0$, and we are done. Otherwise, split the set of all $4n$ rows and columns into two subsets: those that contain more $1$s than $-1$s, and those that contain more $-1$s than $1$s.\n\nOne of these two sets must contain at least $2n$ elements. Without loss of generality, assume that there are at least $2n$ rows and columns (of which $k$ are rows and $\\ell$ columns) that contain more $1$s than $-1$s. In each of these rows and columns, there are at least $n + M/2$ $1$s and at most $n - M/2$ $-1$s. The total number of $1$s in all these rows and columns is therefore at least\n$$\n(k + \\ell) \\left( n + \\frac{M}{2} \\right) - k\\ell,\n$$\nwhere the last term accounts for those $1$s that are possibly double-counted. Hence we have\n$$\n2n^2 \\ge (k+\\ell) \\left(n + \\frac{M}{2}\\right) - k\\ell \\ge (k+\\ell) \\left(n + \\frac{M}{2}\\right) - \\frac{1}{4}(k+\\ell)^2\n$$\nand thus, with $r = k + \\ell$,\n$$\nM \\le \\frac{2n^2 + r^2/4 - rn}{r/2} = \\frac{4n^2}{r} + \\frac{r}{2} - 2n = n - \\frac{(r-2n)(4n-r)}{2r} \\le n,\n$$\nsince $2n \\le r \\le 4n$ by assumption. Hence we have $M \\le n$, which completes the proof in the case that $n$ is even. If $n$ is odd, $M = n$ is impossible, since all row sums and all column sums must be even (sum of an even number of odd numbers), so that we must have $M \\le n - 1$.\n\nWe conclude that the largest possible value of $M$ is $n$ if $n$ is even and $n - 1$ if $n$ is odd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42100,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x_1$, $x_2$ and $x_3$ are the roots of $x^3 - a x^2 + b x - c = 0$, prove that\n$$\n(x_1 - x_2)^2 + (x_2 - x_3)^2 + (x_3 - x_1)^2 = 2a^2 - 6b.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using Vieta's formula, the sum of the roots is $x_1 + x_2 + x_3 = a$, and the sum of the pairwise products is $x_1 x_2 + x_2 x_3 + x_3 x_1 = b$. Starting with the right hand side:\n$$\n\\begin{aligned}\n2a^2 - 6b &= 2(x_1 + x_2 + x_3)^2 - 6(x_1 x_2 + x_2 x_3 + x_3 x_1) \\\\\n&= 2(x_1^2 + x_2^2 + x_3^2 + 2(x_1 x_2 + x_2 x_3 + x_3 x_1)) - 6(x_1 x_2 + x_2 x_3 + x_3 x_1) \\\\\n&= 2(x_1^2 + x_2^2 + x_3^2) - 2(x_1 x_2 + x_2 x_3 + x_3 x_1) \\\\\n&= (x_1 - x_2)^2 + (x_2 - x_3)^2 + (x_3 - x_1)^2.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42101,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{N} \\to \\mathbb{N}$ satisfying $f(n+1)f(n+2) = f(n)^2$ for all $n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "All constant functions f(n) = c for all n, where c is any natural number.",
    "solution": "Suppose that the function is non-constant. Then it will have a least element $f(k)$. Then $f(k+1) \\neq f(k)$, or the function would be constant, and by definition $f(k+1) > f(k)$. But $f(k+2) = \\frac{f(k)^2}{f(k+1)} = f(k)\\frac{f(k)}{f(k+1)} < f(k)$, which is a contradiction. So $f(n)$ constant is the only allowed function.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42102,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, ..., a_n, b_1, b_2, ..., b_n$ be the numbers $1, 2, ..., 2n$ in some order. Suppose that $a_1 < a_2 < ... < a_n$ and that $b_1 > b_2 > ... > b_n$. Prove that\n$$\n\\sum_{i=1}^{n} |a_i - b_i| = n^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that for a given $i$, $a_i$ and $b_i$ are never both in one of $\\{1, 2, ..., n\\}$ or $\\{n+1, ..., 2n\\}$. Suppose, without loss of generality, that $a_i, b_i \\le n$. Then $a_1, a_2, ..., a_i \\le n$, and $b_i, b_{i+1}, ..., b_n \\le n$, giving $n+1$ elements less than or equal to $n$, which is impossible. Then every $|a_i - b_i|$ is a difference between an element in $\\{n+1, ..., 2n\\}$ and an element in $\\{1, 2, ..., n\\}$. Therefore:\n$$\n\\sum_{i=1}^{n} |a_i - b_i| = \\sum_{i=1}^{n} (n + i) - \\sum_{i=1}^{n} i = n^2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42103,
    "subject": "Mathematics (Multi-modal)",
    "question": "An equilateral triangle and a regular hexagon have equal perimeters. What is the ratio of the area of the triangle to the area of the hexagon?",
    "options": [],
    "answer": "2:3",
    "solution": "Let the perimeter be $6k$. The equilateral triangle will have side length $2k$, and will have an area four times the area of a triangle with side length $k$.\n\nThe hexagon will have side length $k$. By joining the vertices of the hexagon to the centre, it can be divided exactly into six equilateral triangles of side length $k$.\n\nThe ratio of the area of the triangle to the area of the hexagon is then $4:6 = 2:3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42104,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence $(a_n)$ is defined by\n$$\na_1 = 1, \\quad a_n = 3a_{n-1} + 2^{n-1}, \\quad \\text{for } n \\ge 2.\n$$\nFind a formula for the general term $a_n$ in terms of $n$.",
    "options": [],
    "answer": "a_n = 3^n - 2^n",
    "solution": "Evaluating the first few terms, one finds $a_1 = 1$, $a_2 = 5$, $a_3 = 19$, $a_4 = 65$, $a_5 = 211$ and $a_6 = 665$. The values always increase by a factor of $3$, plus a little bit. A guess is that the terms are similar to $3^n$, and computing the difference one fits it to be $2^n$. We'll show by induction that $a_n = 3^n - 2^n$.\n\nThe base case ($n=1$) works. Then, for all $n \\in \\mathbb{N}$,\n$$\na_{n+1} = 3a_n + 2^n = 3(3^n - 2^n) + 2^n = 3^{n+1} - 3 \\cdot 2^n + 2^n = 3^{n+1} - 2^{n+1},\n$$\nas required, so the induction holds and $a_n = 3^n - 2^n$ for all $n \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42105,
    "subject": "Mathematics (Multi-modal)",
    "question": "Divide a regular hexagon into exactly 8 identical shapes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The following image is a possible dissection of a hexagon into 8 identical shapes:\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42106,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a convex quadrilateral $ABCD$ the sides $AB$, $BC$, $CD$ are equal and $AC = BD = AD$. Find the angles of $ABCD$.",
    "options": [],
    "answer": "angle A = 72°, angle B = 108°, angle C = 108°, angle D = 72°",
    "solution": "Notice that $\\triangle ABC \\equiv \\triangle BCD$ (s, s, s) and $\\triangle ABD \\equiv \\triangle DCA$ (s, s, s).\nLet $\\alpha = \\angle CBD = \\angle BCA = \\angle CAB = \\angle CDB$\nand $\\beta = \\angle ACD = \\angle ADC = \\angle BAD = \\angle ABD$.\n\n![](attached_image_1.png)\n\nIt can then be seen that $3\\alpha + \\beta = 180^\\circ$ from triangle $ABC$ and $4\\beta + 2\\alpha = 360^\\circ$ from quadrilateral $ABCD$. Solving these two simultaneous equations gives $\\alpha = 36^\\circ$ and $\\beta = 72^\\circ$. Hence $\\angle A = 72^\\circ$, $\\angle B = 108^\\circ$, $\\angle C = 108^\\circ$, $\\angle D = 72^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42107,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the system of equations:\n$xy = x + y;$\n$x^2 + y^2 = 1$",
    "options": [],
    "answer": "(x, y) = ((1 - √2 + √(2√2 - 1))/2, (1 - √2 - √(2√2 - 1))/2) and (x, y) = ((1 - √2 - √(2√2 - 1))/2, (1 - √2 + √(2√2 - 1))/2)",
    "solution": "We are trying to solve the equations\n$$\nxy = x + y\n$$\n$$\n1 = x^2 + y^2.\n$$\n\nSquaring both sides of $xy = x + y$ and using $x^2 + y^2 = 1$ to simplify gives\n$$\nx^2 y^2 = x^2 + 2xy + y^2 = 1 + 2xy \\Rightarrow (xy)^2 - 2(xy) - 1 = 0.\n$$\nThis gives $xy = 1 \\pm \\sqrt{2}$ or $y = \\frac{1\\pm\\sqrt{2}}{x}$.\nNotice that from equation $xy = x + y$, we can write $y = \\frac{x}{x-1}$.\nCombining this with the previous result gives\n$$\nx^2 - x(1 \\pm \\sqrt{2}) + 1 \\pm \\sqrt{2} = 0.\n$$\nIf $x^2 - x(1 + \\sqrt{2}) + 1 + \\sqrt{2} = 0$, then\n$$\n\\begin{aligned}\nx &= \\frac{(1 + \\sqrt{2}) \\pm \\sqrt{(1 + \\sqrt{2})^2 - 4(1 + \\sqrt{2})}}{2} \\\\\n&= \\frac{(1 + \\sqrt{2}) \\pm \\sqrt{-1 - 2\\sqrt{2}}}{2}.\n\\end{aligned}\n$$\nwhich is non-real.\nIf $x^2 - x(1 - \\sqrt{2}) + 1 - \\sqrt{2} = 0$, then\n$$\n\\begin{aligned}\nx &= \\frac{(1 - \\sqrt{2}) \\pm \\sqrt{(1 - \\sqrt{2})^2 - 4(1 - \\sqrt{2})}}{2} \\\\\n&= \\frac{(1 - \\sqrt{2}) \\pm \\sqrt{2\\sqrt{2} - 1}}{2}.\n\\end{aligned}\n$$\nThis presents a viable solution. Notice that if $(x, y)$ is a solution, then $(y, x)$ is also a solution. The two roots of the above equation are the desired values (since they satisfy equation $xy = x + y$ by Vieta's formulas) and we can have $(x, y) = (y, x) = (r_1, r_2)$ where\n$$\nr_1 = \\frac{1 - \\sqrt{2} + \\sqrt{2\\sqrt{2} - 1}}{2} \\quad \\text{and} \\quad r_2 = \\frac{1 - \\sqrt{2} - \\sqrt{2\\sqrt{2} - 1}}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 42108,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the equation $x^5 + y^3 = z^2$ has infinitely many solutions in the natural numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Put $x = k^6$, $y = 2k^{10}$ and $z = 3k^{15}$ for some natural number $k$. Then\n$$\nx^5 + y^3 = (k^6)^5 + (2k^{10})^3 = k^{30} + 8k^{30} = 9k^{30} = (3k^{15})^2 = z^2\n$$\nas required. Since there are infinitely many possible values for $k$, there are infinitely many triples $(x, y, z)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 42109,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that\n$$(x^3 + x^2 + 3)^2 > 4x^3(x - 1)^2$$\nfor all real $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First observe that if $x$ is negative the inequality is trivial since the left hand side is a square and hence larger than or equal to zero. The right hand side would be negative if $x$ is negative.\n\nNow assume $x \\in \\mathbb{R}^+$. Apply the AM-GM inequality in $x^3$ and $(x-1)^2$:\n$$\n\\begin{aligned}\n\\sqrt{x^3(x-1)^2} &\\le \\frac{x^3 + (x-1)^2}{2} \\\\\n\\Leftrightarrow 4x^3(x-1)^2 &\\le (x^3 + (x-1)^2)^2.\n\\end{aligned}\n$$\nNow, notice that $0 < x^3 + (x-1)^2 = x^3 + x^2 - 2x + 1 < x^3 + x^2 + 3$ for positive $x$. This means\n$$\n4x^3(x-1)^2 \\le (x^3 + (x-1)^2)^2 < (x^3 + x^2 + 3)^2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42110,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $10$ numbers written around the circumference of a circle. Some of them are positive and others are negative. During one move we are allowed to change the sign of the numbers in three successive positions, each of the three numbers changes sign. Prove that we can make all the numbers positive after a finite number of moves.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Label the numbers $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ in a clockwise direction around the circle. Notice that if we switch a certain value twice, it remains in its original state. If we switch an element an odd number of times we change its sign. To solve this problem it will suffice to show that we can switch one number's sign and leave all the remaining numbers as they were after a finite number of moves. Without loss of generality, we shall show that the number labelled $0$ can be switched and the remaining numbers left unchanged. If we switch $(9, 0, 1)$, $(0, 1, 2)$, $(8, 9, 0)$, $(2, 3, 4)$, $(8, 7, 6)$, $(7, 6, 5)$, $(5, 4, 3)$ notice that all positions are switched twice except $0$ which is switched $3$ times and hence has its sign changed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42111,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\{a_1, a_2, ..., a_m\\}$ and $\\{b_1, b_2, ..., b_n\\}$ be two sets of reals with the same sum. Prove that you can find an $m \\times n$ array so that the row sums are $a_1, a_2, ..., a_m$ and the column sums are $b_1, b_2, ..., b_n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S = \\sum a_i = \\sum b_i$, and let $\\omega = a_n + b_n - S$. Create the table\n\n| 0   | 0   | ... | 0   | $a_1$ |\n|-----|-----|-----|-----|-------|\n| 0   | 0   | ... | 0   | $a_2$ |\n| ... | ... | ... | ... | ...   |\n| $b_1$ | $b_2$ | ... | $b_{n-1}$ | $\\omega$ |\n\nThe sum of the first $m-1$ rows and $n-1$ columns clearly satisfies the conditions, and the last row sum is $b_1 + b_2 + ... + b_{n-1} + \\omega = \\sum b_i + a_n - S = a_n$ as required. Likewise for the last column sum.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42112,
    "subject": "Mathematics (Multi-modal)",
    "question": "A little boy Mak has drawn a rectangle of size $20 \\times 15$, and divided it by straight lines into unit squares. How many squares (of any size) are there in Mak's picture?",
    "options": [],
    "answer": "1840",
    "solution": "Consider a square of side length $d$, and the position of its bottom-right unit square. It has $d-1$ squares above, and $d-1$ squares to the left of it. Hence there are $(21-d)(16-d)$ possibilities for the bottom-right corner, resulting in $(21-d)(16-d)$ squares of size $d$ appearing in the grid. The total number of squares is\n$$\n\\begin{aligned}\nS &= \\sum_{d=1}^{15} (21 - d)(16 - d) \\\\\n&= \\sum_{d=1}^{15} (336 - 37d + d^2) \\\\\n&= 15 \\cdot 336 - 37 \\sum_{d=1}^{15} d + \\sum_{d=1}^{15} d^2 \\\\\n&= 15 \\cdot 336 - 37 \\frac{15 \\cdot 16}{2} + \\frac{15 \\cdot 16 \\cdot 31}{6} \\\\\n&= 1840.\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42113,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(m, n)$ of positive integers for which $6^m + 2^n + 2$ is a perfect square.",
    "options": [],
    "answer": "(1, 3)",
    "solution": "If both $m$ and $n$ are greater than one, then $6^m + 2^n + 2 \\equiv_4 0 + 0 + 2 \\equiv_4 2$, which is not a perfect square. So at least one of $m$ and $n$ has to be exactly $1$.\n\n* If $m = 1$, then we want $2^n + 8$ to be a square. This has a solution only for $n = 3$. If $n \\ge 4$ the expression will be divisible by $8$ and not $16$, so it cannot be a square.\n\n* If $n = 1$, then we want $6^m + 4$ to be a square. Considering this expression mod $7$, we have $(-1)^m + 4$, which will be $3$ or $5$ (mod $7$). However, perfect squares are only $0$, $1$, $2$ or $4$ (mod $7$), so there are no solutions.\n\nSo the unique solution to the problem is $(m, n) = (1, 3)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42114,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve $k(k + 5)^2 + 12 = n^3$ for positive integers $k$ and $n$.",
    "options": [],
    "answer": "k=5, n=8",
    "solution": "Since $n$ and $k$ are integers, their difference will also be an integer. Since $k(k + 5)^2 = k^3 + 10k^2 + 25k$,\n$$\n\\begin{aligned}\n(k + 2)^3 &= k^3 + 6k^2 + 12k + 8 < k(k + 5)^2 + 12 \\\\\n(k + 4)^3 &= k^3 + 12k^2 + 48k + 64 > k(k + 5)^2 + 12.\n\\end{aligned}\n$$\nThe only way to get equality is if $n = k + 3$. In that case:\n$$\nn^3 = k^3 + 9k^2 + 27k + 27 = k^3 + 10k^2 + 25k + 12,\n$$\nor $0 = k^2 - 2k - 15 = (k - 5)(k + 3)$. As $k$ is a positive integer, $k = 5$ and $n = 8$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42115,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider all the subsets of $\\{1, 2, \\ldots, N\\}$ that do not contain two consecutive numbers. For each subset, calculate the product of the members of the set. What is the sum of the squares of these products?\n\nNote: the product of the elements of an empty set is one.",
    "options": [],
    "answer": "(N + 1)!",
    "solution": "Let $P(N)$ be the sum of squares of the product of these subsets. For $N = 0$ there is only the empty subset, so $P(0) = 1$. For $N = 1$, we add the subset $\\{1\\}$, so $P(1) = 2$. For $N = 2$, we add an extra set $\\{2\\}$, so $P(2) = 6$. Evaluating $P(3) = 24$, one can guess that $P(N) = (N+1)!$. We show this by induction:\n\nTo find $P(N + 1)$, we have all the sets without element $N + 1$, which has sum $P(N)$, and those with $N + 1$, which cannot contain $N$. The product of each of those sets increases by $N + 1$, so the recurrence is $P(N+1) = P(N) + N^2P(N-1)$. The claim holds for small $N$. Suppose it works for some integer $n$. Then\n$$\nP(n + 1) = P(n) + (n + 1)^2 P(n - 1) = (n + 1)! + (n + 1)^2 n! = (n + 2)!,\n$$\nso the induction holds, and $P(N) = (N + 1)!$ for all integers $N$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42116,
    "subject": "Mathematics (Multi-modal)",
    "question": "Can you enumerate the edges of a tetrahedron with numbers from $1$ to $6$ so that all four triangular faces have the same sum, where the sum of a face is calculated by adding the numbers on its edges?",
    "options": [],
    "answer": "No, it cannot be done.",
    "solution": "Label the vertices as shown.\n\n![](attached_image_1.png)\n\nLet $f(XY)$ denote the number on edge $XY$. Notice that\n$$\n\\begin{align*}\nf(BA) + f(AD) + f(DB) &= f(AD) + f(DC) + f(AC) \\\\\n&= f(BC) + f(CD) + f(BD) \\\\\n&= f(AB) + f(BC) + f(AC).\n\\end{align*}\n$$\naccording to the condition. This means that $f(BC) + f(CD) = f(BA) + f(AD)$ and $f(AD) + f(DC) = f(AB) + f(BC)$. Subtracting these two equalities from one another we find that $f(BC) - f(AD) = f(AD) - f(BC)$. Hence $f(AD) = f(BC)$. This means that the numbers cannot be distinct and so it cannot be done.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 42117,
    "subject": "Mathematics (Multi-modal)",
    "question": "Are there infinitely many integers whose square ends in three 4s, i.e. ...444?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that $38^2 = 1444$ which ends in ...444.\n\nMoreover, $(1000k + 38)^2 = 1000000k^2 + 76000k + 1444$ which ends in ...444 for any natural number $k$. Thus there are infinitely many such numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42118,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 10 fencers taking part in a tournament. Each fencer duels with each other exactly once. Is it possible that such a moment exists when no two fencers have participated in the same number of duels?",
    "options": [],
    "answer": "No",
    "solution": "Imagine the fencers as nodes in a graph. There is an edge between two nodes if those fencers have duelled. If everyone is to have duelled a different number of opponents, then since there are only 10 fencers, they must have each duelled a distinct number from $\\{0, 1, 2, \\ldots, 8, 9\\}$. In any graph, the sum of all the degrees of each node is even since each edge is counted twice. However, the sum of all the degrees for the 10 fencers would have to be $1 + 2 + \\dots + 8 + 9 = \\frac{9 \\times 10}{2} = 45$ which is odd. Hence such a state cannot be achieved. Furthermore, note that an easier solution is to just note that you can't simultaneously use $0$ and $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42119,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x^2 + y^2 = 1$ and $x, y > 0$, prove that $x^3 + y^3 \\ge \\sqrt{2}xy$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First notice that by the AM-GM inequality we have\n$$\n\\frac{x^5 y + x^3 y^3}{2} \\geqslant x^4 y^2 \\quad \\text{and} \\quad \\frac{y^5 x + y^3 x^3}{2} \\geqslant y^4 x^2.\n$$\nAlso note that\n$$\n\\begin{align*}\nx^6 + y^6 &\\geqslant x^5 y + x y^5 \\\\\n\\Leftrightarrow x^6 - x^5 y + y^6 - x y^5 &\\geqslant 0 \\\\\n\\Leftrightarrow x^5(x - y) + y^5(y - x) &\\geqslant 0 \\\\\n\\Leftrightarrow (x - y)(x^5 - y^5) &\\geqslant 0,\n\\end{align*}\n$$\nwhich is true when $x > y$ and when $y > x$ and is therefore true. Thus\n$$\n\\begin{align*}\nx^6 + x^3 y^3 + x^3 y^3 + y^6 &\\geqslant (x^5 y + x^3 y^3) + (y^5 x + y^3 x^3) \\geqslant 2x^4 y^2 + 2x^2 y^4 \\\\\n\\Leftrightarrow (x^3 + y^3)^2 &\\geqslant 2x^2 y^2 (x^2 + y^2) = 2x^2 y^2\n\\end{align*}\n$$\nSince $x, y > 0$ we can conclude that $x^3 + y^3 \\geq \\sqrt{2}xy$ as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42120,
    "subject": "Mathematics (Multi-modal)",
    "question": "$B_1$ and $C_1$ are marked on the bisector of angle $A$ in triangle $ABC$ so that $BB_1 \\perp AB$, $CC_1 \\perp AC$. Let $M$ be the midpoint of $B_1C_1$. Prove that $MB = MC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Drop a perpendicular from $C_1$ onto $AB$ meeting $AB$ at $C_2$. Similarly define $B_2$ such that $B_1B_2 \\perp AC$. Let $M_c$ be on $AC$ such that $MM_c \\perp AC$ and $M_b$ on $AB$ such that $MM_b \\perp AB$. Without loss of generality, assume that the points on the angle bisector of $\\angle A$ are in the order as shown:\n$$\n\\text{Now } \\triangle AB_1B_2 \\sim \\triangle AMM_c \\sim \\triangle AC_1C\n$$\nand $\\triangle AC_1C_2 \\sim \\triangle AMM_b \\sim \\triangle AB_1B$.\n![](attached_image_1.png)\nSo by the Midpoint Theorem, $B_2M_c = M_cC$ and $C_2M_b = M_bB$.\nThis means that $B_2M = CM$ and $C_2M = MB$.\nNotice that $\\angle MB_1B_2 = \\angle BB_1M$ and by symmetry, $BB_1 = B_1B_2$.\nThis means that $\\triangle MB_1B_2 \\equiv \\triangle MB_1B$. Hence $BM = MB_2$.\nThus $BM = MB_2 = CM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42121,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $CH$ be the altitude of an acute angled triangle $ABC$, and let $O$ be the centre of its circumcircle. If $T$ is the foot of the perpendicular drawn from vertex $C$ to the line $AO$, prove that the line $TH$ bisects the side $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $TH$ cut $BC$ at $P$. We want to show that $P$ bisects $BC$. Since $\\angle AHC = \\angle ATC = 90^\\circ$, the points $A$, $H$, $T$ and $C$ lie on the same circle.\n\n![](attached_image_1.png)\n\nNow $\\angle PHC = \\angle TAC = \\frac{1}{2}(180^\\circ - \\angle AOC) = 90^\\circ - \\angle ABC = 90^\\circ - \\angle HBC = \\angle BCH = \\angle PCH$, so triangle $PCH$ is isosceles. In addition, $\\angle PBH = 90^\\circ - \\angle BCH = 90^\\circ - \\angle PHC = \\angle PHB$, so the triangle $PBH$ is isosceles as well. Therefore $PC = PH = PB$ and the claim is proved.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 42122,
    "subject": "Mathematics (Multi-modal)",
    "question": "From a point $P$, draw tangents $PB$ and $PT$ to a circle. Let $A$ be the point on the circle such that $AB$ is a diameter, and let $H$ be the foot of the perpendicular from $T$ onto $AB$. Prove that $AP$ bisects $TH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Construct $P'$ on $AT$ such that $OP' \\perp AB$. Then $OP' \\parallel HT$. Since $BT$ is perpendicular to both $OP$ and $AT$, we have $OP \\parallel AT$, and by the midpoint theorem $OP'$ bisects $AP$.\n\n![](attached_image_1.png)\n\nTherefore $OAP'P$ is a parallelogram. Hence $OP'$ is bisected by $AP$, and since triangles $HAT$ and $OAP'$ are similar, $AP$ bisects $HT$ as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42123,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each positive integer $a$ we consider the sequence $(a_n)$ with $a_0 = a$ and $a_n = a_{n-1} + 40^n$ for $n > 0$. Prove that every such sequence contains infinitely many numbers that are divisible by $2009$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\gcd(40, 2009) = 1$, we have $40^{k \\cdot \\varphi(2009)} \\equiv 1 \\pmod{2009}$ for all natural numbers $k$. For $n > \\varphi(2009)$ the exponent $n!$ is certainly a multiple of $\\varphi(2009)$, and therefore $a_{n+1} \\equiv a_n + 1 \\pmod{2009}$. This means that all values modulo $2009$ are taken cyclically and periodically, and all values are obtained infinitely often. Since this also for the value $0$, the proof is complete.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42124,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a regular hexagon, a point inside is chosen and lines to each vertex of the hexagon are drawn, dividing it into six triangles. If the triangles are alternately shaded grey and white, show that the area of the grey triangles is the same as the area of the white triangles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nExpand the hexagon to make a big equilateral triangle, and drop perpendiculars from the point inside to the sides of the triangle. The area of the unshaded region of the hexagon, which is three triangles of equal bases, is equal to $\\frac{1}{2}$ side length multiplied by the sum of the heights. But in an equilateral triangle, no matter where the interior point is chosen, the sum of these dropped perpendiculars is constant (equal to the height of the triangle). So the unshaded and the shaded regions have equal area.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 42125,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d > 0$. Find all possible values of the sum\n$$\nS = \\frac{a}{d+a+b} + \\frac{b}{a+b+c} + \\frac{c}{b+c+d} + \\frac{d}{c+d+a}.\n$$",
    "options": [],
    "answer": "(1, 2)",
    "solution": "Observe that\n$$\n\\begin{aligned}\nS &> \\frac{a}{a+b+c+d} + \\frac{b}{a+b+c+d} + \\frac{c}{a+b+c+d} + \\frac{d}{a+b+c+d} = 1, \\\\\nS &< \\frac{a}{a+b} + \\frac{b}{a+b} + \\frac{c}{c+d} + \\frac{d}{c+d} = 2.\n\\end{aligned}\n$$\nThe function changes smoothly as we vary $a$, $b$, $c$ and $d$. We will prove that it comes arbitrarily close to $1$ and $2$, and therefore it assumes every value in the interval $(1,2)$.\n\nTaking $a = b$ and letting $c = d$ gives\n$$\nS_1(a, c) = \\frac{2a}{2a+c} + \\frac{2c}{a+2c}, \\quad \\Rightarrow \\quad \\lim_{c \\to 0} S_1(a, c) = 1.\n$$\nSimilarly, taking $a = c$ and $b = d$ gives\n$$\nS_2(a, b) = \\frac{2a}{a+2b} + \\frac{2b}{2a+b} \\quad \\Rightarrow \\quad \\lim_{b \\to 0} S_2(a, b) = 2.\n$$\nSo the expression takes on all values between $1$ and $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42126,
    "subject": "Mathematics (Multi-modal)",
    "question": "The cells of a $10 \\times 10$ square are susceptible to infection. In a unit of time, the cells with 2 or more infected neighbours (having a common side) become infected. Is it possible to start the infection in\n\na. nine cells,\nb. ten cells,\n\nin such a way that the infection spreads to the whole square?",
    "options": [],
    "answer": "a) No. b) Yes, for example by infecting all ten cells on the main diagonal.",
    "solution": "It is easy to observe that the infection can never spread outside of a rectangle that bounds it. A less obvious but more useful observation is that the total perimeter of the infected area will never increase.\n\nLet infected squares be black and non-infected squares be white. It is clear that the spread can be considered on a square-by-square basis, as additional infected squares will not hamper the infection of others.\n\nWe need then to consider the cases in which a square may get infected.\nThere are four ways in which it can have two or more black neighbours:\n![](attached_image_1.png)\n\nIf it has exactly two infected neighbours, the black perimeter does not change, if it has three infected neighbours, the black perimeter will decrease by $2$, and if all the neighbours are infected, the perimeter will decrease by $4$.\n\nTo solve the problem, nine cells have a maximal perimeter of $36$, whereas the total $10 \\times 10$ square has a perimeter of $40$, so it is not possible to infect the entire square. If there are ten cells, which all lie along the main diagonal of the square, the infection will spread everywhere.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42127,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triangular numbers which are also perfect squares.",
    "options": [],
    "answer": "All square triangular numbers are given by x^2 = (1/32) ((3 + 2√2)^k − (3 − 2√2)^k)^2 for k ∈ ℕ; equivalently, they correspond to solutions of A^2 − 2B^2 = 1 with A = 2m + 1 and B = 2n.",
    "solution": "Suppose that $n^2 = T_m = \\frac{m(m+1)}{2}$. Multiplying by $8$,\n$$\n8n^2 = 4m^2 + 4m = (2m + 1)^2 - 1 \\Leftrightarrow (2m + 1)^2 - 8n^2 = 1.\n$$\nLetting $A = 2m+1$ and $B = 2n$, this gives Pell's equation, $A^2 - 2B^2 = 1$. The first solution is $A = 3, B = 2$, and the general solution is given by $a_k + \\sqrt{2}b_k = (3 + 2\\sqrt{2})^k$. The solution is\n$$\na_k = \\frac{(3 + 2\\sqrt{2})^k + (3 - 2\\sqrt{2})^k}{2} \\quad \\text{and} \\quad b_k = \\frac{(3 + 2\\sqrt{2})^k - (3 - 2\\sqrt{2})^k}{2\\sqrt{2}}.\n$$\nThe squares themselves are $x^2 = \\frac{1}{32}((3 + 2\\sqrt{2})^k - (3 - 2\\sqrt{2})^k)^2$ for $k \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42128,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a positive integer and $a > 1$. Prove that, for every positive integer $n$, the number\n$$\nn(2n + 1)(3n + 1)\\dots(an + 1)\n$$\nis divisible by all prime numbers smaller than $a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $p < a$ be a prime. If $p \\nmid n$, the claim holds.\nSuppose then that $n$ is not divisible by $p$. Then the numbers $2n + 1, 3n + 1, \\dots, (p+1)n + 1$ give different remainders when divided by $p$, which are denoted, respectively, by $r_1, r_2, \\dots, r_p$. Indeed, if $r_i = r_j$, then\n$$\np|((i+1)n + 1) - ((j+1)n + 1) = (i-j)n,\n$$\nwhich implies that $i = j$. So each of the $p$ possible remainders appears exactly once. In particular, one of these remainders must be zero, meaning that the one of the factors in the product is divisible by $p$, which finishes the problem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42129,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\omega$ be a circle, and let $A$ and $B$ be points on a line not intersecting $\\omega$. Given a point $X_0$ on $\\omega$, define a sequences $X_0, X_1, X_2, \\ldots$ and $Y_0, Y_1, Y_2, \\ldots$ as follows: $Y_n$ is the second intersection of the line $AX_n$ with $\\omega$ and $X_{n+1}$ is the second intersection of the line $BY_n$ with $\\omega$. Prove that if for some integer $k$, it happens that $X_k = X_0$ for some choice of $X_0$ then $X_k = X_0$ for any choice of $X_0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider circles $\\omega_1$ and $\\omega_2$ centred at $A$ and $B$ respectively such that both $\\omega_1$ and $\\omega_2$ are orthogonal to $\\omega$. Let $I$ be one of the intersection points of $\\omega_1$ and $\\omega_2$. Note that $I$ must exist because $AB$ is outside $\\omega$. Then invert the diagram through a circle centred at $I$ with any radius. $\\omega'_1$ and $\\omega'_2$ (the images of $\\omega_1$ and $\\omega_2$) are straight lines passing through the centre of $\\omega'$ (since they are orthogonal and angles between circles are preserved by inversion). $\\omega'_1$ passes through the centre of circle $AX'_iQ'_i$ and $\\omega'_2$ passes through the centre of circle $BX'_iQ'_i$ for the same reason. Hence $Q'_i$ is $X'_i$ reflected in line $\\omega'_1$ and $X'_{i+1}$ is the reflection of $Q'_i$ in the line $\\omega'_2$. Thus $\\angle X'_iOX'_{i+1} = 2\\theta$ where $\\theta$ is the angle between the lines $\\omega'_1$ and $\\omega'_2$. Thus $X'_0 = X'_k$ is equivalent to $k\\theta = 2n\\pi$, for some integer $n$. This means that consecutive terms in the sequence of points $X'_i$ are determined by a rotation through a constant angle. Hence this result will be true irrespective of the choice of $X'_0$. There is a one-to-one correspondence between $X_i$ and $X'_i$ and hence the same is true if $X_0 = X_k$ for some positive integer $k$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42130,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $XYZ$, $L$ is a variable point on a fixed line passing through $X$. $LZ$ meets $XY$ at $P$ and $LY$ meets $XZ$ at $Q$. Show that $PQ$ passes through a fixed point on $YZ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $R$ be the intersection of $YZ$ and $PQ$. We wish to prove that $R$ is constant. Let $R'$ be the intersection of $LQ$ and $XR$. Then it is known that $L, Y, R', Q$ form a harmonic range, i.e. $(L, Y; R', Q) = -1$. The cross ratio depends on the angles $\\angle LXY, \\angle YXR', \\angle R'XZ$. Since $XL, XY$ and $XZ$ are fixed, this implies that $XR'$ is fixed. Hence $R$ is fixed since it is the intersection of $XR'$ and $YZ$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42131,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a $5 \\times 5$ board, two players alternately mark numbers on empty cells. The first player always marks $1$'s, the second $0$'s. One number is marked per turn, until the board is filled. For each of the nine $3 \\times 3$ squares the sum of the nine numbers on its cells is computed. Denote by $A$ the maximum of these sums. How large can the first player make $A$, regardless of the responses of the second player?",
    "options": [],
    "answer": "6",
    "solution": "First, notice that player two can always ensure that $A \\le 6$. The squares of the grid can be partially tiled with $2 \\times 1$ tiles, as shown below. Whenever a $1$ is placed inside a tile, player two can ensure that it also contains a $0$. As every $3 \\times 3$ square contains three complete tiles, it will have at least three zeros in it so, $A \\le 6$.\n\n![](attached_image_1.png)\n\n| 1 | 2 | 3 | 2 | 1 |\n|---|---|---|---|---|\n| 2 | 4 | 6 | 4 | 2 |\n| 3 | 6 | 9 | 6 | 3 |\n| 2 | 4 | 6 | 4 | 2 |\n| 1 | 2 | 3 | 2 | 1 |\n\nLabel each individual cell with the number of $3 \\times 3$ squares that contain it. By going for the highest-placed numbers, player $1$ can get a total of $45$, which spread across $9$ squares gives exactly $5$. Let player $1$ start by placing a $1$ in the middle, and wlog, player two plays in the left side of the board. By placing a $1$ directly right of the middle (value $6$), player two is forced to take the right-most cell in the middle row (else player $1$ will, and there will exist a $3 \\times 3$ square with three $1$s and $6$ empty blocks). By taking the block directly below the middle (value $6$), player one leaves the other stranded: if he does not place a $0$ in the bottom-rightmost $3 \\times 3$ square, player $1$ will get six $1$s in it. If he does, player $1$ will take the last square of value $6$, and will finish with cell totals greater than $45$, which will give $A \\ge 6$. Hence $A = 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42132,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangular grid is obtained by tiling an equilateral triangle of side length $n$ into $n^2$ equilateral triangles of side length $1$. Determine the number of parallelograms bounded by the line segments of the grid.",
    "options": [],
    "answer": "3*binom(n+2,4)",
    "solution": "The parallelograms can be naturally partitioned into three sets, based on the orientation of their sides in the grid. By symmetry there will be the same number of parallelograms of each orientation, so we can restrict ourselves to the 'diamond' case (as in the diagram).\n\n![](attached_image_1.png)\n\nExtend the triangular grid by one. The sides of the parallelogram can then be extended until they cut the new bottom row of the grid. It is easy to see that every parallelogram will correspond to a different set of four intersection points (marked $a$, $b$, $c$ and $d$ for the case of the shaded parallelogram), and that every set of four points will correspond to a parallelogram. This bijection makes it easy to count – there are $n+2$ points on the bottom row, so there are $\\binom{n+2}{4}$ ways of making a parallelogram. Taking into account the orientations, the final answer is $3\\binom{n+2}{4}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42133,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a scalene triangle $ABC$ with $|AB| + |CA| = 2|BC|$, show that the line joining the incenter and the centroid of the triangle is parallel to $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Adopting the usual notation, $2a = b + c$, hence $s = a + b + c = \\frac{3}{2}a$. The area of the triangle is $K = sr = \\frac{3}{2}ar = \\frac{1}{2}a h_a$, where $h_a$ is the altitude from $A$ onto $BC$. Therefore $h_a = 3r$, and so the incenter is a third of the way to the vertex. The centroid is also a third of the way to the vertex; therefore the line joining the incenter and centroid is parallel to $BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42134,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $n \\in \\mathbb{N}$ and $3n+1$ and $4n+1$ are perfect squares, show that $56|n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $3n + 1 = x^2$ and $4n + 1 = y^2$. Then $y^2 - x^2 = n$. As $y$ is odd, $y^2 \\equiv_8 1$, so $n$ has to be even, therefore $x$ is odd, and therefore $y^2 - x^2 \\equiv_8 1 - 1 = 0$, hence $8|n$.\n\nNow notice that $4x^2 - 3y^2 = 1$, so setting $w = 2x$ we obtain Pell's equation $w^2 - 3y^2 = 1$. This has solutions $(2 + \\sqrt{3})^m = w_m + \\sqrt{3}y_m$, but only the odd $m$ solutions correspond to an even $w_m$. Starting with the base solution $2 + \\sqrt{3}$, and multiplying through by $(2 + \\sqrt{3})^2$ we get the recursions\n$$\nw_{m+1} = 2x_{m+1} = 14x_m + 12y_m, \\quad \\text{and} \\quad y_{m+1} = 8x_m + 7y_m.\n$$\nThen $x_{m+1} = 7x_m + 6y_m \\equiv_7 -y_m$, and $y_{m+1} = 8x_m + 7y_m \\equiv_7 x_m$, hence $y_{m+1}^2 - x_{m+1}^2 \\equiv_7 x_m^2 - y_m^2$, and by induction, $x_m^2 - y_m^2 \\equiv_7 0$. Hence 7 divides $n = y_m^2 - x_m^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42135,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two evenly matched teams are engaged in a series of games, which ends as soon as a team wins four games. Is it more likely for the series to end in exactly six games or in exactly seven games?",
    "options": [],
    "answer": "exactly seven games",
    "solution": "Let the two teams be $A$ and $B$, and represent each possible series as a bit string, where a $1$ represents a win by team $A$ and a $0$ a win by team $B$. Let $S$ be the set of all possible bit strings. We need to determine whether there are more bit strings of length $6$ or length $7$ in $S$. For each bit string $s \\in S$, the bit string obtained from $s$ by interchanging $0$ and $1$ is also in $S$, so we may restrict our attention to bit strings ending in a $1$.\n\nThe number of bit strings of length $6$ is equal to $\\binom{5}{3}$ (exactly $3$ of the first $5$ bits must be a $1$ for $A$ to win the tournament in the sixth game), while the number of bit strings of length $7$ is equal to $\\binom{6}{3}$. Since $\\binom{6}{3} > \\binom{5}{3}$, we conclude that the tournament is more likely to end in exactly $7$ games.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42136,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if $x \\le y \\le z$ are real numbers satisfying $xy + yz + zx = 1$, then $xz < \\frac{1}{2}$. Is it possible to replace $\\frac{1}{2}$ with a smaller number?",
    "options": [],
    "answer": "xz < 1/2, and 1/2 is optimal (cannot be replaced by a smaller number).",
    "solution": "If $x$ and $z$ have opposite signs, or if one of them is $0$, then the inequality is obviously true, so $x$, $y$ and $z$ are either all positive or all negative. If they are all negative, we may replace $x$, $y$ and $z$ with $-x$, $-y$ and $-z$ without changing the conditions of the problem, so we can assume that $x$, $y$ and $z$ are all positive.\n\nIn that case, $0 < xy \\le xz \\le yz$, and so $1 = xy + yz + xz \\ge xy + 2xz > 2xz$ which yields $xz < \\frac{1}{2}$.\n\nSetting $x = y = \\frac{1}{n}$ for some natural number $n > 2$, the equation $xy + yz + xz = 1$ implies that $z = \\frac{n^2-1}{2n} > \\frac{1}{n}$ if $n > 2$. Hence $xz = \\frac{n^2-1}{2n^2} = \\frac{1}{2} - \\frac{1}{2n^2}$, which can be made arbitrarily close to $\\frac{1}{2}$. This shows that $\\frac{1}{2}$ cannot be replaced by any smaller number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42137,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a natural number $N$ which is a power of $2$ such that the digits of $N$ can be permuted to form a power of $2$ different from $N$?",
    "options": [],
    "answer": "No",
    "solution": "Suppose that the digits of $2^a$ can be rearranged to form $2^b$, with $a > b$. Then, since the two numbers have the same digit set, it follows that they're congruent modulo $9$. Hence $9 \\mid 2^a - 2^b = 2^b(2^{a-b} - 1)$ and so $2^{a-b} \\equiv_9 1$. However, the smallest positive power of $2$ with this property is $2^6$, so $a-b \\ge 6$. But in that case, $2^a \\ge 2^{b+6} = 64 \\cdot 2^b > 10 \\cdot 2^b$, which means that $2^a$ and $2^b$ can't have the same number of digits, a contradiction. So no such pair exists.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42138,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given that\n$$\n\\frac{1 + 3 + 5 + \\cdots + (2n - 1)}{2 + 4 + 6 + \\cdots + (2n)} = \\frac{2011}{2012},\n$$\ndetermine $n$.",
    "options": [],
    "answer": "2011",
    "solution": "Using the sum formula for arithmetic progressions, we obtain\n$$\n\\frac{1 + 3 + 5 + \\cdots + (2n - 1)}{2 + 4 + 6 + \\cdots + (2n)} = \\frac{n \\cdot \\frac{1+2n-1}{2}}{n \\cdot \\frac{2+2n}{2}} = \\frac{n}{n+1} = \\frac{2011}{2012},\n$$\nfrom which it follows that $n = 2011$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42139,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be real numbers such that\n$$\na_m + a_{m+1} + \\dots + a_n \\ge m + (m+1) + \\dots + n\n$$\nfor every $m = 1, 2, \\dots, n$. Prove that\n$$\na_1^2 + a_2^2 + \\dots + a_n^2 \\ge \\frac{n(n+1)(2n+1)}{6}.\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Make the substitution $a_k = b_k + k$ for $k = 1, 2, \\dots, n$. The conditions of the problem then become $b_m + b_{m+1} + \\dots + b_n \\ge 0$ for every $m = 1, 2, \\dots, n$, while the required inequality becomes\n$$\n\\begin{aligned}\n\\frac{n(n+1)(2n+1)}{6} &\\le (b_1+1)^2 + (b_2+2)^2 + \\dots + (b_n+n)^2 \\\\\n&= (b_1^2 + b_2^2 + \\dots + b_n^2) + (1^2 + 2^2 + \\dots + n^2) + 2(b_1 + 2b_2 + \\dots + n b_n).\n\\end{aligned}\n$$\nBut $b_1^2 + \\dots + b_n^2 \\ge 0$, $1^2 + 2^2 + \\dots + n^2 = \\frac{n(n+1)(2n+1)}{6}$ while\n$b_1+2b_2+\\dots+nb_n = b_n+(b_{n-1}+b_n)+(b_{n-2}+b_{n-1}+b_n)+\\dots+(b_1+b_2+\\dots+b_n) \\ge 0$\nby the (restated) given condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42140,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be two integers with $a > b$. If $ab - 1$ and $a+b$ are relatively prime, and $ab + 1$ and $a-b$ are relatively prime, prove that\n$$\n(ab + 1)^2 + (a - b)^2\n$$\nis not a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We know that\n$$\n(a-b)^2 + (ab+1)^2 = a^2b^2 + a^2 + b^2 + 1 = (a^2+1)(b^2+1),\n$$\nso it will be enough to show that $a^2+1$ and $b^2+1$ are relatively prime. Suppose that there is a prime number $p$ that divides both $a^2+1$ and $b^2+1$. Then $p$ divides $a^2-b^2$, and hence $p$ divides one of $a+b$ or $a-b$.\nIf $p$ divides $a-b$, then $p$ divides $ab-b^2$, and since $p$ divides $b^2+1$, it divides $ab-b^2+b^2+1 = ab+1$, which is impossible, since it is assumed that $a-b$ and $ab+1$ are relatively prime. If $p$ divides $a+b$, a similar contradiction is derived.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42141,
    "subject": "Mathematics (Multi-modal)",
    "question": "The real-valued function $f$ satisfies\n$$\nf(\\tan 2x) = \\tan^4 x + \\cot^4 x\n$$\nfor all real $x$. Prove that, for all real $x$,\n$$\nf(\\sin x) + f(\\cos x) \\ge 196.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that since $\\tan 2\\theta$ attains all real numbers in the interval $\\left(-\\frac{\\pi}{4}, \\frac{\\pi}{4}\\right)$, we may restrict our attention to this domain. Also,\n$$\nf(-\\tan 2\\theta) = f(\\tan 2(-\\theta))\n$$\n$$\n= \\tan^4(-\\theta) + \\cot^4(-\\theta) \\\\\n= \\tan^4 \\theta + \\cot^4 \\theta \\\\\n= f(\\tan 2\\theta),\n$$\nso $f$ is an even function, and we may restrict $\\theta$ to the interval $(0, \\frac{\\pi}{4})$.\n\nNext, let $0 < x < y$, and let $x = \\tan 2\\alpha$, $y = \\tan 2\\beta$, where $0 < \\alpha < \\beta < \\frac{\\pi}{4}$. Then\n$$\n\\begin{aligned}\nf(x) - f(y) &= f(\\tan 2\\alpha) - f(\\tan 2\\beta) \\\\\n&= \\tan^4 \\alpha + \\cot^4 \\alpha - \\tan^4 \\beta - \\cot^4 \\beta \\\\\n&= (\\tan^4 \\alpha - \\tan^4 \\beta) - \\left( \\frac{1}{\\tan^4 \\beta} - \\frac{1}{\\tan^4 \\alpha} \\right) \\\\\n&= (\\tan^4 \\alpha - \\tan^4 \\beta) \\left( 1 - \\frac{1}{\\tan^4 \\alpha \\tan^4 \\beta} \\right).\n\\end{aligned}\n$$\nNow, $\\tan \\theta$ is increasing on the interval $(0, \\frac{\\pi}{4})$, and for $\\theta \\in (0, \\frac{\\pi}{4})$, $0 < \\tan \\theta < 1$, hence $\\tan^4 \\alpha - \\tan^4 \\beta < 0$ and $\\tan^4 \\alpha \\tan^4 \\beta < 1 \\Rightarrow 1 - \\frac{1}{\\tan^4 \\alpha \\tan^4 \\beta} < 0$, hence $f(x) - f(y) > 0$, showing that $f$ is decreasing.\n\nNext, note that both $\\cot \\theta$ and $\\tan \\theta$ are concave up on the interval $(0, \\frac{\\pi}{4})$, so we have that $\\frac{\\tan 2\\alpha + \\tan 2\\beta}{2} \\ge \\tan(\\alpha + \\beta)$, by Jensen's Inequality. Hence, setting $x = \\tan 2\\alpha$ and $y = \\tan 2\\beta$, and recalling that $f$ is decreasing, we have\n$$\n\\begin{aligned}\nf\\left(\\frac{x+y}{2}\\right) &= f\\left(\\frac{\\tan 2\\alpha + \\tan 2\\beta}{2}\\right) \\\\\n&\\le f\\left(\\tan 2\\left(\\frac{\\alpha+\\beta}{2}\\right)\\right) \\\\\n&= \\tan^4\\left(\\frac{\\alpha+\\beta}{2}\\right) + \\cot^4\\left(\\frac{\\alpha+\\beta}{2}\\right) \\\\\n&\\le \\frac{\\tan^4\\alpha + \\tan^4\\beta}{2} + \\frac{\\cot^4\\alpha + \\cot^4\\beta}{2} \\quad \\text{(Jensen's Inequality)} \\\\\n&= \\frac{f(\\tan 2\\alpha) + f(\\tan 2\\beta)}{2} \\\\\n&= \\frac{f(x) + f(y)}{2}.\n\\end{aligned}\n$$\nThis shows that $f$ is concave up.\n\nNow let $x = \\sin \\alpha$ and $y = \\cos \\alpha$. Since $f$ is even, we may assume that both $x$ and $y$ are positive, and hence $\\frac{x+y}{2} \\le \\sqrt{\\frac{x^2+y^2}{2}}$. Using Jensen's Inequality and recalling that $f$ is decreasing, we have that\n$$\nf(x) + f(y) \\ge 2f\\left(\\frac{x+y}{2}\\right) \\ge 2f\\left(\\sqrt{\\frac{x^2+y^2}{2}}\\right) = 2f\\left(\\frac{1}{\\sqrt{2}}\\right).\n$$\nNow, let $\\tan 2\\theta = \\frac{1}{\\sqrt{2}}$. Then $\\tan \\theta = \\sqrt{3} - \\sqrt{2}$ and $\\cot \\theta = \\sqrt{3} + \\sqrt{2}$, and so\n$$\n\\begin{aligned}\nf\\left(\\frac{1}{\\sqrt{2}}\\right) &= f(\\tan 2\\theta) \\\\\n&= \\tan^4 \\theta + \\cot^4 \\theta \\\\\n&= (\\sqrt{3} - \\sqrt{2})^4 + (\\sqrt{3} + \\sqrt{2})^4 \\\\\n&= 2(\\sqrt{3}^4 + 6(\\sqrt{2}\\sqrt{3})^2 + \\sqrt{2}^4) \\\\\n&= 98.\n\\end{aligned}\n$$\nHence $f(\\cos \\theta) + f(\\sin \\theta) \\ge 2(98) = 196$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42142,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each positive integer is coloured red or blue. A function $f$ from the set of positive integers to itself has the following two properties:\n(a) if $x \\le y$, then $f(x) \\le f(y)$; and\n(b) if $x, y$ and $z$ are (not necessarily distinct) positive integers of the same colour and $x + y = z$, then $f(x) + f(y) = f(z)$.\nProve that there exists a positive number $a$ such that $f(x) \\le a x$ for all positive integers $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For integers $x$ and $y$, by a segment $[x, y]$ we mean the set of all integers $t$ such that $x \\le t \\le y$; the length of this segment is $y - x$.\nIf, for every two positive integers $x$ and $y$ of the same colour we have $\\frac{f(x)}{x} = \\frac{f(y)}{y}$, then one can choose $a = \\max\\left\\{\\frac{f(r)}{r}, \\frac{f(b)}{b}\\right\\}$, where $r$ and $b$ are arbitrary red and blue numbers, respectively. So we can assume that there are two red numbers $x$ and $y$ such that $\\frac{f(x)}{x} \\ne \\frac{f(y)}{y}$.\nSet $m = x y$. Then each segment of length $m$ contains a blue number. Indeed, assume that all the numbers on the segment $[k, k+m]$ are red. Then\n$$\nf(k + m) = f(k + x y) = f(k + x(y - 1)) + f(x) = \\dots = f(k) + y f(x), \\\\\nf(k + m) = f(k + x y) = f(k + (x - 1)y) + f(y) = \\dots = f(k) + x f(y),\n$$\nso $y f(x) = x f(y)$, a contradiction. Now we consider two cases.\n\n**Case 1.** Assume that there exists a segment $[k, k+m]$ of length $m$ consisting of blue numbers. Define $D = \\max\\{f(k), \\dots, f(k+m)\\}$. We claim that $f(z) - f(z-1) \\le D$, whenever $z > k$, and the conclusion follows. Consider the largest blue number $b_1$ not exceeding $z$, so $z-b_1 \\le m$, and some blue number $b_2$ in the segment $[b_1+k, b_1+k+m]$, so $b_2 > z$. Write $f(b_2) = f(b_1) + f(b_2-b_1) \\le f(b_1) + D$ to deduce that $f(z+1) - f(z) \\le f(b_2) - f(b_1) \\le D$, as claimed.\n\n**Case 2.** Each segment of length $m$ contains numbers of both colours. Fix any red number $R \\ge 2m$ such that $R+1$ is blue and set $D = \\max\\{f(R), f(R+1)\\}$. Now we claim that $f(z+1)-f(z) \\le D$, whenever $z > 2m$. Consider the largest red number $r$ not exceeding $z$ and the largest blue number $b$ smaller than $r$; then $0 < z-b = (z-r)+(r-b) \\le 2m$, and $b+1$ is red. Let $t = b+R+1$; then $t > z$. If $t$ is blue, then $f(t) = f(b)+f(R+1) \\le f(b)+D$, and $f(z+1)-f(z) \\le f(t)-f(b) \\le D$. Otherwise, $f(t) = f(b+1)+f(R) \\le f(b+1)+D$, hence $f(z+1)-f(z) \\le f(t)-f(b+1) \\le D$, as claimed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42143,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that for every natural number $n$, there exists a natural number $k$ such that the number $kn$ starts with the digits 2011.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose we wish to find a solution where $kn$ has $d+4$ digits. Then we must satisfy $2011 \\cdot 10^d \\le kn < 2012 \\cdot 10^d$, or equivalently\n$$\nk \\in \\left[ \\frac{2011}{n} \\cdot 10^d, \\frac{2012}{n} \\cdot 10^d \\right).\n$$\nThe size of this interval is $\\frac{10^d}{n}$. Clearly, we can choose $d$ sufficiently large such that the interval has size greater than 1, in which case it must contain an integer, and this integer will be a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42144,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a point inside triangle $ABC$. Construct the points $P_1$, $P_2$, $P_3$ such that $PP_1 \\perp BC$, $PP_2 \\perp CA$, $PP_3 \\perp AB$ and $BP_3 = BP_1$, $CP_2 = CP_1$. Prove that $AP_3 = AP_2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Place the points $A$, $B$ and $C$ in the complex plane such that $P$ is at the origin. Then we have the following equations:\n$$\nPP_1 \\perp BC \\implies \\frac{B-C}{P_1} = \\frac{\\bar{C} - \\bar{B}}{\\bar{P}_1} \\qquad (3)\n$$\n$$\nPP_2 \\perp AC \\implies \\frac{A-C}{P_2} = \\frac{\\bar{C} - \\bar{A}}{\\bar{P}_2} \\qquad (4)\n$$\n$$\nPP_3 \\perp AB \\implies \\frac{B-A}{P_3} = \\frac{\\bar{A} - \\bar{B}}{\\bar{P}_3} \\qquad (5)\n$$\n$$\n\\begin{align} BP_1 = BP_3 &\\implies (B - P_1)(\\bar{B} - \\bar{P}_1) = (B - P_3)(\\bar{B} - \\bar{P}_3) \\nonumber \\\\ &\\implies B\\bar{P}_3 + \\bar{B}P_3 - P_3\\bar{P}_3 = B\\bar{P}_1 + \\bar{B}P_1 - P_1\\bar{P}_1 \\tag{6} \\end{align}\n$$\n$$\n\\begin{align} CP_1 = CP_2 &\\implies (C - P_1)(\\bar{C} - \\bar{P}_1) = (C - P_2)(\\bar{C} - \\bar{P}_2) \\nonumber \\\\ &\\implies C\\bar{P}_2 + \\bar{C}P_2 - P_2\\bar{P}_2 = C\\bar{P}_1 + \\bar{C}P_1 - P_1\\bar{P}_1 \\tag{7} \\end{align}\n$$\nThen,\n$$\n\\begin{align*} & AP_2^2 - AP_3^2 \\\\ &= (A - P_2)(\\bar{A} - \\bar{P}_2) - (A - P_3)(\\bar{A} - \\bar{P}_3) \\\\ &= A\\bar{P}_3 + \\bar{A}P_3 - A\\bar{P}_2 - \\bar{A}P_2 + P_2\\bar{P}_2 - P_3\\bar{P}_3 \\\\ &= B\\bar{P}_3 + \\bar{B}P_3 - C\\bar{P}_2 - \\bar{C}P_2 + P_2P_2 - 2P_3\\bar{P}_3 \\quad \\text{(using (4) and (5))} \\\\ &= (B\\bar{P}_3 + \\bar{B}P_3 - P_3\\bar{P}_3) - (C\\bar{P}_2 + \\bar{C}P_2 - P_2\\bar{P}_2) \\end{align*}\n$$\n$$\n\\begin{aligned}\n&= (B\\bar{P}_1 + \\bar{B}P_1 - P_1\\bar{P}_1) - (C\\bar{P}_1 + \\bar{C}P_1 - P_2\\bar{P}_1 \\quad \\text{(using (6) and (7))} \\\\\n&= (B - C)\\bar{P}_1 - (\\bar{C} - \\bar{B})P_1 \\\\\n&= 0, \\quad \\text{(using (3))}\n\\end{aligned}\n$$\nas required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42145,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$, $C$ be three distinct points in the plane for which $AB = AC$. Describe the locus of the points $P$ for which $\\angle APB = \\angle APC$.",
    "options": [],
    "answer": "The perpendicular bisector of BC.",
    "solution": "Clearly every point on the perpendicular bisector of $BC$ satisfies the requirements of the problem. We now show that no other point $P$ does.\nLet $D$ be the midpoint of $BC$, and suppose that there exists a point $P$ not on the line $AD$ such that $\\angle APB = \\angle APC$. Let $Q$ be the reflection of $P$ about the line $AD$; then $Q$ also satisfies $\\angle AQB = \\angle AQC$. Hence $A$, $P$, $Q$ and $C$ are concyclic, and so are $A$, $P$, $Q$ and $B$. However, this means that both $B$ and $C$ lie on the circumcircle of triangle $APQ$, which is clearly impossible.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42146,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the following equation holds for all non-negative integers $n$:\n$$\n\\lfloor \\sqrt{n} + \\sqrt{n+1} + \\sqrt{n+2} \\rfloor = \\lfloor \\sqrt{9n} + 8 \\rfloor.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For each non-negative integer $n$,\n$$\n2\\sqrt{n+1} > \\sqrt{n} + \\sqrt{n+2}.\n$$\nIndeed,\n$$\n\\begin{aligned}\n\\sqrt{n+1} - \\sqrt{n} &= \\frac{1}{\\sqrt{n+1} + \\sqrt{n}} \\\\\n&> \\frac{1}{\\sqrt{n+2} + \\sqrt{n+1}} \\\\\n&= \\sqrt{n+2} - \\sqrt{n+1},\n\\end{aligned}\n$$\nfrom which the result follows.\n\nWe now prove the required equality. It is true for $n = 0$ and $n = 1$, by inspection. Let $n \\ge 2$. By the above result,\n$$\n\\sqrt{n} + \\sqrt{n+1} + \\sqrt{n+2} < 3\\sqrt{n+1} = \\sqrt{9n+9}.\n$$\nWe now show that $\\sqrt{n} + \\sqrt{n+1} + \\sqrt{n+2} > \\sqrt{9n+8}$. First note that, for $n \\ge 2$,\n$$\nn(n+2) - \\left(n + \\frac{7}{9}\\right)^2 = \\frac{4}{9}n - \\frac{49}{81} \\ge \\frac{8}{9} - \\frac{49}{81} > 0\n$$\nwhich implies that\n$$\n\\sqrt{n(n+2)} > n + \\frac{7}{9}.\n$$\nHence,\n$$\n\\begin{aligned}\n[\\sqrt{n} + \\sqrt{n+1} + \\sqrt{n+2}]^2 &> \\left[\\frac{3}{2}(\\sqrt{n} + \\sqrt{n+2})\\right]^2 \\\\\n&= \\frac{9}{4}\\left[n + (n+2) + 2\\sqrt{n(n+2)}\\right] \\\\\n&> \\frac{9}{4}\\left[2n + 2 + 2\\left(n + \\frac{7}{9}\\right)\\right] \\\\\n&= \\frac{9}{4}\\left[4n + \\frac{32}{9}\\right] \\\\\n&= 9n + 8.\n\\end{aligned}\n$$\nHence we have\n$$\n\\sqrt{9n+8} < \\sqrt{n} + \\sqrt{n+1} + \\sqrt{n+2} < \\sqrt{9n+9}.\n$$\nSince $9n+8$ and $9n+9$ are consecutive, no perfect square lies between them, and so no integer lies strictly between $\\sqrt{9n+8}$ and $\\sqrt{9n+9}$. This is enough to conclude the required equality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 42147,
    "subject": "Mathematics (Multi-modal)",
    "question": "The acute-angled triangle $ABC$ has circumcentre $O$ and orthocentre $H$. The perpendicular bisector of $BH$ meets $AB$ at $Q$, the perpendicular bisector of $CH$ meets $AC$ at $P$ and these two bisectors meet at $R$.\n\na. Prove that the circumcircles of $AOP$, $POR$, $ROQ$ and $QOA$ have equal radii. When are they equal in radius to the circumcircle of triangle $ABC$?\n\nb. Prove that if $PQ$ passes through $O$, then it also passes through $H$. What can be said about $\\angle BAC$ in this case?",
    "options": [],
    "answer": "Part (a): The circumcircles of AOP, POR, ROQ, and QOA all have common radius R/(2 sin A), where R is the circumradius of triangle ABC and A is angle BAC. They equal R if and only if angle BAC = 30°. Part (b): If PQ passes through O, then PQ also passes through H; in this case angle BAC = 60°.",
    "solution": "a. Note that $R$ is the circumcentre of triangle $BHC$, hence $OR$ bisects $BC$. Also, $\\angle BRC = 2\\angle(180^\\circ - \\angle BHC) = 2\\angle QRP = 2\\angle A = \\angle BOC$, so it follows that $R$ is the reflection of $O$ across the line $BC$.\n\nNext, since $QR$ is perpendicular to $BH$, which is perpendicular to $AC$, we have that $AC \\parallel QR$, and hence $\\angle BQR = \\angle A = \\angle BOR$, which shows that $BQOR$ is a cyclic quadrilateral. Using the sine rule in triangle $BQR$ we see that the circumradius of triangle $BQR$ is equal to $\\frac{BR}{2\\sin BQR} = \\frac{R}{2\\sin A}$, where $R$ is the circumradius of triangle $ABC$. Similarly, $CPOR$ is a cyclic quadrilateral with circumradius $\\frac{R}{2\\sin A}$.\n\nLet the line through $O$ perpendicular to $BH$ meet $BH$ in $K$ and $AB$ in $L$, let the foot of the altitude from $B$ be $E$ and let the midpoint of $BH$ be $M$. It is well known that $AH = OR$, and hence, since $AH \\parallel OR$ and $QR \\parallel OK \\parallel AC$, it follows that $HE = MK \\implies KE = MH = BM \\implies AL = BQ$. Hence $L$ and $Q$ are symmetric about the midpoint of $AB$, and so $OQ = OL$. Hence, since $OL \\parallel AC$, $\\angle OQA = \\angle OLQ = \\angle A$.\n\nUsing the sine rule in triangle OAQ shows that it has circumradius\n$\\frac{AO}{2 \\sin \\angle AQO} = \\frac{R}{2 \\sin A}$. Similarly, $\\angle OPA = \\angle A$ and triangle OAP has circumradius $\\frac{R}{2 \\sin A}$.\n\nThese radii are equal to $R$ if and only if $\\sin A = \\frac{1}{2}$, i.e. if and only if $\\angle A = 30^\\circ$.\n\n![](attached_image_1.png)\n\nb. Since $QR$ is the perpendicular bisector of $BH$, we have that $\\angle HQR = \\angle BQR = \\angle A$. Considering the opposite angles of these two angles and recalling that $\\angle AQO = \\angle A$, we have that $\\angle HQO + 180^\\circ = 3\\angle A \\Rightarrow \\angle HQO = 3\\angle A - 180^\\circ$. Next, since $\\angle OQA = \\angle A = \\angle OPQ$, it follows that $\\angle POQ = 360^\\circ - 3\\angle A$. Hence $\\angle POQ + HQO = 360^\\circ - 3\\angle A + 3\\angle A - 180^\\circ = 180^\\circ$, which shows that $OP \\parallel QH$. Hence, if $PQ$ passes through $O$, it also passes through $H$. In this case, $\\angle HQO = 0^\\circ = 3\\angle A - 180^\\circ$, implying that $\\angle A = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42148,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Prove that the unit square can be covered by three sets of diameter not exceeding $\\frac{\\sqrt{65}}{8}$.\n\nb. Prove that the unit square can not be covered by three sets of diameter less than $\\frac{\\sqrt{65}}{8}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. Set the square on the coordinate plane with vertices $(\\pm\\frac{1}{2}, \\pm\\frac{1}{2})$. Then the three circles with diameter $\\frac{\\sqrt{65}}{8}$ and centres $(0, \\frac{7}{16})$, $(\\frac{1}{4}, -\\frac{1}{16})$ and $(-\\frac{1}{4}, -\\frac{1}{16})$ covers the unit square.\n\nb. Label the square $ABCD$, and suppose that three sets $S_1$, $S_2$, $S_3$ of diameter less than $\\frac{\\sqrt{65}}{8}$ covers it. Then two of the vertices, without loss of generality $A$ and $B$, belong to the same set, say $S_1$. Then clearly $C$ and $D$ do not belong to $S_1$.\n\nSuppose that $C$ and $D$ are both elements of $S_2$. Consider edge $AD$. Since $a + \\frac{1}{8^2} = \\frac{65}{64}$, $S_1$ can only contain points on $AD$ less than distance $\\frac{1}{8}$ from $A$, and similarly $S_2$ can only contain points on $AD$ less than distance $\\frac{1}{8}$ from $D$, and so neither the midpoints of $AD$ nor $BC$ belongs to $S_1 \\cup S_2$. Hence both midpoints must belong to the set $S_3$, but a similar argument as above shows that $S_3$ can cover at most $2 \\times \\frac{1}{8} = \\frac{1}{4}$ units of segment $AD$. Since $2 \\times \\frac{1}{8} + \\frac{1}{4} = \\frac{3}{8} < 1$, this configuration cannot cover $ABCD$. Hence we may suppose that $C \\in S_2$ and $D \\in S_3$.\n\nAs before, $S_1$ can only contain points at a distance less than $\\frac{1}{8}$ from $A$, $S_3$ must contain the rest of segment $AD$. In particular, if $X$ is the point on $AD$ such that $AX = \\frac{1}{8}$, then $X \\in S_1$. Let $E$ be the midpoint of $CD$. By Pythagoras, the distance $EX = \\frac{65}{8}$, and hence $E$ cannot belong to $S_3$. A similar argument shows that $E$ cannot belong to $S_2$ either, and clearly $E$ does not belong to $S_1$. Hence it is impossible to cover all of $ABCD$ with sets of diameter less than $\\frac{\\sqrt{65}}{8}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42149,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a non-isosceles triangle $ABC$, let $D$, $E$ and $F$ denote the midpoints of the sides $BC$, $CA$ and $AB$ respectively. The circumcircle of triangle $BCF$ and the line $BE$ meet again at $P$, and the circumcircle of $ABE$ and the line $AD$ meet again in $Q$. Finally, the lines $DP$ and $FQ$ meet at $R$. Prove that the centroid $G$ of the triangle $ABC$ lies on the circle $PQR$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will use the following lemma:\n\n**Lemma 1.** Let $AD$ be a median in triangle $ABC$. Then $\\cot \\angle BAD = 2 \\cot A + \\cot B$ and $\\cot \\angle ADC = \\frac{1}{2}(\\cot B - \\cot C)$.\n\n*Proof.* Let $CC_1$ and $DD_1$ be the perpendiculars from $C$ and $D$ to $AB$. Using signed lengths, we write\n$$\n\\begin{align*}\n\\cot BAD &= \\frac{AD_1}{DD_1} \\\\\n&= \\frac{\\frac{1}{2}(AC_1 + AB)}{\\frac{1}{2}CC_1} \\\\\n&= \\frac{CC_1 \\cot A + CC_1(\\cot A + \\cot B)}{CC_1} \\\\\n&= 2 \\cot A + \\cot B.\n\\end{align*}\n$$\nSimilarly, denoting by $A_1$ the projection of $A$ onto $BC$, we get\n$$\n\\cot ADC = \\frac{DA_1}{AA_1}\n$$\n$$\n\\begin{aligned}\n&= \\frac{\\frac{1}{2}BC - A_1C}{AA_1} \\\\\n&= \\frac{\\frac{1}{2}(AA_1 \\cot B + AA_1 \\cot C) - AA_1 \\cot C}{AA_1} \\\\\n&= \\frac{1}{2}(\\cot B - \\cot C).\n\\end{aligned}\n$$\n□\n\nTurning to the given problem, by the lemma we get\n$$\n\\begin{aligned}\n\\cot BPD &= 2 \\cot BPC + \\cot PBC \\\\\n&= 2 \\cot BFC + \\cot PBC \\quad (\\text{from circle } BFPC) \\\\\n&= 2 \\cdot \\frac{1}{2}(\\cot A - \\cot B) + 2 \\cot B + \\cot C \\\\\n&= \\cot A + \\cot B + \\cot C.\n\\end{aligned}\n$$\nSimilarly, $\\cot GQF = \\cot A + \\cot B + \\cot C$, so $\\angle GPR = \\angle GQF$ and $GPRQ$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42150,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(p, q, r)$ of prime numbers which satisfy\n$$\n(p + 1)(q + 2)(r + 3) = 4pqr.\n$$",
    "options": [],
    "answer": "(2, 3, 5), (5, 3, 3), (7, 5, 2)",
    "solution": "Dividing both sides of the equation by $pqr$, we obtain\n$$\n\\left(1 + \\frac{1}{p}\\right) \\left(1 + \\frac{2}{q}\\right) \\left(1 + \\frac{3}{r}\\right) = 4.\n$$\nIf $p, q, r \\ge 5$, then\n$$\n\\left(1 + \\frac{1}{p}\\right) \\left(1 + \\frac{2}{q}\\right) \\left(1 + \\frac{3}{r}\\right) \\le \\frac{6}{5} \\cdot \\frac{7}{5} \\cdot \\frac{8}{5} < 4,\n$$\nhence at least one of $p, q$ or $r$ is less than $5$, and since they're all prime, we have the following cases:\n\n*Case 1:* $p = 2$. Then $3(q+2)(r+3) = 8qr \\implies (5q-6)(5r-9) = 144$ which has the solution $(q, r) = (3, 5)$.\n\n*Case 2:* $p = 3$. Then $4(q+2)(r+3) = 12qr \\implies (q-1)(2r-3) = 9$, which has no solutions.\n\n*Case 3:* $q = 2$. Then $4(p+1)(r+3) = 8pr \\implies (p-1)(r-3) = 6$, which has no solutions.\n\n*Case 4:* $q = 3$. Then $5(p+1)(r+3) = 12pr \\implies (7p-5)(7r-15) = 180$ which has the two solutions $(p, r) = (5, 3), (2, 5)$.\n\n*Case 5:* $r=2$. Then $5(p+1)(q+2) = 8pq \\implies (3p-5)(3q-10) = 80$\nwhich has the solution $(p, q) = (7, 5)$.\n\n*Case 6:* $r = 3$. Then $6(p + 1)(q + 2) = 12pq \\implies (p - 1)(q - 2) = 4$\nwhich has the solution $(p, q) = (5, 3)$.\n\nHence the equation has the three solutions $(p, q, r) = (2, 3, 5), (5, 3, 3)$ and $(7, 5, 2)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42151,
    "subject": "Mathematics (Multi-modal)",
    "question": "A school has $n$ students and there are some extra classes provided for them so that each student can participate in any number of them. We know that there are at least two participants in any class. We also know that if two different classes have two common students, then the numbers of their participants are different. Prove that the total number of classes is not greater than $(n - 1)^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_i$, $2 \\le i \\le n$, be the set of all classes with $i$ participants. We will show that $|A_i| \\le \\frac{n(n-1)}{i(i-1)}$. Since every pair of students is together in at most one class of each $A_i$, we have that $\\binom{i}{2} |A_i| \\le \\binom{n}{2}$, giving\n$$\n|A_i| \\le \\frac{\\binom{n}{2}}{\\binom{i}{2}} = \\frac{n(n-1)}{i(i-1)}.\n$$\nNow, the total number of classes equals\n$$\n\\begin{aligned}\n& |A_2| + |A_3| + \\dots + |A_n| \\\\\n& \\le n(n-1) \\left( \\frac{1}{2(2-1)} + \\frac{1}{3(3-1)} + \\dots + \\frac{1}{n(n-1)} \\right) \\\\\n& = n(n-1) \\left( 1 - \\frac{1}{n} \\right) = (n-1)^2, \\end{aligned}\n$$\nas required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42152,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square and $X$ a point such that $A$ and $X$ are on opposite sides of $CD$. The lines $AX$ and $BX$ intersect $CD$ in $Y$ and $Z$ respectively. If the area of $ABCD$ is $1$ and the area of $XYZ$ is $\\frac{2}{3}$, determine the length of $YZ$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "2/3",
    "solution": "Let the length of $YZ$ be $x$, and let $y$ be the associated height of triangle $XYZ$. Note that triangles $XYZ$ and $XAB$ are similar. Since corresponding sides $YZ$ and $AB$ have lengths $x$ and $1$ respectively, the heights $y$ and $1+y$ have to satisfy\n$$\n\\frac{y}{1+y} = \\frac{x}{1} = x.\n$$\nHence the area of $XYZ$ is\n$$\n\\frac{2}{3} = \\frac{xy}{2} = \\frac{y^2}{2(1+y)}.\n$$\n\nThis yields the quadratic equation\n$$\ny^2 - \\frac{4}{3}y - \\frac{4}{3} = \\left(y + \\frac{2}{3}\\right) (y - 2) = 0,\n$$\nfrom which it follows that $y = 2$ and thus\n$$\nx = \\frac{y}{1+y} = \\frac{2}{3}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42153,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let circles $\\Gamma_1$ and $\\Gamma_2$ intersect at $D$ and $P$. The common tangent of the two circles closest to the point $D$ touches $\\Gamma_1$ at $A$ and $\\Gamma_2$ at $B$. The line $AD$ intersects $\\Gamma_2$ for the second time in $C$. Let $M$ be the middle of the line segment $BC$. Prove that $\\angle DPM = \\angle BDC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nLet $S$ be the intersection of $PD$ and $AB$. Then $S$ lies on the radical axis of the two circles, so $AS = SB$, that is, $PS$ is a median in triangle $PAB$.\nNow, by the tan-chord theorem,\n$$\n\\angle BAP = 180^\\circ - \\angle ADP = \\angle CDP = \\angle CBP.\n$$\nAlso, $\\angle ABP = \\angle BCP$, which implies that $\\triangle PAB \\||\\| \\triangle PBC$. Since $PM$ is a median in triangle $BPC$, it follows that $\\angle DPB = \\angle MPC$. Hence\n$$\n\\angle DPM = \\angle DPB + \\angle BPM = \\angle MPC + \\angle BPM = \\angle BPC = \\angle BDC.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42154,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. A train stops at $2n$ stations, including the first and last ones, numbered in order from the first to the $2n^{th}$. It is known that on a certain car, for each pair of integers $i, j$ such that $1 \\le i < j \\le 2n$, exactly one seat has been reserved for the trip from the $i^{th}$ station to the $j^{th}$. Find the minimum number of seats that must be available in that car.",
    "options": [],
    "answer": "n^2",
    "solution": "When the train leaves station $k$, the passengers on board the train are those that are travelling from station $i$ to $j$, with $i \\le k < j$. Hence there are $k(2n-k)$ people on board the train at that point, with a maximum of $n^2$ between stations $n$ and $n+1$.\n\nMoreover, at station $k$ there are $k-1$ passengers disembarking, and $2n-k$ passengers boarding, so there needs to be\n$$\n(2n - k) - (k - 1) = 2(n - k) + 1\n$$\nseats free in the car for $1 \\le k \\le n$. After station $n$, $2(k-n)-1$ seats are left free. So in order for everyone to have a seat up to station $n$, there needs to be at least $\\sum_{k=1}^{n}[2(n-k)+1] = n^2$ seats on the train. Since the number of passengers never exceed $n^2$, this number will suffice.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42155,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that for every natural number $n$ the product\n$$\n\\left(4 - \\frac{2}{1}\\right) \\left(4 - \\frac{2}{2}\\right) \\left(4 - \\frac{2}{3}\\right) \\dots \\left(4 - \\frac{2}{n}\\right)\n$$\nis an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\begin{align*}\n\\left(4 - \\frac{2}{1}\\right) \\left(4 - \\frac{2}{2}\\right) \\dots \\left(4 - \\frac{2}{n}\\right) &= \\prod_{k=1}^{n} \\left(\\frac{4k-2}{k}\\right) \\\\\n&= \\prod_{k=1}^{n} 2 \\left(\\frac{2k-1}{k}\\right) \\\\\n&= \\prod_{k=1}^{n} \\left(\\frac{2k-1}{k} \\cdot \\frac{2k}{k}\\right) \\\\\n&= \\frac{(2n)!}{(n!)^2} = \\binom{2n}{n},\n\\end{align*}\n$$\nwhich is an integer for each $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42156,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with orthocentre $H$ and let $P$ be a point on its circumcircle. The line through $A$ parallel to $BP$ meets $CH$ at $Q$ and the line through $A$ parallel to $CP$ meets $BH$ at $R$. Prove that $QR$ is parallel to $AP$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet $T$ be the intersection of $BH$ and $PC$ and $F$ be the intersection of $AB$ and $CH$. Then\n$$\n\\begin{aligned}\n\\angle HQA &= 90^\\circ - \\angle BAQ \\\\\n&= 90^\\circ - \\angle ABP \\quad (\\text{since } PB \\parallel AQ) \\\\\n&= 90^\\circ - \\angle ACP \\\\\n&= \\angle RTC \\\\\n&= \\angle ART \\quad (\\text{since } PC \\parallel AR)\n\\end{aligned}\n$$\n\nwhich shows that *AHQR* is a cyclic quadrilateral. Hence\n$$\n\\angle ARQ = \\angle AHF = \\angle ABC = \\angle APC.\n$$\n\nNow, since $AR \\parallel PC$, it follows that $AP \\parallel QR$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42157,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be an odd prime number and let $(a_1, a_2, \\dots, a_p)$ and $(b_1, b_2, \\dots, b_p)$ be arbitrary arrangements of the $p$-tuple $(0, 1, \\dots, p-1)$. For each $i$, let $c_i$ be the non-negative remainder when the product $a_i b_i$ is divided by $p$. Show that $(c_1, c_2, \\dots, c_p)$ cannot be a rearrangement of $(0, 1, \\dots, p-1)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $a_i = b_j = 0$, then $c_i = c_j = 0$. If $(c_1, c_2, \\dots, c_p)$ is a rearrangement of $(0, 1, \\dots, p-1)$, then no entry can appear more than once. Hence $i = j$.\n\nWithout loss of generality we may suppose that $a_p = b_p = 0$ so that $c_p = 0$. Then we have\n$$\n(p-1)! \\equiv_p a_1 a_2 \\dots a_{p-1} \\equiv_p b_1 b_2 \\dots b_{p-1} \\equiv_p c_1 c_2 \\dots c_{p-1}\n$$\nand so\n$$\n\\begin{align*}\n(p-1)! &\\equiv_p c_1 c_2 \\dots c_{p-1} \\\\\n&\\equiv_p (a_1 a_2 \\dots a_{p-1}) (b_1 b_2 \\dots b_{p-1}) \\\\\n&\\equiv_p ((p-1)!)^2,\n\\end{align*}\n$$\nwhich implies that $(p-1)! \\equiv_p 1$. However, since $p$ is odd, this contradicts Wilson's Theorem, and we conclude that $(c_1, c_2, \\dots, c_p)$ is not a rearrangement of $(0, 1, \\dots, p-1)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42158,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer, and let $a_2, a_3, \\dots, a_n$ be positive real numbers such that $a_2 a_3 \\dots a_n = 1$. Prove that\n$$\n(1 + a_2)^2 (1 + a_3)^3 \\dots (1 + a_n)^n > n^n.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\begin{aligned}\n(a_k + 1)^k &= \\left( a_k + \\frac{1}{k-1} + \\frac{1}{k-1} + \\dots + \\frac{1}{k-1} \\right)^k \\\\\n&\\ge k^k \\cdot a_k \\cdot \\frac{1}{(k-1)^{k-1}} \\\\\n&= \\frac{k^k}{(k-1)^{k-1}} \\cdot a_k\n\\end{aligned}\n$$\nThe inequality is strict unless $a_k = \\frac{1}{k-1}$. Multiplying analogous inequalities for $k=2, 3, \\dots, n$ yields\n$$\n\\begin{aligned}\n(a_2 + 1)^2 \\cdot (a_3 + 1)^3 \\cdots (a_n + n)^n &> \\frac{2^2}{1^1} \\cdot \\frac{3^3}{2^2} \\cdot \\frac{4^4}{3^3} \\cdots \\frac{n^n}{(n-1)^{n-1}} \\cdot a_2 \\cdot a_3 \\cdots a_n \\\\\n&= n^n.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42159,
    "subject": "Mathematics (Multi-modal)",
    "question": "a sequence $(a_n)_{n=1}^{\\infty}$ of natural numbers has the property that for any $n \\ge 1$, $a_{n+1} = a_n + b_n$, where $b_n$ is the number having the same digits as $a_n$, but in the reverse order (unlike $a_n$, the number $b_n$ may start with one or more zeroes in the decimal notation). For instance, if $a_1 = 170$ we have $a_2 = 170 + 71 = 241$, $a_3 = 241 + 142 = 383$, and so on.\n\na. Can the number $a_6$ be prime?\n\nb. Can the number $a_7$ be prime?",
    "options": [],
    "answer": "a. No; b. No",
    "solution": "Let $a_n = a_{n,1}a_{n,2}\\dots a_{n,k_n}$ be the decimal representation of $a_n$. Note that if $k_n$ is even, then $a_n \\equiv_{11} a_{n,1} - a_{n,2} + \\dots - a_{n,k_n}$ and $b_n \\equiv_{11} a_{n,k} - a_{n,k-1} + \\dots - a_{n,1} \\equiv -a_n$. Hence $a_{n+1} = a_n + b_n \\equiv_{11} 0$, so $a_{n+1}$ is divisible by $11$. Also note that if $a_n$ is divisible by $11$, then $b_n$ is also divisible by $11$, and hence $a_{n+1}$ is divisible by $11$. In order to prove that $a_7$ cannot be prime it suffices to prove that one of $a_1, a_2, \\dots, a_6$ has an even number of digits.\n\nFor a contradiction, suppose 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 in fact $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, we see that 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+b)$. However, $16(a+d)$ is certainly not less than $10$, which implies that $a_6$ must have more digits than $a_1$, a contradiction. This shows that $a_7$ must be divisible by $11$, and hence cannot be prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42160,
    "subject": "Mathematics (Multi-modal)",
    "question": "The set $T$ consists of 66 points in the plane, and set $P$ consist of 16 lines in the plane. We say that a point $A \\in T$ and a line $\\ell \\in P$ form an *incident pair* if $A \\in \\ell$. Show that the number of incident pairs cannot exceed 159, and that there is such a configuration with exactly 159 incident pairs.",
    "options": [],
    "answer": "159",
    "solution": "Denote by $A_1, A_2, \\dots, A_{66}$ the points from $T$ and by $a_i$ the number of lines from $P$ containing $A_i$. Then the number of pairs of lines intersecting at $A_i$ equals $\\binom{a_i}{2}$, and the number of incident pairs $I = \\sum_{i=1}^{66} a_i$. Since any two lines meet in at most one point, we have $\\sum_{i=1}^{66} \\binom{a_i}{2} \\le \\binom{16}{2} = 120$. Let $b_k$ be the number of points from $T$ which lie on exactly $k$ lines from $P$. Then $\\sum b_k = 66$, $\\sum \\binom{k}{2} b_k \\le 120$ and $I = \\sum k b_k \\le \\sum \\frac{1}{2} \\left(3 + \\binom{k}{2}\\right) = \\frac{1}{2}(3 \\cdot 66 + 120) = 159$, because $3 + \\binom{k}{2} \\ge 2k$. Equality is attained when $b_k = 0$ for $k \\notin \\{2, 3\\}$, $b_2 = 39$ and $b_3 = 27$ - i.e. when the lines from $P$ determine exactly 39 double and 27 triple intersection points.\n\n![](attached_image_1.png)\n\nAn example of a configuration with 159 incident pairs can be constructed using Pappus' Theorem. Take points $A_1, A_2, A_3$ on a line $a$ and $B_1, B_2, B_3$ on line $b \\parallel a$, then draw 9 lines $A_jB_j$, $i, j \\in \\{1, 2, 3\\}$. For instance, in the diagram we set $A_1A_2 : A_2A_3 : B_1B_2 : B_2B_3 = 2 : 2 : 3 : 6$, so among these lines no two are parallel and no three concurrent. By Pappus' Theorem, the 98 lines determine 18 intersection points which are collinear in triples - so these determine another 6 lines. Together with these 6 lines, we have 15 lines and 24 triple intersections. Moreover, three lines obtained by Pappus' Theorem meet in a point (denoted by $K$), which gives us 25 triple intersections. Draw one more line through two double intersection points only. The set $P$ of the 16 drawn lines and set $T$ consisting of the 27 triple intersection points and the 39 remaining double intersection points determine 159 incident pairs.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42161,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the plane there are $2012$ congruent triangles of area $1$ which all have the same orientation. Each of these triangles contains the centroids of all the other triangles.\n\na. Show that the intersection of the $2012$ triangles is a triangle similar to them with area at least $\\frac{1}{9}$.\n\nb. Show that the area of the union of these triangles is less than $\\frac{22}{9}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nLet $A_sB_sC_s$ be the given triangles, with $s \\in \\{1, 2, \\dots, 2012\\}$ and let $\\{XYZ\\}$ stand for the half-plane with the boundary line $XY$ and interior point $Z$. Each of the triangles is the intersection of the three half-planes $\\{A_sB_sC_s\\}$, $\\{B_sC_sA_s\\}$ and $\\{C_sA_sB_s\\}$, hence the intersection of all the triangles is the intersection of all such half-planes. Since the half-planes $\\{A_sB_sC_s\\}$, $1 \\le s \\le 2012$ differ only by translation, their intersection is the half-plane $\\{A_iB_iC_i\\}$ for a certain $i$. Similarly, the intersection of the half-planes $\\{B_sC_sA_s\\}$ is the half-plane $\\{B_jC_jA_j\\}$ for some $j$ and the intersection of the half-planes $\\{C_sA_sB_s\\}$ is the half-plane $\\{C_kA_kB_k\\}$ for some $k$. Hence the intersection of all the half-planes is the intersection of the three half-planes $\\{A_iB_iC_i\\}$, $\\{B_jC_jA_j\\}$ and $\\{C_kA_kB_k\\}$, which is a triangle $ABC$ which is similar to the original triangles, since its sides are parallel to the sides of the original triangles. Let the coefficient of similarity be $\\lambda$. Clearly $0 < \\lambda \\le 1$; we wish to show that $\\lambda \\ge \\frac{1}{3}$.\n\nLet $v$ be the altitude from the vertex $C_i$ to the side $A_iB_i$ in triangle $A_iB_iC_i$. Since the line $A_iB_i$ coincides with the line $AB$, the distance of the centroid $G_i$ of triangle $A_iB_iC_i$ from the line $AB$ equals $\\frac{1}{3}v$. Since every triangle contains the centroid of all the triangles, triangle $ABC$ contains all of them as well, and in particular, $ABC$ contains the centroid $G_i$, which means that the distance between $C$ and the line $AB$ is at least $\\frac{1}{3}v$. Comparing the altitudes from the vertices $C_i$ and $C$ in the similar triangles $A_iB_iC_i$ and $ABC$, we thus obtain immediately that $\\lambda \\ge \\frac{1}{3}$, and so the area of $ABC$ is at least $\\frac{1}{9}$.\n\nb.\nLet $A_sB_sC_s$ be any one of the given triangles. The triangle $ABC$ is contained in it, and so the vertex $A_s$ lies in the half-plane $\\{BCA\\}$ at a distance at most $v$ from the line $BC$, where $v$ is the length of the altitude from the vertex $A_s$ in triangle $A_sB_sC_s$. On the other hand, the distance of the side $B_sC_s$ from $A$ is also at most $v$. Since the triangle $ABC$ contains centroids of all the given triangles, the distance between the side $B_sC_s$ and $BC$ cannot exceed $\\frac{1}{3}v$. The distance of the vertex $A$ from the side $BC$ equals $\\lambda v$, so altogether the distance between the parallel lines $BC$ and $B_sC_s$ is at most $\\min\\{1/3, 1-\\lambda\\} \\cdot v$. All the given triangles thus lie in the belt bounded by the two parallel lines $a_0$ and $a_1$ parallel to $BC$ whose distance from $BC$ is $v$ and $\\min\\{1/3, 1-\\lambda\\} \\cdot v$, respectively. A similar assertion clearly holds for the other two sides $AC$ and $AB$, so the union of the $2012$ triangles lie in the intersection of the three belts.\n\n![](attached_image_1.png)\n\nWe now distinguish two cases, depending on the value of $\\min\\{1/3, 1-\\lambda\\}$.\n\nIf $\\frac{1}{3} \\le \\lambda < \\frac{2}{3}$, then the intersection of the belts is a hexagon which is obtained from the triangle $T$ determined by the triple of lines $(a_1, b_1, c_1)$ upon removing the three small triangles $T_a, T_b$ and $T_c$ determined by the triples of lines $(a_0, b_1, c_1)$, $(a_1, b_0, c_1)$ and $(a_1, b_1, c_0)$ respectively. From the way the lines $a_0, b_0, c_0, a_1, b_1, c_1$ are defined, it follows that the $T$ is similar to the given triangles with coefficient of similarity equal to $1 + \\lambda$ and the triangles $T_a, T_b, T_c$ are also similar to the given triangles with coefficient of similarity $\\lambda - \\frac{1}{3}$.\n\nThe area of the hexagon is therefore given by (recalling that $\\lambda < \\frac{2}{3}$)\n$$\nS = (1 + \\lambda)^2 - 3 \\left( \\lambda - \\frac{1}{3} \\right)^2 = \\frac{8}{3} - 2(\\lambda - 1)^2 < \\frac{8}{3} - \\frac{2}{9} = \\frac{22}{9}.\n$$\n\n![](attached_image_2.png)\n\nIf $\\frac{2}{3} \\le \\lambda \\le 1$, then the intersection of the three belts is again a hexagon, and the corresponding triangle $T$ is similar to the given triangles with coefficient of similarity $3 - 2\\lambda$, and the triangles $T_a, T_b, T_c$ have coefficients of similarity $1 - \\lambda$. The area of the hexagon in this case is given by\n$$\n(3 - 2\\lambda)^2 - 3(1 - \\lambda)^2 = (\\lambda - 3)^2 - 3 \\le \\frac{49}{9} - 3 = \\frac{22}{9},\n$$\nwith equality if $\\lambda = \\frac{2}{3}$.\n\n![](attached_image_3.png)\n\nHowever, since we only start with a finite number of triangles, the side $a_0$ of the hexagon can only contain a finite number of vertices of the given triangles (and cannot coincide with one of these triangles' sides), and so the entire area of the hexagon cannot be covered with only finitely many of these triangles, hence the inequality is strict.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42162,
    "subject": "Mathematics (Multi-modal)",
    "question": "A poker set contains three chips of each of $2n$ different colours for a complete set of $6n$ chips. Let $p_n$ be the number of ways in which the chips can be partitioned into two piles of $3n$ chips each in such a way that no pile contains three chips of the same colour. Prove that $p_n$ is odd if and only if $n$ is a power of 2.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "No three chips of the same colour should be in the same pile, so each pile must contain a chip of each colour. So essentially we need to partition $2n$ chips into two piles of $n$ each. This can be done in $p_n = \\binom{2n}{n}/2$ ways.\n\nNow,\n$$\n\\begin{align*}\np_n &= \\frac{1}{2} \\binom{2n}{n} = \\frac{(2n)!}{2(n!)^2} \\\\\n&= 1 \\cdot 3 \\cdot \\dotsb (2n-1) \\frac{2 \\cdot 4 \\cdot \\dotsb 2n}{2(n!)^2} \\\\\n&= 1 \\cdot 3 \\cdot \\dotsb (2n-1) \\frac{2^n n!}{2(n!)^2} \\\\\n&= 1 \\cdot 3 \\cdot \\dotsb (2n-1) \\frac{2^{n-1}}{n!}.\n\\end{align*}\n$$\n\nNow, let $m$ be a positive integer such that $2^m \\le n < 2^{m+1}$. Then the greatest integer $a$ such that $2^a$ divides $n!$ is\n$$\n\\begin{align*}\na &= \\lfloor \\frac{n}{2} \\rfloor + \\lfloor \\frac{n}{2^2} \\rfloor + \\lfloor \\frac{n}{2^3} \\rfloor + \\dots + \\lfloor \\frac{n}{2^m} \\rfloor \\\\\n&\\le \\frac{n}{2} + \\frac{n}{2^2} + \\frac{n}{2^3} + \\dots + \\frac{n}{2^m} \\\\\n&= n - \\frac{n}{2^m} \\le n-1.\n\\end{align*}\n$$\n\nNow, $p_n$ is odd exactly when $a = n-1$, which happens if only if $n = 2^m$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42163,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square with circumcircle $\\Gamma$. Let $M$ be on minor arc $CD$ of $\\Gamma$. Let $BD$ and $AM$ intersect at $P$, $CD$ and $AM$ intersect at $R$, $BM$ and $AC$ intersect at $Q$ and $BM$ and $DC$ intersect at $S$. Show that $PS \\perp QR$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Place the square in the coordinate plane with $A(-1, 1)$, $B(1, 1)$, $C(1, -1)$ and $D(-1, -1)$. The equation of $\\Gamma$ is $x^2 + y^2 = 2$, the equation of the line $AC$ is $y = -x$ and the equation of the line $BC$ is $y = x$. Let $M$ have coordinates $(m, n)$, where $m^2 + n^2 = 2$. The lines $AM$ and $BM$ have the following equations:\n$$\nAM : y = \\frac{n-1}{m+1}(x+1) + 1\n$$\n$$\nBM : y = \\frac{n-1}{m-1}(x-1) + 1.\n$$\nThe coordinates of the four given intersection points are then\n$$\nP\\left(\\frac{m+n}{2+m-n}, \\frac{m+n}{2+m-n}\\right), \\quad Q\\left(\\frac{n-m}{m+n-2}, \\frac{m-n}{m+n-2}\\right), \\\\\nR\\left(\\frac{1+2m+n}{1-n}, -1\\right), \\quad S\\left(\\frac{2m-n-1}{1-n}, -1\\right).\n$$\nWe now calculate the product of the gradients of the two lines *PS* and *QR*:\n$$\n\\begin{align*} m_{PS}m_{QR} &= \\frac{2(m+1)(1-n)}{(m+n)(1-n) - (2m-n-1)(2+m-n)} \\\\ &\\quad \\times \\frac{2(1-n)(m-1)}{(1-n)(n-m) - (1+2m+n)(m+n-2)} \\\\ &= \\frac{4(m+1)(m-1)(n-1)^2}{4(-n^2-m^2+mn-m+n+1)(-n^2-m^2-mn+m+n+1)} \\\\ &= \\frac{(m+1)(m-1)(n-1)^2}{(m+1)(n-1)(m-1)(1-n)} \\quad \\text{(since } m^2 + n^2 = 2) \\\\ &= -1, \\end{align*}\n$$\nwhich shows that the two lines are perpendicular.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42164,
    "subject": "Mathematics (Multi-modal)",
    "question": "The *liars guessing game* is a game played between two players *A* and *B*. The rules of the game depend on two positive integers *k* and *n* which are known to both players.\n\nAt the start of the game *A* chooses integers $x$ and $N$ with $1 \\le x \\le N$. Player *A* keeps $x$ secret, and truthfully tells $N$ to player *B*. Player *B* now tries to obtain information about $x$ by asking player *A* questions as follows: each question consists of *B* specifying an arbitrary set $S$ of positive integers (possibly one specified in some previous question), and asking *A* whether $x$ belongs to $S$. Player *B* may ask as many such questions as he wishes. After each question, player *A* must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is that, among any $k+1$ consecutive answers, at least one answer must be truthful.\n\nAfter $B$ has asked as many questions as he wants, he must specify a set $X$ of at most $n$ positive integers. If $x$ belongs to $X$, then $B$ wins; otherwise, he loses. Prove that:\n\na. If $n \\ge 2^k$, then $B$ can guarantee a win.\n\nb. For all sufficiently large $k$, there exists an integer $n \\ge 1.99^k$ such that $B$ cannot guarantee a win.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The game can be reformulated in an equivalent one: The player $A$ chooses an element $x$ from the set $S$ (with $|S| = N$) and the player $B$ asks the sequence of questions. The $j$-th question consists of $B$ choosing a set $D_j \\subseteq S$ and player $A$ selecting a set $P_j \\in \\{Q_j, Q_j^C\\}$. The player $A$ has to make sure that for every $j \\ge 1$ the following relation holds:\n$$\nx \\in P_j \\cup P_{j+1} \\cup \\dots \\cup P_{j+k}.\n$$\nThe player $B$ wins if after a finite number of steps he can choose a set $X$ with $|X| \\le n$ such that $x \\in X$.\n\na.\nIt suffices to prove that if $N \\ge 2^k + 1$ then the player $B$ can determine a set $S' \\subseteq S$ with $|S'| \\le N - 1$ such that $x \\in S'$.\n\nAssume that $N \\ge 2^k + 1$. In the first move $B$ selects any set $D_1 \\subseteq S$ such that $|D_1| \\ge 2^{k-1}$ and $|D_1^C| \\ge 2^{k-1}$. After receiving the set $P_1$ from $A$, $B$ makes the second move. The player $B$ selects a set $D_2 \\subseteq S$ such that $|D_2 \\cap P_1^C| \\ge 2^{k-2}$ and $|D_2^C \\cap P_1^C| \\ge 2^{k-2}$. The player $B$ continues this way: in the move $j$ he/she chooses a set $D_j$ such that $|D_j \\cap P_j^C| \\ge 2^{k-j}$ and $|D_j^C \\cap P_j^C| \\ge 2^{k-j}$.\n\nIn this way the player $B$ has obtained the sets $P_1, P_2, \\dots, P_k$ such that $(P_1 \\cup \\dots \\cup P_k)^C \\ge 1$. Then $B$ chooses the set $D_{k+1}$ to be a singleton containing any element outside of $P_1 \\cup \\dots \\cup P_k$. There are two cases now:\n\n*Case 1.* The player $A$ selects $P_{k+1} = D_{k+1}^C$. Then $B$ can take $S' = S \\setminus D_{k+1}$ and the statement is proved.\n\n*Case 2.* The player $A$ selects $P_{k+1} = D_{k+1}$. Now the player $B$ repeats the previous procedure on the set $S_1 = S \\setminus D_{k+1}$ to obtain the sequence of sets $P_{k+2}, P_{k+3}, \\dots, P_{2k+1}$. The following inequality holds:\n$$\n|S_1 \\setminus (P_{k+2} \\cdots P_{2k+1})| \\ge 1,\n$$\nsince $|S_1| \\ge 2^k$. However, now we have\n$$\n|(P_{k+1} \\cup P_{k+2} \\cup \\dots \\cup P_{2k+1})^C| \\ge 1,\n$$\nand we may take $S' = P_{k+1} \\cup \\dots \\cup P_{2k+1}$.\n\nb.\nLet $p$ and $q$ be two real numbers such that $1.99 < p < q < 2$. Let us choose $k_0$ such that\n$$\n\\left(\\frac{p}{q}\\right)^{k_0} \\le 2 \\cdot \\left(1 - \\frac{q}{2}\\right) \\quad \\text{and} \\quad p^k - 1.99^k > 1.\n$$\nWe will prove that for every $k \\ge k_0$ if $|S| \\in (1.99^k, p^k)$ then there is a strategy for the player $A$ to select sets $P_1, P_2, \\dots$ (based on sets $D_1, D_2, \\dots$ provided by $B$) such that for each $j$ the following relation holds:\n$$\nP_j \\cup P_{j+1} \\cup \\dots \\cup P_{j+k} = S.\n$$\nAssuming that $S = \\{1, 2, \\dots, N\\}$, the player $A$ will maintain the following sequence of $N$-tuples: $(\\mathbf{x})_{j=0}^{\\infty} = (x_1^0, x_2^0, \\dots, x_N^0)$. Initially we set $x_1^0 = x_2^0 = \\dots = x_N^0 = 1$. After the set $P_j$ is selected then we define $\\mathbf{x}^{j+1}$ based on $\\mathbf{x}^j$ as follows:\n$$\nx_i^{j+1} = \\begin{cases} 1, & \\text{if } i \\in P_j \\\\ q \\cdot x_i^j, & \\text{if } i \\notin P_j. \\end{cases}\n$$\nThe player $A$ can keep $B$ from winning if $x_i^j \\le q^k$ for each pair $(i, j)$. For a sequence $\\mathbf{x}$, let us define $T(\\mathbf{x}) = \\sum_{i=1}^N x_i$. It suffices for player $A$ to make sure that $T(\\mathbf{x}^j) \\le q^k$ for each $j$.\n\nNotice that $T(\\mathbf{x}^0) = N \\le p^k < q^k$.\n\nWe will now prove that given $\\mathbf{x}^j$ such that $T(\\mathbf{x}^j) \\le q^k$, and a set $D_{j+1}$ the player $A$ can choose $P_{j+1} \\in \\{D_{j+1}, D_{j+1}^C\\}$ such that $T(\\mathbf{x}^{j+1}) \\le q^k$. Let $\\mathbf{y}$ be the sequence that would be obtained if $P_{j+1} = D_{j+1}$, and let $\\mathbf{z}$ be the sequence that would be obtained if $P_{j+1} = D_{j+1}^C$. Then we have\n$$\nT(\\mathbf{y}) = \\sum_{i \\in D_{j+1}^C} q x_i^j + |D_{j+1}|\n$$\n$$\nT(\\mathbf{z}) = \\sum_{i \\in D_{j+1}} q x_i^j + |D_{j+1}^C|.\n$$\nSumming up the previous two equalities gives:\n$$\nT(\\mathbf{y}) + T(\\mathbf{z}) = q \\cdot T(\\mathbf{x}^j) + N \\le q^{k+1} + p^k, \\text{ hence}\n$$\n$$\n\\min \\{T(\\mathbf{y}), T(\\mathbf{z})\\} \\le \\frac{q}{2} \\cdot q^k + \\frac{p^k}{2} \\le q^k,\n$$\nbecause of our choice of $k_0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42165,
    "subject": "Mathematics (Multi-modal)",
    "question": "Sixty points, of which thirty are coloured red, twenty are coloured blue, and ten are coloured green, are marked on a circle. These points divide the circle into sixty arcs. Each of these arcs is assigned a number according to the colours of its endpoints: an arc between a red and a green point is assigned a number 1, an arc between a red and a blue point is assigned a number 2, and an arc between a blue and a green point is assigned a number 3. The arcs between two points of the same colour are assigned a number 0. What is the greatest possible sum of all the numbers assigned to the arcs?",
    "options": [],
    "answer": "100",
    "solution": "Let the score of a red point be $0$, the score of a green point be $1$, and the score of a blue point be $2$. Note that the number assigned to an arc is at most the sum of the scores of the endpoints. This means that the sum of all the numbers assigned to the arcs is at most twice the sum of all the sixty scores, which is\n$$\n2(30 \\cdot 0 + 20 \\cdot 2 + 10 \\cdot 1) = 100.\n$$\nEquality holds if there are no arcs with two green or two blue endpoints. This can be achieved, for instance, by letting red and non-red points alternate. Hence the greatest possible sum is $100$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42166,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$ such that for all $x, y > 0$,\n$$\nf(yf(x))(x + y) = x^2(f(x) + f(y)).\n$$",
    "options": [],
    "answer": "f(x) = 1/x",
    "solution": "Setting $y = x$, we obtain $2x f(xf(x)) = 2x^2 f(x)$, and so $f(xf(x)) = x f(x)$ since $x > 0$.\n\nNow suppose $f(x) = f(y)$. Then\n$$\nx^2(f(x) + f(y)) = f(yf(x))(x + y) = f(yf(y))(x + y) = y f(y)(x + y)\n$$\nand so\n$$\n2x^2 f(x) = (x y + y^2) f(y) \\implies 2x^2 - x y - y^2 = (2x + y)(x - y) = 0 \\implies x = y,\n$$\nsince all the quantities in the equations are positive and so $y \\ne -2x$. So $f$ is injective. Moreover, setting $x = y = 1$, we obtain $f(f(1)) = f(1)$, and so injectivity implies that $f(1) = 1$.\n\nNow set $x = 1$ in the given equation:\n$$\nf(y)(y + 1) = 1 + f(y) \\implies f(y) = \\frac{1}{y},\n$$\nand it is a simple matter to check that the solution $f(x) = \\frac{1}{x}$ satisfies the given equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42167,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all primes $p$ and $q$ such that\n$$\n2q^p - p^q = 7.\n$$",
    "options": [],
    "answer": "[(3, 2), (3, 5)]",
    "solution": "It is easy to see that $p$ is odd and $p \\ne q$, so $p \\ge 3$ and $(p,q) = 1$.\n\nIf $q = 2$, then $2^{p+1} = 7 + p^2$. The only solution is $p = 3$, as $2^{n+1} > 7 + n^2$ for $n \\ge 4$.\n\nFor $q \\ge 3$, by Fermat's Little Theorem we get that $q^p \\equiv_p q$, so $p \\mid 2q^p - 7 \\equiv 2q - 7$ and similarly, $q \\mid p + 7$. Let $p + 7 = kq$ for some positive integer $k$.\n\nIf $2q - 7 \\le 0$, we have $q = 3$ and $p = -1$, which is false.\n\nIf $2q - 7 > 0$, then $2q - 7 \\ge p$, so $2q \\ge p + 7 \\ge kq$, therefore $k = 1$ or $k = 2$.\n\nFor $k = 1$ we obtain $p + 7 = q$, which is impossible for two odd primes $p$ and $q$, so $k = 2$ and $p + 7 = 2q$. Suppose $p > q$. Since $p, q \\ge 3$, we get $q^p \\ge p^q$ and then $7 = 2q^p - p^q \\ge p^q \\ge 27$, a contradiction. Thus\n\n$q > p$ and then $p + 7 = 2q > 2p$ which yields $p = 3$ or $p = 5$. For $p = 3$ we have $q = 5$, while $p = 5$ gives $q \\mid 12$, with no solution.\n\nThe two solutions are thus $(p, q) = (3, 2)$ and $(3, 5)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42168,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, $\\angle A = 60^\\circ$. Let $E$ and $F$ be points on the extensions of $AB$ and $AC$ such that $BE = CF = BC$. The circumcircle of $ACE$ intersects $EF$ in $K$ (different from $E$). Prove that $K$ lies on the bisector of $\\angle BAC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $T$ be the intersection of $BF$ and $CE$. We have that $\\angle BCE + \\angle BEC = \\angle ABC$, and since $BC = BE$, it follows that $\\angle BCE = \\frac{1}{2} \\angle ABC$. Similarly, $\\angle CBF = \\frac{1}{2} \\angle ACB$. Hence\n$$\n\\angle CTF = \\angle BCE + \\angle CBF = \\frac{1}{2}(\\angle ABC + \\angle ACB) = 60^\\circ,\n$$\nand so $ABTC$ is a cyclic quadrilateral. Hence $\\angle EBF = \\angle ACE = \\angle AKE$ and $\\angle ABF = 180^\\circ - \\angle EBF = 180^\\circ - \\angle AKE = \\angle AKF$, which implies that $ABKF$ is cyclic. Hence we have that $\\angle EBK = \\angle CFK$, and since $ACKE$ is cyclic, we have $\\angle BEK = \\angle FCK$. Together with $BE = FC$ it follows that triangles $KBE$ and $KFC$ are congruent. Hence $KC = KE$, which means that $\\angle CAK$ and $\\angle BAK$ are subtended by equal chords in the circumcircle of $ACE$, and hence are equal, as required.\n\nLet $L$ be the intersection of $EF$ and the angle bisector of $\\angle BAC$. Let $BC = a$, $AC = b$ and $AB = c$. Then $EB = FC = a$.\nUsing the cosine rule in triangle $ABC$, we obtain $a^2 = b^2 + c^2 - bc$, and using the cosine rule in triangle $AEF$, we obtain\n$$\nEF^2 = (a+c)^2 + (a+b)^2 - (a+c)(a+b) = a^2 + b^2 + c^2 + ab + ac - bc.\n$$\nSince $AL$ bisects angle $AEF$, it follows that $\\frac{EL}{FL} = \\frac{AE}{FA} = \\frac{a+c}{a+b}$, which yields $FL = \\frac{a+b}{2a+b+c} FE$. Now,\n$$\n\\begin{align*}\nFL \\cdot FE - FC \\cdot FA &= \\frac{a+b}{2a+b+c} FE \\cdot FE - a(a+b) \\\\\n&= \\frac{a+b}{2a+b+c} (a^2 + b^2 + c^2 + ab + ac - bc) - a(a+b) \\\\\n&= \\frac{a+b}{2a+b+c} (b^2 + c^2 - bc - a^2) \\\\\n&= 0 \\quad (\\text{cosine rule in triangle } ABC).\n\\end{align*}\n$$\nHence $FL \\cdot FE = FC \\cdot FA$, which shows that $ACLE$ is a cyclic quadrilateral, i.e. $L = K$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42169,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $n$ points are given on a circle and $nk + 1$ of the chords between these points are drawn, where $2k + 1 < n$. Prove that it is possible to select $k+1$ of the chords such that no two of them intersect.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, we may assume that the $n$ points form a regular $n$-gon. We will partition all the possible chords of the $n$-gon into $n$ different sets such that all the chords in each set are parallel. We distinguish two cases:\n\nIf $n$ is odd, then each chord is parallel to exactly one of the sides of the $n$-gon, so define $A_i$ to be the set of chords parallel to side $i$.\n\nIf $n = 2k$ is even, let the $n$-gon be $P_1P_2\\dots P_{2k}$. Then each chord is parallel to a pair of sides, or parallel to a chord of the form $P_iP_{i+2}$ for some $i$. For $1 \\le i \\le k$, let $A_i$ be the set of chords parallel to side $P_iP_{i+1}$, and for $k < i \\le 2k$, let $A_i$ be the set of chords parallel to $P_{i-2}P_i$.\n\nSince we have $nk + 1$ chords, at least $k + 1$ of these chords must lie in one of the $n$ sets $A_i$. Since the chords in each set $A_i$ are parallel, no two of them intersect.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42170,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ and $k$ be positive integers such that $p$ is prime and $k > 1$. Prove that there is at most one pair $(x, y)$ of positive integers such that\n$$\nx^k + px = y^k.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We distinguish two different cases:\n\n**Case 1:** $\\text{gcd}(x, p) = 1$. In this case, $x$ and $x^{k-1} + p$ do not have a common divisor (other than 1) either, and it follows from the factorisation\n$$\nx(x^{k-1} + p) = y^k\n$$\nthat both $x$ and $x^{k-1} + p$ have to be $k$th powers, say $x = u^k$ and $x^{k-1} + p = v^k$. Then it follows that\n$$\np = v^k - u^{k(k-1)} = (v - u^{k-1}) (v^{k-1} + v^{k-2}u^{k-1} + \\dots + u^{(k-1)^2}).\n$$\nBoth factors have to be positive, and it is clear that\n$$\nv - u^{k-1} < v \\le v^{k-1} \\le v^{k-1} + v^{k-2}u^{k-1} + \\dots + u^{(k-1)^2},\n$$\nso since $p$ is given to be a prime number, we must have $v-u^{k-1} = 1$ and thus $v = u^{k-1}+1$.\nThen\n$$\np = (u^{k-1} + 1)^{k-1} + (u^{k-1} + 1)^{k-2} u^{k-1} + \\dots + u^{(k-1)^2}.\n$$\nThe right hand side is an increasing function of $u$, so there is at most one value of $u$ which satisfies the equation. If there is such an integer $u$, then there is only one corresponding $v$ and thus only one solution $(x, y)$.\n\n**Case 2:** $\\text{gcd}(x, p) = p$. Then $x^k + px$ has to be divisible by $p$, which implies that $y^k$, and thus $y$, is divisible by $p$ as well. It follows that $x^k$ and $y^k$ are divisible by $p^k$, hence this has to be the case for $px$ as well, so $x = p^{k-1}u$ for some integer $u$. Since $y$ is divisible by $p$, we can also set $y = pv$ to obtain\n$$\np^{k(k-1)}u^k + p^k u = p^k v^k\n$$\nso that $v$ would have to lie between the two consecutive integers $p^{k-2}u$ and $p^{k-2}u + 1$, an obvious contradiction.\nWe conclude that there is always at most one solution $(x, y)$.\nSince $x^k + px = y^k$, we define $\\alpha$ as the difference between $x$ and $y$. Then\n$$\nx^k + px = (x + \\alpha)^k > x^k + \\alpha x \\quad \\Rightarrow \\quad \\alpha < p.\n$$\nIn addition\n$$\n\\alpha|(x + \\alpha)^k - x^k = px \\quad \\Rightarrow \\quad \\alpha|x.\n$$\nOn the other hand, by binomially expanding $(x + \\alpha)^k$, every term (except $\\alpha^k$) in the equation is divisible by $x$, therefore $x|\\alpha^k$. Let $\\beta = \\alpha^k/x$ – we will show that $\\beta = 1$.\n$$\n\\begin{aligned}\npx &= (x + \\alpha)^k - x^k \\\\\n\\beta^k px &= (\\beta x + \\beta \\alpha)^k - (\\beta x)^k \\\\\n\\beta^{k-1} p \\alpha^k &= (\\alpha^k + \\beta \\alpha)^k - (\\alpha^k)^k \\\\\n\\beta^{k-1} p &= (\\alpha^{k-1} + \\beta)^k - (\\alpha^{k-1})^k\n\\end{aligned}\n$$\nHowever, since $\\alpha|x$, $x|\\alpha^k$ and $x\\beta = \\alpha^k$, we have $\\beta|\\alpha^{k-1}$. This means that every term on the right hand side is divisible by $\\beta^k$, and the left is only $p\\beta^{k-1}$. So either $p = \\beta$, in which case we require\n$$\ny^k = (x + \\alpha)^k = x^k + px = x^k + \\alpha^k,\n$$\nwhich has no solutions for $x > 0$, or $\\beta = 1$, in which case $x = \\alpha^k$. The equation can then be written as $p = (\\alpha^{k-1} + 1)^k - (\\alpha^{k-1})^k$, and as the right hand side is a strictly increasing function of $\\alpha$, for a given $p$ and $k$ there can be at most one solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42171,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every natural number $n$, there exists a polynomial $p(x)$ with integer coefficients such that $p(1), p(2), \\dots, p(n)$ are all distinct powers of $2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that $p_n(x)$ is a polynomial such that $p_n(1), p_n(2), \\dots, p_n(n)$ are distinct powers of $2$. We claim that $\\gcd(p_n(n+1), n!)$ is a power of $2$. Indeed, if there is a prime number $2 < q \\le n$ that divides $p_n(n+1)$, then $q$ divides $p_n(n+1-q)$ (a power of $2$), which is a contradiction.\n\nLet $2^m$ be the greatest power of $2$ that divides $n!$. So if $a = p_n(n+1)$, then $\\gcd(a, \\frac{n!}{2^m}) = 1$. By Euler's theorem, there exists a natural number $k$ such that $a^k - 1$ is divisible by $\\frac{n!}{2^m}$, so $a^k - 1 = \\frac{t \\cdot n!}{2^m}$ for some integer $t$. Now define\n$$\np_{n+1}(x) = 2^m p_n(x)^k - t(x-1)(x-2)\\dots(x-n).\n$$\nFor $1 \\le i \\le n$, we have $p_{n+1}(i) = 2^m p_n(i)^k$, which is a power of $2$, and\n$$\np_{n+1}(n+1) = 2^m a^k - t \\cdot n! = 2^m \\left( a^k - \\frac{t \\cdot n!}{2^m} \\right) = 2^m.\n$$\nSince the $p_n(i)$, $1 \\le i \\le n$ are distinct powers of $2$ and $k$ is positive, it follows that the $p_{n+1}(i)$, $1 \\le i \\le n+1$ are all distinct powers of $2$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 42172,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the intersection point of the diagonals of the convex quadrilateral $ABCD$, with $AO = OC$. Points $P$ and $Q$ are marked on the segments $AO$ and $CO$, respectively, such that $PO = OQ$. Let $N$ be the intersection of $AB$ and $DP$, and $K$ be the intersection of $CD$ and $BQ$.\nProve that the points $N$, $O$ and $K$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nDraw $NM \\parallel KL \\parallel AC$ as in the figure. From the similarity of the triangles $BOQ$ and $BLK$ it follows that\n$$\n\\frac{LK}{OQ} = \\frac{BL}{BO} = 1 + \\frac{LO}{BO},\n$$\nFrom the similarity of the triangles $DOC$ and $DLK$ it follows that\n$$\n\\frac{LK}{OC} = \\frac{DL}{DO} = 1 - \\frac{LO}{DO}.\n$$\nTherefore\n$$\nLK \\left( \\frac{1}{OQ} - \\frac{1}{OC} \\right) = LO \\left( \\frac{1}{BO} + \\frac{1}{DO} \\right).\n$$\nHence\n$$\n\\frac{LK}{LO} = \\frac{DO + BO}{DO \\cdot BO} \\cdot \\frac{OC \\cdot OQ}{OC - OQ} = \\frac{BD}{DO \\cdot BO} \\cdot \\frac{OC \\cdot OQ}{QC}.\n$$\nSimilarly we obtain\n$$\n\\frac{NM}{OM} = \\frac{BD}{DO \\cdot BO} \\cdot \\frac{OC \\cdot OQ}{QC}.\n$$\nTherefore, since $\\angle NMO = \\angle KLO$ and $\\frac{LK}{LO} = \\frac{NM}{OM}$, triangles $ONM$ and $OKL$ are similar. Hence $\\angle NOM = \\angle KOL$. Since $MOL$ is a straight line, it follows that so is $NOK$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42173,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ such that, for all integers $a$, $b$, and $c$ satisfying $a + b + c = 0$, the following equality holds:\n$$\nf(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).\n$$",
    "options": [],
    "answer": "All such functions are exactly the following families, where c is any integer: (1) f(x) = c x^2 for all integers x; (2) f(x) = 0 for even x and f(x) = c for odd x; (3) f(x) = 0 for x divisible by four, f(x) = c for odd x, and f(x) = 4c for x congruent to two modulo four.",
    "solution": "Substituting $a = b = c = 0$ yields $3f(0)^2 = 6f(0)^2$ which implies $f(0) = 0$. Now we can place $b = -a$, $c = 0$ to obtain $f(a)^2 + f(-a)^2 = 2f(a)f(-a)$, or, equivalently $(f(a) - f(-a))^2 = 0$ which implies $f(a) = f(-a)$.\n\nAssume now that $f(a) = 0$ for some $a \\in \\mathbb{Z}$. Then for any $b$ we have $a+b+(-a-b) = 0$ hence $f(a)^2+f(b)^2+f(a+b)^2 = 2f(b)f(a+b)$, which is equivalent to $(f(b) - f(a+b))^2 = 0$, or $f(a+b) = f(b)$. Therefore if $f(a) = 0$ for some $a \\neq 0$, then $f$ is a periodic function with period $a$.\n\nPlacing $b = a$ and $c = 2a$ in the original equation yields $f(2a) \\cdot (f(2a) - 4f(a)) = 0$. Choosing $a = 1$ we get $f(2) = 0$ or $f(2) = 4f(1)$.\n\nIf $f(2) = 0$, then $f$ is periodic with period $2$ and we must have $f(n) = f(1)$ for all odd $n$. It is easy to verify that for each $c \\in \\mathbb{Z}$ the function\n$$\nf(x) = \\begin{cases} 0, & 2 \\mid n, \\\\ c, & 2 \\nmid n \\end{cases}\n$$\nsatisfies the conditions of the problem.\n\nAssume now that $f(2) = 4f(1)$ and that $f(1) \\neq 0$. Assume that $f(i) = i^2 \\cdot f(1)$ holds for all $i \\in \\{1, 2, \\dots, n\\}$ (it certainly does for $i \\in \\{0, 1, 2\\}$). We place $a = 1$, $b = n$, $c = -n - 1$ in the original equation to obtain:\n$$\nf(1)^2 + n^4 f(1)^2 + f(n+1)^2 = 2n^2 f(1)^2 + 2(n^2 + 1)f(n+1)f(1) \\\\\n\\Leftrightarrow (f(n+1) - (n+1)^2 f(1)) \\cdot (f(n+1) - (n-1)^2 f(1)) = 0.\n$$\nIf $f(n+1) = (n-1)^2 f(1)$ then setting $a = n+1$, $b = 1-n$, and $c = -2$ in the original equation yields\n$$\n2(n - 1)^4 f(1)^2 + 16f(1)^2 = 2 \\cdot 4 \\cdot 2(n - 1)^2 f(1)^2 + 2 \\cdot (n - 1)^4 f(1)\n$$\nwhich implies $(n-1)^2 = 1$ hence $n = 2$. Therefore $f(3) = f(1)$. Placing $a = 1$, $b = 3$, and $c = 4$ into the original equation implies that $f(4) = 0$ or $f(4) = 4f(1) = f(2)$. If $f(4) \\neq 0$ we get\n$$\nf(2)^2 + f(2)^2 + f(4)^2 = 2f(2)^2 + 4f(2)f(4)\n$$\nhence $f(4) = 4f(2)$. We already have that $f(4) = f(2)$ and this implies that $f(2) = 0$, which is impossible according to our assumption.\n\nTherefore $f(4) = 0$ and the function $f$ has period $4$. Then $f(4k) = 0$, $f(4k+1) = f(4k+3) = c$, and $f(4k+2) = 4c$. It is easy to verify that this function satisfies the requirements of the problem.\n\nThe remaining case is $f(n) = n^2 f(1)$ for all $n \\in \\mathbb{N}$, or $f(n) = cn^2$ for some $c \\in \\mathbb{Z}$. This function satisfies the given condition.\n\nThus the solutions are: $f(x) = cx^2$ for some $c \\in \\mathbb{Z}$; $f(x) = \\begin{cases} 0, & 2 \\mid n, \\\\ c, & 2 \\nmid n \\end{cases}$ for some $c \\in \\mathbb{Z}$; and $f(x) = \\begin{cases} 0, & 4 \\mid n, \\\\ c, & 2 \\nmid n, \\\\ 4c, & n \\equiv 2 \\pmod{4} \\end{cases}$ for some $c \\in \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42174,
    "subject": "Mathematics (Multi-modal)",
    "question": "Dylan has a list of all 25-digit numbers consisting of the digits 1, 2, 3 and 4 such that there are an equal number of 1s and 2s. Robert has a list of all 50-digit numbers consisting of 25 digits 1 and 25 digits 2. Show that the number of numbers on each list is the same.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D$ be the set of 25-digit numbers with the same number of 1s and 2s, and let $R$ be the set of 50-digit numbers with 25 digits 1 and 25 digits 2. We define a bijection between the two sets.\n\nLet $d = d_1d_2\\dots d_{25} \\in D$. We define a function on the digits of $d$ as follows:\n$$\nf(d_i) = \\begin{cases} 11 & \\text{if } d_i = 1 \\\\ 22 & \\text{if } d_i = 2 \\\\ 12 & \\text{if } d_i = 3 \\\\ 21 & \\text{if } d_i = 4 \\end{cases}\n$$\nBy replacing each digit $d_i$ in $d$ with $f(d_i)$ we obtain a number in $R$; indeed, if $d_i$ equals 3 or 4, then $f(d_i)$ contains the same number of 1s and 2s, and since there are equal numbers of 1s and 2s in $d$, there will be equal numbers of the digit pairs 11 and 22. Hence each number in $d$ corresponds to a number in $R$ - clearly two different numbers in $d$ will correspond to two different numbers in $R$.\n\nConversely, using the inverse of $f$, we can map each number in $R$ to a number in $D$. A similar argument shows that this mapping is well-defined and injective as well, showing that the two sets have an equal number of elements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42175,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $PQ$ be the diameter of semicircle $H$. Circle $\\omega$ is internally tangent to $H$ and tangent to $PQ$ at $C$. Let $A$ be a point on $H$ and $B$ a point on $PQ$ such that $AB$ is perpendicular to $PQ$ and tangent to $\\omega$. Prove that $AC$ bisects $\\angle PAB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In version centred at $C$ with any radius gives $\\triangle CAP \\sim \\triangle CP'A'$ and $\\triangle CAB \\sim \\triangle CB'A'$. Clearly the line $PQ$ inverts to itself. $\\omega$ inverts to $\\omega'$, a line. $H$ inverts to semicircle $H'$ tangent to $\\omega'$ with diameter $P'Q'$. $AB$ inverts to arc $A'B'$ of a circle with diameter $CB'$ tangent to $\\omega'$. Now observe that arc $A'Q'$ and arc $CA'$ are symmetrical w.r.t the perpendicular bisector of $CQ'$. Which implies $\\angle CP'A' = \\angle CB'A'$. Notice that $AC$ bisects $\\angle PAB$ iff $\\angle CAP = \\angle CAB$ iff $\\angle CP'A' = \\angle CB'A'$ and so we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42176,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle such that $AB \\neq AC$. We denote its orthocentre by $H$, its circumcentre by $O$ and the midpoint of $BC$ by $D$. The extensions of $HD$ and $AO$ meet in $P$. Prove that triangles $AHP$ and $ABC$ have the same centroid.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AO$ intersect the circumcircle at $X$. Since $AX$ is a diameter, $ABX$ is a right angle, hence $BX \\parallel CH$ (both are perpendicular to $AB$). For the same reason, $CX \\parallel BH$, so $BHCX$ is a parallelogram. This means that $XH$ passes through $D$, the midpoint of $BC$. Hence $X$ coincides with $P$ (it lies on both $AO$ and $HD$), and $D$ is also the midpoint of $HX = HP$.\nBut this implies that $AD$ is a median in both triangles $ABC$ and $AHP$. Since the centroid always divides the median in a $2 : 1$ ratio, we conclude that the centroids of $ABC$ and $AHP$ coincide as well.\nIt is well known that the reflection $H'$ of $H$ about the side $BC$ lies on the circumcircle. If we reflect $H'$ again about the perpendicular bisector of $BC$, we obtain another point $H''$ on the circumcircle. The double reflection amounts to a single reflection with centre $D$, since $D$ is the intersection of $BC$ and its perpendicular bisector. Hence $H''$ lies on $HD$. Moreover, since $AH'H''$ is a right angle by construction, $AH''$ is a diameter, so $H''$ lies on $AO$ and must thus coincide with $P$. One could now conclude as in the previous solution, but let us show a different approach instead: the centroid of $ABC$ is known to lie on the Euler line, which also passes through $H$ and $O$. On the other hand, $OH$ is also a median in $AHP$ by our observations. $AD$ is a common median in both triangles, hence the intersection of $OH$ and $AD$ is the centroid of both triangles, which completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42177,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which there exist non-negative integers $a_1, a_2, \\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$$",
    "options": [],
    "answer": "All positive integers n with n ≡ 1 or 2 modulo 4",
    "solution": "Let $M = \\max\\{a_1, \\dots, a_n\\}$. Then we have\n$$\n3^M = 1 \\cdot 3^{M-a_1} + 2 \\cdot 3^{M-a_2} + \\dots + n \\cdot 3^{M-a_n} \\equiv 1 + 2 + \\dots + n = \\frac{n(n+1)}{2} \\pmod n.\n$$\nTherefore, the number $\\frac{n(n+1)}{2}$ must be odd and hence $n \\equiv 1 \\pmod 4$ or $n \\equiv 2 \\pmod 4$.\n\nWe will now prove that each $n \\in \\mathbb{N}$ of the form $4k + 1$ or $4k + 2$ (for some $k \\in \\mathbb{N}$) there exist integers $a_1, \\dots, a_n$ with the described property.\n\nFor a sequence $\\mathbf{a} = (a_1, a_2, \\dots, a_n)$ let us introduce the following notation:\n$$\nL(\\mathbf{a}) = \\frac{1}{2^{a_1}} + \\frac{1}{2^{a_2}} + \\dots + \\frac{1}{2^{a_n}} \\quad \\text{and} \\quad R(\\mathbf{a}) = \\frac{1}{3^{a_1}} + \\frac{2}{3^{a_2}} + \\dots + \\frac{n}{3^{a_n}}.\n$$\nAssume that for $n = 2m + 1$ there exists a sequence $\\mathbf{a} = (a_1, \\dots, a_n)$ of non-negative integers with $L(\\mathbf{a}) = R(\\mathbf{a}) = 1$. Consider the sequence $\\mathbf{a}' = (a'_1, \\dots, a'_{n+1})$ defined in the following way:\n$$\na'_j = \\begin{cases} a_j, & \\text{if } j \\notin \\{m+1, 2m+2\\} \\\\ a_{m+1} + 1, & \\text{if } j \\in \\{m+1, 2m+2\\}. \\end{cases}\n$$\nThen we have\n$$\nL(\\mathbf{a}') = L(\\mathbf{a}) - \\frac{1}{2^{a_{m+1}}} + 2 \\cdot \\frac{1}{2^{a_{m+1}+1}} = 1\n$$\n$$\nR(\\mathbf{a}') = R(\\mathbf{a}) - \\frac{m+1}{3^{a_{m+1}}} + \\frac{m+1}{3^{a_{m+1}+1}} + \\frac{2m+2}{3^{a_{m+1}+1}} = 1.\n$$\nThis implies that if the statement holds for $2m + 1$, then it holds for $2m + 2$.\n\nAssume now that the statement holds for $n = 4m + 2$ for some $m \\ge 2$, and assume that $\\mathbf{a} = (a_1, \\dots, a_{4m+2})$ is the corresponding sequence of $n$ non-negative integers. We will construct a following sequence $\\mathbf{a}' = (a'_1, a'_2, \\dots, a'_{4m+13})$ that satisfies $L(\\mathbf{a}') = R(\\mathbf{a}') = 1$ thus proving that the statement holds for $4m + 13$. Define:\n$$\na'_j = \\begin{cases} a_{m+2} + 2 & \\text{if } j = m + 2 \\\\ a_j + 1 & \\text{if } j \\in \\{2m + 2, 2m + 3, 2m + 4, 2m + 5, 2m + 6\\} \\\\ a_{j/2} + 1 & \\text{if } j \\in \\{4m + 4, 4m + 6, 4m + 8, 4m + 10, 4m + 12\\} \\\\ a_{m+2} + 3 & \\text{if } j \\in \\{4m + 3, 4m + 5, 4m + 7, 4m + 9, 4m + 11, 4m + 13\\} \\\\ a_j & \\text{otherwise.} \\end{cases}\n$$\nWe now have\n$$\nL(\\mathbf{a}') = L(\\mathbf{a}) - \\frac{1}{2^{a_{m+2}}} - \\sum_{j=2}^{6} \\frac{1}{2^{a_{2m+j}}} + \\frac{1}{2^{a_{m+2}+2}} + \\sum_{j=2}^{6} \\frac{1}{2^{a_{2m+j}+1}} + \\sum_{j=2}^{6} \\frac{1}{2^{a_{2m+j}+1}} + 6 \\cdot \\frac{1}{2^{a_{m+2}+3}} = 1.\n$$\nIt remains to verify that $R(\\mathbf{a}') = R(\\mathbf{a}) = 1$. We write\n$$\nR(\\mathbf{a}') - R(\\mathbf{a}) = R\\left(\\begin{array}{ccccccc} a'_{m+2} & a'_{4m+3} & a'_{4m+5} & a'_{4m+7} & a'_{4m+9} & a'_{4m+11} & a'_{4m+13} \\\\ m+2 & 4m+3 & 4m+5 & 4m+7 & 4m+9 & 4m+11 & 4m+13 \\end{array}\\right) - R\\left(\\begin{array}{c} a_{m+2} \\\\ m+2 \\end{array}\\right) + \\sum_{j=2}^{6} \\left(R\\left(\\begin{array}{c} a'_{2m+j} \\\\ 2m+j, 4m+j \\end{array}\\right) - R\\left(\\begin{array}{c} a_{2m+j} \\\\ 2m+j \\end{array}\\right)\\right),\n$$\nwhere\n$$\nR\\left(\\begin{array}{ccc} c_1 & \\dots & c_k \\\\ d_1 & \\dots & d_k \\end{array}\\right)=\\frac{d_1}{3^{c_1}}+\\dots+\\frac{d_k}{3^{c_k}}.\n$$\nFor each $j \\in \\{2, 3, 4, 5, 6\\}$ we have\n$$\nR\\left(\\begin{array}{cc} a'_{2m+j} & a'_{4m+2j} \\\\ 2m+j, & 4m+j \\end{array}\\right)-R\\left(\\begin{array}{c} a_{2m+j} \\\\ 2m+j \\end{array}\\right)=\\frac{2m+j}{3^{a_{2m+j}+1}}+\\frac{4m+2j}{3^{a_{2m+j}+1}}-\\frac{2m+j}{3^{a_{2m+j}}} = 0.\n$$\nThe first term in the expression for $R(\\mathbf{a}') - R(\\mathbf{a})$ is also equal to 0 because\n$$\n\\begin{aligned}\n& R \\begin{pmatrix} a'_{m+2}, & a'_{4m+3}, & a'_{4m+5}, & a'_{4m+7}, & a'_{4m+9}, & a'_{4m+11}, & a'_{4m+13} \\\\ m+2, & 4m+3, & 4m+5, & 4m+7, & 4m+9, & 4m+11, & 4m+13 \\end{pmatrix} \\\n& - R \\begin{pmatrix} a_{m+2} \\\\ m+2 \\end{pmatrix} \\\n&= \\frac{m+2}{3^{a_{m+2}+2}} + \\sum_{j=1}^{6} \\frac{4m+2j+1}{3^{a_{m+2}+3}} - \\frac{m+2}{3^{a_{m+2}}} \\\n&= 0.\n\\end{aligned}\n$$\nThus $R(\\mathbf{a}') = 0$ and the statement holds for $4m + 13$. It remains to verify that there are sequences of lengths 1, 5, 9, 13, and 17. One way to choose these sequences is:\n$$\n(1), \\quad (2, 1, 3, 4, 4), \\quad (2, 3, 3, 3, 3, 4, 4, 4, 4), \\quad (2, 3, 3, 4, 4, 4, 5, 4, 4, 5, 4, 5, 5), \\\\ (3, 2, 2, 4, 4, 5, 5, 6, 5, 6, 6, 6, 6, 6, 6, 6, 5).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42178,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{N} \\to \\mathbb{R}$ ($\\mathbb{N}$ denotes the set of all positive integers, $\\mathbb{R}$ the set of all real numbers) such that\n$$\nf(km) + f(kn) - f(k)f(nm) \\geq 1\n$$\nfor all $k, m, n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "f(n) = 1 for all n",
    "solution": "Plugging in $k = n = m = 1$ yields\n$$\nf(1)^2 - 2f(1) + 1 = (f(1) - 1)^2 \\leq 0,\n$$\nwhich implies $f(1) = 1$. Plugging in $k = 1, n = m$ and $k = n, m = 1$, respectively, we obtain the two inequalities\n$$\n2f(n) - f(n^2) \\geq 1, \\qquad (1)\n$$\n$$\nf(n^2) + f(n) - f(n)^2 \\geq 1. \\qquad (2)\n$$\n\nWe add the two to get\n$$\n-f(n)^2 + 3f(n) \\geq 2\n$$\nor\n$$\n(f(n) - 1)(f(n) - 2) \\leq 0.\n$$\nThis means that $1 \\leq f(n) \\leq 2$ for all $n$. Now assume that there is a positive integer $n$ such that $f(n) = a > 1$. The function $g(x) = \\frac{x^2+1}{x} = x + \\frac{1}{x}$ is increasing on $(1, \\infty)$: for $x > y > 1$,\nwe have\n$$\ng(x) - g(y) = x - y + \\frac{1}{x} - \\frac{1}{y} = (x - y) \\left(1 - \\frac{1}{xy}\\right) > 0.\n$$\nHence, for any $x \\geq a$, we have\n$$\nx^2 + 1 \\geq \\frac{a^2 + 1}{a}x\n$$\nand thus\n$$\nx^2 - x + 1 \\geq \\frac{a^2 - a + 1}{a} \\cdot x.\n$$\nReturning to (2), we now find, by our assumption that $f(n) = a$,\n$$\nf(n^2) \\geq f(n)^2 - f(n) + 1 \\geq \\frac{a^2 - a + 1}{a} \\cdot f(n),\n$$\nand since\n$$\nb = \\frac{a^2 - a + 1}{a} = 1 + \\frac{(a - 1)^2}{a} > 1,\n$$\nwe get\n$$\nf(n^2) \\geq b f(n) \\geq a.\n$$\nIterating this inequality yields $f(n^4) \\geq b^2 f(n) = b^2 a$, $f(n^8) \\geq b^3 f(n) = b^3 a$, etc., and generally (by induction) $f(n^{2^k}) \\geq b^k a$. Since $a > 1$ and $b > 1$, this implies $f(n^{2^k}) > 2$ for sufficiently large $k$, which contradicts the inequality $1 \\leq f(n^{2^k}) \\leq 2$ that was obtained earlier.\nIt follows that there is no $n$ such that $f(n) > 1$, which means that the constant function $f(n) \\equiv 1$ is the only solution (and it is easy to see that this function satisfies the condition, since the left hand side of the inequality is always equal to 1 in this case).\nIt is given that\n$$\nf(km) + f(kn) - f(k)f(mn) \\geq 1 \\quad (3)\n$$\nfor all $k, m, n \\in \\mathbb{N}$.\nPut $m = n = k = 1$ into (3) to get $(f(1) - 1)^2 \\leq 0$, giving\n$$\nf(1) = 1. \\quad (4)\n$$\nPut $m = n = 1$ into (3) to get $2f(k) - f(k) \\geq 1$, giving\n$$\nf(k) \\geq 1 \\quad (5)\n$$\nfor all $k \\in \\mathbb{N}$. Put $k = m = n$ into (3) to get $2f(k^2) - f(k)f(k^2) \\ge 1$, giving\n$$\nf(k^2)(2 - f(k)) \\ge 1, \\qquad (6)\n$$\nagain for all $k \\in \\mathbb{N}$. From (5) and (6) it follows that\n$$\n1 \\le f(k) < 2 \\qquad (7)\n$$\nfor all $k \\in \\mathbb{N}$. We now prove the statement\n$$\nS(M) : 1 \\le f(k) < 1 + \\frac{1}{M} \\text{ for all } k \\in \\mathbb{N}\n$$\nto be true for all $M \\ge 1$, using induction on $M$. The case $M = 1$ is just (7).\nAssume $S(M)$ to be true. Suppose there is a $k_0 \\in \\mathbb{N}$ such that $f(k_0) \\ge 1 + \\frac{1}{M+1}$. Then, using (6), we obtain\n$$\n1 \\le f(k_0^2)(2 - f(k_0)) \\le f(k_0^2) \\left(2 - 1 - \\frac{1}{M+1}\\right) = f(k_0^2) \\left(\\frac{M}{M+1}\\right),\n$$\ngiving $f(k_0^2) \\ge 1 + \\frac{1}{M}$, which violates $S(M)$. So $S(M+1)$ follows.\nIt is now clear that the constant function $f(n) \\equiv 1$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42179,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the sequence $(u_n)$ be defined recursively by\n$$\nu_0 = 0,\\ u_1 = 1,\\ u_n = 2011u_{n-1} - u_{n-2} \\quad \\text{for } n \\ge 2.$$\nFind all the values of $n$ for which $u_n$ is prime.",
    "options": [],
    "answer": "2",
    "solution": "For generality, let $x = 2011$. Expanding the first few terms of the sequence with the recurrence $u_n = x u_{n-1} - u_{n-2}$ we find\n$$\n\\begin{aligned}\nu_0 &= 0, \\quad u_1 = 1, \\quad u_2 = x \\\\\nu_3 &= x^2 - 1 = u_2^2 - u_1^2 \\\\\nu_4 &= x^3 - 2x = 2u_3u_2 - x u_2^2 \\\\\nu_5 &= x^4 - 3x^2 + 1 = u_3^2 - u_2^2 \\\\\nu_6 &= x^5 - 4x^3 + 2x = 2u_4u_3 - x u_3^2.\n\\end{aligned}\n$$\nFrom this we can conjecture that $u_{2n} = 2u_n u_{n+1} - x u_n^2$ and $u_{2n+1} = u_{n+1}^2 - u_n^2$ for $n > 0$. We now prove this by induction. Firstly note that it holds for $n = 1$. Now assume it holds up to $n$, and we will show that it holds for $n + 1$.\n$$\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} + x u_n^2 \\\\\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$$\nand\n$$\nu_{2n+3} = x u_{2n+2} - u_{2n+1}$$\n$$\n\\begin{align*}\n&= x(2u_{n+1}u_{n+2} - x u_{n+1}^2) - (u_{n+1}^2 - u_n^2) \\\\\n&= -x^2 u_{n+1}^2 + x(2u_{n+1}u_{n+2}) + u_n^2 - u_{n+1}^2 \\\\\n&= x^2 u_{n+1}^2 - 2x u_{n+1}u_n + u_n^2 - u_{n+1}^2 \\\\\n&= (x u_{n+1} - u_n)^2 - u_{n+1}^2 \\\\\n&= u_{n+2}^2 - u_{n+1}^2.\n\\end{align*}\n$$\nThe result thus follows by induction.\n\nThe above identities suggest that we may be able to find explicit factorisations, but for a factorisation to prove that a number is not prime it is necessary for the factors to be greater than 1. So let us show that $u_i - u_{i-1} \\ge 2$ for all $i \\ge 2$. For $i = 2$ the result holds by inspection. Suppose it holds for some $i$. Then $u_{i+1} - u_i = 2011u_i - u_{i-1} - u_i = 2010u_i - u_{i-1} \\ge u_i - u_{i-1} \\ge 2$. The result again follows by induction.\n\nInspecting the first few terms, we see that $u_2 = 2011$ is prime. Let us show that there are no other primes. $u_0$ and $u_1$ are non-prime by inspection. For $n \\ge 1$, $u_{2n+1} = u_{n+1}^2 - u_n^2 = (u_{n+1} - u_n)(u_{n+1} + u_n)$. By the inequality above, both factors are at least 2 and so $u_{2n+1}$ is not prime. For $n > 1$, $u_{2n} = u_n(2u_{n+1} - 2011u_n)$. The first term is clearly greater than 1, and the second equals $u_{n+1} - u_{n-1} \\ge u_{n+1} - u_n \\ge 2$.\n\nThus, $u_2 = 2011$ is the only prime in the sequence.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42180,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the positive integer $n$ is the product of exactly two primes that differ by 2 if and only if\n$$\n\\phi(n)\\sigma(n) = (n-3)(n+1).\n$$\n(Recall that $\\phi(n)$ equals the number of positive integers less than or equal to $n$, relatively prime to $n$, and $\\sigma(n)$ equals the sum of the positive divisors of $n$, including 1 and $n$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\phi(n)\\sigma(n) = (n-3)(n+1)$, and let prime $p$ divide $n$ such that $n = p^a m$ for some $m$ not divisible by $p$. Then we have\n$$\n\\begin{align*}\n\\phi(n) &= \\phi(p^a)\\phi(m) = p^{a-1}(p-1)\\phi(m) \\\\\n\\sigma(n) &= \\sigma(p^a)\\sigma(m) = \\left(\\frac{p^{a+1}-1}{p-1}\\right)\\sigma(m),\n\\end{align*}\n$$\nso\n$$\n\\phi(n)\\sigma(n) = p^{a-1}(p^{a+1} - 1)\\phi(m)\\sigma(m).\n$$\nIf $a \\ge 2$, it follows that, modulo $p^{a-1}$,\n$$\n0 \\equiv \\phi(n)\\sigma(n) = (n-3)(n+1) \\equiv -3.\n$$\nHence $a = 2$ and $p = 3$. Thus either $a = 1$ or $a = 2$ and $p = 3$.\nHowever, if $n = 9$ does not satisfy the given equality, and if $n = 9p_1p_2\\dots p_k$, then\n$$\n\\phi(n)\\sigma(n) = 78(p_1^2 - 1)(p_2^2 - 1)\\dots(p_k^2 - 1) \\equiv_9 0,\n$$\nsince $p_1^2 \\equiv_3 1$ for every prime $p_i \\neq 3$, but\n$$\n(n-3)(n+1) \\equiv_9 -3.\n$$\nHence $n = p_1p_2 \\dots p_k$ for some distinct primes $p_i$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42181,
    "subject": "Mathematics (Multi-modal)",
    "question": "An altitude from $A$ to the triangle $ABC$ intersects the side $BC$ in $D$. A circle touches $BC$ in $D$, intersects $AB$ in $M$ and $N$ and intersects $AC$ in $P$ and $Q$. Prove that\n$$\n\\frac{AM + AN}{AC} = \\frac{AP + AQ}{AB}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P(X)$ denote the power of point $X$ with respect to the circle $M, N, D$. Then\n$$\n\\begin{align*}\nAB(AM + AN) &= AB \\cdot AM + AB \\cdot AN \\\\\n&= (AN + NB)AM + AB(AB - BN) \\\\\n&= AN \\cdot AM + BN \\cdot AM + AB^2 - AB \\cdot BN \\\\\n&= P(A) + AD^2 + BD^2 - BN(AB - AM) \\\\\n&= P(A) + AD^2 + P(B) - BN \\cdot BM \\\\\n&= P(A) + AD^2 + P(B) - P(B) \\\\\n&= P(A) + AD^2.\n\\end{align*}\n$$\nA similar argument shows that $AC(AP + AQ) = P(A) + AD^2$, which concludes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42182,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{Q}^{+} \\to \\mathbb{R}^{+}$ be a function that satisfies\n$$\nf(x + y) - f(x - y) = 4\\sqrt{f(x)f(y)}\n$$\nfor all $0 < y < x$.\n(a) Prove that $f(2x) = 4f(x)$ for all $x \\in \\mathbb{Q}^{+}$.\n(b) Find all such functions.",
    "options": [],
    "answer": "All functions f: Q^+ -> R^+ of the form f(x) = c x^2 with c > 0; in particular, f(2x) = 4 f(x).",
    "solution": "Choose any positive rational $p, q$ and let $x = \\frac{p+q}{2}$ and $y = \\frac{p-q}{2}$. Then\n$$\nf(p) - f(q) = 4\\sqrt{f(x)f(y)} > 0,\n$$\nso $f(p) > f(q)$ and hence $f$ is strictly increasing.\n\nNow we show that $f$ has values arbitrarily close to $0$, i.e., for any $\\epsilon \\in \\mathbb{Q}^+$, we can find $x$ such that $f(x) < \\epsilon$. Substituting $x = 2y$ gives\n$$\nf(3y) - f(y) = 4\\sqrt{f(2y)f(y)} > 4\\sqrt{f(y)f(y)} = 4f(y),\n$$\nand hence $f(3y) > 5f(y)$ or $f(y) < \\frac{f(3y)}{5}$. Substituting this into itself gives $f(y) < \\frac{f(3^2y)}{5^2}$, and repeating gives $f(y) < \\frac{f(3^ky)}{5^k}$. Now let $y = \\frac{1}{3^k}$, giving $f(\\frac{1}{3^k}) < \\frac{f(1)}{5^k}$. Since $f(1)$ does not depend on $k$, we can choose $k$ sufficiently large so that $\\frac{f(1)}{5^k}$ is less than $\\epsilon$ and hence $y$ satisfies the condition.\n\nConsider the original equation again, and consider what happens as $f(y)$ tends to zero. $f(x)$ stays fixed so $4\\sqrt{f(x)f(y)}$ can be made arbitrarily small. Also, $f(x)$ lies between $f(x-y)$ and $f(x+y)$, so for any $\\delta$ we can choose a $y$ such that $f(x) - \\delta < f(x-y) < f(x) < f(x+y) < f(x) + \\delta$.\n\nNow let $x = z + \\epsilon$ and $x = z - \\epsilon$. The original equation becomes\n$$\nf(2z) = 4\\sqrt{f(z + \\epsilon)f(z - \\epsilon)} + f(2\\epsilon).\n$$\nBy choosing $\\epsilon$ sufficiently small, we can ensure that $f(\\epsilon)$ is arbitrarily close to zero and $f(z + \\epsilon)$ and $f(z - \\epsilon)$ are arbitrarily close to $f(z)$. The right hand side can thus be made arbitrarily close to $4f(z)$, and it follows that the left hand side must be $4f(z)$.\n\nWe can extend $f(2z) = 4f(z)$ to $f(nz) = n^2 f(z)$ for all positive integers $n$. For $n = 1$ and $n = 2$ we already have it. Suppose it holds up to some $n$. We now show that it also holds for $n + 1$. Let $x = nx, y = x$, giving\n$$\n\\begin{aligned}\nf((n + 1)x) &= 4\\sqrt{f(nx)f(x)} + f((n - 1)x) \\\\\n&= 4\\sqrt{n^2 f(x)f(x)} + (n - 1)^2 f(x) \\\\\n&= 4n f(x) + (n - 1)^2 f(x) \\\\\n&= (n + 1)^2 f(x).\n\\end{aligned}\n$$\nThe result follows by induction.\n\nWe can further extend this to positive rational coefficients. Let $n, m$ be positive integers. $n^2 f(x) = f(m \\cdot \\frac{n}{m} \\cdot x) = m^2 f(\\frac{n}{m} \\cdot x)$, from which it follows that $f(\\frac{n}{m} \\cdot x) = (\\frac{n}{m})^2 f(x)$.\n\nNow let $f(1) = a$. By the above, $f(x) = a x^2$ for all positive rationals $x$. Conversely, it is trivial to substitute this into the original equation to confirm that it is a valid solution, for any $a \\in \\mathbb{Q}^+$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42183,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the last two digits of the product of the squares of all positive odd integers less than $2014$.",
    "options": [],
    "answer": "25",
    "solution": "Since the product of the odd integers less than $2014$ contains $25$ as a factor, it is clearly divisible by $25$. Also, since it is odd, its last two digits have to be $25$ or $75$. So the product is either of the form $100n + 25$ or of the form $100n + 75$. In either case, the last two digits of the squared product (which is the same as the product of the squares) are $25$:\n$$\n(100n + 25)^2 = 10000n^2 + 5000n + 625 = 100(100n^2 + 50n + 6) + 25\n$$\nor\n$$\n(100n + 75)^2 = 10000n^2 + 15000n + 5625 = 100(100n^2 + 150n + 56) + 25.\n$$\nIn fact, we see that the last three digits have to be $625$.\nOur number has $25 = 5^2$ as a factor, so its last two digits are $00$, $25$, $50$ or $75$. Moreover, it is the square of an odd number, which we can write as $(2n+1)^2 = 4n^2 + 4n + 1$. This means that the remainder upon division by $4$ has to be $1$. Now note that $100n + 0$, $100n + 25$, $100n + 50$ and $100n + 75$ leave a remainder of $0$, $1$, $2$, $3$ respectively when divided by $4$. This means that the last two digits are in fact $25$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42184,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given that\n$$\n\\frac{a-b}{c-d} = 2 \\quad \\text{and} \\quad \\frac{a-c}{b-d} = 3\n$$\nfor certain real numbers $a$, $b$, $c$, $d$, determine the value of\n$$\n\\frac{a-d}{b-c}\n$$",
    "options": [],
    "answer": "-5",
    "solution": "Set $x = c - d$ and $y = b - d$. We have\n$$\na - d = (c - d) + (a - c) = x + 3y\n$$\nand\n$$\na - d = (b - d) + (a - b) = y + 2x,\n$$\nhence $x + 3y = y + 2x$, which implies $x = 2y$. Now we get $a - d = 5y$ and\n$$\nb - c = (b - d) - (c - d) = y - x = -y,\n$$\nso\n$$\n\\frac{a-d}{b-c} = -5.\nWe are given that\n$$\na - b = 2c - 2d, \\text{ thus } a + 2d = b + 2c,\n$$\nand\n$$\na - c = 3b - 3d, \\text{ thus } a + 3d = 3b + c.\n$$\nMultiply the first equation by 4 and the second one by 3, and subtract:\n$$\na - d = 4(a + 2d) - 3(a + 3d) = 4(b + 2c) - 3(3b + c) = -5b + 5c,\n$$\nthus\n$$\n\\frac{a-d}{b-c} = -5.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42185,
    "subject": "Mathematics (Multi-modal)",
    "question": "In obtuse triangle $ABC$, with the obtuse angle at $A$, let $D$, $E$, $F$ be the feet of the altitudes through $A$, $B$, $C$ respectively. $DE$ is parallel to $CF$, and $DF$ is parallel to the angle bisector of $\\angle BAC$. Find the angles of the triangle.",
    "options": [],
    "answer": "∠A = 108°, ∠B = 18°, ∠C = 54°",
    "solution": "![](attached_image_1.png)\n\nWe denote the angles of the triangle by $\\alpha = \\angle BAC$, $\\beta = \\angle ABC$ and $\\gamma = \\angle ACB$. $X$ is the intersection of $BC$ with the angle bisector of $\\angle BAC$. Since $\\angle BDA = \\angle BEA = 90^\\circ$, both $D$ and $E$ lie on the circle with diameter $AB$. Thus $AEBD$ is cyclic, which implies that\n$$\n\\angle EDB = \\angle EAB = 180^\\circ - \\alpha.\n$$\nIt is given that $DE$ and $CF$ are parallel, hence\n$$\n\\angle EDB = \\angle FCB = 90^\\circ - \\angle FBC = 90^\\circ - \\beta,\n$$\nso $\\beta = \\alpha - 90^\\circ$. Likewise,\n$$\n\\angle FDC = \\angle FAC = 180^\\circ - \\alpha,\n$$\nand\n$$\n\\angle FDC = \\angle AXC = 180^\\circ - \\angle ACX - \\angle XAC = 180^\\circ - \\gamma - \\alpha/2,\n$$\nso $\\gamma = \\alpha/2$. Since $\\alpha + \\beta + \\gamma = 180^\\circ$, this gives us\n$$\n\\alpha + (\\alpha - 90^\\circ) + \\alpha/2 = 180^\\circ,\n$$\nand thus $\\alpha = 108^\\circ$, $\\beta = 18^\\circ$ and $\\gamma = 54^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42186,
    "subject": "Mathematics (Multi-modal)",
    "question": "(a) Let $a$, $x$, $y$ be positive integers. Prove: if $x \\neq y$, then also\n$$\nax + \\gcd(a, x) + \\text{lcm}(a, x) \\neq ay + \\gcd(a, y) + \\text{lcm}(a, y).\n$$\n(b) Show that there are no two positive integers $a$ and $b$ such that\n$$\nab + \\gcd(a, b) + \\text{lcm}(a, b) = 2014.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(a) Suppose that\n$$\nax + \\gcd(a, x) + \\text{lcm}(a, x) = ay + \\gcd(a, y) + \\text{lcm}(a, y)\n$$\nfor certain positive integers $a$, $x$, $y$. It follows that\n$$\n\\gcd(a, ax + \\gcd(a, x) + \\text{lcm}(a, x)) = \\gcd(a, ay + \\gcd(a, y) + \\text{lcm}(a, y)).\n$$\nSince $a$ divides both $ax$ and $\\text{lcm}(a, x)$, we have\n$$\n\\gcd(a, ax + \\gcd(a, x) + \\text{lcm}(a, x)) = \\gcd(a, \\gcd(a, x)) = \\gcd(a, x)\n$$\nand likewise\n$$\n\\gcd(a, ay + \\gcd(a, y) + \\text{lcm}(a, y)) = \\gcd(a, \\gcd(a, y)) = \\gcd(a, y).\n$$\nTherefore, we must have $\\gcd(a, x) = \\gcd(a, y) = d$ for some positive integer $d$. Since $\\text{lcm}(a, x) = ax/\\gcd(a, x)$ and $\\text{lcm}(a, y) = ay/\\gcd(a, y)$, this gives us\n$$\nax + d + \\frac{ax}{d} = ay + d + \\frac{ay}{d},\n$$\nso\n$$\nax \\left(1 + \\frac{1}{d}\\right) = ay \\left(1 + \\frac{1}{d}\\right),\n$$\nwhich implies $x = y$. This proves the first statement.\n\n(b) Suppose that $ax+\\gcd(a,x)+\\lcm(a,x) = 2014$. Note that the left hand side is divisible by $\\gcd(a,x)$, so $\\gcd(a,x)$ has to be a divisor of 2014, i.e., one of $1, 2, 19, 38, 53, 106, 1007, 2014$. On the other hand,\n$$\n(\\gcd(a, x) + 1)(\\lcm(a, x) + 1) = ax + \\gcd(a, x) + \\lcm(a, x) + 1 = 2015,\n$$\nso $\\gcd(a, x) + 1$ has to divide 2015. Since 2, 3, 20, 39, 54, 107, 1008 are all not divisors of 2015, this leaves us with $\\gcd(a, x) = 2014$. But then $ax + \\lcm(a, x) = 0$, which is clearly impossible since the left hand side is positive.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42187,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 1$ be an integer. An $n \\times n$-square is divided into $n^2$ unit squares. Of these smaller squares, $n$ are coloured green and $n$ are coloured blue. All remaining squares are coloured white. Are there more such colourings for which there are no two green squares in a row, and no two blue squares in a column, or colourings for which there are neither two green squares in a row nor two blue squares in a column?",
    "options": [],
    "answer": "There are more colorings with no two green squares in a row and no two blue squares in a row than with no two green squares in a row and no two blue squares in a column.",
    "solution": "Suppose that $n$ squares have been coloured green, no two of them in the same row. This leaves $n-1$ empty squares in each row, which means that there are $(n-1)^n$ ways to colour $n$ of the remaining $n^2-n$ squares blue such that there are no two blue squares in the same row.\n\nOn the other hand, let $x_i$ be the number of squares in column $i$ that have not been coloured green. Clearly, $x_1 + x_2 + \\dots + x_n = n^2 - n$. The number of ways to colour $n$ of the remaining $n^2 - n$ squares blue in such a way that there are no two blue squares in the same column is\n$$\nx_1 x_2 \\cdots x_n \\le \\left( \\frac{x_1 + x_2 + \\cdots + x_n}{n} \\right)^n = \\left( \\frac{n^2 - n}{n} \\right)^n = (n-1)^n\n$$\nby the inequality between the arithmetic and geometric mean. For some configurations (e.g., all green squares in one column), this holds with strict inequality.\n\nHence we can conclude that there are more possible colourings for which there are no two green squares and no two blue squares in the same row than there are colourings for which there are no two green squares in the same row and no two blue squares in the same column.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 42188,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the centre of a two-dimensional coordinate system, and let $A_1, A_2, \\dots, A_n$ be points in the first quadrant and $B_1, B_2, \\dots, B_m$ points in the second quadrant. We associate numbers $a_1, a_2, \\dots, a_n$ to the points $A_1, A_2, \\dots, A_n$ and numbers $b_1, b_2, \\dots, b_m$ to the points $B_1, B_2, \\dots, B_m$, respectively. It turns out that the area of triangle $OA_jB_k$ is always equal to the product $a_jb_k$, for any $j$ and $k$. Show that either all the $A_j$ or all the $B_k$ lie on a single line through $O$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider first the case that one of the areas is zero, e.g. $\\text{area}(OA_1B_1) = 0$. Then either $a_1 = 0$ or $b_1 = 0$. If $a_1 = 0$, then $\\text{area}(OA_1B_k) = a_1b_k = 0$ for all $k$, which means that all $B_k$ lie on a straight line through $O$ and $A_1$. Likewise, if $b_1 = 0$, then all $A_j$ lie on a straight line through $O$ and $B_1$. So we can assume from now on that none of the areas is zero.\n\nSuppose there are two points among $A_1, A_2, \\dots, A_n$ that are not on the same line through $O$ (since the numbering does not matter, we can assume that $A_1$ and $A_2$ have this property), and at the same time there are two points (again, we can assume them to be $B_1$ and $B_2$) that are not on the same line through $O$. We have\n$$\n\\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$$\nSince triangles $OA_1B_1$ and $OA_1B_2$ have the same base ($OA_1$), their heights must be in a $b_1/b_2$-ratio. The same is true for the heights of triangles $OA_2B_1$ and $OA_2B_2$. If $B_1B_2$ is parallel to $OA_1$, then $b_1 = b_2$, so $B_1B_2$ is parallel to $OA_2$. In this case, $O, A_1$ and $A_2$ lie on one line, contradicting the assumption. The same argument applies if $B_1B_2$ is parallel to $OA_2$.\n\nIf neither $OA_1$ nor $OA_2$ is parallel to $B_1B_2$, let $X$ be the intersection of $OA_1$ and $B_1B_2$, and let $Y$ be the intersection of $OA_2$ and $B_1B_2$.\n\n![](attached_image_1.png)\n\nSince $A_1$ and $A_2$ are in the first quadrant, $X$ and $Y$ are either in the first or third quadrant. In either case, they are not between $B_1$ and $B_2$. Using similar triangles, we see that the ratio of the heights of $OA_1B_1$ and $OA_1B_2$ is $|XB_1|/|XB_2|$, which must be $b_1/b_2$. The same is true (analogously) for the ratio $|YB_1|/|YB_2|$. Hence\n$$\n\\frac{|XB_1|}{|XB_2|} = \\frac{b_1}{b_2} = \\frac{|YB_1|}{|YB_2|}\n$$\nIf $X$ and $Y$ are on different sides of $B_1B_2$, one of these ratios is greater than 1, the other less than 1, a contradiction. Thus we assume that $B_1$ is closer to both $X$ and $Y$ (otherwise, we just interchange the roles of $B_1$ and $B_2$), as in the figure. We get\n$$\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$$\nand thus $|XB_2| = |YB_2|$. This means that $X$ and $Y$ coincide, so $X, Y, O, A_1, A_2$ lie on one line, and we get a contradiction to our assumption again. This completes the proof.\nWe argue as in the first proof and assume again that $A_1$ and $A_2$ do not lie on a common line through $O$, and that $B_1$ and $B_2$ do not lie on a common line through $O$. Let $\\alpha_1, \\alpha_2, \\beta_1, \\beta_2$ be the angles enclosed by $OA_1, OA_2, OB_1$ and $OB_2$ with the x-axis. Then\n$$\n\\text{area}(OA_1B_1) \\cdot \\text{area}(OA_2B_2) = a_1b_1a_2b_2 = \\text{area}(OA_1B_2) \\cdot \\text{area}(OA_2B_1)\n$$\nand thus\n$$\n\\frac{|OA_1||OB_1|\\sin(\\beta_1 - \\alpha_1)}{2} \\cdot \\frac{|OA_2||OB_2|\\sin(\\beta_2 - \\alpha_2)}{2} = \\frac{|OA_1||OB_2|\\sin(\\beta_2 - \\alpha_1)}{2} \\cdot \\frac{|OA_2||OB_1|\\sin(\\beta_1 - \\alpha_2)}{2}\n$$\nIt follows that\n$$\n\\sin(\\beta_1 - \\alpha_1) \\sin(\\beta_2 - \\alpha_2) = \\sin(\\beta_2 - \\alpha_1) \\sin(\\beta_1 - \\alpha_2).\n$$\nNow we use the trigonometric identity $\\sin x \\sin y = \\frac{1}{2}(\\cos(x - y) - \\cos(x + y))$ to get\n$$\n\\frac{1}{2}(\\cos(\\beta_1 - \\alpha_1 + \\alpha_2 - \\beta_2) - \\cos(\\beta_1 - \\alpha_1 + \\beta_2 - \\alpha_2)) = \\frac{1}{2}(\\cos(\\beta_2 - \\alpha_1 + \\alpha_2 - \\beta_1) - \\cos(\\beta_2 - \\alpha_1 + \\beta_1 - \\alpha_2)).\n$$\nThis implies\n$$\n\\cos(\\beta_1 - \\alpha_1 + \\alpha_2 - \\beta_2) = \\cos(\\beta_2 - \\alpha_1 + \\alpha_2 - \\beta_1),\n$$\nand by the addition theorem for the cosine\n$$\n\\cos(\\beta_1 - \\beta_2) \\cos(\\alpha_1 - \\alpha_2) + \\sin(\\beta_1 - \\beta_2) \\sin(\\alpha_1 - \\alpha_2) = \\cos(\\beta_1 - \\beta_2) \\cos(\\alpha_1 - \\alpha_2) - \\sin(\\beta_1 - \\beta_2) \\sin(\\alpha_1 - \\alpha_2),\n$$\nso finally\n$$\n\\sin(\\beta_1 - \\beta_2) \\sin(\\alpha_1 - \\alpha_2) = 0,\n$$\nwhich means that either $\\beta_1 = \\beta_2$ or $\\alpha_1 = \\alpha_2$, contradicting our assumption and thus completing the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42189,
    "subject": "Mathematics (Multi-modal)",
    "question": "I have R4. How many sweets can I buy if one sweet costs 40c?",
    "options": [],
    "answer": "10",
    "solution": "Since total cost = number of sweets × cost of each sweet, it follows that the number of sweets is $\\text{R}4.00 \\div 40\\text{c} = \\text{R}4.00 \\div \\text{R}0.40 = 4 \\div 0.4 = 10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42190,
    "subject": "Mathematics (Multi-modal)",
    "question": "The value of $\\frac{2015 \\times 2 + 4 \\times 4030}{4030}$ is\n(A) 2 (B) 3 (C) 4 (D) 5 (E) 6",
    "options": [],
    "answer": "D",
    "solution": "$$\n\\frac{2015 \\times 2 + 4 \\times 4030}{4030} = \\frac{4030 + 4 \\times 4030}{4030} = \\frac{5 \\times 4030}{4030} = 5.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42191,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $E$ and $F$ lie inside a square $ABCD$ such that the two triangles $ABF$ and $BCE$ are equilateral. Show that $DEF$ is an equilateral triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have $\\angle FAD = \\angle BAD - \\angle BAF = 90^\\circ - 60^\\circ = 30^\\circ$. Since $AF = AB = AD$, triangle $AFD$ is isosceles, which means that $\\angle ADF = \\angle AFD = \\frac{180^\\circ - \\angle FAD}{2} = 75^\\circ$ and $\\angle CDF = \\angle CDA - \\angle ADF = 90^\\circ - 75^\\circ = 15^\\circ$. By symmetry, we also have $\\angle ADE = 15^\\circ$, thus $\\angle EDF = 90^\\circ - \\angle ADE - \\angle CDF = 60^\\circ$.\n\nAgain by symmetry (with respect to the diagonal $BD$), $DE = DF$, so $DEF$ is an isosceles triangle with an angle of $60^\\circ$. Therefore, $DEF$ is indeed an equilateral triangle.\nLet $G$ and $H$ be the midpoints of $AB$ and $CD$ respectively, and let $M$ be the centre of the square. We denote the side length of the square and the two equilateral triangles by $a$. By Pythagoras' Theorem,\n$$\nGF^2 = AF^2 - AG^2 = a^2 - \\left(\\frac{a}{2}\\right)^2 = \\frac{3a^2}{4},\n$$\nso $GF = \\frac{\\sqrt{3}a}{2}$. Next we find $FH = GH - GF = a - \\frac{\\sqrt{3}a}{2} = \\frac{(2-\\sqrt{3})a}{2}$, $FM = GF - GM = \\frac{\\sqrt{3}a}{2} - \\frac{a}{2} = \\frac{(\\sqrt{3}-1)a}{2}$ and by symmetry $EM = FM = \\frac{(\\sqrt{3}-1)a}{2}$.\n\nApplying Pythagoras' Theorem again, we obtain\n$$\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$$\nand\n$$\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$$\nThus $DF = EF$, and by symmetry $DE = EF$. This means that $DEF$ is an equilateral triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42192,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the value of $(5-1) + (0-2)$?",
    "options": [],
    "answer": "2",
    "solution": "$(5-1) + (0-2) = 4 + (-2) = 4 - 2 = 2$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42193,
    "subject": "Mathematics (Multi-modal)",
    "question": "The value of $2 - (0 - (1 - 5))$ is\n(A) 3\n(B) 1\n(C) 0\n(D) -1\n(E) -2",
    "options": [],
    "answer": "E",
    "solution": "**E**\n\n$2 - (0 - (1 - 5)) = 2 - (0 - (-4)) = 2 - (0 + 4) = 2 - 4 = -2$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42194,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players alternate placing $2 \\times 1$ tiles, with no overlap, on a $5 \\times 5$ chessboard until no player can place a tile.\nA number of $1 \\times 1$ squares remain empty.\n![](attached_image_1.png)\n\na) Prove that there must be an odd number of empty squares remaining.\n\nb) Of the empty squares remaining, are more coloured black or white?\n\nc) What is the maximum number of remaining squares?",
    "options": [],
    "answer": "a) odd number of empty squares; b) more black squares; c) 7",
    "solution": "a. The chessboard contains an odd number of squares and every time a $2 \\times 1$ tile is placed, the number of open squares decreases by $2$, an even number. Hence the number of remaining squares is always an odd number (odd minus even is odd).\n\nb. Before any tile is placed, there are $13$ black squares and $12$ white squares. Every time a tile is placed, the tile covers one black square and one white square, so the number of open black squares and open white squares both decrease by $1$. Since there is one more black square than white squares at the start, there will be one more black square than white squares at the end.\n\nc. The following arrangement shows $7$ empty $1 \\times 1$ squares.\n![](attached_image_2.png)\n\nAssume for a contradiction that it is possible to have more than $7$ empty squares. Since there must be an odd number, there are at least $9$ empty squares.\n\nDivide the board into four $2 \\times 3$ rectangles and one $1 \\times 1$ centre square:\n![](attached_image_3.png)\n\nIf the top left $2 \\times 3$ rectangle had $3$ empty squares, then these three empty squares must be touching diagonally, and hence so must the three covered squares. This means that either $A1$ is covered and $B1$ and $A2$ are empty (impossible with $1 \\times 2$ tiles) or $B1$ is covered and $A1$, $B2$ and $C1$ are empty (again impossible). This means that this rectangle can contain at most $2$ empty squares. A similar argument shows that all of the $2 \\times 3$ rectangles have maximum two empty squares. However, this means that to get $9$ empty squares in total, all four rectangles must have exactly two empty squares, and in addition the central square must be empty. It follows that the white squares $C2$, $B3$, $D3$ and $C4$ must all be covered.\n\nNow, from part (b) we know that of the $9$ empty squares, $5$ must be black and $4$ white. The only remaining white squares are $A2$, $A4$, $B1$, $B5$, $D1$, $D5$, $E2$, $E4$, of which four must be empty. None of the pairs $(A2,B1)$, $(A4,B5)$, $(D1,E2)$ and $(D5,E4)$ can be empty, since that would force the corner square also empty, a contradiction. Hence exactly one in each of the above mentioned pairs is empty, and hence the three black squares surrounding it must be covered. For the first white square, this forces three black squares to be covered, and for each of the remaining three white squares, at least two additional black squares are covered, leaving at most $13 - (3 + 2 + 2 + 2) = 4$ black squares empty, which contradicts the fact that $5$ black squares must be empty.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42195,
    "subject": "Mathematics (Multi-modal)",
    "question": "Pyramids with square bases were built by stacking magnetic balls. A pyramid with a $1 \\times 1$ base contains one ball, a pyramid with a $2 \\times 2$ base contains five balls, a pyramid with a $3 \\times 3$ base contains 14 balls, a pyramid with a $4 \\times 4$ base contains 30 balls. How many balls are there in a pyramid with a $10 \\times 10$ base?",
    "options": [],
    "answer": "385",
    "solution": "The $n$-th pyramid is made by putting the $(n - 1)$-st pyramid on an $n \\times n$ square of balls. Thus the number of balls in the $n$-th pyramid is $1^2 + 2^2 + \\ldots + n^2$. The number of balls in the 10th pyramid is therefore $1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 385$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42196,
    "subject": "Mathematics (Multi-modal)",
    "question": "A hiker walks $1$ km East, then $2$ km North, then $3$ km West, then $4$ km South, then $5$ km East and finally $6$ km North. The hiker's straight-line distance in km from the starting point is\n(A) $4$ (B) $5$ (C) $6$ (D) $7$ (E) $8$",
    "options": [],
    "answer": "B",
    "solution": "Horizontal distance travelled:\n$$\n1 - 3 + 5 = 3 \\text{ (i.e. 3 km East)}\n$$\nVertical distance travelled:\n$$\n2 - 4 + 6 = 4 \\text{ (i.e. 4 km North)}\n$$\nBy Pythagoras, the straight-line distance from the starting point is thus $5$ km.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42197,
    "subject": "Mathematics (Multi-modal)",
    "question": "The list of numbers\n16; 19; 24; 25; 30; 31; 32; 46; $x$\nhas the same median and mean. If $x$ is greater than $46$, then the value of $x$ is\n(A) 47 (B) 48 (C) 53 (D) 50 (E) 57",
    "options": [],
    "answer": "A",
    "solution": "The median is the middle number, which is $30$, so the mean is also $30$. The sum of all nine numbers is therefore $9 \\times 30 = 270$, which means that $x = 270 - (16 + 19 + 24 + 25 + 30 + 31 + 32 + 46) = 270 - 223 = 47$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42198,
    "subject": "Mathematics (Multi-modal)",
    "question": "A *new sequence* is formed by deleting numbers from the sequence $1, 2, 3, 4, \\dots, 500$ such that the sum of any two numbers of the *new sequence* is never a multiple of seven.\nWhat is the maximum length of the new sequence?",
    "options": [],
    "answer": "217",
    "solution": "Group the numbers into $7$ groups according to their remainders upon division by $7$: the groups are\n$$\n\\begin{aligned}\n\\{1, 8, 15, \\dots, 498\\}, & \\{2, 9, 16, \\dots, 499\\}, \\{3, 10, 17, \\dots, 500\\}, \\\\\n\\{4, 11, 18, \\dots, 494\\}, & \\{5, 12, 19, \\dots, 495\\}, \\{6, 13, 20, \\dots, 496\\}, \\{7, 14, 21, \\dots, 497\\}.\n\\end{aligned}\n$$\nNote that if a number from a group that leaves remainder $x$ is not deleted, then all the numbers from the group that leaves remainder $7-x$ must be deleted, otherwise two numbers will sum to a multiple of $7$. In particular, only one number from the last group above may be selected.\nNote also that if one number from a group (besides the last one) is selected, we may select all numbers in that group, since they all leave the same remainder when divided by $7$. The first three groups contain $72$ numbers each, while the last four contain $71$ numbers each. If we select the first three groups, we may not select any number in the next three groups, and finally we can select one number from the last group. This yields $72 \\times 3 + 1 = 217$ numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42199,
    "subject": "Mathematics (Multi-modal)",
    "question": "When a water-tank is $30\\%$ empty it has $30$ litres more in it than when it is $30\\%$ full. How many litres can the tank hold when it is full?",
    "options": [],
    "answer": "75",
    "solution": "$75$\n\nIf $W$ is the number of litres the tank can hold, then when it is $30\\%$ empty it is holding $0.7W$ litres, and so we have $0.7W = 0.3W + 30$, i.e. $0.4W = 30$ and then $W = 30 \\div 0.4 = 300 \\div 4 = 75$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42200,
    "subject": "Mathematics (Multi-modal)",
    "question": "Lewis Hamilton completes a 72-lap race travelling at an average speed of 288 km/h. Each lap is 6 km in length. The time taken, in hours, for him to complete the race is\n(A) 1 (B) 2 (C) 2.5 (D) 3 (E) 1.5",
    "options": [],
    "answer": "E",
    "solution": "Time = Distance/Speed, so the time taken is equal to $$(72 \\times 6)/288 = 1.5$$ hours.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42201,
    "subject": "Mathematics (Multi-modal)",
    "question": "When $a$ is increased by $20\\%$ and $b$ is decreased by $20\\%$ the resulting values are equal. The ratio of $a$ to $b$ is\n\n(A) $\\frac{1}{2}$ (B) $1$ (C) $\\frac{2}{3}$ (D) $\\frac{3}{4}$ (E) $\\frac{4}{3}$",
    "options": [],
    "answer": "C",
    "solution": "$120\\%$ of $a$ is $\\frac{6}{5}a$ while $80\\%$ of $b$ is $\\frac{4}{5}b$.\nWe thus have $\\frac{6}{5}a = \\frac{4}{5}b$ from which it follows that $\\frac{a}{b} = \\frac{4}{5} \\times \\frac{5}{6} = \\frac{2}{3}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42202,
    "subject": "Mathematics (Multi-modal)",
    "question": "I buy a number of chocolates which cost R25 each and cool-drinks which cost R9 each. I buy more chocolates than cool-drinks. How many cool-drinks do I buy if I pay R839?",
    "options": [],
    "answer": "21",
    "solution": "If I buy $x$ chocolate bars and $y$ cool-drinks, then $25x + 9y = 839$ with $x > y$. If we put $x = y$ as a first guess, then $x = y = 839/34 \\approx 24$, so we can write $x = 24 + a$ and $y = 24 - b$. The equation becomes $25a - 9b = 23$, which by easy trial and error (guess and check) has a solution $a = 2$ and $b = 3$. Thus $x = 24 + 2 = 26$ and $y = 24 - 3 = 21$, which is the required answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42203,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find three prime factors of: $5^{2015} + 5^{2016} + 5^{2017} + 31^{2016}$\nNB: Show all your working!",
    "options": [],
    "answer": "2, 3, 31",
    "solution": "Let us consider the expression:\n$$\n5^{2015} + 5^{2016} + 5^{2017} + 31^{2016}\n$$\nFirst, factor $5^{2015}$ from the first three terms:\n$$\n5^{2015} + 5^{2016} + 5^{2017} = 5^{2015}(1 + 5 + 5^2) = 5^{2015}(1 + 5 + 25) = 5^{2015} \\times 31\n$$\nSo the expression becomes:\n$$\n5^{2015} \\times 31 + 31^{2016}\n$$\nNow, factor $31$:\n$$\n= 31(5^{2015} + 31^{2015})\n$$\nSo $31$ is a prime factor.\n\nNow, consider $5^{2015} + 31^{2015}$.\nNotice that both $5$ and $31$ are odd, so their powers are odd, and their sum is even. Thus, $2$ is a factor.\n\nLet us check if $2$ is a factor:\n- $5^{2015}$ is odd\n- $31^{2015}$ is odd\n- Odd + Odd = Even\nSo $2$ is a factor.\n\nNow, check if $5 + 31 = 36$ is divisible by $3$.\nLet us check modulo $3$:\n- $5 \\equiv 2 \\pmod{3}$\n- $31 \\equiv 1 \\pmod{3}$\nSo $5^{2015} \\equiv 2^{2015} \\pmod{3}$\n$31^{2015} \\equiv 1^{2015} = 1 \\pmod{3}$\nNow, $2^{2015}$ is $2$ if $2015$ is odd, which it is.\nSo $2^{2015} \\equiv 2 \\pmod{3}$\nThus,\n$$\n5^{2015} + 31^{2015} \\equiv 2 + 1 = 3 \\equiv 0 \\pmod{3}\n$$\nSo $3$ is a factor.\n\nTherefore, the three prime factors are $2$, $3$, and $31$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42204,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $A_1$, $A_2$, $A_3$ ... are constructed as follows: the length $OA_1$ is $4$, $\\angle OA_1A_2 = 90^\\circ$ and the length $A_1A_2 = 1$; then a right angle is constructed at $A_2$ to find $A_3$, and so on as shown in the diagram.\nThe length of $OA_{21}$ is\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "6",
    "solution": "**6** By Pythagoras, $OA_2 = \\sqrt{17}$. Then $OA_3 = \\sqrt{18}$, $OA_4 = \\sqrt{19}$ and so on, with $OA_n = \\sqrt{n+15}$, and thus $OA_{21} = \\sqrt{36} = 6$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42205,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a + 2b = 13$ and $5a - 2b = 5$, the value of $b$ is\n(A) 1\n(B) 2\n(C) 3\n(D) 4\n(E) 5",
    "options": [],
    "answer": "E",
    "solution": "Given $a + 2b = 13$ and $5a - 2b = 5$, we can add both left-hand sides and both right-hand sides to find $6a = 18$. Thus $a = 3$, and then since $a + 2b = 13$, we must have $2b = 10$, i.e. $b = 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42206,
    "subject": "Mathematics (Multi-modal)",
    "question": "The last digit of $2^{2015} + 5^{2015}$ is\n(A) 1   (B) 3   (C) 5   (D) 7   (E) 9",
    "options": [],
    "answer": "B",
    "solution": "The last digits of the first few powers of $2$ are $2$, $4$, $8$, $6$, $2$, $4$, ..., so the digits repeat in cycles of length $4$. Since the remainder after dividing $2015$ by $4$ is $3$, it follows that the last digit of $2^{2015}$ is the same as the last digit of $2^3$, which is $8$. Finally, the last digit of any power of $5$ is $5$, so the last digit of $2^{2015} + 5^{2015}$ is equal to the last digit of $8 + 5$, which is $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42207,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of the squares of 3 consecutive positive integers is 770. The largest of these integers is\n(A) 15 (B) 16 (C) 17 (D) 18 (E) 19",
    "options": [],
    "answer": "C",
    "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 = 768/3 = 256 = 16^2$, so $n = 16$ (since $n > 0$), and the largest number is $n+1 = 17$.\n\nWithout doing any algebra, it is clear that $n^2 \\approx 770/3 = 256\\frac{2}{3}$, and the nearest perfect square is 256.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42208,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some balls were distributed into $2015$ boxes which were arranged in a row as indicated below. Any four consecutive boxes always had a total of $30$ balls. How many balls were there in the $2015$th box?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "7",
    "solution": "If the number of balls in the boxes are denoted $x_1, x_2, \\dots$, then\n\n$$\nx_n + x_{n+1} + x_{n+2} + x_{n+3} = 30 = x_{n+1} + x_{n+2} + x_{n+3} + x_{n+4}.\n$$\n\nThus $x_n = x_{n+4}$ for all $n$, which means that the numbers recur in cycles of length four. Since $2015 = 4 \\times 503 + 3$, it follows that $x_{2015} = x_3 = 7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42209,
    "subject": "Mathematics (Multi-modal)",
    "question": "$\\triangle ABC$ is right-angled at $A$, and $\\triangle ABP$ is equilateral with $AB = 2$. The length of $AC$ is\n![](attached_image_1.png)\n(A) $\\sqrt{6}$\n(B) $\\sqrt{8}$\n(C) $\\sqrt{10}$\n(D) $\\sqrt{12}$\n(E) $4$",
    "options": [],
    "answer": "D",
    "solution": "$BAP = 60^\\circ$ while $BAC = 90^\\circ$, so $PAC = 30^\\circ$. With $BPA = 60^\\circ$ that means $C = 30^\\circ$, and so $\\angle 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": 42210,
    "subject": "Mathematics (Multi-modal)",
    "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![](attached_image_1.png)\n(A) $5 : 2$\n(B) $3 : 2$\n(C) $2 : 1$\n(D) $3 : 1$\n(E) $9 : 4$",
    "options": [],
    "answer": "C",
    "solution": "$\\frac{\\text{area } \\triangle ABP}{\\text{area } ABCD} = \\frac{1}{3}$ so $\\frac{\\frac{1}{2} \\cdot BP \\cdot AB}{BC \\cdot AB} = \\frac{1}{3}$ so $\\frac{\\frac{1}{2} BP}{BC} = \\frac{1}{3}$ and therefore $\\frac{BP}{BC} = \\frac{2}{3}$\nand then $BP : PC$ is $\\frac{2}{3} : \\frac{1}{3} = 2 : 1$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42211,
    "subject": "Mathematics (Multi-modal)",
    "question": "$\\triangle ABE$ and $\\triangle BCF$ are equilateral triangles and $ABCD$ is a square.\nProve that $\\triangle DEF$ is an equilateral triangle.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First note that reflecting the figure around the line $BD$ doesn't change the figure (the equilateral triangle $AEB$ reflects to the triangle $CBF$). This means that $DE$ and $DF$ are symmetric about the line $BD$, which means that $DF = DE$. If we can show that $\\angle FDE = 60^\\circ$, then triangle $DEF$ is an isosceles triangle with one $60^\\circ$ angle, which means it's equilateral.\n\nNow, since $DA = AB = AE$, triangle $ADE$ is isosceles. Next,\n$$\n\\angle DAE = 90^\\circ - \\angle EAB = 90^\\circ - 60^\\circ = 30^\\circ \\text{ and so } \\angle ADE = \\frac{1}{2}(180^\\circ - \\angle DAE) = \\frac{1}{2}(180^\\circ - 30^\\circ) = 75^\\circ.\n$$\nSimilarly, $\\angle DFC = 75^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42212,
    "subject": "Mathematics (Multi-modal)",
    "question": "A die has 20 identical equilateral triangular faces numbered from $1$ to $20$. If two such dice are rolled the most probable sum of the numbers showing on the top faces is\n(A) $18$   (B) $19$   (C) $20$   (D) $21$   (E) $2$",
    "options": [],
    "answer": "D",
    "solution": "It is easy to see that for $n = 1, 2, \\ldots, 20$ there are $n$ equally probable ways to obtain a total of $n+1$. (One die shows any number $x$ between $1$ and $n$, and the other die shows $n+1-x$.) In particular, a total of $21$ can be obtained with $20$ different throws. Beyond that, the number of possibilities decreases again from $19$ down to $1$. Thus the most probable total is $21$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42213,
    "subject": "Mathematics (Multi-modal)",
    "question": "Five squares (labelled $A$, $B$, $C$, $D$ and $E$) are drawn along a straight line, each touching the next square at a vertex, as shown in the diagram. Square $B$ has an area of $20$. Square $C$ has a side length of $4$. Square $D$ has double the area of square $B$. What is the area of square $E$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "24",
    "solution": "Denote the side lengths of squares $A$, $B$, $C$, $D$, $E$ by $a$, $b$, $c$, $d$, $e$ respectively. The two right-angled triangles enclosed by squares $C$, $D$, $E$ and the line are congruent to each other and have sides $c$, $d$, $e$, so by Pythagoras' theorem $d^2 = c^2 + e^2$. We are given that $c^2 = 4^2 = 16$ and $d^2 = 2 \\times 20 = 40$, so $e^2 = 40 - 16 = 24$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 42214,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sums of three out of four numbers (omitting each of the four numbers in turn) are $20$, $22$, $24$ and $27$, respectively. What is the sum of the four numbers?",
    "options": [],
    "answer": "31",
    "solution": "Since each of the given sums involves exactly three of the numbers, it follows that the sum of the four sums is three times the sum of the four numbers. Thus the required sum is equal to $(20 + 22 + 24 + 27)/3 = 93/3 = 31$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42215,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a line segment with the points $R$, $A$, $T$ and $E$, the ratio $RA : AT = 1 : 3$ and $AT : TE = 5 : 2$. The ratio $RA : TE$ is\n(A) $1 : 2$ (B) $1 : 3$ (C) $3 : 5$ (D) $5 : 6$ (E) $3 : 2$",
    "options": [],
    "answer": "D",
    "solution": "We are given $\\frac{RA}{AT} = \\frac{1}{3}$ and $\\frac{AT}{TE} = \\frac{5}{2}$, so\n$$\n\\frac{RA}{TE} = \\frac{RA}{AT} \\times \\frac{AT}{TE} = \\frac{1}{3} \\times \\frac{5}{2} = \\frac{5}{6}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42216,
    "subject": "Mathematics (Multi-modal)",
    "question": "The number of different positive integers such that $\\frac{n}{n+4} < \\frac{4}{9}$ is\n(A) 0\n(B) 1\n(C) 2\n(D) 3\n(E) 4",
    "options": [],
    "answer": "D",
    "solution": "Multiplying gives us $9n < 4n + 16$, i.e. $5n < 16$, so $n = 1$ or $2$ or $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42217,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many pairs of non-negative integers $x$ and $y$ are solutions of $\\frac{x}{20} + \\frac{y}{15} = 1$?",
    "options": [],
    "answer": "6",
    "solution": "**6**\n\nWe can re-write $\\frac{x}{20} + \\frac{y}{15} = 1$ in the form $3x + 4y = 60$. Then we can see that $3x$ must be divisible by $4$, so $x$ must be, and trying successive possible values we see that only the following combinations of $x$- and $y$-values are acceptable: $(0; 15)$, $(4; 12)$, $(8; 9)$, $(12; 6)$, $(16; 3)$, $(20; 0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42218,
    "subject": "Mathematics (Multi-modal)",
    "question": "The number of squares that have $(-1; -1)$ as a vertex and at least one of the coordinate axes as an axis of symmetry is\n(A) 1 (B) 2 (C) 3 (D) 4 (E) 5",
    "options": [],
    "answer": "E",
    "solution": "Answer E.\nIf we reflect $P(-1; -1)$ in the $x$-axis to map to $Q(-1; 1)$ we obtain 3 squares with the $x$-axis as a line of symmetry, as shown in the diagram. Two of the cases have $PQ$ as an edge/side and in one case $PQ$ is a diagonal of the square. Similarly, if we reflect $P$ in the $y$-axis to $R$, then there are three possible squares. However, one of these is the same as before, so in total there are exactly five squares possible.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 42219,
    "subject": "Mathematics (Multi-modal)",
    "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![](attached_image_1.png)\n(A) $5 : 2$ \n(B) $3 : 2$ \n(C) $2 : 1$ \n(D) $3 : 1$ \n(E) $9 : 4$",
    "options": [],
    "answer": "C",
    "solution": "$\\frac{\\text{area } \\triangle ABP}{\\text{area } ABCD} = \\frac{1}{3}$ so $\\frac{\\frac{1}{2} \\text{BP} \\cdot \\text{AB}}{\\text{BC} \\cdot \\text{AB}} = \\frac{1}{3}$ so $\\frac{\\frac{1}{2} \\text{BP}}{\\text{BC}} = \\frac{1}{3}$ and therefore $\\frac{\\text{BP}}{\\text{BC}} = \\frac{2}{3}$\nand then $BP : PC$ is $\\frac{2}{3} : \\frac{1}{3} = 2 : 1$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42220,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f$ be a function satisfying\n$$\nf(xy) = \\frac{f(x)}{y}\n$$\nfor all positive real numbers $x$ and $y$. If $f(500) = 3$, what is the value of $f(100)$?",
    "options": [],
    "answer": "15",
    "solution": "We are given $3 = f(500) = f(100 \\times 5) = \\frac{f(100)}{5}$, so $f(100) = 3 \\times 5 = 15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42221,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many of the integers between $97$ and $199$ are multiples of $2$ or $3$?\n(A) $33$ (B) $40$ (C) $55$ (D) $60$ (E) $68$",
    "options": [],
    "answer": "E",
    "solution": "**E** Between (and including) $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$. The number we seek is $51 + 34 - 17 = 68$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42222,
    "subject": "Mathematics (Multi-modal)",
    "question": "Calculate $2015 - (2015 - (2015 - (2015 - 1)))$.",
    "options": [],
    "answer": "1",
    "solution": "$2015 - (2015 - (2015 - (2015 - 1))) = 2015 - (2015 - (2015 - 2014)) = 2015 - (2015 - 1) = 2015 - 2014 = 1.$\n\nAlternatively,\n$2015 - (2015 - (2015 - (2015 - 1))) = 2015 - 2015 + 2015 - 2015 + 1 = 1.$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42223,
    "subject": "Mathematics (Multi-modal)",
    "question": "Human hair grows at a rate of about $1$ centimetre per month. This is equivalent to about how many millimetres every ten years?\n(A) $12$\n(B) $120$\n(C) $1\\,200$\n(D) $12\\,000$\n(E) $120\\,000$",
    "options": [],
    "answer": "C",
    "solution": "**C** $1$ cm per month $= 10$ mm per month $= 120$ mm per year $= 1\\,200$ mm in ten years",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42224,
    "subject": "Mathematics (Multi-modal)",
    "question": "The largest number below is\n(A) $2^3$ (B) $\\frac{15}{2}$ (C) $\\sqrt{81}$ (D) $4^2$ (E) $\\frac{31}{4}$",
    "options": [],
    "answer": "D",
    "solution": "Answer D.\nAfter simplification, the numbers are (A) $8$, (B) $7.5$, (C) $9$, (D) $16$, (E) $7.75$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42225,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many positive factors of $128$ are not factors of $120$?",
    "options": [],
    "answer": "4",
    "solution": "$128$ has factors that are all powers of $2$, i.e. $2^0, 2^1, \\ldots, 2^7$.\n\nThe largest power of $2$ that divides $120$ is $2^3$.\n\nThe powers of $2$ dividing $128$ and exceeding $8$ are $2^4, 2^5, 2^6$ and $2^7$.\n\nSo, there are $4$ positive factors of $128$ that are not factors of $120$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42226,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs of real numbers $a$ and $x$ that satisfy the simultaneous equations\n$$\n5x^3 + ax^2 + 8 = 0\n$$\nand\n$$\n5x^3 + 8x^2 + a = 0.\n$$",
    "options": [],
    "answer": "(8, -2), (-3, -1), (-13, 1)",
    "solution": "**Solution 1:** If we subtract the two equations, we obtain\n$$\nax^2 + 8 - 8x^2 - a = (a-8)(x^2-1) = (a-8)(x+1)(x-1) = 0,\n$$\nthus either $a = 8$ or $x = -1$ or $x = 1$. If $a = 8$, we are left with\n$$\n5x^3 + 8x^2 + 8 = (x+2)(5x^2 - 2x + 4) = 0.\n$$\nThe second factor has no real roots, since its discriminant $2^2 - 4 \\cdot 5 \\cdot 4 = -76$ is negative.\nThus $x = -2$ in this case.\nIf $x = -1$, we get $a = -5x^3 - 8x^2 = -3$, and if $x = 1$, we get $a = -5x^3 - 8x^2 = -13$.\nIn summary, there are three possible pairs: $(a, x) = (8, -2)$, $(a, x) = (-3, -1)$ and $(a, x) = (-13, 1)$.\n\n\n**Solution 2:** Solving the second equation for $a$ gives us $a = -5x^3 - 8x^2$, thus\n$$\n5x^3 + (-5x^3 - 8x^2)x^2 + 8 = -5x^5 - 8x^4 + 5x^3 + 8 = 0.\n$$\nThe polynomial can be factorised:\n$$\n\\begin{aligned}\n-5x^5 - 8x^4 + 5x^3 + 8 &= 5x^3(1 - x^2) + 8(1 - x^4) = 5x^3(1 - x^2) + 8(1 + x^2)(1 - x^2) \\\\\n&= (1 - x^2)(5x^3 + 8x^2 + 8) = (1 - x)(1 + x)(x + 2)(5x^2 - 2x + 4).\n\\end{aligned}\n$$\nAgain, we find that $x = 1$, $x = -1$ or $x = -2$, since the quadratic factor has no real roots.\nThe value of $a$ is obtained from the equation $a = -5x^3 - 8x^2$, so we end up with the three possibilities $(a, x) = (8, -2)$, $(a, x) = (-3, -1)$ and $(a, x) = (-13, 1)$ again.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 42227,
    "subject": "Mathematics (Multi-modal)",
    "question": "If the average of four different positive integers is $8$, what is the largest possible value of any one of these integers?",
    "options": [],
    "answer": "26",
    "solution": "If the average of four numbers is $8$, their sum must be $8 \\times 4 = 32$. To maximize the largest one of these numbers, we choose the smallest possible values for the other three, which are $1$, $2$ and $3$ (since the integers must be positive and different). The remaining number is then $32 - (1 + 2 + 3) = 26$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42228,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many different numbers can be written as the product of two or more of the numbers $3$, $4$, $5$, $6$, $7$, $7$, $7$?",
    "options": [],
    "answer": "138",
    "solution": "The required numbers are of the form $3^a4^b5^c6^d7^e$, where\n$$\n0 \\le a \\le 1, \\quad 0 \\le b \\le 2, \\quad 0 \\le c \\le 2, \\quad 0 \\le d \\le 1, \\quad 0 \\le e \\le 3,\n$$\nand $a+b+c+d+e \\ge 2$. This gives two possible values of $a$, three of $b$, three of $c$, two of $d$, and four of $e$, making a total of $2 \\times 3 \\times 3 \\times 2 \\times 4 = 144$ possible numbers if we ignore the last restriction. To satisfy it, we must exclude one possibility with $a+b+c+d+e = 0$ and five possibilities with $a+b+c+d+e = 1$, so the final total is $144 - 1 - 5 = 138$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42229,
    "subject": "Mathematics (Multi-modal)",
    "question": "Anelize put $120$ identical cubes together to form a rectangular prism. She then painted all $6$ faces of the prism. Once the paint had dried she disassembled the cubes and found that $24$ of the cubes had not been painted on any face. The surface area of the prism she formed was\n(A) $148$ (B) $150$ (C) $152$ (D) $154$ (E) $156$",
    "options": [],
    "answer": "A",
    "solution": "Let $l$, $b$, $h$ denote the numbers of cubes in the length, breadth and height respectively. Then $lbh = 120$. To find the unpainted cubes we must remove a layer at both ends of the length, breadth and height, giving $(l-2)(b-2)(h-2) = 24$. By trying various factorizations of $120$, we soon find that $l$, $b$, $h$ are equal to $4$, $5$, $6$ in some order. The surface area is $2(lb + lh + bh) = 2(20 + 30 + 24) = 148$. (Trying to solve the equations algebraically is harder than using trial and error.)",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 42230,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $t$ toffees cost $c$ cents, the number of toffees that can be bought for $r$ rands is\n(A) $\\frac{100rc}{t}$ (B) $\\frac{100rt}{c}$ (C) $\\frac{100r}{ct}$ (D) $\\frac{rt}{100c}$ (E) $\\frac{100c}{rt}$",
    "options": [],
    "answer": "B",
    "solution": "Since $t$ toffees cost $c$ cents, each toffee costs $\\frac{c}{t}$ cents.\n\n$r$ rands equals $100r$ cents.\n\nThe number of toffees that can be bought for $100r$ cents is thus $100r \\div \\frac{c}{t} = \\frac{100rt}{c}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42231,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call a divisor $d$ of a positive integer $n$ *special* if $d + 1$ is also a divisor of $n$. Prove: 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": "2, 6, 12",
    "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 $n/d$ and $n/(d + 1)$ are both integers, which means that their difference is at least 1. Thus\n$$\n\\frac{n}{d} \\geq \\frac{n}{d+1} + 1,\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": 42232,
    "subject": "Mathematics (Multi-modal)",
    "question": "One of the factors of $48$ is chosen at random. What is the probability that the chosen factor is NOT a multiple of $4$?\n\n(A) $40\\%$\n(B) $30\\%$\n(C) $25\\%$\n(D) $20\\%$\n(E) $10\\%$",
    "options": [],
    "answer": "A",
    "solution": "There are ten factors: $1$, $2$, $3$, $4$, $6$, $8$, $12$, $16$, $24$, $48$. Four of them are not multiples of $4$, viz. $1$, $2$, $3$, $6$. The probability is therefore $\\frac{4}{10}$ or $40\\%$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42233,
    "subject": "Mathematics (Multi-modal)",
    "question": "Marks are drawn on a thin strip of paper dividing the strip into 4 equal lengths. Marks are also drawn dividing the strip into 3 equal lengths. After the strip has been cut at each mark, how many pieces will there be?",
    "options": [],
    "answer": "6",
    "solution": "6\nIf the length of the strip is 12 units, marks will be made at 3 units, 6 units and 9 units for four equal lengths, and at 4 and 8 units for the three equal lengths. This gives 5 marks in all, meaning there will be 6 pieces when the cuts are made.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42234,
    "subject": "Mathematics (Multi-modal)",
    "question": "At a certain school, break starts at 11:45 and ends at 12:12. How long is break, in minutes?",
    "options": [],
    "answer": "27",
    "solution": "Since $11{:}45$ is $45$ minutes after $11{:}00$ and $12{:}12$ is $60 + 12 = 72$ minutes after $11{:}00$, the length of break is $72 - 45 = 27$ minutes.\n\nAlternatively, there are $15$ minutes from $11{:}45$ to $12{:}00$ and $12$ minutes from $12{:}00$ to $12{:}12$, so the length of break is $15 + 12 = 27$ minutes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42235,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $\\frac{12}{x}$ is a natural number and $x$ is a natural number, how many possible values are there for $x$?\n(A) 4\n(B) 5\n(C) 6\n(D) 7\n(E) 8",
    "options": [],
    "answer": "C",
    "solution": "$x$ must be a divisor of $12$, and the possibilities are $1$, $2$, $3$, $4$, $6$, $12$, which is $6$ in all.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42236,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of three different positive integers is $7$. Their product is\n(A) $12$ (B) $10$ (C) $9$ (D) $8$ (E) $5$",
    "options": [],
    "answer": "D",
    "solution": "Suppose the numbers are $x$, $y$, $z$, with $0 < x < y < z$ and $x + y + z = 7$. If $x \\ge 2$, then $y \\ge 3$ and $z \\ge 4$, so $x + y + z \\ge 9$, which is too large. Therefore $x = 1$, which leaves $y + z = 6$. Since $2 \\le y < z$, the only possibility is $y = 2$ and $z = 4$. The product $xyz = 1 \\times 2 \\times 4 = 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42237,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many zeroes will there be in the answer if the following number is divided by $111$?\n$1110222222003333333330004444444444440000$",
    "options": [],
    "answer": "28",
    "solution": "$\\frac{1110222222003333333330004444444444440000}{111} = 100020020000300300300000400400400400000$, which contains $28$ zeroes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42238,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $\\frac{20}{3} = \\frac{120}{y}$ then $y$ equals\n(A) 9\n(B) 12\n(C) 15\n(D) 18\n(E) 20",
    "options": [],
    "answer": "D",
    "solution": "If the numerator has been multiplied by $6$ and the fraction stays the same, the denominator is also multiplied by $6$ and becomes $3 \\times 6 = 18$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42239,
    "subject": "Mathematics (Multi-modal)",
    "question": "For how many positive values of $n$ are both $\\frac{1}{2}n$ and $2n$ two-digit integers?",
    "options": [],
    "answer": "15",
    "solution": "15\n\nThe smallest value of $n$ is 20 (for which $\\frac{1}{2}n$ is 10) and the largest value of $n$ is 49 (for which $2n$ is 98). From 20 to 49 inclusive represents 30 integers, half of which are even (so that $\\frac{1}{2}n$ is an integer).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42240,
    "subject": "Mathematics (Multi-modal)",
    "question": "Steve was sorting $1\\,000$ eggs into sizes. He got paid $20$ cents for each egg that he sorted. For each egg that he broke while sorting he did not get paid and had to pay his employer $R1$. Steve was paid $R176$. How many eggs did Steve break?",
    "options": [],
    "answer": "20",
    "solution": "Suppose Steve breaks $n$ eggs, so $1\\,000 - n$ eggs are unbroken. He receives $R0.20 \\times (1\\,000 - n)$, but has to repay $R1 \\times n$. Thus $0.2(1\\,000 - n) - n = 176$, giving $200 - 1.2n = 176$, so $1.2n = 200 - 176 = 24$ and $n = 20$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42241,
    "subject": "Mathematics (Multi-modal)",
    "question": "Eskom announces that there is a 60% chance of Stage 1 load shedding for a specific week.\nIn a specific suburb Stage 1 load shedding is from 10:00 to 12:30 on a Monday, Wednesday, Friday and Sunday. What is the probability that in that specific suburb there will be load shedding at a moment in the week, without knowing the day, or time or whether or not it is day or night?",
    "options": [],
    "answer": "1/28",
    "solution": "There are $7 \\times 24 = 168$ hours in a week and there are $4 \\times 2.5 = 10$ hours during which load shedding may occur. Since there is a 60% chance of load shedding, the probability that there is load shedding at a particular moment during the week is\n$$\n\\frac{4 \\times 2.5}{7 \\times 24} \\times 60\\% = \\frac{10}{168} \\times \\frac{60}{100} = \\frac{1}{28}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42242,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle with $AB < AC$, and let points $D$ and $E$ be chosen on the sides $AC$ and $BC$ respectively in such a way that $AD = AE = AB$. The circumcircle of $ABE$ intersects the line $AC$ at $A$ and $F$ and the line $DE$ at $E$ and $P$. Prove that $P$ is the circumcentre of $BDF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $AD = AE$, $AED$ is isosceles, so we also have $\\angle ADE = \\angle AED$. Combining this with the fact that $ABPE$ and $ABPF$ are cyclic, we obtain\n\n$$\n\\angle ABP = 180^{\\circ} - \\angle AEP = \\angle AED = \\angle ADE = \\angle ADP\n$$\nas well as\n$$\n\\angle ABP = 180^{\\circ} - \\angle AFP = \\angle DFP,\n$$\nso $\\angle FDP = \\angle DFP$, which implies that $PD = PF$. Likewise, $\\angle ABE = \\angle AEB$ since $AB = AE$, which means that\n$$\n\\angle APB = \\angle AEB = \\angle ABE = \\angle APE = \\angle APD.\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": 42243,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $4350$ is written as a product of its prime factors, then the largest prime factor is\n(A) $5$   (B) $13$   (C) $19$   (D) $29$   (E) $31$",
    "options": [],
    "answer": "D",
    "solution": "Clearly $4350$ is divisible by $5$, since the last two digits are $50$. It is also divisible by $3$, since the sum of the digits is divisible by $3$. Dividing by $150$ gives a quotient of $29$, which is prime, so the prime factorization is $4350 = 2 \\times 3 \\times 5^2 \\times 29$, and the largest prime factor is $29$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42244,
    "subject": "Mathematics (Multi-modal)",
    "question": "The last digit of $2011 \\times 2013 \\times 2015 - 2010 \\times 2012 \\times 2014$ is\n(A) 0\n(B) 1\n(C) 2\n(D) 4\n(E) 5",
    "options": [],
    "answer": "E",
    "solution": "The last digit of $2011 \\times 2013 \\times 2015$ is a five, and the last digit of $2010 \\times 2012 \\times 2014$ is zero, so the last digit of the difference will be $5 - 0 = 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42245,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $2x + 2y = 14$ and $x^2 - y^2 = 21$ then $x - y$ is equal to\n(A) 1 (B) 2 (C) 3 (D) 4 (E) -1",
    "options": [],
    "answer": "C",
    "solution": "Since $2x + 2y = 14$, we have $x + y = 7$. Also since $x^2 - y^2 = (x + y)(x - y)$, we have $7(x - y) = 21$, so $x - y = 21/7 = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42246,
    "subject": "Mathematics (Multi-modal)",
    "question": "Several small villages are situated on the banks of a straight river. On one side, there are $20$ villages in a row, and on the other there are $15$ villages in a row. We would like to build bridges, each of which connects a village on the one side with a village on the other side. The bridges must be straight, must not cross, and it should be possible to get from any village to any other village using only those bridges (and not any roads that might exist between villages on the same side of the river). How many different ways are there to build the bridges?",
    "options": [],
    "answer": "818809200",
    "solution": "We show that the answer is generally $\\binom{a+b-2}{a-1} = \\frac{(a+b-2)!}{(a-1)!(b-1)!}$ if there are $a$ towns on one side and $b$ on the other. In our particular instance, there are thus $\\binom{33}{14} = 818\\,809\\,200$ ways to build the bridges. We prove our general formula by induction on the total number of villages. Note that the formula always holds if either $a=1$ or $b=1$. In this case, $\\binom{a+b-2}{a-1} = 1$, and indeed there is only one possibility: to build bridges from the single village on the one side of the river to all other villages. This forms the base of our induction.\n\nWe call the villages one the one side $A_1, A_2, \\dots, A_\\alpha$, and the villages on the other side $B_1, B_2, \\dots, B_\\beta$ (in this order). Note first that $A_1$ and $B_1$ cannot both be connected by a bridge to villages other than each other: if there is a bridge between $A_1$ and $B_k$ and a bridge between $B_1$ and $A_l$, where $k, l > 1$, then these bridges cross, which is impossible.\nThis leaves us with two possibilities:\n\n* $A_1$ is directly connected to $B_1$ only, while $B_1$ is connected to the rest. In this case, we can ignore $A_1$ and only count the possibilities to build bridges between $A_2, A_3, \\dots, A_\\alpha$ and $B_1, B_2, \\dots, B_\\beta$. By the induction hypothesis, this can be done in $\\binom{a+b-3}{a-2}$ ways.\n* $B_1$ is directly connected to $A_1$ only, while $A_1$ is connected to the rest. In this case, we can ignore $B_1$, and the induction hypothesis shows that there are $\\binom{a+b-3}{a-1}$ possibilities.\n\nAltogether, this gives us\n$$\n\\binom{a+b-3}{a-2} + \\binom{a+b-3}{a-1} = \\binom{a+b-2}{a-1}\n$$\npossibilities to build the bridges, which completes our induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42247,
    "subject": "Mathematics (Multi-modal)",
    "question": "When simplified, the fraction $\\frac{2}{0 + \\frac{1}{5 + \\frac{2}{0 + \\frac{1}{5}}}}$ equals",
    "options": [],
    "answer": "30",
    "solution": "$$\n\\frac{2}{0 + \\frac{1}{5 + \\frac{2}{0 + \\frac{1}{5}}}} = \\frac{2}{0 + \\frac{1}{5 + \\frac{2}{5 + \\frac{1}{5}}}} = \\frac{2}{0 + \\frac{1}{5 + 10}} = \\frac{2}{\\frac{1}{15}} = 2 \\times 15 = 30\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42248,
    "subject": "Mathematics (Multi-modal)",
    "question": "The dimensions of a box and a book are, respectively, $36\\ \\text{cm} \\times 30\\ \\text{cm} \\times 20\\ \\text{cm}$ and $20\\ \\text{cm} \\times 15\\ \\text{cm} \\times 6\\ \\text{cm}$, as shown in the diagram. What is the maximum number of books that can be packed into the box with no books sticking out?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "12",
    "solution": "We must decide whether the books are to be packed flat, on their spines, or upright (or possibly a combination). In this case it is easy, because if they are packed upright, then their height is $20\\ \\text{cm}$, the same as the depth of the box. The width of each book is $15\\ \\text{cm}$, which goes twice into the width of the box; the thickness of each book is $6\\ \\text{cm}$, which goes six times into the length of the box. Thus $2 \\times 6 = 12$ upright books will fill the box completely, so no higher number is possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42249,
    "subject": "Mathematics (Multi-modal)",
    "question": "If we place a $3$ at both ends of a number, its value is increased by $3372$.\nFind the original number.",
    "options": [],
    "answer": "41",
    "solution": "Placing a $3$ at both ends of the number increases it by $3372$, a four-digit number. Hence the original number is a two-digit number. Suppose the original number is $x$. Then the number formed by placing a $3$ at both ends is equal to $3003 + 10x$. This is $3372$ more than $x$, that is,\n$$\n3003 + 10x - x = 3372 \\implies 9x = 369 \\implies x = 41.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42250,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $a$ is an integer, and that $n! + a$ divides $(2n)!$ for infinitely many positive integers $n$. Prove that $a = 0$.",
    "options": [],
    "answer": "0",
    "solution": "$$(2n)! = \\binom{2n}{n} \\cdot n!^2 \\equiv \\binom{2n}{n} \\cdot (-a)^2 \\mod (n! + a),$$\nso if $n! + a$ divides $(2n)!$, then it also divides $a^2 \\binom{2n}{n}$. We will show that when $n$ is large, $n! + a$ is greater than $a^2 \\binom{2n}{n}$ and therefore does not divide it (unless $a = 0$). Assume in the following that $a \\neq 0$ and $n > \\frac{4^5}{12}a^2$. First of all, by the binomial theorem,\n$$\n\\binom{2n}{n} \\le \\sum_{k=0}^{2n} \\binom{2n}{k} = 2^{2n} = 4^n.\n$$\nIf $a > 0$, we have\n$$\n0 < \\frac{a^2}{n! + a} \\binom{2n}{n} < \\frac{a^2}{n!} \\binom{2n}{n} \\le \\frac{a^2}{n!} 4^n = \\left( \\frac{4^{n-5}}{5 \\cdot 6 \\cdots (n-1)} \\right) \\frac{4^5 a^2}{24n} < 1.\n$$\nIf $a < 0$, then we observe that $n! > n > 2|a|$, so that\n$$\n0 < \\frac{a^2}{n! + a} \\binom{2n}{n} = \\frac{a^2}{n! - |a|} \\binom{2n}{n} < \\frac{a^2}{\\frac{1}{2}n!} \\binom{2n}{n} \\le \\frac{2a^2}{n!} 4^n = \\left( \\frac{4^{n-5}}{5 \\cdot 6 \\cdots (n-1)} \\right) \\frac{4^5 a^2}{12n} < 1.\n$$\n\nThus $\\frac{a^2 \\binom{2n}{n}}{n!+a}$ is not an integer for all such $n$, which contradicts our assumption that $n! + a$ divides $(2n)!$ (and thus $a^2 \\binom{2n}{n}$) for infinitely many $n$.\nSo $a = 0$ is the only possibility (and indeed $n!$ divides $(2n)!$ for all $n$).\nWe show that $n! + a$ cannot divide $(2n)!$ if $a \\neq 0$, $n \\ge 2|a|$ and $n \\ge 9$. In this case, $\\frac{n!}{a}$ is an integer. If $n! + a$ divides $(2n)!$, then so does $\\frac{n!}{a} + 1$. Next note that every prime $p \\le n$ divides $\\frac{n!}{a}$: if $p \\nmid a$, this is obvious (since $p \\nmid n!$), and if $p \\mid a$, then we see that $p$ divides $2|a|$, which occurs as a factor of $\\frac{n!}{a}$. In either case, we find that $\\frac{n!}{a} + 1$ is not divisible by $p$. So $\\frac{n!}{a} + 1$ does not have any prime factors $\\le n$, which means that it is coprime to all numbers $\\le n$ as well as all the even numbers $\\le 2n$.\nTherefore $\\frac{n!}{a} + 1$ must divide the product of all odd numbers greater than $n$ and less than $2n$. By our assumption, $\\left|\\frac{n!}{a} + 1\\right| \\ge \\frac{n}{a} \\cdot (n-1)! - 1 \\ge 2(n-1)! - 1$. We show that the product of the odd numbers between $n$ and $2n$ is less than $2(n-1)! - 1$ for $n \\ge 9$, which yields the desired contradiction.\nFor $n = 9$ and $n = 10$, we have the two inequalities\n$$\n2 \\cdot 8! - 1 = 80639 > 36465 = 11 \\cdot 13 \\cdot 15 \\cdot 17\n$$\nand\n$$\n2 \\cdot 9! - 1 = 725759 > 692835 = 11 \\cdot 13 \\cdot 15 \\cdot 17 \\cdot 19\n$$\nrespectively. Now we proceed by induction. Assume that\n$$\n2(n-1)! - 1 > \\prod_{\\substack{n<k<2n \\\\ k \\text{ odd}}} k.\n$$\nIf $n > 10$ is even, we use the induction hypothesis to obtain\n$$\n\\prod_{\\substack{n<k<2n \\\\ k \\text{ odd}}} k = \\frac{(2n-3)(2n-1)}{n-1} \\prod_{\\substack{n-2<k<2n-4 \\\\ k \\text{ odd}}} k < 2(2n-1) \\prod_{\\substack{n-2<k<2n-4 \\\\ k \\text{ odd}}} k < (4n-2)(2(n-3)!-1) \\\\\n< (n-1)(n-2)(2(n-3)!-1) = 2(n-1)! - (n-1)(n-2) < 2(n-1)!-1,\n$$\nand if $n > 10$ is odd, we get analogously\n$$\n\\prod_{\\substack{n<k<2n \\\\ k \\text{ odd}}} k = \\frac{(2n-3)(2n-1)}{n} \\prod_{\\substack{n-2<k<2n-4 \\\\ k \\text{ odd}}} k < 2(2n-3) \\prod_{\\substack{n-2<k<2n-4 \\\\ k \\text{ odd}}} k < (4n-6)(2(n-3)!-1) \\\\\n< (n-1)(n-2)(2(n-3)!-1) = 2(n-1)! - (n-1)(n-2) < 2(n-1)!-1.\n$$\n\nThis completes the induction and thus the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42251,
    "subject": "Mathematics (Multi-modal)",
    "question": "If today is Thursday, what day of the week will it be in 150 days from now?\n(A) Sunday\n(B) Monday\n(C) Tuesday\n(D) Wednesday\n(E) Thursday",
    "options": [],
    "answer": "A",
    "solution": "Every multiple of $7$ represents a full week. Since today is Thursday, in one day's time it will be Friday. Thus each full week after today starts on a Friday and ends on a Thursday. $150$ days divided by $7$ equals $21$ full weeks with a remainder of $3$ days. The $147$th day from now will thus be a Thursday (end of $21$st full week), and consequently the $150$th day from now will be a Sunday.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42252,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a group of $108$ people, $1$ in $4$ of them has a pen and $1$ in $3$ has a pencil. What is the minimum possible number of people that have something to write with?",
    "options": [],
    "answer": "36",
    "solution": "$36$ people must have a pen and $36$ must have a pencil. If all the people with pens also have pencils, this could be as few as $36$ people in all.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 42253,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x = \\sqrt[3]{900}$ then\n(A) $7 < x < 8$ (B) $9 < x < 10$ (C) $11 < x < 12$ (D) $10 < x < 11$ (E) $12 < x < 13$",
    "options": [],
    "answer": "B",
    "solution": "Answer B.\nSince $9^3 = 729 < 900$ and $10^3 = 1000 > 900$, it follows that $9 < \\sqrt[3]{900} < 10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42254,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is said that you can only fold a piece of paper in half 7 times. Harold folds a sheet of paper in half 5 times and then makes a hole in the folded paper. How many holes does the sheet of paper have after it is unfolded again?\n(A) 32   (B) 36   (C) 81   (D) 50   (E) 64",
    "options": [],
    "answer": "A",
    "solution": "Each fold doubles the number of layers that will be pierced. There will be $2^5$ layers and therefore $2^5 = 32$ holes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42255,
    "subject": "Mathematics (Multi-modal)",
    "question": "If 8 athletes run a race and no two athletes finish exactly together, the number of different possible results for the first, second and third positions is\n(A) 360 (B) 300 (C) 56 (D) 336 (E) 512",
    "options": [],
    "answer": "D",
    "solution": "There are 8 possibilities for first place, each of which can be combined with 7 possibilities for second place and 6 possibilities for third place. Thus the number of possible results for all three places is $8 \\times 7 \\times 6 = 336$. (This assumes that no two athletes finish exactly together.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42256,
    "subject": "Mathematics (Multi-modal)",
    "question": "Mollie, Alfred and four other people want to be in a group photograph. In how many different ways can they be arranged in a row with Mollie and Alfred together in the middle?\n\n(A) 8 (B) 16 (C) 24 (D) 48 (E) 80",
    "options": [],
    "answer": "D",
    "solution": "The person on the extreme left can be any one of the four people that is neither Alfred nor Mollie; the second left can be any one of the remaining three; the first person on the right of centre... and so on. For every arrangement of the people around them, Alfred and Mollie can swap places to make a new arrangement. So the number of possibilities is $(4 \\times 3 \\times 2 \\times 1) \\times 2 = 48$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 42257,
    "subject": "Mathematics (Multi-modal)",
    "question": "The smallest number bigger than $2015$ that is divisible by all of $2$, $3$, $4$, $5$ and $6$ is",
    "options": [],
    "answer": "2040",
    "solution": "$2040$\n\nIn order to be divisible by $2$, $3$, $4$, $5$ and $6$, the number only needs to be divisible by $2^2 \\times 3 \\times 5$, i.e. it must be a multiple of $60$. The smallest multiple of $60$ bigger than $2015$ is $2040$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42258,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$, and $C$ be distinct points on a straight line with $AB = AC = 1$.\nSquare $ABDE$ and equilateral triangle $ACF$ are drawn on the same side of line $BC$.\nLines $EC$ and $BF$ cut in $G$.\nWhat is the size of $\\angle EGB$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "75°",
    "solution": "We will compute the angle sizes of quadrilateral $DEGB$ and use the fact that the sum of the interior angles of a quadrilateral is equal to $360^\\circ$.\n\nSince $AFC$ is an equilateral triangle, $\\angle FAC = 60^\\circ$ and so $\\angle EAF = 90^\\circ - 60^\\circ = 30^\\circ$, which yields $\\angle BAE = 90^\\circ + 30^\\circ = 120^\\circ$. Since $AB = AC = AF$, triangle $BAF$ is isosceles, and so\n$$\n\\angle GBC = \\angle FBA = \\frac{1}{2}(180^\\circ - \\angle BAF) = \\frac{1}{2}(180^\\circ - 120^\\circ) = 30^\\circ.\n$$\nNext, $AE = AB = AC$, and so triangle $EAC$ is a right-angled isosceles triangle, which means that $\\angle BCG = \\angle ACE = 45^\\circ$. Hence,\n$$\n\\angle BGE = \\angle GBC + \\angle GCB = 30^\\circ + 45^\\circ = 75^\\circ.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42259,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is said that you can only fold a piece of paper in half 7 times. Harold folds a sheet of paper in half 5 times and then makes a hole in the folded paper. How many holes does the sheet of paper have after it is unfolded again?\n(A) 32\n(B) 36\n(C) 81\n(D) 50\n(E) 64",
    "options": [],
    "answer": "A",
    "solution": "Each fold doubles the number of layers that will be pierced. There will be $2^5$ layers and therefore $2^5 = 32$ holes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42260,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $\\frac{n+3}{13}$ is a positive integer, then the remainder when $n$ is divided by $13$ is",
    "options": [],
    "answer": "10",
    "solution": "10 Suppose $\\frac{n+3}{13} = k$; then $n = 13k - 3$ or $n = 13(k-1) + 10$. This shows that $n$ leaves remainder $10$ on division by $13$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42261,
    "subject": "Mathematics (Multi-modal)",
    "question": "The digits from $1$ to $9$ are added, in order, over and over again until the total is $460$.\n\n$1+2+3+4+5+6+7+8+9+1+2+3+\\ldots$\n\nThe last digit that was added is\n\n(A) 2   (B) 4   (C) 6   (D) 8   (E) 9",
    "options": [],
    "answer": "B",
    "solution": "The digits $1$ to $9$ total $45$, and $460 = 10 \\times 45 + 10$. So the complete set $1\\ldots9$ will appear ten times, and then as many more digits as total $10$, viz. $1+2+3+4$. So the last digit is $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42262,
    "subject": "Mathematics (Multi-modal)",
    "question": "The values of $x$ satisfying the equation $(x - 7)(x + 12) = -48$ are\n(A) 7 or 12\n(B) -7 or 12\n(C) 4 or -9\n(D) -4 or 9\n(E) -12 or 7",
    "options": [],
    "answer": "C",
    "solution": "Multiplying out the equation gives $x^2 + 5x - 84 = -48$, so $x^2 + 5x - 36 = 0$. This factorizes to $(x + 9)(x - 4) = 0$, so $x = -9$ or $x = +4$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 42263,
    "subject": "Mathematics (Multi-modal)",
    "question": "Alfred and four other people want to be in a group photograph. In how many different ways can they be arranged in a row with Alfred in the middle?\n(A) 4\n(B) 8\n(C) 12\n(D) 24\n(E) 40",
    "options": [],
    "answer": "D",
    "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": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 42264,
    "subject": "Mathematics (Multi-modal)",
    "question": "A rectangular sheet of paper can be used to form a cylinder by joining two opposite sides together:\n![](attached_image_1.png)\nShould the short edges or the long edges be joined together to obtain the largest volume of the cylinder?\nNB: Show all your working!",
    "options": [],
    "answer": "Join the short edges.",
    "solution": "Suppose that the rectangular sheet of paper has dimensions $a$ and $b$, with $b$ being the longer side. We calculate the volume of the two cylinders formed by joining the long sides and short sides, respectively.\n\n* Suppose the short sides are glued together. Then the height of the cylinder is $a$ and the circumference of the cylinder is $b$. If $r$ is the radius of the cylinder, it means that $2\\pi r = b$, or $r = \\frac{b}{2\\pi}$. Hence the cylinder's volume is $\\pi r^2 h = \\pi (\\frac{b}{2\\pi})^2 \\cdot a = \\frac{ab^2}{4\\pi}$.\n\n* Suppose the long sides are glued together. Then the height of the cylinder is $b$ and the circumference of the cylinder is $a$. If $r$ is the radius of the cylinder, it means that $2\\pi r = a$, or $r = \\frac{a}{2\\pi}$. Hence the cylinder's volume is $\\pi r^2 h = \\pi (\\frac{a}{2\\pi})^2 \\cdot b = \\frac{a^2b}{4\\pi}$.\n\nComparing these two numbers, we see that the first volume can be written as $b(\\frac{ab}{4\\pi})$, while the second volume is $a(\\frac{ab}{4\\pi})$. Since the number in brackets is the same for both and $b > a$, it means that the first volume $b(\\frac{ab}{4\\pi})$ is the largest. Hence the largest volume is obtained when the short sides of the sheet of paper are glued together.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 42265,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each interior angle of a regular 180 sided polygon is equal to\n(A) 175° (B) 180° (C) 157.5° (D) 120° (E) 178°",
    "options": [],
    "answer": "E",
    "solution": "It is easier to work with exterior angles, since the sum of the exterior angles of a polygon is $360°$. There are $180$ exterior angles, which are all equal since the polygon is regular. Thus each exterior angle is equal to $360°/180 = 2°$, and each interior angle is equal to $180° - 2° = 178°$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42266,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circle centre $O$ has radius $4$ cm. If the area of sector $OPQ$ is $\\frac{1}{2}\\pi$, and if the probability that a point chosen randomly in the circle lies in the shaded sector is $\\frac{1}{n}$, then the value of $n$ is\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "32",
    "solution": "The area of the whole circle is $\\pi \\cdot 4^2 = 16\\pi$, but the area of the sector is $\\frac{1}{2}\\pi$, so the probability is $\\frac{\\frac{1}{2}\\pi}{16\\pi} = \\frac{1}{32}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42267,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a \\bullet b$ means the value of $ab + a + b$, and $5 \\bullet x = 35$, the value of $x$ must be\n\n(A) 5\n(B) 7\n(C) 9\n(D) 12\n(E) 15",
    "options": [],
    "answer": "A",
    "solution": "By definition of $\\bullet$, we know $5 \\bullet x = 5x + 5 + x = 6x + 5$. If this is $35$, then $6x = 30$, so $x = 5$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42268,
    "subject": "Mathematics (Multi-modal)",
    "question": "As a decimal number $6.28\\%$ is equal to\n(A) $0.0628$ (B) $0.628$ (C) $6.28$ (D) $62.8$ (E) $628$",
    "options": [],
    "answer": "A",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42269,
    "subject": "Mathematics (Multi-modal)",
    "question": "The value of $1 + \\frac{1}{3 + \\frac{1}{2}}$ is\n(A) $\\frac{6}{5}$ (B) $\\frac{7}{6}$ (C) $\\frac{9}{2}$ (D) $\\frac{6}{7}$ (E) $\\frac{9}{7}$",
    "options": [],
    "answer": "E",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42270,
    "subject": "Mathematics (Multi-modal)",
    "question": "The tens digit of the product $1 \\times 2 \\times 3 \\times \\cdots \\times 98 \\times 99$ is\n(A) 0 (B) 1 (C) 2 (D) 4 (E) 9",
    "options": [],
    "answer": "A",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42271,
    "subject": "Mathematics (Multi-modal)",
    "question": "A palindromic number is a number that reads the same backwards as it does forwards.\nFor example: $15751$ and $909$.\n\na) What is the smallest palindromic number greater than $2016$?\n\nb) What is the largest palindromic number less than $2016$?",
    "options": [],
    "answer": "a) 2112; b) 2002",
    "solution": "a) $2112$, since the next smaller palindrome is $2002$, which is less than $2016$.\n\nb) $2002$, since $2112$ is the next palindrome.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42272,
    "subject": "Mathematics (Multi-modal)",
    "question": "The last digit when $8\\,045 - 4\\,058$ is calculated is\n(A) 1   (B) 3   (C) 5   (D) 7   (E) 9",
    "options": [],
    "answer": "D",
    "solution": "The last digit will be found by subtracting $8$ from $15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42273,
    "subject": "Mathematics (Multi-modal)",
    "question": "1. 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?\n\n2. If $2016 \\times x$ is a perfect square, what is the smallest positive integer value for $x$?",
    "options": [],
    "answer": "25%; 14",
    "solution": "1.\n$$\n\\frac{10}{40} \\times 100 = 25.\n$$\nSo, $25\\%$ of the bulbs are blue.\n\n2.\nThe prime factorisation of $2016$ is $2^5 \\times 3^2 \\times 7$. In a perfect square, each prime factor has an even exponent, so we need one more power of $2$ and one more power of $7$. The answer is $2 \\times 7 = 14$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42274,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the value of $2 + 2 \\times 3^2$",
    "options": [],
    "answer": "20",
    "solution": "The value is $2 + 2 \\times 9 = 2 + 18 = 20$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42275,
    "subject": "Mathematics (Multi-modal)",
    "question": "$\\frac{3 \\times 2016 + 13 \\times 2016}{1008} =$\n\n(A) 2   (B) 16   (C) 32   (D) $19 \\times 2016$   (E) $6 + 13 \\times 2016$",
    "options": [],
    "answer": "C",
    "solution": "$\\frac{3 \\times 2016 + 13 \\times 2016}{1008} = \\frac{(3 + 13) \\times 2016}{1008} = (3 + 13) \\times \\frac{2016}{1008} = 16 \\times 2 = 32.$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42276,
    "subject": "Mathematics (Multi-modal)",
    "question": "Together, the two positive integers $a$ and $b$ have 9 digits and contain each of the digits $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$ exactly once. For which possible values of $a$ and $b$ is the fraction $a/b$ closest to $1$?",
    "options": [],
    "answer": "a = 9876, b = 12345",
    "solution": "If $a > b$, then $a$ has at least five digits, so $a \\ge 12345$, and $b$ has at most four digits, so $b \\le 9876$. In this case, we have\n$$\n\\frac{a}{b} \\ge \\frac{12345}{9876} > 1,\n$$\nso the value of $a/b$ that is closest to $1$ is\n$$\n\\frac{12345}{9876} = 1 + \\frac{2469}{9876}\n$$\nin this case.\n\nOn the other hand, if $a < b$ ($a = b$ is impossible, since $a$ and $b$ cannot have the same number of digits), then $a$ has at most four digits and $b$ at least five, so $a \\le 9876$ and $b \\ge 12345$, which means that\n$$\n\\frac{a}{b} \\le \\frac{9876}{12345} < 1.\n$$\nHence in this case the value that is closest to $1$ is\n$$\n\\frac{9876}{12345} = 1 - \\frac{2469}{12345}.\n$$\nSince the denominator $12345$ is greater than the denominator $9876$, we see that the value of $9876/12345$ is closer to $1$ than that of $12345/9876$ (in fact, we have $9876/12345 = 4/5$ and $12345/9876 = 5/4$, so the distances are $1/5$ and $1/4$ respectively). Thus the values of $a$ and $b$ for which $a/b$ is closest to $1$ are $a = 9876$ and $b = 12345$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42277,
    "subject": "Mathematics (Multi-modal)",
    "question": "One horse eats $40\\%$ of a bale of hay and another horse eats $P\\%$ of what is left. If both horses ate the same amount, the value of $P$ is\n\n(A) $43$\n(B) $66\\frac{2}{3}$\n(C) $50$\n(D) $75$\n(E) $80$",
    "options": [],
    "answer": "B",
    "solution": "The second horse eats $P\\%$ of $60\\%$ and this is the same as $40\\%$: so $P\\%$ is $\\frac{2}{3}$, i.e. $P = 66\\frac{2}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42278,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the school library there are $1\\,024$ books of which $64$ are detective stories. Twenty-four of these detective stories are about Sherlock Holmes. What is the probability that a book you selected randomly is a detective story which is not about Sherlock Holmes?\n\n(A) $\\frac{1}{16}$ (B) $\\frac{3}{8}$ (C) $\\frac{3}{128}$ (D) $\\frac{5}{128}$ (E) $\\frac{1}{24}$",
    "options": [],
    "answer": "D",
    "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 $1\\,024$ books is $\\frac{40}{1\\,024} = \\frac{5}{128}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42279,
    "subject": "Mathematics (Multi-modal)",
    "question": "If the difference between two prime numbers is also prime, what is the smallest value of the sum of those two primes?\n(A) 9\n(B) 7\n(C) 6\n(D) 4\n(E) 3",
    "options": [],
    "answer": "B",
    "solution": "Looking at the difference of successive pairs of small primes, we note that $3 - 2 = 1$ which is not prime, $5 - 2 = 3$ and $5 - 3 = 2$, and then larger numbers differing by $2$ (or more). So the smallest sum comes from $5 + 2 = 7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42280,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the difference between the largest and the second largest odd factors of $2016$?",
    "options": [],
    "answer": "42",
    "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": 42281,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be the product of any three consecutive positive odd integers. What is the highest common factor (greatest common divisor) of all such numbers $P$?",
    "options": [],
    "answer": "3",
    "solution": "For any three consecutive odd integers, exactly one is divisible by $3$, so $3$ is a common factor of all such numbers $P$. Even the first two non-overlapping values $P = 1 \\times 3 \\times 5$ and $P = 7 \\times 9 \\times 11$ have highest common factor $3$, which is therefore the highest common factor of all values of $P$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42282,
    "subject": "Mathematics (Multi-modal)",
    "question": "Zola and Ron play a game by alternately moving a single ten cent coin on a circular board. The game starts with the ten cent coin already on the board as shown. A player may move the coin either clockwise one position or one position toward the centre, but cannot move to a position that has been previously occupied.\n\nThe last person who is able to move wins the game.\n\nIf Zola starts, which player can play in a way that guarantees a win?\n\nExplain this player's winning strategy.",
    "options": [],
    "answer": "Zola",
    "solution": "Since the game must end (finite number of blocks and can't play on a previously occupied block) and only one person can move last, there will be a winner and hence a winning strategy.\n\nWe claim that the first player (Zola) has the winning strategy and it is to always move clockwise. There are $7$ positions left in the outer ring so only player $1$ (Zola) can complete the outer ring and then player $2$ (Ron) will be forced to go inwards. Note at any stage player $2$ (Ron) can opt to go inwards, but if he moves inwards, there will be $7$ positions left in that ring and again only player $1$ (Zola) can complete that ring. This is the same for each ring. Hence only player $1$ (Zola) can complete the inner ring and move last and hence win.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42283,
    "subject": "Mathematics (Multi-modal)",
    "question": "Water pours into a tank at a constant rate. After 2 hours the tank was 36% full and after 10 hours it is 100% full. How full was the tank to begin with?\n(A) 10%\n(B) 12%\n(C) 15%\n(D) 18%\n(E) 20%",
    "options": [],
    "answer": "E",
    "solution": "In 8 hours (from 2 hours to 10 hours), the tank filled up a further $64\\%$ (from $36\\%$ to $100\\%$). That is $8\\%$ per hour.\n\nSo in the first two hours it filled up $16\\%$, meaning it started with $36 - 16 = 20\\%$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42284,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are twice as many girls as boys at a school. If $30\\%$ of the girls and $45\\%$ of the boys have already completed their holiday project, what percentage of the learners still needs to complete their project?",
    "options": [],
    "answer": "65%",
    "solution": "For convenience assume there are $300$ learners, i.e. $200$ girls and $100$ boys.\n\n$30\\%$ of $200$ is $60$ and $45\\%$ of $100$ is $45$.\n\nThus, $60 + 45 = 105$ learners have completed their project.\n\nThe percentage of learners who still need to complete their project is thus\n$$\n\\frac{300 - 105}{300} \\times 100 = \\frac{195}{3} = 65.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42285,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two sequences of real numbers are defined as follows:\n$$\nu_1 = 0, \\quad u_{n+1} = \\frac{1}{2}(u_n + v_n)$$\n$$v_1 = 1, \\quad v_{n+1} = \\frac{1}{4}(u_n + 3v_n)$$\nFind the value of $v_{2016} - u_{2016}$.",
    "options": [],
    "answer": "1/4^{2015}",
    "solution": "$u_1 = 0$\n$v_1 = 1$\n$$\nu_{n+1} = \\frac{1}{2}(u_n + v_n)$$\n$$v_{n+1} = \\frac{1}{4}(u_n + 3v_n)$$\n$$\n\\begin{aligned}\nv_{2016} - u_{2016} &= \\frac{1}{4}(u_{2015} + 3v_{2015}) - \\frac{1}{2}(u_{2015} + v_{2015}) \\\\\n&= \\frac{1}{4}v_{2015} - \\frac{1}{4}u_{2015} \\\\\n&= \\frac{1}{4}(v_{2015} - u_{2015}) \\\\\n&= \\frac{1}{4}\\left(\\frac{1}{4}(v_{2014} - u_{2014})\\right) \\\\\n&= \\frac{1}{4}\\left(\\frac{1}{4}\\left(\\frac{1}{4}\\right)^{2013}(v_1 - u_1)\\right) \\\\\n&= \\frac{1}{4^{2015}}(v_1 - u_1) \\\\\n&= \\frac{1}{4^{2015}}(1 - 0) \\\\\n&= \\frac{1}{4^{2015}}.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42286,
    "subject": "Mathematics (Multi-modal)",
    "question": "$\\left(\\frac{1}{8}\\right)^2 \\div \\left(\\frac{1}{2}\\right)^8$ is equal to\n(A) 2\n(B) 4\n(C) 8\n(D) 16\n(E) 32",
    "options": [],
    "answer": "B",
    "solution": "$\\frac{\\left(\\frac{1}{8}\\right)^2}{\\left(\\frac{1}{2}\\right)^8} = \\frac{\\frac{1}{64}}{\\frac{1}{256}} = \\frac{1}{64} \\times \\frac{256}{1} = 4$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42287,
    "subject": "Mathematics (Multi-modal)",
    "question": "At a height of *h* metres above sea level, the distance $d$, in kilometres, to the visible horizon is given approximately by the formula $d = 8\\sqrt{\\frac{h}{5}}$. A mountaineer on the side of Table Mountain can see a distance of $80$ kilometres to the horizon. How high, in metres, is the mountaineer above sea level?\n\n(A) 1 000   (B) 500   (C) 650   (D) 250   (E) 400",
    "options": [],
    "answer": "B",
    "solution": "Since $d = 8\\sqrt{\\frac{h}{5}}$, it follows that $d^2 = \\frac{64h}{5}$, so $h = \\frac{5}{64}d^2$. With $d = 80$, this gives $d^2 = 6400$, so $h = \\frac{5}{64} \\times 6400 = 500$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42288,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the figure, $ABCD$ is a rectangle.\nThe area of triangle $DEC$ is $9\\ \\text{cm}^2$ and $BE = \\frac{2}{5} BC$.\nWhat is the area of $ABCD$ in $\\text{cm}^2$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "30",
    "solution": "Let $BE = 2x$; then with $BE = \\frac{2}{5} BC$ we must have $EC = 3x$.\nNow for $\\triangle DEC$, $\\frac{1}{2}(3x)(DC) = 9$, giving $DC = \\frac{6}{x}$, and since $BC = 5x$,\nthe area of rectangle $ABCD$ is $5x \\cdot \\frac{6}{x} = 30$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42289,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $f(x) = 2^x$, what is the value of $\\frac{f(x+1) + f(x)}{f(x)}$?",
    "options": [],
    "answer": "3",
    "solution": "$$\n\\frac{f(x + 1) + f(x)}{f(x)} = \\frac{f(x + 1)}{f(x)} + 1 = \\frac{2^{x+1}}{2^x} + 1 = 2^1 + 1 = 3.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42290,
    "subject": "Mathematics (Multi-modal)",
    "question": "Mrs. Habana turns $45$ years old in $2016$ and her son $17$. In which year will Mrs. Habana turn double her son's age?\n(A) $2033$ (B) $2044$ (C) $2031$ (D) $2040$ (E) $2027$",
    "options": [],
    "answer": "E",
    "solution": "Mrs Habana will be $45 + x$ years old and her son will be $17 + x$ old where $x$ is the number of years from $2016$. Therefore $45 + x = 2(17 + x)$, giving $45 + x = 34 + 2x$, so $x = 45 - 34 = 11$. The year will then be $2016 + 11 = 2027$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42291,
    "subject": "Mathematics (Multi-modal)",
    "question": "Boris takes a taxi to his home but falls asleep when the taxi is halfway to his house. He wakes up when the remaining part of his journey is equal to half the distance covered by the taxi while he was asleep.\nIf the fraction of the journey for which he slept is $\\frac{1}{n}$, what is the value of $n$?",
    "options": [],
    "answer": "3",
    "solution": "3 Assume that the remaining distance home is 1 unit. Then he slept for 2 units. Hence, the whole journey is 6 units and the fraction of the ride that he slept for is $\\frac{2}{6} = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42292,
    "subject": "Mathematics (Multi-modal)",
    "question": "A certain pen can draw a line $0.4$ mm wide and $1$ km long before it runs out of ink. The largest square you can colour in with this pen has an area, in square metres, of\n(A) $4\\,000$   (B) $400$   (C) $40$   (D) $4$   (E) $0.4$",
    "options": [],
    "answer": "E",
    "solution": "Since $0.4 \\text{ mm} = 0.4 \\times 10^{-3} \\text{ m}$ and $1 \\text{ km} = 10^3 \\text{ m}$, it follows that the area of the line is $(0.4 \\times 10^{-3}) \\times 10^3 = 0.4 \\text{ m}^2$, and the area of the square will be the same.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42293,
    "subject": "Mathematics (Multi-modal)",
    "question": "$1 \\div 13$ is a recurring decimal that begins $0.076923076923076923\\ldots$. The 100th digit after the decimal comma is\n(A) $0$\n(B) $2$\n(C) $6$\n(D) $7$\n(E) $9$",
    "options": [],
    "answer": "E",
    "solution": "There are $6$ digits in the repeating part of the decimal form. $100 = 6 \\times 16 + 4$, so the $100$th digit will be the $4$th in the repeating part $076923$, which is $9$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 42294,
    "subject": "Mathematics (Multi-modal)",
    "question": "14. I accidentally decreased a number by $60\\%$ instead of increasing it by $60\\%$. This incorrect value now needs to be increased by $k\\%$ to get to the correct value. What is the value of $k$?\n\n15. 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": "k = 300; adopted child's present age = 11",
    "solution": "14. $300$\nI ended up with $40\\%$ of the number instead of $160\\%$. So I need to multiply this new result by $4$, or add it three times to itself, which is an increase of $300\\%$.\n\n15. $11$\n2 years ago the sum of all the family's ages was $4 \\times 19 = 76$. That should have increased by $2 \\times 4 = 8$, but has actually become $5 \\times 19 = 95$, i.e. increased by $19$. So the new person is now $19 - 8 = 11$ years old.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42295,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are infinitely many terms of the arithmetic sequence $1, 14, 27, 40, \\dots, 1+13k, \\dots$ which are of the form $222\\ldots22$. In other words a number that is made up using only the digit $2$.\n\n(Hint: $1001 = 7 \\times 11 \\times 13$)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $221 = 13 \\times 17$ and therefore $222 = 13 \\times 17 + 1$ is in the arithmetic sequence. Furthermore, $222\\,222\\,221$ is divisible by $13$ and hence $222\\,222\\,222$ is in the sequence, since\n$$\n222222 = 222 \\times 1001 = 222 \\times 7 \\times 11 \\times 13.\n$$\nIn fact, any number consisting of $6m + 3$ $2$s will be one more than a multiple of $13$, since the number consisting of $6m$ $2$s will be a multiple of $1001$ and hence a multiple of $13$. Thus there are infinitely many terms in the sequence that only uses the digit $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42296,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle RUG$, $RU = 10$ cm, $UG = 21$ cm, $RG = 17$ cm and $RY$ is perpendicular to $UG$ at point $Y$. The length of $RY$, in cm, is\n![](attached_image_1.png)\n(A) 8 (B) 5 (C) 7 (D) 6 (E) 8.1",
    "options": [],
    "answer": "A",
    "solution": "Let $h = RY$ and $x = UY$, so $GY = 21 - x$. By Pythagoras' theorem, $h^2 = 10^2 - x^2$ and also $h^2 = 17^2 - (21-x)^2$, so $100 - x^2 = 289 - 441 + 42x - x^2$. This gives $42x = 100 - 289 + 441 = 252$ and therefore $x = 252 \\div 42 = 6$. Finally, $h^2 = 100 - 6^2 = 64$, so $h = 8$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42297,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the minimum number of integers which must be removed from the first 50 positive even integers so that the sum of the remaining integers is $2016$?",
    "options": [],
    "answer": "6",
    "solution": "The first $50$ even positive integers are $2, 4, 6, \\ldots, 96, 98, 100$, and their sum is $50 \\times \\frac{1}{2}(2 + 100) = 50 \\times 51 = 2550$. To reduce the sum to $2016$ we must subtract $534$. The average of the numbers removed must be less than $100$, so we need to remove at least six. To use as few numbers as possible, we first remove the five largest numbers from $100$ down. Now $100 + 98 + 96 + 94 + 92 = 480$, which is nearly there, so to bring the total subtracted to $534$ we finally need to remove $54$.\n\n[There are, of course, many other combinations of five integers with a sum of $534$ that can be removed. ]",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42298,
    "subject": "Mathematics (Multi-modal)",
    "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 on the field (11 players per team) were all the numbers from $1$ to $22$. At half-time, the coach substituted her goal-keeper (who had the number $1$ on her jersey) for a reserve player. The coach then noticed that after the substitution, no two players on the field had the same jersey number and that the sum of the jersey numbers of each of the teams were exactly equal.\n\na) What is the smallest (positive) possible jersey number of the reserve player?\n\nb) What is the greatest (positive) possible jersey number of the reserve player?\n\n[10]",
    "options": [],
    "answer": "a) 24; b) 122",
    "solution": "The sum from $2$ to $22$ is $2 + 3 + \\cdots + 22 = \\frac{22(23)}{2} - 1 = 11 \\times 23 - 1 = 252$.\n\nTherefore the new number must be even or otherwise the sum can't be exactly divisible by $2$.\n\nThe smallest possible even number we can use is $24$. To see that this is indeed possible take\n$$\n24 + 22 + 21 + 20 + 19 + 12 + 6 + 5 + 4 + 3 + 2 = 138 = 18 + 17 + 16 + 15 + 14 + 13 + 11 + 10 + 9 + 8 + 7.\n$$\n\nTo find the greatest number possible, we try and put the $10$ smallest numbers with the greatest number. The sum of the $11$ numbers from $12$ to $22$ is\n$$\n12 + 13 + \\cdots + 22 = \\frac{22(23)}{2} - \\frac{11(12)}{2} = 11 \\times 23 - 11 \\times 6 = 11 \\times 17 = 187.\n$$\n\nHence the largest total can be $187$. Subtracting the numbers $2$ to $11$ from this gives us\n$$\n187 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 = 122,\n$$\nwhich is an even number. Thus $122$ is the largest possible jersey number to achieve this.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 42299,
    "subject": "Mathematics (Multi-modal)",
    "question": "Devious Steve either lies for the whole day or tells the truth for the whole day. Which one of the following statements can he never say?\n(A) \"Yesterday I told the truth\"\n(B) \"Yesterday I lied\"\n(C) \"Today I am telling the truth\"\n(D) \"Today I am lying\"\n(E) \"Tomorrow I shall lie\"",
    "options": [],
    "answer": "D",
    "solution": "Statements A, B and E can be made irrespective of whether Devious Steve is lying or telling the truth. Statement C can also be made whether he is telling the truth or lying. However, if statement D is true, then Steve is a liar, which contradicts the truth of statement D. Similarly if statement D is false, then Steve is a truth-teller, which contradicts the falsity of statement D.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42300,
    "subject": "Mathematics (Multi-modal)",
    "question": "$OABC$ is a rectangle of length $16$ cm. $M$ is the midpoint of $CB$ and $P$ lies on $AB$ so that $BP = 3$ cm. The length $OM$ is $17$ cm. The length of $OP$, in cm, is\n![](attached_image_1.png)\n(A) $18$\n(B) $19$\n(C) $20$\n(D) $21$\n(E) $22$",
    "options": [],
    "answer": "C",
    "solution": "$CM = MB = 8$. Then by Pythagoras in $\\triangle OMC$, $OC = 15$. But then $AP = 15 - 3 = 12$, and Pythagoras in $\\triangle OAP$ shows $OP = 20$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42301,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence of $2016$ terms is constructed as follows: The first two terms of the sequence are both equal to $3$. Starting from the third term, each subsequent term is the sum of the preceding two terms. Each of the terms of this sequence $3$, $3$, $6$, $9$, $...$ is now divided by $2$ and the remainders are added. What is the sum of all the remainders of the $2016$ terms?\n(A) $504$ (B) $1\\,008$ (C) $1\\,344$ (D) $1\\,512$ (E) $2\\,016$",
    "options": [],
    "answer": "C",
    "solution": "The first six terms of the sequence are $3$, $3$, $6$, $9$, $15$, $24$, $...$, which are Odd, Odd, Even, Odd, Odd, Even, $...$, since Odd $+$ Odd $=$ Even and Odd $+$ Even $=$ Odd. The pattern continues in the same way in cycles of length $3$, with two odd numbers and one even number in each cycle. After division by $2$, the remainders in each cycle are $1$, $1$, $0$, so the sum of the three remainders is $2$. The sum of the first $2016$ remainders is therefore $\\frac{2}{3} \\times 2016 = 1344$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42302,
    "subject": "Mathematics (Multi-modal)",
    "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": "11",
    "solution": "2 years ago the sum of all the family's ages was $4 \\times 19 = 76$. That should have increased by $2 \\times 4 = 8$, but has actually become $5 \\times 19 = 95$, i.e. increased by $19$. So the new person is now $19 - 8 = 11$ years old.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42303,
    "subject": "Mathematics (Multi-modal)",
    "question": "A box contains $5$ red pens, $6$ blue pens and $4$ green pens. Three pens are consecutively taken from the box, at random and without replacement. The probability that the first pen is red, the second pen blue and the third pen green, is\n(A) $\\frac{7}{225}$ (B) $\\frac{1}{5}$ (C) $\\frac{9}{82}$ (D) $\\frac{4}{91}$ (E) $\\frac{3}{14}$",
    "options": [],
    "answer": "D",
    "solution": "There are $5 + 6 + 4 = 15$ pens in the box, so the probability of first picking a red pen is $\\frac{5}{15} = \\frac{1}{3}$. This leaves $14$ pens (since the red pen is not replaced), and the probability of next picking a blue pen is $\\frac{6}{14} = \\frac{3}{7}$. Finally, the probability of picking a green pen is $\\frac{4}{13}$. The events are not independent, and the probability is $\\frac{1}{3} \\times \\frac{3}{7} \\times \\frac{4}{13} = \\frac{4}{7 \\times 13} = \\frac{4}{91}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42304,
    "subject": "Mathematics (Multi-modal)",
    "question": "A palindrome is a whole number that reads the same forwards and backwards (e.g. $47274$). How many $3$-digit palindromes are there?\n(A) $95$ (B) $90$ (C) $85$ (D) $80$ (E) $70$",
    "options": [],
    "answer": "B",
    "solution": "There are $9$ ways to select the nonzero hundreds' digit, then $1$ way to select the units' digit (same as the hundreds' digit), and then $10$ ways to select the tens' digit. Hence there are $9 \\times 1 \\times 10 = 90$ three-digit palindromes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42305,
    "subject": "Mathematics (Multi-modal)",
    "question": "$D$ is a point on side $AB$ of $\\triangle ABC$, $E$ is a point on $CD$ and $F$ is a point on $CE$. The areas of triangles $AED$, $AEC$, $BFD$ and $BFC$ are $6$, $10$, $17$ and $7$, respectively. What is the area of $\\triangle BEF$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "8",
    "solution": "If two triangles have the same height, then the ratio of their areas is equal to the ratio of their bases. It follows that $\\frac{DE}{EC} = \\frac{6}{10} = \\frac{3}{5}$, so $\\frac{DE}{DC} = \\frac{3}{3+5} = \\frac{3}{8}$. Similarly, $\\frac{DF}{FC} = \\frac{17}{7}$, so $\\frac{DF}{DC} = \\frac{17}{17+7} = \\frac{17}{24}$. Therefore $\\frac{EF}{DC} = \\frac{DF - DE}{DC} = \\frac{17}{24} - \\frac{3}{8} = \\frac{17-9}{24} = \\frac{8}{24} = \\frac{1}{3}$. It follows that $\\triangle BEF = \\frac{1}{3}\\triangle BDC = \\frac{1}{3}(17+7) = 8$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42306,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $169! = 1 \\times 2 \\times 3 \\times \\cdots \\times 169$ is written as the product of prime numbers, how many times would $13$ appear as a factor?\n(A) 12 (B) 13 (C) 14 (D) 15 (E) 16",
    "options": [],
    "answer": "C",
    "solution": "Since $13$ is a prime number, the only factors in $169!$ supplying powers of $13$ to the product are $13$, $26$, $39$, $\\ldots$, $156$, $169$. The first $12$ of these provide one power of $13$ each, but since $169 = 13^2$, it follows that the total power of $13$ is $12 + 2 = 14$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42307,
    "subject": "Mathematics (Multi-modal)",
    "question": "A and $B$ are opposite vertices of a regular hexagon. $C$ and $D$ are midpoints of two opposite sides. If the area of the hexagon is $126$, then $AB \\times CD$ is\n![](attached_image_1.png)\n(A) $129$ (B) $132$ (C) $84\\sqrt{3}$ (D) $168$ (E) $248$",
    "options": [],
    "answer": "D",
    "solution": "By joining opposite vertices, the hexagon can be divided into six congruent equilateral triangles. If we form a rectangle around the hexagon by drawing lines through $A$ and $B$ parallel to $CD$, then the area of the rectangle is $AB \\times CD$. Next, the portion of the rectangle outside the hexagon is composed of four right-angled triangles, which can be combined into two equilateral triangles congruent to the first six. Thus the area of the rectangle is $\\frac{8}{6} \\times 126 = 168$, which is also equal to $AB \\times CD$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42308,
    "subject": "Mathematics (Multi-modal)",
    "question": "All my Maths tests are out of the same total, and the average of my marks in the four tests so far is $75\\%$. What percentage do I need in a fifth test so that my overall average will be $80\\%$?\n\n(A) $100$\n(B) $90$\n(C) $85$\n(D) $80$\n(E) $75$",
    "options": [],
    "answer": "A",
    "solution": "Let the five marks be $a$, $b$, $c$, $d$ and $e$.\n\n$$\n\\frac{a+b+c+d+e}{5} = 80 \\therefore a+b+c+d+e = 400\n$$\nBut $\\frac{a+b+c+d}{4} = 75 \\therefore a+b+c+d = 300$, and so $e = 100$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42309,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the remainder when $2^{2016}$ is divided by $13$?",
    "options": [],
    "answer": "1",
    "solution": "(We do this by inspection, by trying out some values until we can see the pattern.) Draw up a list of the remainders left by the powers of $2$ after division by $13$:\n\n| $n$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |\n|-----|---|---|---|---|---|---|----|----|----|----|----|----|----|\n| $2^n \\bmod 13$ | 1 | 2 | 4 | 8 | 3 | 6 | 12 | 11 | 9 | 5 | 10 | 7 | 1 |\n\nWe see that the remainders repeat every $12$ terms, since $2^{12} \\equiv 1 \\pmod{13}$.\n\nNow, $2016 \\div 12 = 168$ with remainder $0$, so $2016 = 12 \\times 168 + 0$.\n\nTherefore,\n$$\n2^{2016} \\equiv 2^0 \\equiv 1 \\pmod{13}.\n$$\n\nSo the remainder is $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42310,
    "subject": "Mathematics (Multi-modal)",
    "question": "$\\left(\\frac{1}{8}\\right)^2 \\times 2^8$ is equal to\n(A) 2\n(B) 4\n(C) 8\n(D) 16\n(E) 32",
    "options": [],
    "answer": "B",
    "solution": "$\\left(\\frac{1}{8}\\right)^2 \\times 2^8 = \\frac{1}{64} \\times 256 = 4$\n\nOR\n\n$\\left(\\frac{1}{2^3}\\right)^2 \\times 2^8 = \\frac{1}{64} \\times 2^8 = 2^2 = 4$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42311,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many grams of pure gold must be added to five grams of a mixture that consists of 10% gold and 90% of some other metal so that the mixture will contain 20% gold?\n(A) 0.125 (B) 0.25 (C) 0.375 (D) 0.5 (E) 0.625",
    "options": [],
    "answer": "E",
    "solution": "The initial $5$ g of mixture contains $10\\% \\times 5 = 0.5$ g of gold. If we add $x$ g of pure gold, then the mass of gold is $(0.5 + x)$ g and the total mass is $(5 + x)$ g. Therefore $0.5 + x = 20\\% \\times (5 + x) = 1 + 0.2x$, so $0.8x = 0.5$ and $x = 5/8 = 0.625$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42312,
    "subject": "Mathematics (Multi-modal)",
    "question": "A target consists of two circles with the same centre. The larger circle has radius $12$ cm. When shots hit the target randomly, $403$ out of $900$ land in the inner circle. The radius of the inner circle is closest to\n![](attached_image_1.png)\n(A) $6$ cm\n(B) $7$ cm\n(C) $8$ cm\n(D) $9$ cm\n(E) $10$ cm",
    "options": [],
    "answer": "C",
    "solution": "If the radius is $r$ cm then\n$$\n\\frac{\\text{area smaller circle}}{\\text{area larger circle}} = \\frac{\\pi r^2}{\\pi \\cdot 12^2} = \\frac{r^2}{12^2} = \\frac{403}{900} \\approx \\frac{4}{9}\n$$\nSo $r^2$ is close to $\\frac{2^2 \\cdot 12^2}{3^2}$ which means $r$ is $\\frac{2 \\cdot 12}{3} = 8$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42313,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs of real numbers $a$ and $b$, $b > 0$, such that the solutions to the two equations\n$$\nx^2 + a x + a = b\n$$\nand\n$$\nx^2 + a x + a = -b\n$$\nare four consecutive integers.",
    "options": [],
    "answer": "a = -1, b = 1 and a = 5, b = 1",
    "solution": "The quadratic formula gives us\n$$\n\\frac{-a \\pm \\sqrt{a^2 - 4a + 4b}}{2}\n$$\nand\n$$\n\\frac{-a \\pm \\sqrt{a^2 - 4a - 4b}}{2}\n$$\n\nSuppose that the four consecutive numbers are $n - 1, n, n + 1, n + 2$. The parabola $y = x^2 + a x + a$ reaches its minimum at $x = -a/2$, and the line $x = -a/2$ is its axis of symmetry. The two pairs of solutions both have to have this axis of symmetry, and the solutions to the second equation have to lie closer to the minimum at $x = -a/2$. Thus the only possibility is that $n - 1$ and $n + 2$ are the solutions to the first equation, while $n$ and $n + 1$ are the solutions to the second equation. This gives us the equations\n$$\n(n - 1)^2 + a(n - 1) + a = b, \\quad (1)\n$$\n$$\nn^2 + a n + a = -b, \\quad (2)\n$$\n$$\n(n + 1)^2 + a(n + 1) + a = -b. \\quad (3)\n$$\nSubtract (1) from (3) to obtain\n$$\n4n + 2a = -2b. \\quad (4)\n$$\nSubtract (2) from (3) to obtain\n$$\n2n + 1 + a = 0. \\quad (5)\n$$\nMultiply (5) by 2 and subtract from (4):\n$$\n-2 = -2b,\n$$\nso $b = 1$. Now we know that the two roots of $x^2 + a x + a = 1$ differ by 3:\n$$\n\\frac{-a + \\sqrt{a^2 - 4a + 4}}{2} = \\frac{-a - \\sqrt{a^2 - 4a + 4}}{2} + 3,\n$$\nso\n$$\n\\sqrt{a^2 - 4a + 4} = \\sqrt{(a - 2)^2} = 3,\n$$\nwhich means that $a - 2 = \\pm 3$. Once again, we find that the two possibilities are $a = -1, b = 1$ and $a = 5, b = 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 42314,
    "subject": "Mathematics (Multi-modal)",
    "question": "The value of $\\sqrt{2 \\times 0 \\times 1 \\times 6}$ is\n(A) 6 (B) 3 (C) 2 (D) 1 (E) 0",
    "options": [],
    "answer": "E",
    "solution": "$2 \\times 0 \\times 1 \\times 6 = 0$, and $\\sqrt{0} = 0$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42315,
    "subject": "Mathematics (Multi-modal)",
    "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 in such a way 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": "sqrt(65)",
    "solution": "![](attached_image_1.png)\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$$\nx^2 + (5-x)^2 = 13,\n$$\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$$\n7^2 + y^2 = 8^2 + (5-y)^2.\n$$\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": 42316,
    "subject": "Mathematics (Multi-modal)",
    "question": "All four-digit positive integers which are rearrangements of the number $2316$ are written in increasing order. What is the largest possible difference between two adjacent numbers in this list?",
    "options": [],
    "answer": "2502",
    "solution": "$6123$ is the smallest number starting with a $6$ and $3621$ is the largest number starting with a $3$. The difference between these is $6123 - 3621 = 2502$. Adjacent numbers between $1236$ and $3621$ will be less than $2500$ apart, while adjacent numbers larger than $6123$ will be less than $1000$ apart. Thus $2502$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42317,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the value of $\\frac{13 \\times 13 \\times 12}{1 + 13 + 13 + 12}$",
    "options": [],
    "answer": "52",
    "solution": "The fraction is $\\frac{13 \\times 13 \\times 12}{39} = \\frac{13 \\times 13 \\times 12}{3 \\times 13} = \\frac{13 \\times 12}{3} = 13 \\times 4 = 52$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42318,
    "subject": "Mathematics (Multi-modal)",
    "question": "Which term must be removed from the sequence $\\frac{1}{2}$, $\\frac{1}{4}$, $\\frac{1}{6}$, $\\frac{1}{10}$ and $\\frac{1}{12}$ so that the sum of the remaining terms is equal to 1?\n(A) $\\frac{1}{2}$ (B) $\\frac{1}{4}$ (C) $\\frac{1}{6}$ (D) $\\frac{1}{10}$ (E) $\\frac{1}{12}$",
    "options": [],
    "answer": "D",
    "solution": "If all terms are brought to $60$, the least common multiple (LCM), then all the numerators except $6$ have a factor $5$. Since the required numerator is $60$, which also has a factor $5$, the term that must be dropped is $\\frac{1}{10}$.\n\nAlternatively,\n$$\n\\frac{1}{2} + \\frac{1}{4} + \\frac{1}{6} + \\frac{1}{10} + \\frac{1}{12} = \\frac{30 + 15 + 10 + 6 + 5}{60} = \\frac{66}{60} = 1 + \\frac{1}{10}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42319,
    "subject": "Mathematics (Multi-modal)",
    "question": "For two positive real numbers $a$ and $b$, which may be equal, what is the smallest possible value of $\\frac{a}{b} + \\frac{b}{a}$?",
    "options": [],
    "answer": "2",
    "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\nLet $x = \\frac{a}{b}$, so $x > 0$. Then $\\frac{a}{b} + \\frac{b}{a} = x + \\frac{1}{x}$.\n\nBy the AM-GM inequality:\n\n$$\nx + \\frac{1}{x} \\geq 2\\sqrt{x \\cdot \\frac{1}{x}} = 2\\sqrt{1} = 2.\n$$\n\nEquality holds when $x = 1$, i.e., $a = b$.\n\nTherefore, the smallest possible value is $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42320,
    "subject": "Mathematics (Multi-modal)",
    "question": "For how many positive three-digit numbers is the hundreds digit smaller than the units digit?",
    "options": [],
    "answer": "360",
    "solution": "360\n\nSystematic counting soon reveals a pattern.\n\nIn the 900s there are clearly no cases.\n\nIn the 800s there are 10: 809, 819, 829, ..., 899.\n\nIn the 700s there are 20: 708, 718, 728, ..., 798\n709, 719, 729, ..., 799.\n\nIn the 600s there are 30: 607, 617, 627, ..., 697\n608, 618, 628, ..., 698\n609, 619, 629, ..., 699.\n\nThis pattern continues until the 100s, where there are 80.\n\nThe desired total is thus $0 + 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 = 360$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42321,
    "subject": "Mathematics (Multi-modal)",
    "question": "The inscribed circle of triangle $ABC$, with centre $I$, touches sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$, respectively. Let $P$ be a point, on the same side of $FE$ as $A$, for which $\\angle PFE = \\angle BCA$ and $\\angle PEF = \\angle ABC$. Prove that $P$, $I$ and $D$ lie on a straight line.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nSince $\\angle AEI = \\angle AFI = 90^\\circ$, the points $A$, $E$, $F$, $I$ lie on a circle with diameter $AI$. Moreover, since $\\angle PFE = \\angle BCA$ and $\\angle PEF = \\angle ABC$ by our assumptions on $P$, triangles $ABC$ and $PEF$ are similar, so $\\angle EPF = \\angle BAC = \\angle EAF$. $P$ was assumed to lie on the same side of $EF$ as $A$, thus it follows that $P$ also lies on the same circle as $A$, $E$, $F$ and $I$.\nWe also know that $BDIF$ is a cyclic quadrilateral (using the same reasoning as before, namely that $\\angle BDI = \\angle BFI = 90^\\circ$), so $\\angle FID + \\angle FBD = 180^\\circ$.\nNow we can conclude that $\\angle PIF = \\angle PEF = \\angle ABC = \\angle FBD = 180^\\circ - \\angle FID$, so $\\angle PIF + \\angle FID = 180^\\circ$, which means that $P$, $I$, $D$ lie on a straight line.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42322,
    "subject": "Mathematics (Multi-modal)",
    "question": "Jess is standing in a queue of people. She is 18th from the front and 35th from the back.\nHow many people are in the queue?",
    "options": [],
    "answer": "52",
    "solution": "There are $17$ people in front of Jess and $34$ behind her. Including herself this makes $17 + 1 + 34 = 52$ people.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42323,
    "subject": "Mathematics (Multi-modal)",
    "question": "One of the numbers below is multiplied by $2016$ and the product is a perfect square. The number is\n\n(A) $10$ (B) $14$ (C) $32$ (D) $100$ (E) $108$",
    "options": [],
    "answer": "B",
    "solution": "Factorise $2016$. It is divisible by $2$, because it is an even number and it is also divisible by $9$ since the sum of the digits is $9$. The full factorisation is $2^5 \\times 3^2 \\times 7$. Each prime factor of a complete square must have an even number as the exponent (i.e. every prime factor must appear an even number of times) so that $2016$ multiplied by a number $n$ will be a complete square. Therefore, $n$ must be equal to $2 \\times 7$ which is equal to $14$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42324,
    "subject": "Mathematics (Multi-modal)",
    "question": "$0.125$ divided by $0.25$ is\n(A) $0.75$ (B) $0.65$ (C) $0.55$ (D) $0.50$ (E) $0.40$",
    "options": [],
    "answer": "D",
    "solution": "(D) $0.125$ is half of $0.25$, so result is $0.5$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42325,
    "subject": "Mathematics (Multi-modal)",
    "question": "A representation of $\\frac{17}{20}$ as a sum of reciprocals\n$$\n\\frac{17}{20} = \\frac{1}{a_1} + \\frac{1}{a_2} + \\cdots + \\frac{1}{a_k}\n$$\nis called a *calm representation* with $k$ terms if the $a_i$ are distinct positive integers and at most one of them is not a power of two.\n\na. Find the smallest value of $k$ for which $\\frac{17}{20}$ has a calm representation with $k$ terms.\n\nb. Prove that there are infinitely many calm representations of $\\frac{17}{20}$.",
    "options": [],
    "answer": "k = 3; infinitely many calm representations exist.",
    "solution": "Note first that there is no calm representation with 2 terms: if either $a_1$ or $a_2$ is 1 or $a_1 = a_2 = \\frac{1}{2}$, then the sum is greater than $\\frac{17}{20}$. Otherwise, the sum is at most $\\frac{1}{2} + \\frac{1}{3} = \\frac{5}{6} < \\frac{17}{20}$, thus too small.\n\nOn the other hand, there is a representation with 3 terms, namely\n$$\n\\frac{17}{20} = \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{10},\n$$\nshowing that the smallest possible value of $k$ is 3.\n\nNow we show that there are infinitely many calm representations: take $a_1 = 5 \\cdot 2^{4n+1}$ and consider the difference\n$$\n\\frac{17}{20} - \\frac{1}{5 \\cdot 2^{4n+1}} = \\frac{17 \\cdot 2^{4n-1} - 1}{5 \\cdot 2^{4n+1}}.\n$$\nSince $2^4 = 16 \\equiv 1 \\mod 5$, the numerator is $17 \\cdot 2^{4n-1} - 1 \\equiv 17 \\cdot 8 - 1 = 135 \\equiv 0 \\mod 5$. Thus the factor 5 cancels, and we have\n$$\n\\frac{17}{20} - \\frac{1}{5 \\cdot 2^{4n+1}} = \\frac{A}{2^{4n+1}}\n$$\nfor some positive integer $A < 2^{4n+1}$. Since $A$ has a binary representation as $A = 2^{b_1} + 2^{b_2} + \\dots + 2^{b_r}$ with distinct nonnegative integers $b_1, b_2, \\dots, b_r$, we get\n$$\n\\frac{17}{20} = \\frac{1}{5 \\cdot 2^{4n+1}} + \\frac{1}{2^{4n+1-b_1}} + \\frac{1}{2^{4n+1-b_2}} + \\dots + \\frac{1}{2^{4n+1-b_r}},\n$$\nwhich is a calm representation for every $n$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42326,
    "subject": "Mathematics (Multi-modal)",
    "question": "The value of $2 - 0 + 1 \\times 6$ is\n(A) $-4$\n(B) $4$\n(C) $8$\n(D) $9$\n(E) $18$",
    "options": [],
    "answer": "C",
    "solution": "$2 - 0 + 1 \\times 6 = 2 - 0 + 6 = 8$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42327,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\n\\sqrt{5050^2 - 4950^2} =\n$$\n(A) 10 (B) $10^2$ (C) $10^3$ (D) $10^4$ (E) $10^5$",
    "options": [],
    "answer": "C",
    "solution": "$$\n\\sqrt{5050^2 - 4950^2} = \\sqrt{(5050 + 4950)(5050 - 4950)} = \\sqrt{10000 \\times 100} = \\sqrt{10^6} = 10^3.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42328,
    "subject": "Mathematics (Multi-modal)",
    "question": "$p$, $q$ and $r$ represent the numbers $2$, $3$, $4$ in some order. What is the greatest possible value of $p^q \\times r$?",
    "options": [],
    "answer": "162",
    "solution": "$3^4 \\times 2 = 81 \\times 2 = 162$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42329,
    "subject": "Mathematics (Multi-modal)",
    "question": "Andile and Zandre play a game on a $2017 \\times 2017$ board. At the beginning, Andile declares some of the squares *forbidden*, meaning that nothing may be placed on such a square. After that, they take turns to place coins on the board, with Zandre placing the first coin. It is not allowed to place a coin on a forbidden square or in the same row or column where another coin has already been placed. The player who places the last coin wins the game.\nWhat is the least number of squares Andile needs to declare as forbidden at the beginning to ensure a win? (Assume that both players use an optimal strategy.)",
    "options": [],
    "answer": "2017",
    "solution": "The minimum number is $2017$. For example, Andile can achieve a win by declaring all squares of the last row forbidden, so that $2016$ rows remain. After that, there will be exactly $2016$ moves possible, no matter how the two play, since placing a coin always eliminates exactly one row and one column from further use. This means that Andile gets the last move.\n\nOn the other hand, we prove that $2016$ or fewer forbidden squares are not sufficient, no matter how they are placed. Generally, we show by induction that Zandre has a winning strategy on a $(2n-1) \\times (2n-1)$ board if he gets to place a coin first and no more than $2n-2$ squares have been forbidden. This is trivial for $n=1$: Zandre can simply place a coin on the only square.\n\nFor the induction step, consider a $(2n+1) \\times (2n+1)$ board with at most $2n$ forbidden squares. If there are two forbidden squares in the same row or column somewhere, then Zandre places a coin in this row/column. This is possible since there are fewer forbidden squares than squares in a row or column. If there are at least two forbidden squares but no two of them in the same row or column, Zandre chooses any two of them, then places a coin on the intersection of the row of the first forbidden square and the column of the second forbidden square. This is possible because of the assumption that there are no two forbidden squares in the same row or column. Finally, if there is only one forbidden square, Zandre places a coin anywhere in the same row or column, and if there are no forbidden squares, he just places it on an arbitrary square.\n\nAfter Andile's move, the rows and columns where Andile and Zandre placed their coins can be removed, since no further coins can be placed there anymore. This leaves us with a $(2n-1) \\times (2n-1)$ board, and by the choice of Zandre's move there are at most $2n-2$ forbidden squares on it. So by the induction hypothesis, Zandre has a winning strategy for the remaining position, which completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42330,
    "subject": "Mathematics (Multi-modal)",
    "question": "John spends a total of $3$ hours weeding, planting and watering in his garden. He spends $25$ minutes more on weeding than planting, and $55$ minutes less on watering than planting. How many minutes did he spend planting?\n\n(A) $45$\n(B) $50$\n(C) $65$\n(D) $70$\n(E) $80$",
    "options": [],
    "answer": "D",
    "solution": "If he spends $x$ minutes on planting, then his total time spent in the garden is $x + 25 + x + x - 55$ and is $180$ minutes. This means $3x - 30 = 180$, so $x = 70$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 42331,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which integers $n \\ge 2$ 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": "all integers n ≥ 2 with n ≠ 4, 5",
    "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 $n \\ge 6$: 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": 42332,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every positive integer $n$, determine the greatest possible value of the quotient\n$$\n\\frac{1 - x^n - (1-x)^n}{x(1-x)^n + (1-x)x^n}\n$$\nwhere $0 < x < 1$.",
    "options": [],
    "answer": "2^n - 2",
    "solution": "It is convenient to write $y = 1 - x$, so that $x + y = 1$. We show that the maximum of the resulting expression\n$$\n\\frac{1 - x^n - y^n}{xy^n + yx^n}\n$$\nis attained when $x = y = \\frac{1}{2}$. The value in this case is $2^n - 2$. We first rewrite the expression as follows:\n$$\n\\begin{aligned} \\frac{1 - x^n - y^n}{xy^n + yx^n} &= \\frac{x + y - x^n - y^n}{xy^n + yx^n} = \\frac{x(1 - x^{n-1}) + y(1 - y^{n-1})}{xy(x^{n-1} + y^{n-1})} \\\\ &= \\frac{x(1-x)(1+x+\\cdots+x^{n-2}) + y(1-y)(1+y+\\cdots+y^{n-2})}{xy(x^{n-1} + y^{n-1})} \\\\ &= \\frac{xy(1+x+\\cdots+x^{n-2}+1+y+\\cdots+y^{n-2})}{xy(x^{n-1} + y^{n-1})} \\\\ &= \\frac{1+1}{x^{n-1} + y^{n-1}} + \\frac{x+y}{x^{n-1} + y^{n-1}} + \\cdots + \\frac{x^{n-2} + y^{n-2}}{x^{n-1} + y^{n-1}}. \\end{aligned}\n$$\n\nWe will therefore be done if we can show that\n$$\n\\frac{x^a + y^a}{x^b + y^b}\n$$\nattains its maximum for $x = y = \\frac{1}{2}$ whenever $0 \\le a < b$. This can be achieved in many ways, for example as follows: the general mean inequality yields\n$$\n\\left( \\frac{x^a + y^a}{2} \\right) \\le \\left( \\frac{x^b + y^b}{2} \\right)^{a/b}\n$$\nand\n$$\n\\frac{1}{2} = \\frac{x+y}{2} \\le \\left( \\frac{x^b + y^b}{2} \\right)^{1/b},\n$$\nUsing the same notation as in the previous solution, the binomial theorem gives us\n$$\n\\frac{1 - x^n - y^n}{xy^n + yx^n} = \\frac{(x + y)^n - x^n - y^n}{xy^n + yx^n} = \\frac{\\sum_{k=1}^{n-1} \\binom{n}{k} x^k y^{n-k}}{xy(x^{n-1} + y^{n-1})} = \\frac{\\sum_{k=1}^{n-1} \\binom{n}{k} x^{k-1} y^{n-k-1}}{x^{n-1} + y^{n-1}}.\n$$\nWe group terms pairwise ($k$ and $n-k$ forming a pair):\n$$\n\\frac{1 - x^n - y^n}{xy^n + yx^n} = \\sum_{k=1}^{\\lfloor(n-1)/2\\rfloor} \\binom{n}{k} \\frac{x^{k-1}y^{n-k-1} + x^{n-k-1}y^{k-1}}{x^{n-1} + y^{n-1}} + \\begin{cases} \\binom{n}{n/2} \\frac{x^{n/2-1}y^{n/2-1}}{x^{n-1}+y^{n-1}} & n \\text{ even,} \\\\ 0 & \\text{otherwise.} \\end{cases}\n$$\nNow we only need to show that\n$$\n\\frac{x^{k-1}y^{n-k-1} + x^{n-k-1}y^{k-1}}{x^{n-1} + y^{n-1}} = \\frac{(xy)^{k-1}(x^{n-2k} + y^{n-2k})}{x^{n-1} + y^{n-1}}\n$$\nattains its maximum when $x = y = \\frac{1}{2}$. Note that the potential extra term for even $n$ is also of this form, up to a factor 2:\n$$\n\\frac{x^{n/2-1}y^{n/2-1}}{x^{n-1} + y^{n-1}} = \\frac{1}{2} \\cdot \\frac{(xy)^{n/2-1}(x^0 + y^0)}{x^{n-1} + y^{n-1}}.\n$$\nSince $xy \\le \\left(\\frac{x+y}{2}\\right)^2 = \\frac{1}{4}$ by the inequality between the arithmetic and geometric mean (with equality for $x = y = \\frac{1}{2}$), this can again be achieved by showing that\n$$\n\\frac{x^a + y^a}{x^b + y^b}\n$$\nattains its maximum for $x = y = \\frac{1}{2}$ whenever $0 \\le a < b$, as in the first solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42333,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $\\hat{A} = 120^\\circ$ and $\\hat{B}$ is five times $\\hat{C}$. The number of degrees in $\\hat{C}$ is\n(A) 10 (B) 12 (C) 15 (D) 18 (E) 20",
    "options": [],
    "answer": "A",
    "solution": "The angles of the triangle add up to $180^\\circ$, so $\\hat{C} + 5\\hat{C} + 120^\\circ = 180^\\circ$, i.e. $6\\hat{C} = 60^\\circ$ so $\\hat{C} = 10^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42334,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle with circumcircle $\\Gamma$. Let $D$ be a point on segment $BC$ such that $\\angle BAD = \\angle DAC$, and let $M$ and $N$ be points on segments $BD$ and $CD$, respectively, such that $\\angle MAD = \\angle DAN$. Let $S, P$ and $Q$ (all different from $A$) be the intersections of the rays $AD, AM$ and $AN$ with $\\Gamma$, respectively. Show that the intersection of $SM$ and $QD$ lies on $\\Gamma$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nLet $X$ be the intersection of $SM$ with the circumcircle. Note that $\\angle MXA = \\angle SXA = \\angle SBA = \\angle SBC + \\angle CBA$. Since $AS$ bisects the angle $\\angle BAC$, we have $\\angle SBC = \\angle SAC = \\angle BAS$. It follows that\n$$\n\\angle MXA = \\angle BAS + \\angle CBA = \\angle BAD + \\angle DBA = 180^\\{\\circ\\} - \\angle ADB = 180^\\{\\circ\\} - \\angle ADM,\n$$\nso $AXMD$ is a cyclic quadrilateral. Consequently, $\\angle SXD = \\angle MXD = \\angle MAD = \\angle DAN = \\angle SAQ = \\angle SXQ$, which means that $X, D$ and $Q$ lie on a common straight line. Hence $X$ is also the intersection of $SM$ and $QD$, which completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42335,
    "subject": "Mathematics (Multi-modal)",
    "question": "The lengths of all sides of a triangle are prime numbers. If two sides are $5$ and $7$ what is the longest possible length of the third side?",
    "options": [],
    "answer": "11",
    "solution": "The length of the third side of a triangle cannot exceed the sum of the lengths of the first two sides, which is $5 + 7 = 12$. The largest prime number less than $12$ is $11$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42336,
    "subject": "Mathematics (Multi-modal)",
    "question": "Water freezes at $0^\\circ$ Celsius and boils at $100^\\circ$ Celsius. On the SAMO scale, water freezes at $20^\\circ$ and boils at $170^\\circ$. The two scales are linearly related. What is the temperature on the SAMO scale if it is $80^\\circ$ on the Celsius scale?",
    "options": [],
    "answer": "140",
    "solution": "A temperature of $80^\\circ$ C is $80\\%$ of the distance from freezing point to boiling point, so on the SAMO scale the reading is\n$$\n20 + \\frac{80}{100}(170 - 20) = 20 + 120 = 140^\\circ.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42337,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(P, d)$ of a polynomial $P$ with integer coefficients and an integer $d$ such that the equation $P(x) - P(y) = d$, where $x$ and $y$ are integers and $x \\neq y$, has infinitely many solutions.",
    "options": [],
    "answer": "All such pairs are exactly:\n1) For any integer d, any linear polynomial with integer coefficients P(x) = (d/a) x + b, where a is a nonzero integer divisor of d and b is any integer (including the case d = 0, which yields constant polynomials).\n2) For d = 0, any polynomial that is even after a shift, i.e., P(x) = R((x − c)^2) for some polynomial R, where 2c is an integer (and P has integer coefficients).",
    "solution": "Note first that $x - y$ divides $P(x) - P(y)$. So if $d \\neq 0$, there are only finitely many possibilities for $x - y$. For one of these possible values, let us denote it by $a$, there must be\n\ninfinitely many pairs $(x, y)$ such that $x - y = a$ and $P(x) - P(y) = d$. Note that\n$$\nP(x) - P(x - a) - d\n$$\nis still a polynomial. The only way it can have infinitely many zeros is that it is identically zero. But then $P(x) = dx/a + b$ for some number $b$.\nThis gives us the first set of solutions, consisting of an arbitrary integer $d$ and a polynomial $P(x) = dx/a + b$, where $a$ is a (positive or negative) divisor of $d$ and $b$ an arbitrary integer. This includes the solution where $d = 0$ and the polynomial $P(x)$ is constant.\nFor the rest of this solution, we can assume that $d = 0$. If the degree of the polynomial is odd, then there exist integers $A$ and $B$ such that $P(x)$ is either increasing for $x \\ge A$ and for $x \\le B$, or decreasing for $x \\ge A$ and $x \\le B$. If $P(x)$ is increasing for $x \\ge A$, then $P(A)$, $P(A + 1)$, $P(A + 2)$, ... are a strictly increasing sequence of integers, thus distinct. Moreover, from some point on these numbers are all greater than the maximum of $P(x)$ for $x \\le A$. Likewise, $P(B)$, $P(B - 1)$, $P(B - 2)$, ... are a strictly decreasing sequence of integers, and from some point on they are all less than the minimum of $P(x)$ for $x \\ge B$. This means that there cannot be infinitely many pairs $x, y$ with $x \\ne y$ such that $P(x) - P(y) = 0$. The same argument applies if $P(x)$ is decreasing for $x \\ge A$ and $x \\le B$.\nThus $P(x)$ must be a polynomial of even degree. Now there exist integers $A$ and $B$ such that $P(x)$ is increasing for $x \\ge A$ and decreasing for $x \\le B$, or vice versa. Thus there can only be infinitely many pairs $x, y$ with $x \\ne y$ such that $P(x) - P(y) = 0$ if one of the two $(x, y)$ is $\\ge A$ while the other is $\\le B$. We show that $x + y$ has to be constant for these solutions. Let the first terms of the polynomial be as follows:\n$$\nP(x) = ax^n + bx^{n-1} + \\dots\n$$\nAssume that $a$ is positive, for otherwise one can replace $P(x)$ by $-P(x)$. If $an(x+y) \\ge -2b+1$, then we have\n$$\nP(x) - P\\left(\\frac{-2b+1}{an} - x\\right) = x^{n-1} + \\dots,\n$$\nwhere the dots stand for terms involving lower powers of $x$. For large enough $x$, this will be strictly positive, so $P(x) - P(y) \\ne 0$. Likewise, if $an(x + y) \\le -2b - 1$, then we have\n$$\nP(x) - P\\left(\\frac{-2b-1}{an} - x\\right) = -x^{n-1} + \\dots,\n$$\nwhich is negative for large enough $x$. Hence $P(x) - P(y) \\ne 0$ in this case as well. Thus there must be infinitely many solutions of $P(x) - P(y) = 0$ with $an(x + y) = -2b$. In this case,\n$$\nP(x) - P\\left(\\frac{-2b}{an} - x\\right)\n$$\nis a polynomial with infinitely many zeros, hence it is constant. Thus\n$$\nP(x) = P\\left(\\frac{-2b}{an} - x\\right)\n$$\nfor all $x$, which gives\n$$\nQ(x) = P\\left(\\frac{-b}{an} + x\\right) = P\\left(\\frac{-b}{an} - x\\right) = Q(-x)\n$$\nfor all $x$. It follows that the polynomial $Q(x)$ can only contain even powers of $x$, i.e. $Q(x) = R(x^2)$. This gives us the second set of solutions, consisting of a polynomial of the form $P(x) = R((x-c)^2)$, where $2c$ is an integer, and $d=0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 42338,
    "subject": "Mathematics (Multi-modal)",
    "question": "The square of the difference between the largest and smallest of three consecutive integers is\n(A) 1 (B) 9 (C) 4 (D) 100 (E) 2",
    "options": [],
    "answer": "C",
    "solution": "Any three consecutive integers are of the form $n$, $n+1$, $n+2$. The difference between the largest and the smallest is $(n+2) - n = 2$, and the square of the difference is $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42339,
    "subject": "Mathematics (Multi-modal)",
    "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 $L_i \\cdot \\left[\\binom{m}{i} - \\binom{m-k}{i}\\right]$ for all $i \\ge 1$. [Here, $\\binom{n}{i} = \\frac{n!}{i!(n-i)!}$ denotes a binomial coefficient. Note that $\\binom{n}{i} = 0$ if $n < i$.]",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove the statement by induction on $m$. When $m = k$, we have\n$$\nL_i \\left[ \\binom{k}{i} - \\binom{0}{i} \\right] = L_i \\binom{k}{i} = L_i \\cdot \\frac{k!}{i!(k-i)!} = k \\cdot \\frac{L_i}{i} \\cdot \\frac{(k-1)!}{(i-1)!(k-i)!} = k \\cdot \\frac{L_i}{i} \\cdot \\binom{k-1}{i-1}.\n$$\nSince $\\frac{L_i}{i}$ is an integer (by definition of $L_i$), and $\\binom{k-1}{i-1}$ is a binomial coefficient and thus also an integer, we see that $k$ is indeed a divisor.\n\nFor the induction step, assume that the statement holds for a specific value of $m$. We use the recursion $\\binom{m+1}{i} = \\binom{m}{i} + \\binom{m}{i-1}$ to show that it holds for $m+1$ as well:\n$$\n\\begin{aligned}\nL_i \\left[ \\binom{m+1}{i} - \\binom{m+1-k}{i} \\right] &= L_i \\left[ \\binom{m}{i} - \\binom{m-k}{i} \\right] + L_i \\left[ \\binom{m}{i-1} - \\binom{m-k}{i-1} \\right] \\\\\n&= L_i \\left[ \\binom{m}{i} - \\binom{m-k}{i} \\right] + \\frac{L_i}{L_{i-1}} \\cdot L_{i-1} \\left[ \\binom{m}{i-1} - \\binom{m-k}{i-1} \\right].\n\\end{aligned}\n$$\nNote that $k$ divides both $L_i[\\binom{m}{i} - \\binom{m-k}{i}]$ and $L_{i-1}[\\binom{m}{i-1} - \\binom{m-k}{i-1}]$ by the induction hypothesis (if $i=1$, the latter term is simply zero), and $L_{i-1}$ (the least common multiple of $1, 2, \\dots, i-1$) divides $L_i$ (the least common multiple of $1, 2, \\dots, i$, or equivalently the least common multiple of $L_{i-1}$ and $i$). It follows that $k$ is also a divisor of $L_i[\\binom{m+1}{i} - \\binom{m+1-k}{i}]$, which completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42340,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x - y > x$, then which of the sentences MUST be true?\n(A) $x > 0$ (B) $y < 0$ (C) $x > y$ (D) $y > 0$ (E) $x < 0$",
    "options": [],
    "answer": "B",
    "solution": "If $x - y > x$ then $-y > 0$, so $y < 0$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42341,
    "subject": "Mathematics (Multi-modal)",
    "question": "Which one of the following is not a rational number?\n(A) $\\sqrt{4}$\n(B) $\\pi^0$\n(C) $\\sqrt{6.25}$\n(D) $\\sqrt{14}$\n(E) $\\sqrt{0.49}$",
    "options": [],
    "answer": "D",
    "solution": "D The others are, in order, $2$; $1$; $2.5$ and $0.7$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42342,
    "subject": "Mathematics (Multi-modal)",
    "question": "A shop has a special offer on pumpkins: \"Buy one, get one at half price\". For this offer, the average cost per pumpkin is the same as in\n(A) Two for the price of one\n(B) Three for the price of one\n(C) Three for the price of two\n(D) Four for the price of three\n(E) Five for the price of four",
    "options": [],
    "answer": "D",
    "solution": "If you buy one item and get one at half price you are getting two items for the price of one and a half items. Since $2 : 1\\frac{1}{2}$ is equivalent to $4 : 3$, the average cost per item is the same as four items for the price of three.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42343,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the value of $(2+4+6+...+198+200) - (1+3+5+...+197+199)$?",
    "options": [],
    "answer": "100",
    "solution": "Rearranging, the value is $(2-1) + (4-3) + (6-5) + \\dots + (200-199)$ which has $100$ brackets and therefore totals $100 \\times 1 = 100$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42344,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let one dollar cost $R12$, and let one British pound cost $R21$. What does one pound cost in dollars?\n(A) 1.25   (B) 1.30   (C) 1.50   (D) 1.60   (E) 1.75",
    "options": [],
    "answer": "E",
    "solution": "One rand is $\\frac{1}{12}$ of a dollar, so one pound is $\\frac{21}{12} = \\frac{7}{4} = 1.75$ dollars.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42345,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many integers between $100$ and $1000$ are multiples of $7$?\n(A) $120$ (B) $125$ (C) $128$ (D) $132$ (E) $140$",
    "options": [],
    "answer": "C",
    "solution": "The multiples of $7$ include $98$ ($= 7 \\times 14$), $105$, $112$, $\\ldots$, $994$ ($= 7 \\times 142$), $1001$, $\\ldots$ Of these, $142 - 14 = 128$ are between $100$ and $1000$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42346,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is estimated that there are $3\\,000\\,000$ km² of rain forest in Brazil, and it is being reduced at the rate of $7\\,400$ km² per year. If this rate stays the same, then the number of years before there is no rain forest left is approximately\n(A) $400$ (B) $300$ (C) $200$ (D) $150$ (E) $100$",
    "options": [],
    "answer": "A",
    "solution": "The number of years is $\\frac{3\\,000\\,000}{7\\,400} = \\frac{30 \\times 100}{7.4} \\approx 400$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42347,
    "subject": "Mathematics (Multi-modal)",
    "question": "In rugby it is possible to score 3, 5 or 7 points at a time. If South Africa wins the next Rugby World Cup and scores 34 points in the final, the highest number of times 3 points could have been scored by South Africa is\n(A) 6 (B) 7 (C) 8 (D) 9 (E) 10",
    "options": [],
    "answer": "D",
    "solution": "We need $34 - 5x - 7y$ to be a multiple of $3$, so $5x + 7y$, like $34$, must have remainder $1$ after division by $3$. For the largest possible multiple of $3$ we need the smallest such value of $5x + 7y$, which is $0 \\times 5 + 1 \\times 7 = 7$. Then $(34 - 7) \\div 3 = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42348,
    "subject": "Mathematics (Multi-modal)",
    "question": "Bee, Cee and Vee live in Microphyllia where there are only two types of creatures – those that consistently tell the truth and those that consistently lie. The former creatures are Trudees and the latter Falsees. When I last visited Microphyllia I asked Bee: “Who of you all are Trudees?”\nBee mumbled its answer so I didn't quite catch what she said.\n“Bee said that only one of the three of us is a Trudee”, Cee noted.\nVee turned to me and said: “Don’t believe Cee, he’s not telling the truth”.\nWho of them are Trudees and who are Falsees?",
    "options": [],
    "answer": "Cee is a Falsee, Vee is a Trudee, and Bee’s status cannot be determined from the given information.",
    "solution": "Suppose Cee tells the truth, then Bee said that only one of the three tells the truth, so Bee can't tell the truth or otherwise there will be at least two truth tellers which would be a contradiction. Hence, if Cee tells the truth Bee lies and there must be more than one truth teller, so Vee must be telling the truth. However, Vee says Cee lies contradicting our assumption. Hence Cee can't tell the truth and lies.\nWe don't know what Bee said, so we don't know if he tells the truth or lies. Thus Cee lies, Vee tells the truth and we don't know about Bee.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42349,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a$ and $b$ are positive integers such that $\\frac{a}{b} = \\frac{2}{3}$ and $a + b = 80$, then the product $ab$ is equal to\n\n(A) 1 500   (B) 1 599   (C) 1 667   (D) 1 596   (E) 1 536",
    "options": [],
    "answer": "E",
    "solution": "Since $a = \\frac{2}{3}b$ and $a + b = 80$, we have\n$$\n80 = \\frac{2}{3}b + b = \\frac{5}{3}b\n$$\nso\n$$\nb = \\frac{3}{5} \\times 80 = 48.\n$$\nThen $a = 80 - 48 = 32$ and $ab = 32 \\times 48 = 1536$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42350,
    "subject": "Mathematics (Multi-modal)",
    "question": "If the product of four consecutive integers is equal to the value of one of those integers, then what is the largest possible value for any of the integers?",
    "options": [],
    "answer": "3",
    "solution": "3\n\nWith a little trial and error it should soon be clear that this scenario is only possible if one of the integers is zero. The options are $-3$; $-2$; $-1$; $0$ or $-2$; $-1$; $0$; $1$ or $-1$; $0$; $1$; $2$ or $0$; $1$; $2$; $3$. The largest possible integer is thus $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42351,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for all natural numbers $n$, $\\frac{n}{3} + \\frac{n^2}{2} + \\frac{n^3}{6}$ is also a natural number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\begin{aligned}\n\\frac{n}{3} + \\frac{n^2}{2} + \\frac{n^3}{6} &= \\frac{2n + 3n^2 + n^3}{6} \\\\\n&= \\frac{n(2 + 3n + n^2)}{6} \\\\\n&= \\frac{n(n + 1)(n + 2)}{6}\n\\end{aligned}\n$$\nSince $n$, $n + 1$, $n + 2$ are three consecutive integers, at least one is divisible by $2$ and one is divisible by $3$. Hence $n(n + 1)(n + 2)$ is divisible by $6$ and therefore $\\frac{n}{3} + \\frac{n^2}{2} + \\frac{n^3}{6}$ must be an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42352,
    "subject": "Mathematics (Multi-modal)",
    "question": "One hundred glasses are arranged in a $10 \\times 10$ array. Now we pick $a$ of the rows and pour blue liquid into all glasses in these rows, so that they are half full. The remaining rows are filled halfway with yellow liquid. Afterwards, we pick $b$ of the columns and fill them up with blue liquid. The remaining columns are filled up with yellow liquid. The mixture of blue and yellow liquid turns green. If both halves in a glass have the same colour, then that colour remains as it is.\n\na. Determine all possible combinations of values for $a$ and $b$ so that exactly half of the glasses contain green liquid at the end.\n\nb. Is it possible that precisely one quarter of the glasses contain green liquid at the end?",
    "options": [],
    "answer": "a. Exactly half are green if and only if either a = 5 or b = 5 (the other can be any integer from 0 to 10). b. No, one quarter green is impossible.",
    "solution": "The total number of glasses that are green at the end of the procedure is\n$$\na(10 - b) + b(10 - a) = 10a + 10b - 2ab = 2(5a + 5b - ab).\n$$\nWe immediately observe that this number is always even, so the number of green glasses cannot be $25$ (i.e., one quarter). Hence the answer to the second question is no.\n\nFor the first question, we have to solve the equation\n$$\n10a + 10b - 2ab = 50,\n$$\nwhich is equivalent to\n$$\n2ab - 10a - 10b + 50 = 2(a - 5)(b - 5) = 0.\n$$\nThus exactly half of the glasses contain green liquid if either $a = 5$ or $b = 5$ (and the other is arbitrary).",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 42353,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, $AB = AC$, and $D$ is on $BC$. A point $E$ is chosen on $AC$, and a point $F$ is chosen on $AB$, such that $DE = DC$ and $DF = DB$. It is given that $\\frac{DC}{BD} = 2$ and $\\frac{AF}{AE} = 5$. Determine the value of $\\frac{AB}{BC}$.",
    "options": [],
    "answer": "sqrt(3)/2",
    "solution": "Note that triangles $ABC$ and $DFB$ are both isosceles, and they share the angle at $B$. Therefore, they must be similar. Likewise, triangle $DCE$ is similar to these two. Let $BF = x$ and $BD = FD = y$. Then we have $DC = 2BD = 2y$, and since $DFB$ and $DCE$ are similar, it also follows that $CE = 2x$. Next, we let $AB = AC = a$, so $AF = AB - BF = a - x$ and $AE = AC - CE = a - 2x$. We are given that $AF = 5AE$, so\n$$\na - x = 5(a - 2x),\n$$\nwhich is equivalent to $4a = 9x$, i.e. $a = \\frac{9x}{4}$. Finally, since $ABC$ and $DFB$ are similar, we have\n$$\n\\frac{\\frac{9x}{4}}{3y} = \\frac{AB}{BC} = \\frac{BD}{BF} = \\frac{y}{x}.\n$$\nThus\n$$\n\\left(\\frac{y}{x}\\right)^2 = \\frac{3}{4},\n$$\nand finally\n$$\n\\frac{AB}{BC} = \\frac{y}{x} = \\frac{\\sqrt{3}}{2}.\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42354,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest positive integer $n$ whose prime factors are all greater than $18$, and that can be expressed as $n = a^3 + b^3$ with positive integers $a$ and $b$.",
    "options": [],
    "answer": "1843",
    "solution": "We can factorise $n$ as\n$$\nn = a^3 + b^3 = (a + b)(a^2 - ab + b^2).\n$$\nThe first factor $a + b$ has to be at least $19$, since $n$ would otherwise contain a prime factor that is smaller than $18$. Setting $a + b = s$, we obtain\n$$\na^2 - ab + b^2 = a^2 - a(s - a) + (s - a)^2 = 3a^2 - 3as + s^2 = 3\\left(a - \\frac{s}{2}\\right)^2 + \\frac{s^2}{4}\n$$\nby completing the square. Hence the second factor is greater or equal to $\\frac{s^2}{4}$, and becomes smaller the closer $a$ is to $\\frac{s}{2}$. If $s = 19$, then for $a = 9$ or $a = 10$, the second factor is $91 = 7 \\cdot 13$, which contains a prime factor smaller than $18$. For $a = 8$ or $a = 11$, however, it is equal to $97$, which is prime. In this case, $n = 19 \\cdot 97 = 1843 = 11^3 + 8^3$ satisfies the conditions.\n\nIf $s = 19$ and $a < 8$ or $a > 11$, then\n$$\na^2 - ab + b^2 = 3\\left(a - \\frac{19}{2}\\right)^2 + \\frac{19^2}{4} > 3\\left(\\frac{3}{2}\\right)^2 + \\frac{19^2}{4} = 97,\n$$\nthus $n > 19 \\cdot 97 = 1843$. If $s > 19$, then $s$ must be at least $20$ (in fact at least $23$, so that it does not contain prime factors smaller than $18$), so\n$$\nn = s\\left(3\\left(a - \\frac{s}{2}\\right)^2 + \\frac{s^2}{4}\\right) \\geq s \\cdot \\frac{s^2}{4} \\geq \\frac{20^3}{4} = 2000.\n$$\nThis means that $1843$ is indeed the smallest number with the desired properties.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42355,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle with circumradius $R$, and let $\\ell_A, \\ell_B, \\ell_C$ be the altitudes through $A, B, C$ respectively. The altitudes meet at $H$. Let $P$ be an arbitrary point in the same plane as $ABC$. The feet of the perpendicular lines through $P$ onto $\\ell_A, \\ell_B, \\ell_C$ are $D, E, F$ respectively. Prove that the areas of $DEF$ and $ABC$ satisfy the following equation:\n$$\n\\text{area}(DEF) = \\frac{PH^2}{4R^2} \\text{area}(ABC).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nNote that $\\angle PDH = \\angle PFH = \\angle PEH = 90^\\circ$ by construction, so by Thales's Theorem, $D, E$ and $F$ lie on a circle whose diameter is $PH$. Therefore, the angle between lines $DE$ and $FE$ is the same as the angle between lines $DH$ and $FH$ (angles subtended by the chord $DF$), which in turn is the same as the angle between lines $AB$ and $BC$ (angles between pairwise perpendicular lines). Repeating the argument, we find that the angles between lines $DE, DF, EF$ coincide with those between lines $AB, AC, BC$, so triangles $ABC$ and $DEF$ are similar. The diameter of the circumcircle of $ABC$ is $2R$, the diameter of the circumcircle of $DEF$ is $PH$ as observed before. Since areas of similar triangles are proportional to squared lengths, we obtain\n$$\n\\text{area}(DEF) = \\frac{PH^2}{4R^2} \\text{area}(ABC)\n$$\nas required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42356,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all sequences $a_1, a_2, a_3, \\dots$ of nonnegative integers such that $a_1 < a_2 < a_3 < \\dots$ and $a_n$ divides $a_{n-1} + n$ for all $n \\ge 2$.",
    "options": [],
    "answer": "All such sequences are exactly the following three families:\n1) a_n = n − 1 for all n.\n2) a_n = (n^2 + n)/2 + k for all n, where k is a fixed nonnegative integer.\n3) For a fixed nonnegative integer N,\n   a_n = n − 1 for n ≤ N, and a_n = (n^2 + n)/2 − (N^2 − N + 2)/2 for n > N.",
    "solution": "We claim that the only possible sequences are the following:\n* $a_n = n - 1$ for all $n$, or\n* $a_n = \\frac{n^2+n}{2} + k$ for all $n$, where $k$ is a fixed nonnegative integer, or\n* $a_n = \\begin{cases} n-1 & n \\le N, \\\\ \\frac{n^2+n}{2} - \\frac{N^2-N+2}{2} & n > N, \\end{cases}$ where $N$ is a fixed nonnegative integer.\n\nLet us first verify that each of these sequences satisfies the conditions:\n* If $a_n = n - 1$ for all $n$, then $a_{n-1} + n = 2n - 2 = 2a_n$ is indeed divisible by $a_n$.\n* If $a_n = \\frac{n^2+n}{2} + k$ for all $n$, then $a_{n-1} + n = \\frac{n^2-n}{2} + k + n = \\frac{n^2+n}{2} + k = a_n$ is also divisible by $a_n$.\n* In the third case, $a_n$ divides $a_{n-1} + n$ for $n \\le N$ as in the first case. Next note that $a_{N+1} = \\frac{(N+1)^2+(N+1)}{2} - \\frac{N^2-N+2}{2} = 2N$ divides $a_N + (N+1) = 2N$. Finally, for $n > N+1$, we have $a_n = a_{n-1} + n$ as in the second case, so $a_n$ again divides $a_{n-1} + n$.\n\nNow we prove that these are the only such sequences. First, let $a_k$ be an element of the sequence such that $a_k \\ge k$ (if such an element exists). Recall that $a_k + k + 1$ has to be a multiple of $a_{k+1}$. However, since $a_{k+1} > a_k$, we have\n$$\n2a_{k+1} \\ge 2(a_k + 1) > 2a_k + 1 \\ge a_k + k + 1.\n$$\nSo the only possible multiple of $a_{k+1}$ that $a_k + k + 1$ could be is $1 \\cdot a_{k+1}$, and it follows that $a_{k+1} = a_k + k + 1$. But then $a_{k+1} \\ge k + k + 1 \\ge k + 1$, so we can repeat the argument with $k+1$ instead of $k$ to show that $a_{k+2} = a_{k+1} + k + 2$, etc. Generally, we get $a_{n+1} = a_n + n + 1$ for all $n \\ge k$.\n\nIf $a_1 \\ge 1$, then we can invoke this observation immediately: $a_{n+1} = a_n + n + 1$ for all $n \\ge 1$, so\n$$\na_n = a_{n-1} + n = a_{n-2} + (n-1) + n = \\dots = a_1 + 2 + 3 + \\dots + (n-1) + n = \\frac{n^2+n}{2} + (a_1 - 1),\n$$\nwhich is exactly our second solution.\n\nSuppose finally that $a_1 = 0$, and let $N$ be the largest index for which $a_N = N - 1$; if there is no largest index, then $a_n = n - 1$ for all $n$, and we obtain the first solution. Next note that $a_{N+1}$ has to divide $a_N + N + 1 = 2N$. By our choice of $N$, we have $a_{N+1} \\ne N$, and since $a_{N+1} > a_N = N - 1$, the only possible value (the only divisor of $2N$) for $a_{N+1}$ is $2N$. But then $a_{N+1} = 2N \\ge N + 1$, and we can apply the same observation as before: $a_{m+1} = a_m + m + 1$ for all $m \\ge N$, thus\n$$\na_n = a_{n-1} + n = \\dots = a_N + (N+1) + (N+2) + \\dots + (N-1) + n = \\\\\n= (N-1) + \\frac{n^2+n}{2} - \\frac{N^2+N}{2} = \\frac{n^2+n}{2} - \\frac{N^2-N+2}{2}\n$$\nfor all $n > N$, which is indeed the third solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42357,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, and let $x_1, x_2, \\dots, x_n$ be distinct positive integers with $x_1 = 1$. Construct an $n \\times 3$ table where the entries of the $k$-th row are $x_k, 2x_k, 3x_k$ for $k = 1, 2, \\dots, n$. Now follow a procedure where, in each step, two identical entries are removed from the table. This continues until there are no more identical entries in the table.\n\na. Prove that at least three entries remain at the end of the procedure.\n\nb. Prove that there are infinitely many possible triples that can remain at the end of the procedure for suitable choices of $n$ and $x_1, x_2, \\dots, x_n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nWe write $X$ as a shorthand for the set $\\{x_1, x_2, \\dots, x_n\\}$. Note that the number $1$ can only appear in the first row, so it remains at the end of the procedure. Next, let $x_k$ be the largest power of $2$ that occurs in $X$. This is possible since $x_1 = 1$ is a power of $2$. The number $2x_k$ in the second column of row $k$ is a power of two, and it does not occur in any of the other rows: it is not one of the numbers in the first column (i.e., one of the other $x_j$) by our choice of $x_k$, it is not anywhere else in the second column (i.e., equal to $2x_j$ for some other $j$) since the $x_j$ are all distinct, and it is not in the third column (i.e., of the form $3x_j$) since the numbers there are multiples of $3$. Hence $2x_k$ remains in the table until the end of the procedure.\n\nBy the same argument, if we choose $x_\\ell$ to be the largest power of $3$ that occurs among the numbers $x_1, x_2, \\dots, x_n$, then $3x_\\ell$ remains in the table at the end. Thus there are at least three numbers left at the end, namely $1, 2x_k, 3x_\\ell$.\n\nb.\nWe construct a sequence of sets $X_r$ in the following way: $X_0 = \\{1\\}$, and\n$$\nX_{r+1} = X_r \\cup \\{2^{2^r}x : x \\in X_r\\} \\cup \\{3^{2^r}x : x \\in X_r\\}\n$$\nfor all $r \\ge 0$. In words: $X_{r+1}$ consists of $X_r$, all elements of $X_r$ multiplied by $2^{2^r}$, and all elements of $X_r$ multiplied by $3^{2^r}$. We claim that the final table, if $x_1, x_2, \\dots, x_n$ are chosen to be the elements of $X_r$, consists of the three numbers $1, 2^{2^r}$ and $3^{2^r}$.\n\nClearly, this is true for $r = 0$: there is only one row, and the initial numbers $1, 2, 3$ remain in the table.\n\nFor the induction step, suppose the statement is true for $X_r$; we prove it for $X_{r+1}$. By the induction hypothesis, we can perform the reduction procedure in such a way that the rows corresponding to $X_r$ are reduced to $1, 2^{2^r}$ and $3^{2^r}$. The rows corresponding to $\\{2^{2^r}x : x \\in X_r\\}$ can be reduced to $2^{2^r}, 2^{2^r} \\cdot 2^{2^r} = 2^{2^{r+1}}$ and $2^{2^r} \\cdot 3^{2^r} = 6^{2^r}$, and the rows corresponding to $\\{3^{2^r}x : x \\in X_r\\}$ can be reduced to $3^{2^r}, 3^{2^r} \\cdot 2^{2^r} = 6^{2^r}$ and $3^{2^r} \\cdot 3^{2^r} = 3^{2^{r+1}}$.\n\nApplying the reduction procedure three more times to $2^{2^r}, 3^{2^r}$ and $6^{2^r}$, we end up with the three numbers $1, 2^{2^{r+1}}$ and $3^{2^{r+1}}$. This completes the induction and thus the proof of the second part.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42358,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $a$ for which $a^a$ is divisible by $20^{19}$.",
    "options": [],
    "answer": "All positive integers a with a = 20 or a = 10k for integer k ≥ 4.",
    "solution": "If $20^{19} \\mid a^a$, then $10 \\mid a$, hence $a \\in \\{10, 20, 30, \\dots\\}$. Clearly, $a = 10$ is not possible ($10^{10}$ is smaller than $20^{19}$), while $a = 20$ is possible ($20^{20} = 20 \\cdot 20^{19}$). Also, all $a = 10k$, $k \\ge 4$, are possible (for $q = 5^{10k-19} \\cdot 2^{10k-38} \\cdot k^{10k}$ we have $(10k)^{10k} = q \\cdot 20^{19}$). What about $a = 30$? If $30^{30} = q \\cdot 20^{19}$ for some $q \\in \\mathbb{Z}$, then $3^{30} \\cdot 5^{11} = q \\cdot 2^8$, with the left-hand side odd and the right-hand side even, an impossibility. Thus $a = 30$ is not possible. The solution is therefore all $a \\in \\{20\\} \\cup \\{10k : k \\ge 4\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42359,
    "subject": "Mathematics (Multi-modal)",
    "question": "We have a deck of 90 cards that are numbered from 10 to 99 (all two-digit numbers). How many sets of three or more different cards in this deck are there such that the number on one of them is the sum of the other numbers, and those other numbers are consecutive?",
    "options": [],
    "answer": "92",
    "solution": "We are looking at sums of the form\n$$\ns(10 + \\ell, k) = \\sum_{i=0}^{k-1} (10 + \\ell + i) = k\\left(10 + \\ell + \\frac{1}{2}(k-1)\\right),\n$$\nwhere $\\ell \\ge 0$ ($10 + \\ell$ represents the starting number of the consecutive numbers in a set) and $k \\ge 2$ represents the number of consecutive numbers in a set. Each of these sums (the $(k+1)$-th number in the set) must be less than, or equal to, $99$. By taking $\\ell = 0$, we see that the maximum possible value for $k$ is $7$.\n\nFor the case $k = 2$ (i.e., where three cards are used), we must have $2 \\cdot (10+\\ell+\\frac{1}{2}(2-1)) \\le 99$, so that $\\ell \\le 39$. The $39+1=40$ possibilities of three cards are: $10+11=21, 11+12=23, \\dots, 49+50=99$.\n\nFor the case $k = 3$ (i.e., where four cards are used), we get $3 \\cdot (10+\\ell+\\frac{1}{2}(3-1)) \\le 99$, so that $\\ell \\le 22$, representing the $22+1=23$ groups of four cards: $10+11+12=33, \\dots, 32+33+34=99$.\n\nContinuing in this way, we find that $\\ell \\le 13$ when $k = 4$, $\\ell \\le 7$ when $k = 5$, $\\ell \\le 4$ when $k = 6$, and $\\ell \\le 1$ when $k = 7$. All in all, there are $40+23+14+8+5+2=92$ sets of cards satisfying the conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42360,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$, $C$ be points on a circle whose centre is $O$ and whose radius is $1$, such that $\\angle BAC = 45^\\circ$. Lines $AC$ and $BO$ (possibly extended) intersect at $D$, and lines $AB$ and $CO$ (possibly extended) intersect at $E$. Prove that $BD \\cdot CE = 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nFigure 1\n\nPut $x = OD$ and $y = OE$. Let $BO$ extended meet the circle again in $F$ (so that $DF = 1-x$). Since $\\angle BFC = 45^\\circ$ and $BF$ is a diameter of the circle, we have $\\angle BCF = 90^\\circ$ and $BC = CF = \\sqrt{2}$. Similarly, if $CE$ extended meets the circle again in $G$, we have $GE = 1-y$, and $CG$ and $BF$ are the diagonals of square $BCFG$ that intersect perpendicularly at $O$. Therefore, from the right triangles $BOE$ and $COD$, we have $CD = \\sqrt{1+x^2}$ and $BE = \\sqrt{1+y^2}$.\nBy the similarity of triangles $BDA$ and $CDF$, we have $\\frac{AD}{DF} = \\frac{BD}{CD}$, i.e., $\\frac{AD}{1-x} = \\frac{1+x}{\\sqrt{1+x^2}}$, so that\n$$AD = \\frac{1-x^2}{\\sqrt{1+x^2}}.$$\nThen, by the similarity of triangles $BEG$ and $CEA$, $\\frac{AD+DC}{GB} = \\frac{CE}{BE}$, i.e.,\n$$\n\\frac{\\frac{1-x^2}{\\sqrt{1+x^2}} + \\sqrt{1+x^2}}{\\sqrt{2}} = \\frac{1+y}{\\sqrt{1+y^2}}.\n$$\nSimplifying and solving for $x$ gives $x = \\frac{1-y}{1+y}$, so that\n$$\nBD \\cdot CE = (1+x)(1+y) = \\left(1 + \\frac{1-y}{1+y}\\right) (1+y) = 2.\n$$\nNote that we have solved here the case where $A$ lies strictly between $G$ and $F$ on the short arc $\\widehat{FG}$. The case where $A = F$ or $A = G$ follows trivially, because here $BD \\cdot CE$ reduces to either $BF \\cdot CO = 2 \\cdot 1 = 2$ or $CG \\cdot BO = 2$. The case where $A$ lies between $C$ and $F$ on the short arc $\\widehat{CF}$ is shown in Figure 2.\n\n![](attached_image_2.png)\n\nFigure 2\n\nIn a way similar to the first case, using the similarity between triangles $BDA$ and $CDF$, we see that $\\frac{CD}{2+x} = \\frac{\\sqrt{1+(1+x)^2}}{2+x} = \\frac{x}{AD}$, so that $AD = \\frac{x(2+x)}{\\sqrt{1+(1+x)^2}}$. The similarity between triangles $BEG$ and $CEA$ gives\n$$\n\\frac{1-y}{\\sqrt{1+y^2}} = \\frac{CA}{\\sqrt{2}} = \\frac{CD-AD}{\\sqrt{2}} = \\frac{\\sqrt{1+(1+x)^2} - \\frac{x(2+x)}{\\sqrt{1+(1+x)^2}}}{\\sqrt{2}}\n$$\nFrom this equation we solve $x = \\frac{2y}{1-y}$ and it follows that $BD \\cdot CE = (2+x)(1-y) = (2+\\frac{2y}{1-y})(1-y) = 2$.\nFinally, the case where $A$ lies between $B$ and $G$ on the short arc $\\overset{\\sim}{B}G$ follows symmetrically from the above.\nWe can also follow a trigonometric approach: We first consider the case where $O$ is inside triangle $ABC$. See Figure 3.\n\n![](attached_image_3.png)\n\nFigure 3\n\nWe have $\\angle BOC = 90^\\circ$ (twice $\\angle BAC = 45^\\circ$). Put $x = OD$, $y = OE$, $\\alpha = \\angle ACE$, $\\beta = \\angle ABD$. Then $\\tan \\alpha = x$ and $\\tan \\beta = y$. Furthermore, $\\alpha + \\beta = 45^\\circ$ (using $180^\\circ = \\angle BAD + \\angle ABD + \\angle BDA = 45^\\circ + \\beta + (90^\\circ + \\alpha)$), so that\n$$\n1 = \\tan(\\alpha + \\beta) = \\frac{\\tan \\alpha + \\tan \\beta}{1 - \\tan \\alpha \\tan \\beta} = \\frac{x+y}{1-xy}.\n$$\nHence, $x+y = 1-xy$, and we have $BD \\cdot CE = (1+x)(1+y) = 1+x+y+xy = 1+1 = 2$.\n\nIn Figure 4 we have $O$ outside triangle $ABC$.\n\n![](attached_image_4.png)\n\nFigure 4\n\nIn this case we have $\\beta - \\alpha = 45^\\circ$ (using $180^\\circ = \\angle BAC + \\angle ABC + \\angle BCA = 45^\\circ + (\\beta + 45^\\circ) + (45^\\circ - \\alpha)$), so that\n$$\n1 = \\tan(\\beta - \\alpha) = \\frac{\\tan \\beta - \\tan \\alpha}{1 + \\tan \\alpha \\tan \\beta} = \\frac{y-x}{1+xy}.\n$$\nHence, $y-x = 1+xy$, and we have $BD \\cdot CE = (1-x)(1+y) = 1-x+y-xy = 1+1 = 2$.\nFinally, there are the cases where $O$ is either on $AB$ or on $AC$. In the former case we have $D = A$ and $E = O$, so that $BD \\cdot CE = BA \\cdot CO = 2 \\cdot 1 = 2$, and in the latter case, $D = O$ and $E = A$, so that $BD \\cdot CE = BO \\cdot CA = 1 \\cdot 2 = 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42361,
    "subject": "Mathematics (Multi-modal)",
    "question": "The squares of an $8 \\times 8$ board are coloured alternatingly black and white. A rectangle consisting of some of the squares of the board is called *important* if its sides are parallel to the sides of the board and all its corner squares are coloured black. The side lengths can be anything from $1$ to $8$ squares. On each of the $64$ squares of the board, we write the number of important rectangles in which it is contained. The sum of the numbers on black squares is $B$, and the sum of the numbers on white squares is $W$. Determine the difference $B-W$.",
    "options": [],
    "answer": "200",
    "solution": "In each important rectangle, the number of black squares is one more than the number of white squares. Hence, each important rectangle contributes $+1$ to the difference $B-W$. The value of $B-W$ is thus the same as the number of important rectangles on the board.\n\nLet us number the rows on the board $1, 2, \\ldots, 8$ from the top downwards and the columns $1, 2, \\ldots, 8$ from the left to the right. So $(1, 1)$ is the upper left square and $(8, 8)$ denotes the lower right square. Assume $(1, 1)$ is a black square. Then all $(i, j)$ with both $i$ and $j$ odd, as well as all those with both $i$ and $j$ even, are black squares. All other squares are white.\n\nBy focusing only on the four odd-numbered rows and the four odd-numbered columns, we find that they determine $(4 + \\binom{4}{2})^2 = 100$ important rectangles. Similarly, the four even-numbered rows and the four even-numbered columns determine another $100$ important rectangles, giving a total of $200$ important rectangles on the board. It follows that $B-W = 200$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42362,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ such that\n$$\nf(a^3) + f(b^3) + f(c^3) + 3f(a+b)f(b+c)f(c+a) = (f(a+b+c))^3\n$$\nfor all $a, b, c \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "All solutions f: Z -> Z are exactly the following five functions:\n1) f(n) = 0 for all integers n.\n2) f(n) = n for all integers n.\n3) f(n) = -n for all integers n.\n4) f(n) = 0 if n is divisible by 3, f(n) = 1 if n ≡ 1 mod 3, and f(n) = -1 if n ≡ 2 mod 3.\n5) The negation of (4): f(n) = 0 if n is divisible by 3, f(n) = -1 if n ≡ 1 mod 3, and f(n) = 1 if n ≡ 2 mod 3.",
    "solution": "Suppose $f$ satisfies the condition.\nBy taking $(a, b, c) = (0, 0, 0)$, we get $3f(0) + 3f(0)^3 = f(0)^3$, so that either $f(0) = 0$, or $3 = -2f(0)^2$. The latter is not possible in $\\mathbb{Z}$, so we must have $f(0) = 0$.\nBy taking $(a, b, c) = (n, -n, 0)$, we get $f(n^3) + f(-n^3) = 0$, resulting in\n$$\nf(-n^3) = -f(n^3) \\text{ for all } n \\in \\mathbb{Z}. \\qquad (1)\n$$\nBy taking $(a, b, c) = (n, 0, 0)$, we get\n$$\nf(n^3) = f(n)^3 \\text{ for all } n \\in \\mathbb{Z}. \\qquad (2)\n$$\nBy combining (1) and (2), we see that, for any $n \\in \\mathbb{Z}$,\n$$\nf(-n)^3 = f((-n)^3) = f(-n^3) = -f(n^3) = -f(n)^3 = (-f(n))^3,\n$$\nso that $f(-n) = -f(n)$, i.e., $f$ is an odd function.\nNow take $(a, b, c) = (k, 1-k, 0)$, so that\n$$\nf(k)^3 + f(1-k)^3 + 3f(k)f(1-k)f(1) = f(1)^3 \\text{ for all } k \\in \\mathbb{Z}. \\qquad (3)\n$$\nAlso, from (2), $f(1) = f(1)^3$, so that $f(1) \\in \\{-1, 0, 1\\}$.\nFirst, if $f(1) = 0$, then from (3), $f(k) = -f(1-k) = f(k-1)$ for all $k \\in \\mathbb{Z}$, so that $f(n) = 0$ for all $n \\in \\mathbb{Z}$ (using induction and the fact that $f$ is odd).\nSecond, if $f(1) = 1$, then from (3), $f(k)^3 + f(1-k)^3 + 3f(k)f(1-k) = 1$, for all $k \\in \\mathbb{Z}$, i.e., the Diophantine equation $X^3 + Y^3 + 3XY = 1$ is satisfied by $(X, Y) = (f(k), f(1-k))$. This equation can be rewritten as $(X + Y - 1)(X^2 - XY + Y^2 + X + Y + 1) = 0$. Note that $X^2 - XY + Y^2 + X + Y + 1 = 0$ is only solvable in $\\mathbb{Z}$ if $X = Y = -1$ (otherwise the discriminant is negative when considered as a quadratic in $X$). So there are two options here:\n\n1. $f(k) = 1 - f(1-k) = 1 + f(k-1)$ for all $k \\in \\mathbb{Z}$. Then it follows immediately by induction and the fact that $f$ is odd that $f(n) = n$ for all $n \\in \\mathbb{Z}$.\n2. There is some $k_0 \\in \\mathbb{Z}$ such that $f(k_0) = f(1-k_0) = -1$. Then $f(-k_0) = 1 = 1 - f(1+k_0)$ implies that $f(1+k_0) = 0$. Hence, by using $(a, b, c) = (1+k_0, 1-k_0, -1)$ in the functional equation, we get that $0 - 1 - 1 + 3f(2)(-1)(1) = 1$, so that $f(2) = -1$. Thus $k_0 = 2$ is the smallest positive value of $k_0$ with the property that $f(k_0) = -1 = f(1-k_0)$. We have therefore established the base case of the proof by induction that $(f(3k), f(3k+1), f(3k+2)) = (0, 1, -1)$ for all $k \\ge 0$. Assume now that this statement is true for some $k \\ge 0$. Then, from $f(-3k-2) = 1 = 1 - f(1+3k+2)$, we find that $f(3+3k) = f(3(k+1)) = 0$. Using $(a, b, c) = (3k, 1, 3)$ in the functional equation (and recalling that $f(3) = f(1+2) = f(1+k_0) = 0$), we get $0 + 1 + 0 + 3(0)(1)(0) = f(3k+4)^3$, so that $f(3k+4) = f(3(k+1)+1) = 1$. Similarly, with $(a, b, c) = (3k, 2, 3)$, we see that $f(3k+5) = f(3(k+1)+2) = -1$, and the induction is complete. It follows now from the fact that $f$ is odd that $(f(3k), f(3k+1), f(3k+2)) = (0, 1, -1)$ for all $k \\in \\mathbb{Z}$, i.e.,\n$$\nf(n) = \\begin{cases} 0 & \\text{if } n \\equiv 0 \\pmod{3} \\\\ 1 & \\text{if } n \\equiv 1 \\pmod{3}, & \\text{for all } n \\in \\mathbb{Z}. \\\\ -1 & \\text{if } n \\equiv 2 \\pmod{3} \\end{cases} \\quad (4)\n$$\nFinally, the observation that whenever $f$ satisfies the equation then $-f$ also satisfies the equation, takes care of the case $f(1) = -1$.\nIt is straightforward to check that the three functions $f(n) = 0$, $f(n) = n$ and $f(n) = -n$ (for all $n \\in \\mathbb{Z}$) satisfy the given functional equation. The function (4) requires a little more effort (and time) to check, but offers no difficulty. Its negative then follows automatically. We conclude that there are five functions that solve the equation, as described above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42363,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(m, n)$ of non-negative integers that satisfy the equation\n$$\n20^m - 10m^2 + 1 = 19^n.\n$$",
    "options": [],
    "answer": "(2, 2)",
    "solution": "Let the pair $(m, n)$ satisfy the equation. $m = 0$ implies that $2 = 19^n$, an impossibility. So we must have $m > 0$. Taking both sides modulo $10$ gives $1 \\equiv (-1)^n \\pmod{10}$, implying that $n$ must be even. Taking both sides modulo $20$ gives $-10m^2 + 1 \\equiv (-1)^n \\pmod{20}$, so that $2 \\mid m^2$, whence $m$ is even too. So let us put $m = 2k$ and $n = 2l$ for a positive integer $k$ and a non-negative integer $l$. The original equation can now be written as\n$$\n10m^2 - 1 = 20^{2k} - 19^{2l} = (20^k - 19^l)(20^k + 19^l).\n$$\nAs both $20^k + 19^l$ and $10m^2 - 1$ are positive (recall that $m \\ge 1$), we must have that the integer $20^k - 19^l \\ge 1$. It follows that\n$$\n10m^2 - 1 = (20^k - 19^l)(20^k + 19^l) \\ge 20^k + 19^l \\ge 20^k + 1,\n$$\nand we see that $20^k \\le 10(2k)^2 - 2 = 40k^2 - 2$. This inequality is not true for $k = 2$, and it can be easily verified by induction that it also fails to hold for any $k > 2$. The inequality holds for $k = 1$, and it follows (and is easily checked) that the only pair that solves the original equation, is $(m, n) = (2, 2)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 42364,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive multiple of $20$ with exactly $20$ positive divisors.",
    "options": [],
    "answer": "240",
    "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 $M_k$.\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| $a_1 + 3$ | $a_2 + 1$ | $a_3 + 2$ | $a_1$ | $a_2$ | $a_3$ | $k = 2^{a_1} \\cdot 3^{a_2} \\cdot 5^{a_3}$ |\n|---|---|---|---|---|---|---|\n| 4 | 1 | 5 | 1 | 0 | 3 | 250 |\n| 5 | 1 | 4 | 2 | 0 | 2 | 100 |\n| 5 | 2 | 2 | 2 | 1 | 0 | 12 |\n| 10 | 1 | 2 | 7 | 0 | 0 | 128 |\n\nWe conclude that the smallest positive multiple of $20$ with exactly $20$ positive divisors is $12 \\times 20 = 240$.\nOf course, one can try to be lucky, and simply start to determine the numbers of divisors of $1 \\times 20, 2 \\times 20, 3 \\times 20, \\dots$, and soon end up with $12 \\times 20$ having $20$ positive divisors (for the first time in this sequence). Hence, $12 \\times 20$ is the smallest such multiple.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42365,
    "subject": "Mathematics (Multi-modal)",
    "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": "4 - 2/sqrt(3)",
    "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 $BC$. See Figure 1.\n\n![](attached_image_1.png)\nFigure 1\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$$\n4 - 4 \\cdot \\frac{1}{2} \\cdot 1 \\cdot \\frac{1}{\\sqrt{3}} = 4 \\cdot \\left(1 - \\frac{1}{2\\sqrt{3}}\\right).\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42366,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x$, $y$, $z$ are real numbers satisfying\n$$\n\\begin{aligned}\n(x + 1)(y + 1)(z + 1) &= 3 \\\\\n(x + 2)(y + 2)(z + 2) &= -2 \\\\\n(x + 3)(y + 3)(z + 3) &= -1,\n\\end{aligned}\n$$\nfind the value of\n$(x + 20)(y + 20)(z + 20)$.",
    "options": [],
    "answer": "6748",
    "solution": "By putting $X = x+2$, $Y = y+2$ and $Z = z+2$, the system of equations translates to\n\n$$\n(X - 1)(Y - 1)(Z - 1) = 3 \\quad \\dots (1)\n$$\n\n$$\nXYZ = -2 \\quad \\dots (2)\n$$\n\n$$\n(X + 1)(Y + 1)(Z + 1) = -1. \\quad \\dots (3)\n$$\n\nBy expanding the left hand sides of (1) and (3), and substituting $XYZ = -2$, we obtain the system\n\n$$\n\\begin{aligned}\nT_1 - T_2 &= 6 \\\\\nT_1 + T_2 &= 0,\n\\end{aligned}\n$$\n\nwhere $T_1 = X + Y + Z$ and $T_2 = XY + YZ + ZX$. From this, we solve $T_1 = 3$ and $T_2 = -3$, so that\n\n$$\n\\begin{aligned}\n(x + 20)(y + 20)(z + 20) &= (X + 18)(Y + 18)(Z + 18) \\\\\n&= -2 + 18 \\cdot T_2 + 18^2 \\cdot T_1 + 18^3 \\\\\n&= -2 + 18 \\cdot (-3) + 18^2 \\cdot (3) + 18^3 \\\\\n&= 6748.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42367,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $k$ is said to be *visionary* if there are integers $a > 0$ and $b \\ge 0$ such that $a \\cdot k + b \\cdot (k+1) = 2020$. How many visionary integers are there?",
    "options": [],
    "answer": "88",
    "solution": "All lower case variables in this solution denote integers. Let $X$ denote the set of visionary integers. We show that $X = \\{\\lfloor \\frac{2020}{n} \\rfloor : 1 \\le n \\le 2020\\}$.\n\nIf $k$ is a visionary integer, then there exist $a > 0$ and $b \\ge 0$ such that $a k + b(k + 1) = 2020$, i.e., $2020 = k(a + b) + b$, where $0 \\le b < a + b$. This implies that $k = \\lfloor \\frac{2020}{a+b} \\rfloor$ and we also have $1 \\le a + b \\le 2020$, since $a \\ge 1$ and $k \\ge 1$. Hence $k \\in X$. Conversely, let $k \\in X$, i.e., $k = \\lfloor \\frac{2020}{n} \\rfloor$ for some $1 \\le n \\le 2020$. Then $2020 = k n + r$, where $0 \\le r < n$. Put $b = r$ and $a = n - r$. Then $a > 0$ and $b \\ge 0$, and we see that $a k + b(k + 1) = 2020$, i.e., $k$ is visionary.\n\nIn order to solve the problem, we need to find the cardinality of the set $X$. To this end, write $X$ as the disjoint union of $X_1 = \\{\\lfloor \\frac{2020}{n} \\rfloor : 1 \\le n < \\sqrt{2020}\\}$ and $X_2 = \\{\\lfloor \\frac{2020}{n} \\rfloor : \\sqrt{2020} < n \\le 2020\\}$. (Note that 2020 is not a square.) Also, since the smallest element of $X_1$ is $\\lfloor \\frac{2020}{\\lfloor \\sqrt{2020} \\rfloor} \\rfloor = \\lfloor \\frac{2020}{44} \\rfloor = 45$, and the largest element of $X_2$ is $\\lfloor \\frac{2020}{\\lfloor \\sqrt{2020} \\rfloor+1} \\rfloor = \\lfloor \\frac{2020}{45} \\rfloor = 44$, the sets $X_1$ and $X_2$ are indeed disjoint.\n\nLet $1 \\le n_1 < n_2 \\le \\lfloor \\sqrt{2020} \\rfloor$. If $\\lfloor \\frac{2020}{n_1} \\rfloor = \\lfloor \\frac{2020}{n_2} \\rfloor$, then $0 < \\frac{2020}{n_1} - \\frac{2020}{n_2} < 1$, implying that $0 < n_2 - n_1 < 1$, an impossibility. This shows that $X_1$ has exactly $\\lfloor \\sqrt{2020} \\rfloor = 44$ elements.\n\nNext, consider any $n$ such that $1 \\le n \\le \\lfloor \\sqrt{2020} \\rfloor$. We show that there exists a $q$, where $\\sqrt{2020} < q \\le 2020$, such that $\\lfloor \\frac{2020}{q} \\rfloor = n$. By the Division Algorithm, there exist (unique) $q$ and $r$ such that $2020 = q n + r$, where $0 \\le r < n$. Now if $q \\le \\lfloor \\sqrt{2020} \\rfloor$, then $2020 = q n + r < q n + n = n(q+1) \\le \\lfloor \\sqrt{2020} \\rfloor \\cdot (\\lfloor \\sqrt{2020} \\rfloor + 1) = 44 \\cdot 45 = 1980$, a contradiction. So we have $\\sqrt{2020} < q \\le 2020$. Moreover, $n = \\lfloor \\frac{2020}{q} \\rfloor$, as $2020 = n q + r$, where $0 \\le r < n < q$. Finally, if $\\sqrt{2020} < n \\le 2020$, then $1 \\le \\frac{2020}{n} < \\sqrt{2020}$, so that also $1 \\le \\lfloor \\frac{2020}{n} \\rfloor < \\sqrt{2020}$, i.e., all elements of $X_2$ lie in the interval $[1, \\lfloor \\sqrt{2020} \\rfloor]$. This shows that $|X_2| = |X_1|$, and we conclude that $|X| = |X_1| + |X_2| = 44 + 44 = 88$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42368,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, and let $T$ be a point on the extension of $AB$ beyond $B$, and $U$ a point on the extension of $AC$ beyond $C$, such that $BT = CU$. Moreover, let $R$ and $S$ be points on the extensions of $AB$ and $AC$ beyond $A$ such that $AS = AT$ and $AR = AU$. Prove that $R, S, T, U$ lie on a circle whose centre lies on the circumcircle of $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nFigure 2\n\nSince $AS = AT$, we have $\\angle AST = \\angle ATS = \\frac{1}{2}\\angle BAC$. Similarly, $\\angle ARU = \\angle AUR = \\frac{1}{2}\\angle BAC$, so that $\\angle S = \\angle R$. This implies that $R, S, T$ and $U$ are concylic.\n\nWe now show that the centre $O$ of the circle $\\omega$ through $R, S, T$ and $U$, lies on the circumcircle of $ABC$. We know that $O$ lies on the perpendicular bisector of $ST$ (which is also the perpendicular bisector of $RU$, since $ST \\parallel RU$). This perpendicular bisector forms a diameter of $\\omega$, and contains $A$.\n\nIf $O = A$, then we are finished, since $A$ is certainly on the circumcircle of $ABC$. So suppose that the points $O$ and $A$ are different, as shown in Figure 2. Drop perpendiculars from $O$ to $BR$ (which is $BA$ extended), to $AC$, and to $CB$. Let the feet of these perpendiculars be $I, J$ and $K$, respectively. It is known that $O$ is on the circumcircle of $ABC$ if and only if the points $I, J$ and $K$ are collinear. (In case this happens, the line through $I, J, K$ is called the *Simson line* of $ABC$ determined by $O$.)\n\nIn order to show that $I, J, K$ are collinear, it is sufficient to show that $\\angle JKO = \\angle IKO$. Since $\\angle OJC = \\angle OKC = 90^\\circ$, $OJKC$ is a cyclic quadrilateral, so that $\\angle JKO = \\angle JCO$. Our next observation is that $IT = JU$. This follows from the fact that $OI = OJ$ (from symmetry – recall that triangle $RAU$ is isosceles, and $AO$ is on the perpendicular bisector of $RU$), and $OT = OU$, giving $IT^2 = OT^2 - OI^2 = OU^2 - OJ^2 = JU^2$. Hence, $IB = IT - BT = JU - CU = JC$, from which we get that triangles $OIB$ and $OJC$ are congruent. We therefore have $\\angle IBO = \\angle JCO$. Finally, since $IOKB$ is a cyclic quadrilateral ($\\angle BIO = \\angle BKO = 90^\\circ$), we also have $\\angle IBO = \\angle IKO$.\n\nPutting everything together, we conclude that $\\angle JKO = \\angle JCO = \\angle IBO = \\angle IKO$, and we are done.\nThe same strategy as in Solution 1 shows that $RSTU$ is cyclic.\n\nThe centre $O$ must lie on the four perpendicular bisectors of $RS, ST, SU$ and $RU$. But since triangles $RAU$ and $SAT$ are isosceles, the perpendicular bisectors of $ST$ and $RU$ pass through $A$ and bisect the angle $RAU$.\n\nNow $\\angle TOU = 2\\angle TSU$, since $O$ is the centre of circle $RSTU$, which equals $\\angle TSU + \\angle ATS$, since $AS=AT$, which in turn equals $\\angle TAU$, the exterior angle in triangle $AST$. This proves that $AOUT$ is cyclic and hence $\\angle BTO = \\angle ATO = \\angle AUO = \\angle CUO$.\n\nAlong with $OT=OU$ (radii) and the given $BT=CU$, we conclude that triangles $BTO$ and $CUO$ are congruent, so $\\angle BOT = \\angle COU$. Finally $\\angle BAC = \\angle TAU = \\angle TOU = \\angle BOC - \\angle BOT + \\angle COU = \\angle BOC$ and thus $ABCO$ is cyclic.\n\n(The problem could also be finished by noting that $O$ lies on the perpendicular bisector of $BC$ and the exterior angle bisector of $A$ and thus lies on the circumcircle, but this assumes some extra knowledge that the above presentation doesn't.)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42369,
    "subject": "Mathematics (Multi-modal)",
    "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": "1000977",
    "solution": "The answer is $1000977$. Let $M$ 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, $0 \\le b < 2^a$) or $(2^a + b)^2 + 2^a + 3b + 2$ ($a, b$ nonnegative integers, $0 \\le b < 2^a - 1$), 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 $(n-1)^2 < M \\le n^2$. Then the $M$ members will stand in an $n \\times n$ square, and if $M \\ne n^2$, 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 $0 \\le b < 2^a$. 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 $M \\le n^2 - n + 1$, so that $M - n \\le (n-1)^2 = (2^a + b)^2$. 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$$\nM = (2^a + b - 1)^2 + 2^a + 3b = (2^a + b)^2 - (2^a - b - 1) \\le (n - 1)^2.\n$$\n\nThe latter equation yields\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$$\nM = (2^a + b)^2 + 2^a + 3b + 2,\n$$\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$$\nM = (2^a + b)^2 + 2^{a+1} + 4b + 3 = (2^a + b + 1)^2 + 2(b + 1) > n^2,\n$$\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": 42370,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest and largest integers with decimal representation of the form $ababa$ ($a \\neq 0$) that are divisible by 11.",
    "options": [],
    "answer": "17171 and 98989",
    "solution": "The number $N = \\underline{ababa}$ is divisible by 11 if and only if $a-b+a-b+a = 3a-2b$ is divisible by 11. To find the smallest possible $N$ with this property, we observe that $a=1$ (the smallest possible value for $a$) and $b=7$ offer a solution: $3 \\cdot 1 - 2 \\cdot 7 = -11$ is divisible by 11 (and no smaller $b \\in \\{0, 1, \\dots, 9\\}$ works). So $N = 17171$ is the smallest number of the required form that is divisible by 11. Similarly, the largest possible value for $a$ is $a=9$, and we do have a solution in this case, with $b=8$: $3 \\cdot 9 - 2 \\cdot 8 = 11$ is divisible by 11 (and no larger $b \\in \\{0, 1, \\dots, 9\\}$ works). So $N = 98989$ is the largest number of the required form that is divisible by 11.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42371,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $PAB$ and $PBC$ 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": "sqrt(2)",
    "solution": "![](attached_image_1.png)\nFigure 1\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 $AC = PC = x + \\frac{1}{x}$.\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$$\n(x + \\frac{1}{x})^2 = 1^2 + \\frac{x^2+1}{x^2} - 2 \\cdot \\frac{\\sqrt{x^2+1}}{x} \\cdot \\cos(180^\\circ - \\alpha) = 1 + \\frac{x^2+1}{x^2} + 2 \\cdot \\frac{\\sqrt{x^2+1}}{x} \\cdot \\cos \\alpha = 3 + \\frac{x^2+1}{x^2}.\n$$\n\nWe now easily solve for $x$, and find that $\\frac{PA}{AB} = x = \\sqrt{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42372,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest integer $k > 1$ such that there exist $k$ distinct primes whose squares sum to a power of $2$.",
    "options": [],
    "answer": "5",
    "solution": "For $p_1^2 + p_2^2$ (where $p_1$ and $p_2$ are two distinct primes) to be equal to $2^n$ (where $n \\ge 2$), we must have both $p_1$ and $p_2$ odd. This would give $p_1^2 + p_2^2 \\equiv 2 \\pmod 4$, while $2^n \\equiv 0 \\pmod 4$, a contradiction.\n\nFor $p_1^2 + p_2^2 + p_3^2$ (where $p_1, p_2$ and $p_3$ are three distinct primes) to be equal to $2^n$ (where $n \\ge 2$), we must have (say) $p_1 = 2$ and both $p_2$ and $p_3$ odd. This would give $p_1^2 + p_2^2 + p_3^2 \\equiv 2 \\pmod 4$, while $2^n \\equiv 0 \\pmod 4$, a contradiction.\n\nFor $p_1^2+p_2^2+p_3^2+p_4^2$ (where $p_1, p_2, p_3$ and $p_4$ are four distinct primes) to be equal to $2^n$ (where $n \\ge 3$), we must have all four of these primes odd, say $p_i = 2m_i+1$, $i = 1, 2, 3, 4$. This would give $\\sum_{i=1}^4 p_i^2 = \\sum_{i=1}^4 (2m_i+1)^2 = \\sum_{i=1}^4 (4m_i(m_i+1)+1) = 8L+4 = 2^n$ for some integer $L$, since each $m_i(m_i+1)$ is even. Then $2L+1 = 2^{n-2}$ would be a contradiction, as $2^{n-2}$ is even if $n \\ge 3$.\n\nWe see that the smallest $k > 1$ that we are looking for must satisfy $k \\ge 5$. Now, for the sum of the squares of five distinct primes to be a power of $2$, one of them must be $2$. The sum of the squares of the five smallest primes is not a power of two, but the very next choice of smallest primes does indeed satisfy the requirement: $2^2 + 3^2 + 5^2 + 7^2 + 13^2 = 256 = 2^8$. Hence $k = 5$ is the smallest $k$ we are looking for.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42373,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $\\angle ABC \\ne 90^\\circ$ and $AB$ its shortest side. Denote by $H$ the intersection of the altitudes of triangle $ABC$. Let $K$ be the circle through $A$ with centre $B$. Let $D$ be the other intersection of $K$ and $AC$. Let $K$ intersect the circumcircle of $BCD$ again at $E$. If $F$ is the intersection of $DE$ and $BH$, show that $BD$ is tangent to the circle through $D, F$ and $H$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider Figure 2:\n\n![](attached_image_1.png)\nFigure 2\n\nWe note that $H$ must also be on $V$, the circumcircle of triangle $BDC$. This is because triangles $BLH$ and $BMD$ are similar (note that triangle $ABD$ is iscoceles, with $BA = BD$, and $BM$ is a perpendicular bisector of $AD$, that also bisects $\\angle ABD$), implying that angles $BHC$ and $BDC$ are equal, showing that $BHDC$ is a cyclic quadrilateral.\n\nLet $O$ be the centre of $W$, the circumcircle of triangle $HDF$ and drop the perpendicular from $O$ to $DF$, with foot $P$. Put $\\theta = \\angle DOP$, so that $\\angle DOF = 2\\theta$, giving $\\angle DHF = \\theta$. Since $HM$ is a perpendicular bisector of $AD$, and $HA = HD$, it follows that $\\angle HAC = 90^\\circ - \\theta$. Thus, $\\angle ACN = \\theta$. But the chords $BD$ and $BE$ of circle $V$ have equal length (since they are both radii of circle $K$), hence subtend the same angle in circle $V$, giving $\\angle BCE = \\theta$. But then $\\angle BDE = \\theta$, so that $\\angle FDR = \\theta$, where $R$ is an arbitrary point on the line through $B$ and $D$, with $R$ and $B$ on opposites sides of $D$.\n\nFinally, since $\\angle ODP = 90^\\circ - \\theta$, we conclude that $\\angle ODR = 90^\\circ$, i.e. the line through $B$ and $D$ must be tangent to $W$, at the point $D$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42374,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $a(x), b(x), c(x), d(x)$ with real coefficients satisfying the simultaneous equations\n\n$$\n\\begin{aligned}\nb(x)c(x) + a(x)d(x) &= 0 \\\\\na(x)c(x) + (1 - x^2)b(x)d(x) &= x + 1\n\\end{aligned}\n$$",
    "options": [],
    "answer": "{(a(x), b(x), c(x), d(x)) : b(x) = 0, d(x) = 0, and a(x)c(x) = x + 1} = {(k, 0, (x+1)/k, 0) : k ∈ ℝ \\ {0}} ∪ {((x+1)/k, 0, k, 0) : k ∈ ℝ \\ {0}}",
    "solution": "We first show that it is not possible for all four polynomials to be non-zero. Suppose they are. Denote the leading coefficients of the polynomials $a(x), b(x), c(x), d(x)$ (which exist, because the polynomials are non-zero) by $A, B, C, D$, respectively. Then the first equation implies $BC = -AD$ and thus $ABCD = -(BC)^2 < 0$. In the second equation, the leading coefficient of $a(x)c(x)$ is $AC$ and the leading coefficient of $(1-x^2)b(x)d(x)$ is $-BD$. Since the degree of $(1-x^2)b(x)d(x)$ is at least 2, these leading coefficients must cancel if we wish to end up with $x+1$ on the right. Thus $AC - BD = 0$, which implies that $ABCD = (AC)^2 > 0$. We have two contradictory inequalities, proving that this system has no solution when all four polynomials are non-zero.\n\nNow suppose that $a(x) = 0$. Then, by the first equation, $b(x) = 0$ or $c(x) = 0$. If $b(x) = 0$, we see that the second equation is not satisfied. If $c(x) = 0$, we see that, again, the second equation is not satisfied, due to differences in degrees. Similarly, when $c(x) = 0$, we find, in a symmetrical way, that there are no solutions.\n\nHenceforth, we assume that both $a(x)$ and $c(x)$ are non-zero. So either $b(x)$ or $d(x)$ (or both) must be zero. From the first equation, $b(x) = 0$ if and only if $d(x) = 0$. It follows that the complete set of solutions is given by all $(a(x), b(x), c(x), d(x))$ where $a(x)c(x) = x+1$ and $b(x) = d(x) = 0$, i.e.,\n\n$$\n\\{(k, 0, \\frac{x}{k} + \\frac{1}{k}, 0) : k \\text{ a non-zero real number}\\} \\cup \\{(\\frac{x}{k} + \\frac{1}{k}, 0, k, 0) : k \\text{ a non-zero real number}\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 42375,
    "subject": "Mathematics (Multi-modal)",
    "question": "Jacob and Laban take turns playing a game. Each of them starts with a list of square numbers $1, 4, 9, \\dots, 2021^2$, and there is a whiteboard in front of them with the number $0$ on it. Jacob chooses a number $x^2$ from his list, removes it from his list, and replaces the number $W$ on the whiteboard with $W + x^2$. Laban then does the same with a number from his list, and they repeat back and forth until both of them have no more numbers in their list. Now every time that the number on the whiteboard is divisible by $4$ after a player has taken their turn, Jacob gets a sheep. Jacob wants to have as many sheep as possible by the end of the game, whereas Laban wants Jacob to have as few sheep as possible. What is the greatest number $K$ such that Jacob can guarantee to get at least $K$ sheep by the end of the game, no matter how Laban plays?",
    "options": [],
    "answer": "506",
    "solution": "Since $n^2 \\equiv 0 \\pmod{4}$ if and only if $n$ is even, and $n^2 \\equiv 1 \\pmod{4}$ if and only if $n$ is odd, we can simplify notation by replacing the even squares by $0$ and the odd squares by $1$, in each of the two lists. Thus Jacob and Laban each has a pool of $1010$ zeros and $1011$ ones to choose from when the game starts.\n\nOur first observation is that, no matter in which order they write down their ones and zeros on the board, after every fourth one, Jacob will earn another sheep. Since there are in total $2022$ ones between the two players, and $2022 = 4 \\times 505 + 2$, Jacob is guaranteed to have at least $505$ sheep after the game. But since Jacob has the privilege of playing the first move, he can start with one of his zeros, which will secure a further sheep for himself, leaving him with a total of $506$ (or more) sheep after the game.\n\nWe now argue that if Laban plays intelligently, he can make sure that Jacob only get these $506$ sheep, and nothing more. The secret for him is to make sure that no zeros are played (by anybody) immediately after every fourth one. Such zeros would add more sheep to Jacob's collection. So, immediately after every $4k$-th one (where $1 \\le k \\le 505$), another one should follow. Laban can easily achieve this by writing down a one after the zero with which Jacob has started the game, and after that simply copying each move of Jacob. This will ensure that Jacob must write down every $4k$-th one (for all $1 \\le k \\le 505$), and Laban then follows with another one immediately after each of these ones, achieving his goal.\n\nBy following this rule, Laban will not run out of zeros or ones before the $2020$-th one has been used. After the first two moves (a zero by Jacob and a one by Laban), Laban has more zeros than Jacob, and one one less than Jacob in his pool, but since there are two spare ones left after $2020$ ones have been used, this offers no problem for Laban. Any zeros or ones played after the $2021$-th one will not contribute to a sum divisible by four, hence no further sheep for Jacob.\n\nThere is, of course, also the possibility that Jacob decides (for whatever reason) to start with a one, and not a zero. If this is the case, then Laban simply copies each move by Jacob right from the start, except the second one played by Jacob (which will be the third one of the game). After this one, Laban should proceed with a zero (if there are any zeros left in his list), and then continue to copy each move of Jacob after that. In doing so, no zero will be played directly after the $4k$-th one (for any $1 \\le k \\le 505$). If there are no zeros left in Laban's list after the third one of the game, then there will also be no zeros left in Jacob's list. So Laban simply continues with the ones from his list and Jacob cannot earn any more sheep than the guaranteed $505$ produced by every fourth one. Hence, with this option where Jacob starts with a one, he is only guaranteed to earn $505$ sheep.\n\nWe conclude that Jacob can guarantee (by starting with a zero) to collect $506$ sheep for himself, but no more.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42376,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider $16$ points arranged as shown, with horizontal and vertical distances of $1$ between consecutive rows and columns. In how many ways can one choose four of these points such that the distance between every two of those four points is strictly greater than $2$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "7",
    "solution": "Label the sixteen points $A$, $B$, $C$, $\\ldots$, $P$, as shown below in Figure 1:\n\n![](attached_image_1.png)\n\nA selection of four points satisfying the condition that the distance between every two of the four points is greater than $2$, will be called a valid selection.\n\nLet us first consider a valid selection which includes one of the points of the inner square $FGKJ$. Clearly, only one of these points ($F$, $G$, $K$, $J$) can be used; say we select $F$. The only points from the outer square $ABCDHLPONMIE$ at distance more than $2$ from $F$, are $D$, $L$, $P$, $O$ and $M$. If we choose either $L$ or $O$, then there are not enough points left among the remaining ones on the outer square to form a valid selection. We are therefore forced to select $D$, $P$ and $M$, together with $F$, to obtain a valid selection. Similarly, by symmetry, there are three further valid selections that contain points from the inner square: $\\{G, P, M, A\\}$, $\\{K, M, A, D\\}$ and $\\{J, A, D, P\\}$.\n\nAll that remains, is to consider valid selections using only points from the outer square. We note that the only way to choose two points in a valid selection from the same side of the outer square, is to choose two corner points, such as $A$ and $D$. But then the only option for the other two points in the valid selection would be to choose the other two corner points of the outer square, $P$ and $M$, giving the valid selection $\\{A, D, P, M\\}$. Moving away from corner points, leaves us with the two remaining valid selections (where only one point from each of the sides of the outer square is selected), namely $\\{B, H, O, I\\}$ and $\\{C, L, N, E\\}$.\n\nHence, there are seven possible valid selections.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42377,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of real numbers $x$ and $y$ which satisfy the following equations:\n$$\n\\begin{aligned}\nx^2 + y^2 - 48x - 29y + 714 &= 0 \\\\\n2xy - 29x - 48y + 756 &= 0\n\\end{aligned}\n$$",
    "options": [],
    "answer": "[(28, 7), (20, 22), (33/2, 37/2), (63/2, 21/2)]",
    "solution": "The two equations can be rewritten as\n$$\n\\begin{aligned}\n(x - 24)^2 + \\left(y - \\frac{29}{2}\\right)^2 &= \\frac{289}{4} \\\\\n(x - 24)\\left(y - \\frac{29}{2}\\right) &= -30.\n\\end{aligned}\n$$\nBy putting $X = x - 24$ and $Y = y - \\frac{29}{2}$, we obtain\n$$\nX^2 + Y^2 = \\frac{289}{4} \\tag{1}\n$$\n$$\nXY = -30. \\tag{2}\n$$\nFrom (2) we have $Y = -30/X$. Plug this into (1) to get $X^2 + \\frac{900}{X^2} = \\frac{289}{4}$, which simplifies to\n$$\nX^4 - \\frac{289}{4}X^2 + 900 = 0.\n$$\nWe can now solve for $X^2$, using the quadratic formula:\n$$\n\\begin{aligned}\nX^2 &= \\frac{1}{2} \\left( \\frac{289}{4} \\pm \\sqrt{\\left(\\frac{289}{4}\\right)^2 - 60^2} \\right) \\\\\n&= \\frac{1}{2} \\left( \\frac{289}{4} \\pm \\sqrt{\\left(\\frac{289}{4} - 60\\right) \\left(\\frac{289}{4} + 60\\right)} \\right) \\\\\n&= \\frac{1}{2} \\left( \\frac{289}{4} \\pm \\sqrt{\\frac{49}{4} \\cdot \\frac{529}{4}} \\right) \\\\\n&= \\frac{1}{2} \\left( \\frac{289 \\pm 161}{4} \\right),\n\\end{aligned}\n$$\nso that $X^2 = \\frac{225}{4}$ or $X^2 = 16$. Thus, $X = \\pm\\frac{15}{2}$ or $X = \\pm 4$, and correspondingly, $Y = \\mp 4$ or $Y = \\mp\\frac{15}{2}$. Using $x = X + 24$ and $y = Y + \\frac{29}{2}$, we find that $(x, y)$ solves the original set of equations if and only if $(x, y) \\in \\{(28, 7), (20, 22), (\\frac{33}{2}, \\frac{37}{2}), (\\frac{63}{2}, \\frac{21}{2})\\}$.\nProceed as in Solution 1 up to the two equations\n$$\nX^2 + Y^2 = \\frac{289}{4} \\tag{1}\n$$\n$$\nXY = -30. \\tag{2}\n$$\nBy either adding $2XY = -60$ to (1), or subtracting it from (1), we get, respectively,\n$$\n\\begin{aligned}\nX^2 + 2XY + Y^2 &= (X + Y)^2 = \\frac{49}{4} = \\left(\\frac{7}{2}\\right)^2 \\\\\nX^2 - 2XY + Y^2 &= (X - Y)^2 = \\frac{529}{4} = \\left(\\frac{23}{2}\\right)^2.\n\\end{aligned}\n$$\nFrom this, we have to solve four systems of equations:\n$$\n\\begin{aligned}\nX + Y &= \\pm\\frac{7}{2} \\\\\nX - Y &= \\pm\\frac{23}{2}\n\\end{aligned}\n$$\n\n| X + Y | X - Y | X | Y | x | y |\n|-------|-------|----|----|----|----|\n| $\\frac{7}{2}$ | $\\frac{23}{2}$ | $\\frac{15}{2}$ | $-4$ | $\\frac{63}{2}$ | $\\frac{21}{2}$ |\n| $\\frac{7}{2}$ | $-\\frac{23}{2}$ | $-4$ | $\\frac{15}{2}$ | $20$ | $22$ |\n| $-\\frac{7}{2}$ | $\\frac{23}{2}$ | $4$ | $-\\frac{15}{2}$ | $28$ | $7$ |\n| $-\\frac{7}{2}$ | $-\\frac{23}{2}$ | $-\\frac{15}{2}$ | $4$ | $\\frac{33}{2}$ | $\\frac{37}{2}$ |\n\nWe conclude that there are exactly four pairs $(x, y)$ of real numbers that solve the original two equations, namely $(\\frac{63}{2}, \\frac{21}{2})$, $(20, 22)$, $(28, 7)$ and $(\\frac{33}{2}, \\frac{37}{2})$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 42378,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, and $c$ be nonzero integers. Show that there exists an integer $k$ such that\n$$\n\\gcd(a + kb, c) = \\gcd(a, b, c).\n$$\n(Note: 'gcd' stands for 'greatest common divisor')",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We may assume that $a$, $b$, $c$ are all positive, since if $a$, $b$, $c$ are all positive, and $\\gcd(a + kb, c) = \\gcd(a, b, c)$ 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 $a$, $b$ and $c$ 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, $\\gcd(a + kb, c) = \\gcd(a, b, c) = 1$.\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$$\na = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_n^{\\alpha_n}\n$$\n$$\nb = p_1^{\\beta_1} p_2^{\\beta_2} \\cdots p_n^{\\beta_n}\n$$\n$$\nc = p_1^{\\gamma_1} p_2^{\\gamma_2} \\cdots p_n^{\\gamma_n}.\n$$\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$$\nk = \\begin{cases} p_{i_1} p_{i_2} \\cdots p_{i_t} & \\text{if } E \\neq \\emptyset \\\\ 1 & \\text{if } E = \\emptyset. \\end{cases}\n$$\nWe show that, for this $k$, $\\gcd(a + kb, c) = \\gcd(a, b, c) = 1$.\nSince $\\gcd(a + kb, c)$ is a divisor of $c$, the only possible way that $\\gcd(a + kb, c) > 1$ 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 $a + kb$ (where $k = p_{i_1}p_{i_2}\\cdots p_{i_t}$).\n\nSo $\\gcd(a+kb, c)$ is not divisible by any of the one-primes.\n\nSecondly, 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 $a+kb$ (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 $a+kb$.\n\nIt follows that $\\gcd(a+kb, c) = 1 = \\gcd(a, b, c)$.\n\nFinally, if $\\gcd(a, b, c) = d > 1$, then $\\gcd(\\frac{a}{d}, \\frac{b}{d}, \\frac{c}{d}) = 1$, and from the above, there exists an integer $k$ such that $\\gcd(\\frac{a}{d} + k \\cdot \\frac{b}{d}, \\frac{c}{d}) = 1 = \\gcd(\\frac{a}{d}, \\frac{b}{d}, \\frac{c}{d})$. Multiplying both sides by $d$ gives $\\gcd(a + kb, c) = d = \\gcd(a, b, c)$, and we are done.\nAs in Solution 1, we may assume that $\\gcd(a, b, c) = 1$. By Dirichlet's Theorem, the sequence $s_n = a/\\gcd(a, b) + n \\cdot b/\\gcd(a, b)$, $n \\in \\mathbb{N}$, contains infinitely many primes. Let $p = a/\\gcd(a, b) + k \\cdot b/\\gcd(a, b)$, $k \\in \\mathbb{N}$, be prime, with $p > c$. Then, as $\\gcd(a, b)$ and $c$ are relatively prime,\n$$\n1 = \\gcd(p, c) = \\gcd(p \\cdot \\gcd(a, b), c) = \\gcd(a + kb, c).\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42379,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB < AC$. A point $P$ on the circumcircle of $ABC$ (on the same side of $BC$ as $A$) is chosen in such a way that $BP = CP$. Let $BP$ and the angle bisector of $\\angle BAC$ intersect at $Q$, and let the line through $Q$ and parallel to $BC$ intersect $AC$ at $R$. Prove that $BR = CR$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AQ$ extended meet the circumcircle of triangle $ABC$ in $T$, and let the line through $Q$ and $R$ intersect $PC$ in $D$. See Figure 2:\n![](attached_image_1.png)\nFigure 2\nPut $\\angle BAQ = \\angle CAQ = \\alpha$, so that also $\\angle BCT = \\alpha$ and $\\angle TBC = \\alpha$. Put $\\angle BCA = \\theta$, so that also $\\angle BTA = \\theta$. Put $\\angle ACP = \\phi$, so that also $\\angle ABP = \\phi$.\n\nThen $\\angle PBC = \\angle PCB$ (since $BP = CP) = \\theta + \\phi$. From $\\angle PBT + \\angle TCP = 180^{\\circ}$ (since $BPCT$ is a cyclic quadrilateral), we therefore have $2(\\alpha + \\theta + \\phi) = 180^{\\circ}$, giving $\\alpha + \\theta + \\phi = 90^{\\circ}$. Since triangle $QPD$ is isosceles ($QD \\parallel BC$, so that $\\angle PQD = \\angle PBC = \\angle PCB = \\angle PDQ$), we have $PQ = PD$, forcing $BQ = CD$. Furthermore, chords $BT$ and $CT$ both subtend the same angle $\\alpha$ at $A$, hence are equal in length. It follows that $\\triangle TBQ \\equiv \\triangle TCD$, hence $\\angle DTC = \\angle QTB = \\theta$. But since $\\angle DRC = \\angle BCR = \\theta$ (recall that $QD \\parallel BC$), we see that $RDCT$ is a cyclic quadrilateral. This implies that $\\angle TRD = 90^{\\circ}$ (because $\\angle TCD = \\alpha+\\theta+\\phi = 90^{\\circ}$). So $\\angle TRQ = 90^{\\circ}$, and we also have that $RQBT$ is a cyclic quadrilateral. (Recall that $\\angle QBT = 90^{\\circ}$.) It follows that $\\angle QRB = \\angle QTB = \\theta$, and we thus have $\\angle RBC = \\theta$, since $QD \\parallel BC$. Finally, $BR = RC$ follows from the fact that $\\angle RBC = \\angle RCB = \\theta$.\nLet the line through $R$ and $Q$ intersect $AB$ and $PC$ in $S$ and $T$, respectively. See Figure 3:\n![](attached_image_2.png)\nFigure 3\nPut $\\gamma = \\angle PBC = \\angle PCB$. Then also $\\angle PAC = \\gamma$. From the fact that $ST \\parallel BC$, it follows that $\\angle BQS = \\angle QBC = \\gamma$, so that $\\angle PQR = \\gamma$. Thus $PAQR$ is a cyclic quadrilateral. Hence $\\angle RPQ = \\angle RAQ = \\alpha$, and since $\\angle CPB = \\angle CAP = 2\\alpha$, we have $\\angle RPT = \\alpha$. So $QR = TR$.\n\nWe see that $\\triangle BQR \\equiv \\triangle CTR$ ($\\angle BQR = \\angle CTR = 180^{\\circ} - \\gamma$, $QR = RT$ and $BQ = CT$ (from $PB = PC$ and $PQ = PT$)). Consequently, $BR = CR$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42380,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer, and consider a set of $n$ points in three-dimensional space such that:\n(i) every two distinct points are connected by a string which is either red, green, blue, or yellow;\n(ii) for every three distinct points, if the three strings between them are not all of the same colour, then they are of three different colours;\n(iii) not all the strings have the same colour.\nFind the maximum possible value of $n$.",
    "options": [],
    "answer": "9",
    "solution": "Let us fix a notation: Number the points $1, 2, \\dots, n$; The colour of the string between points $i$ and $j$ is denoted by $c(i, j)$, so that $c(i, j) \\in \\{\\text{B}, \\text{G}, \\text{R}, \\text{Y}\\}$ for all $1 \\le i, j \\le n$, $i \\ne j$, and where $\\text{B} = \\text{Blue}$, $\\text{G} = \\text{Green}$, $\\text{R} = \\text{Red}$, and $\\text{Y} = \\text{Yellow}$.\n\nFor $n = 3$, there is a trivial solution, by putting, say, $c(1,2) = \\text{B}$, $c(2,3) = \\text{G}$, $c(3,1) = \\text{R}$. Also, for $n = 4$ it is fairly straightforward to see a solution, say $c(1,2) = c(3,4) = \\text{B}$, $c(1,3) = c(2,4) = \\text{G}$, and $c(1,4) = c(2,3) = \\text{R}$.\n\nHenceforth, assume that $n > 4$.\n\nWe first observe that no point can have three strings of the same colour attached to it. For suppose (without loss of generality) that $c(1,2) = c(1,3) = c(1,4) = \\text{B}$. Then $c(2,3) = c(2,4) = c(3,4) = \\text{B}$. This implies that there must be a fifth point, $5$, say, such that $c(1,5) \\neq \\text{B}$ (otherwise all strings will have the same colour $\\text{B}$). Assume $c(1,5) = \\text{G}$. Then $c(2,5), c(3,5), c(4,5) \\notin \\{\\text{B}, \\text{G}\\}$. Assume that $c(2,5) = \\text{R}$, say. Then $c(3,5) \\neq \\text{R}$, forcing $c(3,5) = \\text{Y}$. But then $c(4,5) \\notin \\{\\text{R}, \\text{Y}\\}$, by condition (ii) applied to the two sets $\\{2,4,5\\}$ and $\\{3,4,5\\}$ of three points each. The contradiction $c(4,5) \\notin \\{\\text{B}, \\text{G}, \\text{R}, \\text{Y}\\}$ proves that our claim is valid. Consequently, there can be at most two strings of each colour attached to each point, implying that there can be no more than nine points in total.\n\nWe now show that it is indeed possible that nine points can be mutually connected by strings such that all conditions (i) to (iii) are satisfied.\n\nPartition the points $1, 2, \\dots, 9$ into three subsets of three points each: $X = \\{1, 2, 3\\}$, $Y = \\{4, 5, 6\\}$, $Z = \\{7, 8, 9\\}$. First, consider the three 'internal' monochromatic triangles (all sides Blue) where the points in each of these subsets are connected with Blue strings. Then we construct nine further monochromatic triangles $KLM$, where $K \\in X$, $L \\in Y$, and $M \\in Z$ (three of each of the colours Green, Red and Yellow). This can be done in several ways (six ways, to be precise). We show here a nice symmetrical one: Three Green triangles $\\{1, 4, 7\\}$, $\\{2, 5, 8\\}$, $\\{3, 6, 9\\}$; Three Red triangles $\\{1, 5, 9\\}$, $\\{2, 6, 7\\}$, $\\{3, 4, 8\\}$; Three Yellow triangles $\\{1, 6, 8\\}$, $\\{2, 4, 9\\}$, $\\{3, 5, 7\\}$. See Figure 4.\n\n![](attached_image_1.png)\n\nThe motivation behind the method used here was to write a $3 \\times 3$ table:\n$$\nT = \\begin{array}{|c|c|c|}\n\\hline\n1 & 2 & 3 \\\\\n\\hline\n4 & 5 & 6 \\\\\n\\hline\n7 & 8 & 9 \\\\\n\\hline\n\\end{array}\n$$\nThe rows of $T$ ($\\{1, 2, 3\\}$, $\\{4, 5, 6\\}$, $\\{7, 8, 9\\}$) are used to determine the three Blue triangles. The columns ($\\{1, 4, 7\\}$, $\\{2, 5, 8\\}$, $\\{3, 6, 9\\}$) are used to determine the three Green triangles. For the final two colours we use one element from each row and one element from each column, in two possible ways: For the Red triangles: $\\{1, 5, 9\\}$, $\\{2, 6, 7\\}$, $\\{3, 4, 8\\}$; and for the Yellow triangles: $\\{3, 5, 7\\}$, $\\{2, 4, 9\\}$, $\\{1, 6, 8\\}$.\n\nNow we have a string between each two points, and if a triangle contains two sides/strings of the same colour, the third side must also be of that colour.\n\nWe conclude that the maximum possible value of $n$ is $n = 9$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 42381,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there are infinitely many polynomials $P$ with real coefficients such that if $x$, $y$, and $z$ are real numbers such that $x^2 + y^2 + z^2 + 2xyz = 1$, then\n$$\nP(x)^2 + P(y)^2 + P(z)^2 + 2P(x)P(y)P(z) = 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us call a triple $(x, y, z)$ of real numbers a *-triple if it satisfies\n$$\nx^2 + y^2 + z^2 + 2xyz = 1.\n$$\nLet us call a polynomial $p(x)$ with real coefficients a *-polynomial if $(x, y, z)$ a *-triple implies $(p(x), p(y), p(z))$ a *-triple.\nWe first investigate polynomials of degree at most 1. Hence, assume that $p(x) = ax + b$ ($a, b \\in \\mathbb{R}$), and suppose that $p(x)$ is a *-polynomial. Since $(x, -x, 1)$ and $(x, x, -1)$ are *-triples for all $x \\in \\mathbb{R}$, we have\n$$\np(x)^2 + p(-x)^2 + p(1)^2 + 2p(x)p(-x)p(1) = 1 \\text{ and } p(x)^2 + p(x)^2 + p(-1)^2 + 2p(x)p(x)p(-1) = 1\n$$\nfor all $x \\in \\mathbb{R}$. This simplifies, respectively, to\n$$\n2a^2(1 - a - b)x^2 + 2b^2(1 + a + b) + (a + b)^2 = 1 \\quad (1)\n$$\n$$\n\\text{and}\\quad 2a^2(1 - a + b)x^2 + 4ab(1 - a + b)x + 2b^2(1 - a + b) + (b - a)^2 = 1. \\quad (2)\n$$\nSince these equations are valid for all $x \\in \\mathbb{R}$, we must have both the coefficients $2a^2(1-a-b)$ and $2a^2(1-a+b)$ equal to 0. Therefore, $a=0$, or $1-a-b=0=1-a+b$.\nIf $a = 0$, then, from (1), $2b^2(1+b) + b^2 = 1$. It is clear that $b = -1$ is a solution to this equation, and it follows readily that $b = \\frac{1}{2}$ is the only other solution. It is straightforward to check that the constant polynomials $p(x) = -1$ and $p(x) = \\frac{1}{2}$ are indeed *-polynomials.\nIn case $a \\neq 0$, then $1 - a - b = 0 = 1 - a + b$, so that $b = 0$. From (1), the coefficient $2a^2(1 - a - b) = 2a^2(1 - a) = 0$, giving $a = 1$. From this we get the trivial *-polynomial $p(x) = x$.\n\nOur next observation is that when $p(x)$ is a *-polynomial, then $p(p(x)) = p^2(x)$ is also a *-polynomial — and, in fact, it follows by an easy induction that $p^n(x)$ are *-polynomials for all $n \\ge 1$, where $p^n(x)$ means the composition of $p(x)$ with itself, $n$ times. So if we can find a *-polynomial $p(x)$ such that infinitely many polynomials in the sequence $p(x), p^2(x), p^3(x), \\dots$ are different from each other, the problem will be solved. Unfortunately, none of the three *-polynomials we have found so far has this property. We therefore look for a possible second degree *-polynomial $p(x) = ax^2 + bx + c$, $a, b, c \\in \\mathbb{R}$, $a \\neq 0$. In this case, assuming that $p(x)$ is a *-polynomial, and again using the *-triples $(x, -x, 1)$ and $(x, x, -1)$ (for all $x \\in \\mathbb{R}$), we obtain, after simplification:\n$$\n2a^2(1 + a + b + c)x^4 + 2[2ac + b^2 + (2ac - b^2)(a + b + c)]x^2 + 2c^2(1 + a + b + c) + (a + b + c)^2 = 1 \\quad (3)\n$$\nand\n$$\n2a^2(1 + a - b + c)x^4 + 4ab(1 + a - b + c)x^3 + 2(2ac + b^2)(1 + a - b + c)x^2 + 4bc(1 + a - b + c)x + 2c^2(1 + a - b + c) + (a - b + c)^2 = 1 \\quad (4)\n$$\nSince the coefficients of $x^4$ in (3) and (4) must be 0, and we have $a \\neq 0$, we must have $a + b + c = -1 = a - b + c$, so that $b = 0$. Thus $a + c = -1$ and we conclude that\n$$\np(x) = ax^2 + c = (-1-c)x^2 + c\n$$\nfor some $c \\in \\mathbb{R}$. But $p(x)$ is assumed to be a *-polynomial, and since $(x, \\sqrt{1-x^2}, 0)$ are *-triples for all $-1 \\le x \\le 1$, we get that\n$$\n((-1-c)x^2+c)^2 + ((-1-c)(1-x^2)+c)^2 + c^2 + 2((-1-c)x^2+c)((-1-c)(1-x^2)+c)c = 1,\n$$\nwhich simplifies to\n$$\n2(1+c)^2(1-c)x^4 - 2(1+c)^2(1-c)x^2 + 1 = 1.\n$$\nAs before, the coefficients of $x^4$ and $x^2$ must be 0, and we see that $c \\in \\{-1, 1\\}$. The case $c = -1$ gives $p(x) = -1$, which we have already dealt with. Hence, the only possible candidate at this stage for a second degree *-polynomial, is $p(x) = -2x^2 + 1$. We now verify that $p(x) = -2x^2 + 1$ is indeed a *-polynomial:\nLet $(x, y, z)$ be an arbitrary *-triple. Then\n$$\n\\begin{aligned}\n& (-2x^2 + 1)^2 + (-2y^2 + 1)^2 + (-2z^2 + 1)^2 + 2(-2x^2 + 1)(-2y^2 + 1)(-2z^2 + 1) \\\\\n&= -16x^2y^2z^2 + 4(x^4 + y^4 + z^4) + 8(x^2y^2 + y^2z^2 + z^2x^2) - 8(x^2 + y^2 + z^2) + 5 \\\\\n&= -4(1 - (x^2 + y^2 + z^2))^2 + 4(x^2 + y^2 + z^2)^2 - 8(x^2 + y^2 + z^2) + 5 \\\\\n&= 1,\n\\end{aligned}\n$$\nand we conclude that $p(x) = -2x^2 + 1$ is indeed a *-polynomial.\nThis solves the problem, since we now have an infinite sequence $p(x), p^2(x), p^3(x), \\dots$ of *-polynomials, and they are all different, since $\\deg(p^n(x)) = 2^n$ for each $n \\ge 1$, a fact that can easily be verified by induction.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 42382,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x$ and $y$ are positive real numbers such that\n$$\n\\sqrt{2x} + \\sqrt{y} = 13 \\text{ and } \\sqrt{8x} + \\sqrt{9y} = 35,\n$$\ncalculate $20x + 23y$.",
    "options": [],
    "answer": "2023",
    "solution": "The second equation can be written as $2\\sqrt{2x} + 3\\sqrt{y} = 35$. Together with the first equation, we obtain a system of two linear equations in the two unknowns $\\sqrt{2x}$ and $\\sqrt{y}$, which can easily be solved to give $\\sqrt{2x} = 4$ and $\\sqrt{y} = 9$. From this follows $x = 8$ and $y = 81$, so that $20x + 23y = 20 \\cdot 8 + 23 \\cdot 81 = 2023$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42383,
    "subject": "Mathematics (Multi-modal)",
    "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": "24",
    "solution": "Extend $AE$ and $DC$ until they meet in $F$, say. (These two lines will definitely meet: since $ABCD$ is a cyclic quadrilateral, $\\angle ABC > 90^\\circ$ implies that $\\angle ADC < 90^\\circ$.) See Figure 1.\n\n![](attached_image_1.png)\n\nFigure 1\n\nWe have $\\angle FBC = \\angle FDA = \\angle FCE$ (note that $EC \\parallel AD$), so that triangles $FAD$, $FCB$ and $FEC$ are all similar to each other. Put $x = AD$ and $y = EF$. From the similarity between triangles $FAD$ and $FEC$, it follows that\n$$\n\\frac{x}{y+16} = \\frac{12}{y}. \\qquad (1)\n$$\nFurthermore, $BC = \\sqrt{9^2 + 12^2} = 15$, so that from the similarity between triangles $FCB$ and $FEC$, it follows that\n$$\n\\frac{15}{\\sqrt{12^2 + y^2}} = \\frac{12}{y}. \\qquad (2)\n$$\nFrom (2) we get $y = 16$, and then from (1) we find that $x = AD = 24$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42384,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider two positive integers $a$ and $b$ such that $a + 2b = 2020$.\n\na. Determine the largest possible value of the greatest common divisor of $a$ and $b$.\n\nb. Determine the smallest possible value of the least common multiple of $a$ and $b$.",
    "options": [],
    "answer": "a) 505; b) 808",
    "solution": "We observe that $2020 = 2^2 \\cdot 5 \\cdot 101$. Whenever $a + 2b = 2020$, $d = \\gcd(a, b)$ must divide $2020$. So, to find the largest possible value of $d$, we have to look at the large divisors of $2020$. $d = 2020$ is not possible, since in this case $a + 2b \\ge 2020 + 2 \\cdot 2020 > 2020$. The next largest divisor is $1010$. Since $3 \\cdot 1010 > 2020$, $d = 1010$ is also not possible. Next is $505$, and here we find that $a = 1010$ and $b = 505$ work, since $\\gcd(1010, 505) = 505$, and $1010 + 2 \\cdot 505 = 2020$. It follows that the largest possible value for $\\gcd(a, b)$ is $505$. This solves part (a).\n\nTo solve part (b), we first consider two cases for a pair $(a, b)$ such that $a + 2b = 2020$:\n\n(i) $a \\mid b$: Let $b = k a$ for some integer $k \\ge 1$. Then $a + 2k a = 2020$, i.e., $a = \\frac{2020}{1 + 2k}$, and we have $\\operatorname{lcm}(a, b) = k a = \\frac{2020 k}{1 + 2k} = \\frac{2020}{2 + 1/k}$. Hence $k$ should be minimized in order to minimize $\\operatorname{lcm}(a, b)$. Since $1 + 2k \\mid 2020$, we must take $k = 2$ (note that $3 \\nmid 2020$). It follows that $a + 4a = 5a = 2020$, giving $a = 404$ and $b = 2a = 808$, so that the smallest least common multiple of $a$ and $b$ in this case is given by $\\operatorname{lcm}(404, 808) = 808$.\n\n(ii) $b \\mid a$: Let $a = k b$ for some integer $k \\ge 1$. Then $k b + 2b = 2020$, i.e., $b = \\frac{2020}{2 + k}$, and we have $\\operatorname{lcm}(a, b) = k b = \\frac{2020 k}{2 + k} = \\frac{2020}{1 + 2/k}$. Hence $k$ should be minimized in order to minimize $\\operatorname{lcm}(a, b)$. Since $2 + k \\mid 2020$, we must take $k = 2$ (again, $3 \\nmid 2020$). It follows that $2b + 2b = 4b = 2020$, giving $b = 505$ and $a = 2b = 1010$, so that the smallest least common multiple of $a$ and $b$ in this case is given by $\\operatorname{lcm}(1010, 505) = 1010$.\n\nWe now show that we may assume that either (i) or (ii) holds in order to solve the problem: Among all possible greatest common divisors of $a$ and $b$, where $a + 2b = 2020$, let us fix one of them, say $d$. Let $(a, b)$ be any pair such that $a + 2b = 2020$ with $\\gcd(a, b) = d$. Put $a = \\hat{a} d$ and $b = \\hat{b} d$. Then $\\hat{a}$ and $\\hat{b}$ are relatively prime, so that $\\operatorname{lcm}(\\hat{a}, \\hat{b}) = \\hat{a} \\hat{b} > \\max\\{\\hat{a}, \\hat{b}\\}$. The smallest possible value for $\\operatorname{lcm}(\\hat{a}, \\hat{b})$ (and hence also for $\\operatorname{lcm}(a, b)$) is thus obtained if either $\\hat{a} = 1$ or $\\hat{b} = 1$ (in which case $\\operatorname{lcm}(\\hat{a}, \\hat{b})$ is equal to either $\\hat{b}$ or $\\hat{a}$). But $\\hat{a} = 1$ implies $a \\mid b$ and $\\hat{b} = 1$ implies $b \\mid a$. Since we can repeat this argument for any possible value of $d$, we may assume that (i) or (ii) holds in order to solve the problem.\n\nWe conclude that the smallest possible least common multiple of $a$ and $b$, where $a + 2b = 2020$, is $\\operatorname{lcm}(404, 808) = 808$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42385,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a set of real numbers satisfying the following:\n(a) $\\sqrt{n^2+1} \\in A$ for all positive integers $n$;\n(b) if $x \\in A$ and $y \\in A$, then $x - y \\in A$.\nProve that every integer can be written as a product of two different elements in $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first observe some natural consequences of property (b):\n(i) $0 \\in A$, since if $x \\in A$ is arbitrary, then $0 = x - x \\in A$.\n(ii) If $x \\in A$, then $-x = 0 - x \\in A$.\n(iii) If $x, y \\in A$, then $x + y = x - (-y) \\in A$.\n(iv) If $x \\in A$, then $kx \\in A$ for any integer $k$. This follows easily by induction on $k$ (for positive $k$), using (iii). For negative $k$, apply (ii). For $k = 0$, use (i).\n\nSo now we know that $k\\sqrt{m^2+1} + l\\sqrt{n^2+1} \\in A$ for all integers $k$ and $l$, and all positive integers $m$ and $n$.\nLet us first try to write the integer 1 as a product of two different elements in $A$: We can find (by inspection) positive integers $k, l, m, n$ such that $1 = k^2(m^2+1) - l^2(n^2+1)$, for example, $k=3, m=1, l=1, n=4$. (There are many other possibilities as well.) Hence,\n$$\n\\begin{aligned}\n1 &= 3^2(1^2 + 1) - 1^2(4^2 + 1) \\\\\n&= (3\\sqrt{1^2 + 1} - 1\\sqrt{4^2 + 1})(3\\sqrt{1^2 + 1} + 1\\sqrt{4^2 + 1}),\n\\end{aligned}\n$$\na product of two different elements of $A$.\nIf $p$ is an arbitrary integer, then we have\n$$\n\\begin{aligned}\np = p \\cdot 1 &= p \\cdot (3\\sqrt{1^2+1} - 1\\sqrt{4^2+1})(3\\sqrt{1^2+1} + 1\\sqrt{4^2+1}) \\\\\n&= (3p\\sqrt{1^2+1} - p\\sqrt{4^2+1})(3\\sqrt{1^2+1} + 1\\sqrt{4^2+1}),\n\\end{aligned}\n$$\nagain a product of two elements of $A$. These two elements of $A$ are indeed different, since if $3p\\sqrt{1^2+1} - p\\sqrt{4^2+1} = 3\\sqrt{1^2+1} + 1\\sqrt{4^2+1}$, then we would have $p = 35 + 6\\sqrt{34}$, which is not an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42386,
    "subject": "Mathematics (Multi-modal)",
    "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$$\n\\frac{t!}{k!(t-k)!} \\le C \\le \\frac{(t+k-1)!}{k!(t-1)!},\n$$\nwhere $n! = 1 \\times 2 \\times \\dots \\times n$ for any positive integer $n$, and $0! = 1$.",
    "options": [],
    "answer": "Detailed 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![](attached_image_1.png)\nFigure 2\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": 42387,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABIH$, $BDEC$, and $ACFG$ be arbitrary rectangles constructed (externally) on the sides of triangle $ABC$. Choose point $S$ outside rectangle $ABIH$ (on the opposite side as triangle $ABC$) such that $\\angle SHI = \\angle FAC$ and $\\angle HIS = \\angle EBC$. Prove that the lines $FI$, $EH$, and $CS$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $T = EH \\cap FI$. Let $K$ and $L$ be the feet of the perpendiculars from $A$ to $FI$ and from $B$ to $EH$, respectively. Then $K$ and $L$ are on the circumcircle $\\Gamma$ of $ABIH$. Note that $K$ is also on the circumcircle of $ACFG$ and $L$ is also on the circumcircle of $BDEC$. Let $CK$ extended intersect $\\Gamma$ in $M$, and let $CL$ extended intersect $\\Gamma$ in $N$. Then the lines $HM$ and $IN$ intersect in $S$. $[\\angle EBC = \\angle ELC = \\angle HLN = \\angle HIN]$. But $\\angle HIS = \\angle EBC$, so $S$ is on $IN$. Similarly, $S$ is on $HM$.\n\n![](attached_image_1.png)\n\nConsider the self-crossing hexagon $IKMHLN$ inscribed in the circle $\\Gamma$. By Pascal's Theorem, the points $IK \\cap HL = T$, $KM \\cap LN = C$, and $MH \\cap NI = S$ are collinear. This shows that $T$ is a point that belongs to all three lines $FI$, $EH$ and $CS$, i.e., these lines are concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42388,
    "subject": "Mathematics (Multi-modal)",
    "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.\nDetermine the value(s) of $n$ and how many faces of the large cube were painted.",
    "options": [],
    "answer": "n = 7; two faces painted, and they are opposite each other",
    "solution": "Let us first determine how the number of painted faces influences the formula for the total number of unpainted unit cubes.\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.\nIf one face is painted, the number of unpainted unit cubes is $n^3 - n^2$.\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).\nIf three faces are painted, the number of unpainted unit cubes is either $(n - 1)^3$ (if they share a vertex) or $n(n - 1)(n - 2)$.\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).\nIf five faces are painted, the number of unpainted unit cubes is $(n - 1)(n - 2)^2$.\nNow we factorise $245 = 5 \\times 7 \\times 7$. Since these are three prime factors, the only possibility from the expressions above is if $5 \\cdot 7^2 = (n - 2)n^2$.\nThus $n = 7$ and two faces opposite each other are painted.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42389,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine which of the following is larger:\n$$\n\\sqrt{2 + \\sqrt[3]{5}} \\quad \\text{or} \\quad \\sqrt[3]{5 + \\sqrt{2}}\n$$\nFully explain your reasoning.",
    "options": [],
    "answer": "sqrt(2 + cube_root(5)) is larger than cube_root(5 + sqrt(2)).",
    "solution": "Let us denote $x = \\sqrt{2 + \\sqrt[3]{5}}$ and $y = \\sqrt[3]{5 + \\sqrt{2}}$. Since both are positive, $x - y$ has the same sign as $x^6 - y^6$. We have\n$$\nx^6 = (2 + \\sqrt[3]{5})^3 = 8 + 12\\sqrt[3]{5} + 6\\sqrt[3]{25} + 5 = 13 + 12\\sqrt[3]{5} + 6\\sqrt[3]{25}\n$$\nand\n$$\ny^6 = (5 + \\sqrt{2})^2 = 25 + 10\\sqrt{2} + 2 = 27 + 10\\sqrt{2}\n$$\nNow we do a rough estimate on the terms in $x^6$:\n$$\n12\\sqrt[3]{5} = \\sqrt[3]{12^3 \\cdot 5} = \\sqrt[3]{8640} > \\sqrt[3]{8000} = 20\n$$\n\nand\n$$\n6\\sqrt[3]{25} = \\sqrt[3]{6^3 \\cdot 25} = \\sqrt[3]{5400} > \\sqrt[3]{1000} = 10.\n$$\nThus $x^6 > 13 + 20 + 10 = 43$. On the other hand\n$$\ny^6 = 27 + 10\\sqrt{2} = 27 + \\sqrt{200} < 27 + \\sqrt{225} = 27 + 15 = 42.\n$$\nThus $x^6 > y^6$, and thus $x > y$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42390,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each of the lattice points $(x, y)$ ($x$ and $y$ integers) in the plane can be coloured black or white. A single strike by an L-shaped punch changes the colour of the four lattice points $(a, b)$, $(a+1, b)$, $(a, b+1)$ and $(a, b+2)$. All lattice points are initially coloured white. Prove that after any number of strikes, the number of black lattice points will be either zero or greater than or equal to four.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "When the lattice points $(a, b)$, $(a+1, b)$, $(a, b+1)$ and $(a, b+2)$ change colour after a single strike by the punch, let us call the point $(a, b)$ the corner point of the strike.\n\nWe first show (by induction on the number of strikes) that, after any finite number of strikes, there will always be an even number of black lattice points. After the first strike, four points (an even number) will be black. Suppose that after $n$ strikes, for some $n \\ge 1$, an even number, say $2m$, of black lattice points $X = \\{(x_1, y_1), (x_2, y_2), \\dots, (x_{2m}, y_{2m})\\}$ remain. For the $(n+1)$-th strike, with corner point $(a, b)$, put $Y = \\{(a, b), (a+1, b), (a, b+1), (a, b+2)\\}$. Let $|X \\cap Y| = k \\in \\{0, 1, 2, 3, 4\\}$, where $|S|$ denotes the size of a set $S$. Then, after strike number $n+1$, the number of black points will be $|(X \\cup Y) \\setminus (X \\cap Y)|$, which is $|X| + |Y| - 2k = 2m + 4 - 2k$, an even number. This completes the induction.\n\nWe now show that it never happens that exactly two black points remain after any finite number of strikes, which will solve the problem.\n\nFirst, we observe that the moment the same corner point $(a, b)$ appears twice in a series of strikes, we can just as well ignore these two strikes, since the colours of the points in the set $\\{(a, b), (a+1, b), (a, b+1), (a, b+2)\\}$ will remain the same after two strikes with corner point $(a, b)$, irrespective of where they appear in the series of strikes. Hence, we may assume that all the corner points in the finite series of strikes are different (each appearing exactly once).\n\nLet $X = \\{(x_1, y_1), (x_2, y_2), \\dots, (x_{2m}, y_{2m})\\}$ be the total number of black lattice points that remain after these strikes.\n\nAmongst the $y$-coordinates $y_1, y_2, \\dots, y_{2m}$, let $y'$ be the smallest. Say $y_{i_1}, y_{i_2}, \\dots, y_{i_k}$ are all the $y$-coordinates equal to $y'$. We assume them to be ordered such that $x_{i_1} < x_{i_2} < \\dots < x_{i_k}$. It follows that, amongst these points, the point $(x_{i_1}, y')$ is the corner point of a strike, while $(x_{i_k}, y')$ is not (with $(x_{i_k} - 1, y')$ the corner point of the strike producing $(x_{i_k}, y')$). So the two points $(x_{i_1}, y')$ and $(x_{i_k}, y')$ cannot be the same, since $(x_{i_1}, y')$ is a corner point of a strike, while $(x_{i_k}, y')$ is not. Hence, there are at least two black points with $y$-coordinates equal to $y'$.\n\nAnalogously, amongst the $x$-coordinates $x_1, x_2, \\dots, x_{2m}$, let $x'$ be the smallest. Say $x_{j_1}, x_{j_2}, \\dots, x_{j_m}$ are all the $x$-coordinates equal to $x'$. We assume them to be ordered such that\n\n$y_{j_1} < y_{j_2} < \\cdots < y_{j_m}$.\n\nIt follows that, amongst these points, the point $(x', y_{j_1})$ is the corner point of a strike, while $(x', y_{j_m})$ is not (with $(x', y_{j_m} - 2)$ the corner point of the strike producing $(x', y_{j_m})$). So the two points $(x', y_{j_1})$ and $(x', y_{j_m})$ cannot be the same, since $(x', y_{j_1})$ is a corner point of a strike, while $(x', y_{j_m})$ is not. Hence, there are at least two black points with $x$-coordinates equal to $x'$.\n\nWe see that there are at least three black points in the subset $\\{(x_{i_1}, y'), (x_{i_k}, y'), (x', y_{j_1}), (x', y_{j_m})\\}$ of $X$. (There is a possibility that $(x_{i_1}, y') = (x', y_{j_1})$.) Therefore, since $X$ contains an even number of points, we conclude that $|X| \\ge 4$.\n\nRecall that we can strike the punch twice at corner point $(0,0)$, say, to obtain a scenario where there are no black points after a series of strikes.\n\nOur conclusion is that the number of black lattice points after a finite number of strikes is an even number that is either $0$, or greater than or equal to $4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42391,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f$ from integers to integers such that\n$$\nf(m+n) + f(m-n) - 2f(m) = 6mn^2\n$$\nfor all integers $m$ and $n$.",
    "options": [],
    "answer": "All solutions are f(n) = n^3 + a n + c for integers a and c.",
    "solution": "First note that if we replace $f(x)$ by $f(x) + ax + c$ for some constants $a$ and $c$, then we get another solution. Thus we may assume for now that $f(0) = 0$ and $f(1) = 1$.\nLet $P(m, n)$ be the statement that was given. Then $P(0, n)$ gives\n$$\nf(n) + f(-n) - 2f(0) = 0\n$$\nand since $f(0) = 0$, we get $f(-n) = -f(n)$. Then taking $P(n, m)$,\n$$\nf(m+n) + f(n-m) - 2f(n) = 6m^2n\n$$\nand subtracting this from $P(m, n)$ gives\n$$\n2f(m-n) + 2f(n) - 2f(m) = 6mn(n-m).\n$$\nSetting $n = -1$, dividing by 2 and rearranging gives\n$$\nf(m+1) = f(m) + f(1) + 3m(m+1).\n$$\nThe first few values of $m$ tell us:\n$$\nf(2) = 8, \\quad f(3) = 27, \\quad f(4) = 64.\n$$\nWe hypothesise for an inductive argument that $f(n) = n^3$ and show that\n$$\nf(n+1) = f(1) + f(n) + 3n(n+1) = n^3 + 3n^2 + 3n + 1 = (n+1)^3.\n$$\nThus $f(n) = n^3$ for all positive integers, and hence by $f(-n) = -f(n)$ also for all negative integers.\nNow incorporating our initial remark, we find that all solutions are of the form $f(n) = n^3 + an + c$ for some integer constants $a$ and $c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42392,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider three circles $\\Gamma_1, \\Gamma_2$ and $\\Gamma_3$, with centres $O_1, O_2$ and $O_3$, respectively, such that each pair of circles is externally tangent. Suppose we have another circle $\\Gamma$ with centre $O$ on the line segment $O_1O_3$ such that $\\Gamma_1, \\Gamma_2$ and $\\Gamma_3$ are each internally tangent to $\\Gamma$.\nShow that $\\angle O_1O_2O_3$ measures less than $90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the points of tangency of $\\Gamma$ with $\\Gamma_1, \\Gamma_2$ and $\\Gamma_3$ be $A, B$ and $C$, respectively.\n\n![](attached_image_1.png)\n\nThen $AOB$ is a straight line, and since it is a diameter, $\\angle ACB = 90^\\circ$. Let $E$ be the other intersection point of $AC$ with $\\Gamma_3$ and $F$ be the other intersection point of $BC$ with $\\Gamma_3$. Then $C, O_3$ and $O$ lie on a straight line, so $\\angle CEO_3 = \\angle ECO_3 = \\angle ACO = \\angle CAO$, the first equality following from radii $EO_3 = O_3C$, and the last from radii $AO = CO$. Thus $EO_3 \\parallel AO$ and similarly $O_3F \\parallel OB$.\n\nWe now claim that $EF < AO_1 + O_2B$ and that this finishes the problem. To see that it finishes the problem construct points $X$ and $Y$ on segment $AB$ such that $AX = EO_3$ and $YB = O_3F$. Then $\\angle O_1O_3O_2 < \\angle XO_3Y = \\angle XO_3O + \\angle OO_3Y = \\angle ACO + \\angle OCB = 90^\\circ$.\n\nFor the claim $EF < AO_1 + O_2B$, we introduce the values $r_1, r_2, r_3, r$ for the radii of circles $\\Gamma_1, \\Gamma_2, \\Gamma_3, \\Gamma$, respectively. Then $2r_1 + 2r_2 = 2r$ by comparing radii along the diameter $AB$, so $r_1 + r_2 = r$. To show the claim we need to show $2r_3 < r_1 + r_2$, or equivalently $2r_3 < r$. But this is clear by looking at the segment $CO$, where two segments of length $r_3$ are contained within one segment of length $r$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42393,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{N} \\to \\mathbb{N}_0$ be a function that satisfies\n$$\nf(mn) = mf(n) + nf(m)\n$$\nfor all positive integers $m, n$ and $f(2024) = 10120$.\nProve that there are two integers $m, n$ with $m \\neq n$ such that $f(m) = f(n)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By setting $n = 1$, we obtain $mf(1) = 0$ for all $m$, so $f(1) = 0$. Next we notice that since every positive integer can be written as a product of prime numbers, the function $f$ is determined completely by its value on the primes. If there exists $m \\ge 2$ such that $f(m) = 0$, then $f(m) = f(1)$ and we are done. Thus assume for all $m \\ge 2$ that $f(m) \\ge 1$.\nThus\n$$\n\\begin{aligned}\nf(2024) &= f(8 \\cdot 11 \\cdot 23) = 23f(8 \\cdot 11) + 88f(23) = 23(11f(8) + 8f(11)) + 88f(23) \\\\\n&= 253f(8) + 184f(11) + 88f(23).\n\\end{aligned}\n$$\nTo study $f(8)$, we instead derive a general formula for perfect powers:\n$$\nf(a^2) = af(a) + af(a) = 2af(a),\n$$\n$$\nf(a^3) = a^2f(a) + af(a^2) = 3a^2f(a).\n$$\nWe assume as inductive hypothesis that\n$$\nf(a^n) = na^{n-1}f(a),\n$$\nand prove it for $n+1$ by\n$$\nf(a^{n+1}) = f(a^n \\cdot a) = af(a^n) + a^n f(a) = a(na^{n-1})f(a) + a^n f(a) = (n+1)a^n f(a).\n$$\nThus\n$$\n10120 = f(2024) = 253(3 \\cdot 2^2)f(2) + 184f(11) + 88f(23) = 3036f(2) + 184f(11) + 88f(23). \\quad (1)\n$$\nConsidering this equation modulo 11, we conclude that $f(11)$ is divisible by 11, and by the assumption that $f(11) \\ge 1$, we conclude that $f(11) \\ge 11$. Similarly, working modulo 23 tells us that $f(23)$ is divisible by 23 and thus $f(23) \\ge 23$. Finally, working modulo 8 tells us that $4f(2)$ is divisible by 8, so $f(2) \\ge 2$. Thus the right hand-side of (1) is greater than or equal to\n$$\n3036 \\cdot 2 + 184 \\cdot 11 + 88 \\cdot 23 = 10120.\n$$\nSince we have equality, we must conclude that\n$$\nf(2) = 2, \\quad f(11) = 11 \\quad \\text{and} \\quad f(23) = 23.\n$$\nOur only hope to find two integers $m$ and $n$ for which $f(m) = f(n)$ is to use the information just obtained. Thus, we set\n$$\nm = 2^a 11^b 23^c \\quad n = 2^d 11^e 23^f.\n$$\nIntuitively we hope to find values of **d**, **e**, **f** that are close to **a**, **b**, **c** (e.g. differ by 1) and then solve for values **a**, **b**, **c** that happen to work. We notice that $2 \\cdot 11 = 22$ is close to 23, so let us study\n$$\n\\begin{align*} \nf(22^a 23^c) &= 23^c f(22^a) + 22^a f(23^c) \\\\ \n&= 23^c (a \\cdot 22^{a-1} \\cdot f(22)) + 22^a (c \\cdot 23^{c-1} \\cdot f(23)) \\\\ \n&= 23^c (a \\cdot 22^{a-1} \\cdot (2f(11) + 11f(2))) + 22^a (c \\cdot 23^{c-1} \\cdot 23) \\\\ \n&= 23^c (a \\cdot 22^{a-1} \\cdot 44) + 22^a (c \\cdot 23^{c-1} \\cdot 23) \\\\ \n&= (2a + c) 22^a 23^c \n\\end{align*}\n$$\nNow we compare $f(22^a 23^{b+1})$ with $f(22^{a+1} 23^b)$ and hope to find values of **a**, **b** for which they are equal.\n$$\n\\begin{aligned} \nf(22^a 23^{b+1}) &= (2a + b + 1) 22^a 23^{b+1}, \\\\ \nf(22^{a+1} 23^b) &= (2(a + 1) + b) 22^{a+1} 23^b. \n\\end{aligned}\n$$\nThus if we can find values of **a**, **b** such that\n$$\n2a + b + 1 = 22 \\quad \\text{and} \\quad 2a + b + 2 = 23\n$$\nwe would be done. Luckily there are many such pairs, for example\n$(\\mathbf{a}, \\mathbf{b}) \\in \\{(1, 19), (2, 17), (3, 15), (4, 13), (5, 11), (6, 9), (7, 7), (8, 5), (9, 3), (10, 1)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42394,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle with $\\angle B \\neq \\angle C$. The incircle $I$ of a triangle $ABC$ touches the sides $BC, CA, AB$ at the points $D, E, F$, respectively. Let $P$ be the intersection of $AD$ and the incircle $I$, which is different from $D$.\nLet $Q$ be the intersection of the line $EF$ and the line passing $P$ and perpendicular to $AD$, and let $X, Y$ be intersections of the line $AQ$ and $DE, DF$, respectively. Show that the point $A$ is the midpoint of $XY$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $Q'$ be the intersection of the line passing $A$ and parallel to $BC$ and the line passing $P$ and perpendicular to $AD$. Let $U$ be the intersection of $DI$ and $AQ'$, and $V$ be the intersection of $PQ'$ and the circle $I$. ($V \\neq P$)\nSince $\\angle VPD = 90^\\circ$, four points $D, I, V, U$ lie on a line.\nSince $\\angle BDI = 90^\\circ$ we have $\\angle AUI = 90^\\circ$ and since $\\angle AFI = \\angle AEI = 90^\\circ$ we have that five points $A, F, I, E, U$ lie on a circle, say, $C_1$.\nFour points $A, P, V, U$ lie on a circle, say, $C_2$, because $\\angle APV = \\angle AUV = 90^\\circ$.\nFor given three circles, three perpendicular bisectors of the line segments joining two centers of two circles meet at one point. Considering three circles $C_1, C_2$, and the incircle $I$, we have that three lines $AU, PV, EF$ meet at one point $Q'$. Thus we have $Q' = Q$.\nSince $\\triangle AXE \\sim \\triangle CDE$, we have $\\frac{AX}{AE} = \\frac{CD}{CE}$. So\n$$\nAX = \\frac{CD}{CE} \\times AE = AE.\n$$\nSince $\\triangle AYF \\sim \\triangle BDF$, we have $\\frac{AY}{AF} = \\frac{BD}{BF}$. So\n$$\nAY = \\frac{BD}{BF} \\times AF = AF.\n$$\nSince $AE = AF$ we have $AX = AY$, which completes the proof. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42395,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $a$, let $S_a$ be the set of primes satisfying the following condition:\n(condition) For any $p \\in S_a$, there exists an odd integer $b$ such that $(2^{2^a})^b - 1$ is divisible by $p$.\nFor any positive integer $a$, prove that there exist infinitely many primes that are not contained in $S_a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any positive integer $n$, define $a_n := 2^{2^n} + 1$. Since\n$$\na_1 a_2 \\cdots a_{n-1} = a_n - 2,\n$$\n$\\gcd(a_n, a_m) = 1$ for any $n \\neq m$. Now, for any positive integer $n$ such that $n \\ge a$, let $p_n$ be any prime dividing $a_n$. Note that $p_n$ is different from $p_m$ for any $m \\ne n$.\n\nSuppose that $2^{2^n}b - 1$ is divisible by $p_n$ for some odd integer $b$. Since\n$$\n2^{2^{2n}} \\equiv 1 \\pmod{p_n}, \\quad 2^{2^{2n+1}} \\equiv 1 \\pmod{p_n},\n$$\n$2^{2^n} \\equiv 1 \\pmod{p_n}$. This is a contradiction to the fact that $2^{2^n} \\equiv -1 \\pmod{p_n}$. Therefore, $p_n \\notin S_a$ for any integer $n$ such that $n \\ge a$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42396,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three schools $A$, $B$ and $C$ participate in a chess tournament with five students from each school. Let $a_1, a_2, \\dots, a_5; b_1, b_2, \\dots, b_5; c_1, c_2, \\dots, c_5$ be the list of the players from $A$, $B$, $C$ respectively. Let $P_A$, $P_B$, $P_C$ be the scores that schools $A$, $B$, $C$ make, respectively, when the tournament is over following the rules described below. Find the remainder of the number of possible triples $(P_A, P_B, P_C)$ when it is divided by $8$.\n\n1. Players from each school have matches in order from the first one, and if a player is once defeated then he (or she) is eliminated from the tournament. The first match of the tournament is between the players $a_1$ and $b_1$.\n\n2. When $x_i$ from school $X$ is defeated by $y_i$ from $Y$, if there are players left from the school $Z$ other than $X$, $Y$ then $y_i$ have another match with a player from $Z$, if there is no player left in $Z$ but there are players from $X$ then $y_j$ have a match with $x_{i+1}$. The tournament is over when no player is left in two schools.\n\n3. School $X$ adds $10^{i-1}$ points to its score when $x_i$ from $X$ wins a match.",
    "options": [],
    "answer": "4",
    "solution": "(i) The triple $(10, 0, 0)$ can be made only when $a_1$ wins exactly $10$ times, and $(0, 10, 0)$ is only when $b_1$ wins $10$ games. The number of games that a player $x \\notin \\{a_1, b_1\\}$ can win is strictly less than $10$. Therefore, the number of possible triples $(P_A, P_B, P_C)$ is exactly the same as the number of different processes of the tournament.\n\n(ii) Let $f(l, m, n)$ be the number of possible triples $(P_A, P_B, P_C)$ when $l$, $m$, $n$ students from schools $A$, $B$, $C$, respectively, participate in the tournament. Two possible results of the first match will give us the following:\n$$\nf(l, m, n) = f(n, m, l-1) + f(l, n, m-1).\n$$\nIf there is no player left for a school, then we have a binomial number, and we have\n$$\nf(l, m, 0) = \\binom{l+m}{m} \\qquad f(l, m, n) = 0 \\quad \\text{if } n < 0.\n$$\nMoreover, we have $f(l, m, n) = f(m, l, n)$.\nSince the number we are looking for is $f(5, 5, 5)$, let us consider the number $f(n, n, n)$:\n$$\nf(n, n, n) = 2f(n, n, n-1) \\\\\nf(n, b, k) = f(k, n, n-1) + f(n, k, n-1) = 2f(n, k, n-1) \\\\\nf(n, n, n) = 4f(n, n-1, n-1).\n$$\nTherefore, the remainder of $f(5, 5, 5)$ divided by $8$ is $4$ if $f(5, 4, 4)$ is odd and $0$ if $f(5, 4, 4)$ is even.\n\n(iii) Let us calculate $f(5, 4, 4)$ modulo $2$.\n$$\n\\begin{align*}\n& f(5, 4, 4) \\\\\n= & f(4, 4, 4) + f(5, 4, 3) \\\\\n= & 0 + f(3, 4, 4) + f(5, 3, 3) \\\\\n= & f(4, 4, 2) + f(3, 4, 3) + f(3, 3, 4) + f(5, 3, 2) \\\\\n= & 0 + f(3, 4, 3) + 0 + f(5, 3, 2) \\\\\n= & f(3, 4, 2) + f(3, 3, 3) + f(2, 3, 4) + f(5, 2, 2) \\\\\n= & f(3, 4, 2) + 0 + f(2, 3, 4) + f(5, 2, 2) \\\\\n= & f(2, 4, 2) + f(3, 2, 3) + f(4, 3, 1) + f(2, 4, 2) + f(2, 2, 4) + f(5, 2, 1) \\\\\n= & f(3, 2, 3) + f(4, 3, 1) + f(5, 2, 1) \\\\\n= & f(3, 2, 2) + f(3, 3, 1) + f(1, 3, 3) + f(4, 1, 2) + f(1, 2, 4) + f(5, 1, 1) \\\\\n= & f(3, 2, 2) + 0 + f(1, 3, 3) + f(4, 1, 2) + f(1, 2, 4) + f(5, 1, 1) \\\\\n= & f(2, 2, 2) + f(3, 2, 1) + f(3, 3, 0) + f(1, 3, 2) + f(2, 1, 3) + f(4, 2, 0) \\\\\n& + f(4, 2, 0) + f(1, 4, 1) + f(1, 1, 4) + f(5, 1, 0) \\\\\n= & 0 + f(3, 2, 1) + 0 + f(1, 3, 2) + f(2, 1, 3) + 0 + 0 + f(1, 4, 1) + 0 + 0 \\\\\n= & f(1, 2, 2) + f(3, 1, 1) + f(2, 3, 0) + f(1, 2, 2) + f(3, 1, 1) + f(2, 3, 0) \\\\\n& + f(1, 4, 0) + f(1, 1, 3) \\\\\n= & f(1, 4, 0) = 1\n\\end{align*}\n$$\n$f(5, 4, 4)$ is odd and the answer is $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42397,
    "subject": "Mathematics (Multi-modal)",
    "question": "For three mutually distinct real numbers $a_1$, $a_2$, $a_3$, define three real numbers $b_1$, $b_2$, $b_3$ as follows:\n$$\nb_j = \\left( 1 + \\frac{a_j a_i}{a_j - a_i} \\right) \\left( 1 + \\frac{a_j a_k}{a_j - a_k} \\right), \\quad \\{i, j, k\\} = \\{1, 2, 3\\}.\n$$\nProve the inequality\n$$\n1 + |a_1 b_1 + a_2 b_2 + a_3 b_3| \\le (1 + |a_1|)(1 + |a_2|)(1 + |a_3|).\n$$\nWhen does the equality hold?",
    "options": [],
    "answer": "Equality holds if and only if all three numbers are nonnegative.",
    "solution": "Define\n$$\nA = \\frac{a_1 a_2}{a_1 - a_2}, \\quad B = \\frac{a_1 a_3}{a_1 - a_3}, \\quad C = \\frac{a_2 a_3}{a_2 - a_3}.\n$$\nThen\n$$\n\\begin{align*}\n& a_1 b_1 + a_2 b_2 + a_3 b_3 \\\\\n= & a_1 (1+A) (1+B) + a_2 (1-A) (1+C) + a_3 (1-B) (1-C) \\\\\n= & a_1 + a_2 + a_3 + (a_1 - a_2)A + (a_1 - a_3)B \\\\\n& + (a_2 - a_3)C + a_1 AB - a_2 AC + a_3 BC.\n\\end{align*}\n$$\nSome computation shows that\n$$\n(a_1 - a_2)A + (a_1 - a_3)B + (a_2 - a_3)C = a_1 a_2 + a_1 a_3 + a_2 a_3,\n$$\n\n\\begin{aligned}\n& a_1AB - a_2AC + a_3BC \\\\\n&= a_1a_2a_3 \\frac{a_1^2(a_2 - a_3) + a_2^2(a_3 - a_1) + a_3^2(a_1 - a_2)}{(a_1 - a_2)(a_2 - a_3)(a_1 - a_3)} \\\\\n&= a_1a_2a_3.\n\\end{aligned}\n$$\nThus\n$$\n\\begin{aligned}\n& 1 + |a_1b_1 + a_2b_2 + a_3b_3| \\\\\n&= 1 + |a_1 + a_2 + a_3 + a_1a_2 + a_1a_3 + a_2a_3 + a_1a_2a_3| \\\\\n&\\le 1 + |a_1| + |a_2| + |a_3| + |a_1a_2| + |a_1a_3| + |a_2a_3| + |a_1a_2a_3| \\\\\n&= (1 + |a_1|)(1 + |a_2|)(1 + |a_3|).\n\\end{aligned}\n$$\nThe equality holds if and only if 7 real numbers $a_1$, $a_2$, $a_3$, $a_1a_2$, $a_1a_3$, $a_2a_3$, $a_1a_2a_3$ are either all non-negative or all non-positive. Note that at most one of the three numbers $a_1$, $a_2$, $a_3$ can be zero. Thus, the equality holds if and only if the three real numbers $a_1$, $a_2$, $a_3$ are all non-negative. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 42398,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a convex hexagon $ABCDEFG$, three triangles $ABC$, $CDE$, $EFA$ are similar. That is,\n$$\n\\angle BAC = \\angle DCE = \\angle FEA\n$$\n$$\n\\angle BCA = \\angle DEC = \\angle FAE.\n$$\nFind conditions on these three triangles under which $\\triangle ACE$ is an equilateral triangle if and only if $\\triangle BDF$ is an equilateral triangle.",
    "options": [],
    "answer": "The equivalence holds if and only if the three similar triangles are not the isosceles type with two equal angles of thirty degrees (i.e., not 30–30–120).",
    "solution": "Suppose $\\triangle ACE$ is equilateral, then $\\triangle BDF$ is always equilateral, because three triangles $ACE$, $CDE$, $EFA$ are identical and shape of the hexagon $ABCDEF$ is invariable under the $120^\\circ$ rotation around the center of $\\triangle ACE$. That is, since $B \\to D \\to F$ is obtained from the $120^\\circ$ rotation, $\\triangle BDF$ is an equilateral triangle.\n\nWe now show that if three similar triangles $ABC$, $CDE$, $EFA$ are not of a isosceles triangle whose two same angles are $30^\\circ$, the converse is true. First, let $\\angle BAC = \\angle DCE = \\angle FEA = \\alpha$, $\\angle BCA = \\angle DEC = \\angle FAE = \\beta$. Suppose that $\\triangle BDF$ is equilateral. Let\n$$\n\\begin{aligned}\n\\frac{AB}{AC} &= \\frac{CD}{CE} = \\frac{EF}{EA} = s \\\\\n\\frac{BC}{AC} &= \\frac{DE}{CE} = \\frac{FA}{EA} = t,\n\\end{aligned}\n$$\n$AC = a$, $CE = b$, $EA = c$. For $\\triangle ABF$, we have\n$$\n\\begin{aligned}\nBF^2 &= AB^2 + FA^2 - 2AB \\cdot FA \\cdot \\cos(\\angle BAF) \\\\\n&= (sa)^2 + (tc)^2 - 2stac \\cdot \\cos(\\alpha + \\beta + A) \\\\\n&= s^2a^2 + t^2c^2 - 2stac[\\cos(\\alpha + \\beta) \\cos A - \\sin(\\alpha + \\beta) \\sin A].\n\\end{aligned}\n$$\n\nSimilarly we have,\n$$BD^2 = s^2b^2 + t^2a^2 - 2stab[\\cos(\\alpha + \\beta) \\cos C - \\sin(\\alpha + \\beta) \\sin C]$$\n$$DF^2 = s^2b^2 + t^2a^2 - 2stab[\\cos(\\alpha + \\beta) \\cos C - \\sin(\\alpha + \\beta) \\sin C].$$\n$\\frac{1}{2}ac \\sin A = \\frac{1}{2}ab \\sin C = \\frac{1}{2}bc \\sin E$ equals the area of $\\triangle ACE$, and $BF^2 = BD^2 = DF^2$. Moreover, we have $2ac \\cos A = a^2 + c^2 - b^2$, $2ab \\cos C = a^2 + b^2 - c^2$, $2bc \\cos E = b^2 + c^2 - a^2$. Let $\\cos(\\alpha+\\beta) = K$, then we have\n$$\n\\begin{aligned}\n& s^2a^2 + t^2c^2 - stK(a^2 + c^2 - b^2) \\\\\n&= s^2b^2 + t^2c^2 - stK(a^2 + b^2 - c^2) \\\\\n&= s^2c^2 + t^2b^2 - stK(b^2 + c^2 - a^2)\n\\end{aligned}\n$$\nNow let\n$$\ns^2 - t^2 = x, \\quad 2stK - s^2 = y, \\quad t^2 - 2stK = z,\n$$\nthen we have\n$$\nxa^2 + yb^2 + zc^2 = ya^2 + zb^2 + xc^2 = za^2 + xb^2 + yc^2 \\quad (1)\n$$\nFrom the equations (1), suppose first that $x = y = z$, then we have $s = t$ and $K = \\frac{1}{2}$; hence $\\alpha + \\beta = 60^\\circ$. That is, three similar triangles $ABC$, $CDE$, $EFA$ are of a isosceles triangle whose two same angles are $30^\\circ$. We exclude this case. Second, suppose $x, y, z$ are either all distinct or two of them are equal, then we get the equations $a = b = c$ by some calculations, which means that $\\triangle ACE$ is equilateral. We conclude that $\\triangle ACE$ is equilateral unless three similar triangles $ABC$, $CDE$, $EFA$ are of a isosceles triangle whose two same angles are $30^\\circ$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42399,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $N$ is said to be an \"*n*-good number\" if it satisfies the following two properties:\n(Property 1) $N$ is divisible by at least $n$ distinct primes\n(Property 2) There exist distinct positive divisors $1, x_2, \\dots, x_n$ of $N$ such that\n$$\n1 + x_2 + \\dots + x_n = N.\n$$\nShow that there exists an \"*n*-good number\" for each $n \\ge 6$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use an induction on $n$.\n\na. For $n = 6$, put $N_6 = 2 \\cdot 3 \\cdot 7 \\cdot 43 \\cdot 1807 = 1806 \\cdot 1807$. Since $1807 = 13 \\cdot 139$, $N_6$ is divisible by 6 distinct primes.\nMoreover, since\n$$\n\\frac{1}{m} = \\frac{1}{m+1} + \\frac{1}{m(m+1)},\n$$\nwe have\n\n$$\n\\begin{align*}\n1 &= \\frac{1}{2} + \\frac{1}{2} \\\\\n  &= \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{6} \\\\\n  &= \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{7} + \\frac{1}{42} \\\\\n  &= \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{7} + \\frac{1}{43} + \\frac{1}{1806}, \\quad (42 \\cdot 43 = 1806) \\\\\n  &= \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{7} + \\frac{1}{43} + \\frac{1}{1807} + \\frac{1}{N_6}.\n\\end{align*}\n$$\nHence, by multiplying both sides of above equality by $N_6$, we get\n$$\nN_6 = 1 + \\frac{N_6}{2} + \\frac{N_6}{3} + \\frac{N_6}{7} + \\frac{N_6}{43} + \\frac{N_6}{1807},\n$$\nwhere each term of the right hand side is a divisor of $N_6$.\n\nb. Suppose there is an $n$-good number $N_n$. Put\n$$\nN_{n+1} = N_n(N_n + 1).\n$$\nThen\n$$\n\\begin{align*}\nN_{n+1} &= (1 + x_2 + \\cdots + x_n)(N_n + 1) \\\\\n         &= 1 + N_n + x_2(N_n + 1) + \\cdots + x_n(N_n + 1).\n\\end{align*}\n$$\nHence $N_{n+1}$ is a sum of $n+1$ distinct divisors.\nSince $(N_n, N_n + 1) = 1$, prime divisors of $N_n + 1$ are different from those of $N_n$. Since $N_n + 1$ has at least one prime divisor, and since $N_n$ has at least $n$ distinct prime divisors, $N_{n+1}$ has at least $n+1$ distinct prime divisors. Therefore $N_{n+1}$ is a $(n+1)$-good number. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42400,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle and $O$ be its circumcircle. Let $O'$ be the circle that is tangent to $O$ at $A$ and tangent to the side $BC$ at $D$. $O'$ intersects the lines $AB$ and $AC$ at $E$ and $F$, respectively. $O'$ intersects the lines $OO'$ and $EO'$ at $A' (\\neq A)$ and $G (\\neq E)$, respectively. The lines $BO$ and $A'G$ intersect at $H$. Prove that $DF^2 = AF \\cdot GH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $X$ be the intersection of the line $BO$ with the circle $O$. Since the line segments $EG$ and $BX$ are diameters of the circles $O$ and $O'$, respectively, $\\angle EAG = \\angle BAX = 90^\\circ$ and hence three points $A, G, X$ are collinear.\n\nSince $O'$ is tangent to $O$ at $A$, we have $\\angle AEG = \\angle ABX = \\angle ABH$ and thus the lines $EG$ and $BH$ are parallel. Since the segment $AA'$ is a diameter of the circle $O'$, $\\angle AGA' = \\angle AGE + \\angle EGA' = 90^\\circ$. So from $\\angle AGE + \\angle AEG = 90^\\circ$ follows $\\angle AEG = \\angle EGA' = \\angle EGH$. Thus, the quadrilateral $BEGH$ is parallelogram and therefore, we get\n$$\nGH = BE. \\tag{1}\n$$\n\nOn the other hand, since the lines $EF$ and $BC$ are also parallel, we have $\\angle EFD = \\angle FDC$. Thus, we get\n$$\n\\angle EFD = \\angle FDC = \\angle DAF = \\angle DEF,\n$$\nwhich implies that\n$$\nDF = DE. \\tag{2}\n$$\nIn particular, since $\\angle ADB = \\angle AFD$ and\n$$\n\\angle BAD = \\angle BDF = \\angle DEF = \\angle EFD = \\angle CDF = \\angle CAD,\n$$\nthe two triangles $ABD$ and $ADF$ are similar and therefore, we get\n$$\n\\frac{AF}{AD} = \\frac{DF}{BD}. \\tag{3}\n$$\nSince the triangles $ABD$ and $DBE$ are also similar, we get\n$$\n\\frac{DE}{AD} = \\frac{BE}{BD}. \\tag{4}\n$$\n\nCombining the equations (1), (2), (3) and (4), we get\n$$\n\\frac{GH}{DF} = \\frac{BE}{DE} = \\frac{BD}{AD} = \\frac{DF}{AF},\n$$\nwhich yields $DF^2 = AF \\cdot GH$ as desired. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42401,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the sixteen tiles fixed on a wall as shown below. How many ways are there to write either $0$ or $1$ on each tile so that the product of the two numbers written on every neighboring pair of tiles (sharing a common side) is always $0$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "1234",
    "solution": "When $0$ or $1$ is written on each tile in such a way that the product of the two numbers written on every neighboring pair of tiles is always $0$, we'll call the status a *z-pattern*. First, we prove a couple of lemmas.\n\n**Lemma 1.** Let $a_n$ be the number of z-patterns for $n$ tiles laid in a row. Then\n$$\na_n = F_{n+2}\n$$\nfor all $n \\ge 1$, where $F_n$ is the $n$-th Fibonacci number defined by $F_1 = 1$, $F_2 = 1$, and $F_{n+2} = F_{n+1} + F_n$ for all $n \\ge 1$.\n\n*Proof.* We prove by induction on $n \\ge 1$. It is clear that the lemma holds for $n=1$ and $n=2$. Assume the lemma for all $k$, $1 \\le k < n$, where $n \\ge 3$. If $0$ is written on the tile at one end, then any number ($0$ or $1$) can be written on the next tile and hence there are $a_{n-1}$ such z-patterns. If $1$ is written there, then $0$ should be written on the next tile and hence there are $a_{n-2}$ such z-patterns. Thus we get\n$$\na_n = a_{n-1} + a_{n-2} = F_{n+1} + F_n = F_{n+2},\n$$\nwhich proves the lemma.\n\n**Lemma 2.** Let $b_n$ be the number of z-patterns for $n$ tiles fixed on a wall in a ring shape. Then\n$$\nb_n = a_{n-1} + a_{n-3} = F_{n+1} + F_{n-1}\n$$\nfor all $n \\ge 4$.\n\n*Proof.* Let $n \\ge 4$ and choose any tile among the $n$ tiles fixed on a wall in a ring shape. If $0$ is written on the tile, then any number can be written on the neighboring tiles and hence there are $a_{n-1}$ such z-patterns. If $1$ is written on the tile, then $0$ should be written on the neighboring two tiles and hence there are $a_{n-3}$ such z-patterns. Thus we get\n$$\nb_n = a_{n-1} + a_{n-3} = F_{n+1} + F_{n-1},\n$$\nwhich proves the lemma.\n\nWe now return to the problem. Consider the $2 \\times 2$ tiles in the center. Observe that the number of $0$'s that can be written on these four tiles equals $4$, $3$ or $2$.\n\nCase 1) Four $0$'s:\n$$\n\\begin{bmatrix} 0 & 0 \\\\ 0 & 0 \\end{bmatrix}\n$$\nWe may apply Lemma 2 to the $12$ tiles surrounding the four tiles in the center because any number can be written on the $12$ tiles. Therefore, the number of z-patterns in this case equals\n$$\nb_{12} = F_{13} + F_{11} = 233 + 89 = 322.\n$$\n\nCase 2) Three $0$'s:\n$$\n\\begin{bmatrix} 1 & 0 \\\\ 0 & 0 \\end{bmatrix}\n$$\n$$\n\\begin{bmatrix} 0 & 1 \\\\ 0 & 0 \\end{bmatrix}\n$$\n$$\n\\begin{bmatrix} 0 & 0 \\\\ 1 & 0 \\end{bmatrix}\n$$\n$$\n\\begin{bmatrix} 0 & 0 \\\\ 0 & 1 \\end{bmatrix}\n$$\nIn each of the four subcases, only $0$ can be written on two neighboring tiles of the tile marked by $1$. For the remaining $10$ tiles, it is clear that the number of z-patterns equals to $a_9 \\times 2 = F_{11} \\times 2 = 178$. Therefore, the total number of z-patterns in this case equals\n$$\n178 \\times 4 = 712.\n$$\n\nCase 3) Two $0$'s:\n$$\n\\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}\n$$\n$$\n\\begin{bmatrix} 0 & 1 \\\\ 1 & 0 \\end{bmatrix}\n$$\nIn each of the two subcases, only $0$ can be written on four neighboring tiles of the two tiles marked by $1$. For the remaining $6$ tiles, it is clear that the number of z-patterns equals to $a_3^2 \\times 2^2 = F_5^2 \\times 4 = 100$. Therefore, the total number of z-patterns in this case equals\n$$\n100 \\times 2 = 200.\n$$\n\nCombining the three cases, the answer is: $322 + 712 + 200 = 1234$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 42402,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triplets $(x, y, z)$ of positive integers satisfying $1 + 4^x + 4^y = z^2$.",
    "options": [],
    "answer": "(x, y, z) = (t, 2t - 1, 1 + 2^{2t - 1}) or (2t - 1, t, 1 + 2^{2t - 1}) for any positive integer t",
    "solution": "Without loss of generality, we may assume that $x \\le y$. Suppose that $2x < y + 1$. Then\n$$\n(2^y)^2 < 1 + 4^x + 4^y < (1 + 2^y)^2,\n$$\nwhich implies that $1 + 4^x + 4^y$ is not a square of an integer. If $2x = y + 1$, then $1 + 4^x + 4^y = 1 + 2^{y+1} + 4^y = (1 + 2^y)^2$. Hence\n$$\n(x, y, z) = (x, 2x-1, 1 + 2^{2x-1})\n$$\nis a solution of the equation for any positive integer $x$. Suppose that $2x > y + 1$. Note that\n$$\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$$\nfor any $x > 1$. Therefore the solutions are:\n$$\n(x, y, z) = (x, 2x-1, 1 + 2^{2x-1}) \\text{ or } (2x-1, x, 1 + 2^{2x-1})\n$$\nfor any positive integer $x$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42403,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pair $(p, q)$ of primes such that $p^p + q^q + 1$ is divisible by $pq$.",
    "options": [],
    "answer": "(2, 5) and (5, 2)",
    "solution": "Clearly, $p \\ne q$. So we may assume that $p < q$ without loss of generality. Assume that $p = 2$. Then since\n$$\nq^q + 5 \\equiv 5 \\equiv 0 \\pmod{q},\n$$\nthe only possible prime for $q$ is $5$. Furthermore, $(p, q) = (2, 5)$ satisfies the above condition. Now we assume that both $p$ and $q$ are odd primes. Since $p^p + 1 \\equiv 0 \\pmod{q}$, $p^{p-1} - p^{p-2} + \\dots - p + 1$ is divisible by $q$ and $p^{2p} \\equiv 1 \\pmod{q}$. On the other hand, Fermat's Little Theorem says\n$$\np^{q-1} \\equiv 1 \\pmod{q}.\n$$\nIf $\\gcd(2p, q-1) = 2$, then $p^2 \\equiv 1 \\pmod{q}$. Hence $p \\equiv 1 \\pmod{q}$ or $p \\equiv -1 \\pmod{q}$. Therefore\n$$\n0 \\equiv p^{p-1} - p^{p-2} + \\dots - p + 1 \\equiv 1 \\pmod{q},\n$$\nwhich is a contradiction. Assume that $\\gcd(2p, q - 1) = 2p$, that is, $q \\equiv 1 \\pmod p$. In this case we have\n$$\n0 \\equiv p^p + q^q + 1 \\equiv p^p + 1 + 1 \\equiv 2 \\pmod{p},\n$$\nwhich is again a contradiction.\nTherefore $(p, q) = (2, 5)$ or $(5, 2)$ is the only pair of primes satisfying the above condition. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42404,
    "subject": "Mathematics (Multi-modal)",
    "question": "For the vertex $A$ of $\\triangle ABC$, define $\\ell_A$ by the distance between the feet of perpendiculars drawn to the sides $AB$ and $AC$ from the intersection of the angle bisector of $\\angle A$ and the side $BC$. Similarly, define $\\ell_B$ and $\\ell_C$ for the vertices $B$ and $C$, respectively. Prove the inequality\n$$\n\\frac{\\ell_A \\ell_B \\ell_C}{\\ell^3} \\le \\frac{1}{64}\n$$\nwhere $\\ell$ is the perimeter of $\\triangle ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a$, $b$ and $c$ be the lengths of the opposite sides of vertices $A$, $B$ and $C$, respectively. Also, let $D$ be the intersection of $BC$ and the angle bisector of $\\angle A$. Put $p = BD$ and $q = CD$. Since $AD$ is the angle bisector of $\\angle A$, we have $bp = cq$ and thus a little manipulation yields\n$$\np = \\frac{ac}{b+c} \\quad \\text{and} \\quad q = \\frac{ab}{b+c}. \\qquad (1)\n$$\nMeanwhile, since $\\cos(\\angle ADB) + \\cos(\\angle ADC) = 0$, we have by the Cosine Law\n$$\n\\frac{x^2 + p^2 - c^2}{2px} + \\frac{x^2 + q^2 - b^2}{2qx} = 0\n$$\nwhere $x = AD$. This, together with (1), yields\n$$\nx^2 = bc - pq = bc \\left[ 1 - \\left( \\frac{a}{b+c} \\right)^2 \\right] = \\frac{bc(b+c-a)}{(b+c)^2} \\ell. \\qquad (2)\n$$\nNow, let $E$ and $F$ be the feet of perpendiculars drawn from $D$ to $AB$ and $AC$, respectively. Then, since $\\square AEDF$ is inscribed in a circle, we have $\\angle DEF = \\angle DAF$. Accordingly, applying the Sine Law, we have\n$$\n\\frac{\\ell_A}{\\sin A} = \\frac{\\ell_A}{\\sin(\\angle EDF)} = \\frac{DF}{\\sin(\\angle DEF)} = \\frac{DF}{\\sin(\\angle DAF)} = x.\n$$\nThus, denoting the area of $\\triangle ABC$ by $T$, we have by (2)\n$$\n\\ell_A = x \\sin A = \\frac{2xT}{bc} = \\frac{2T}{(b+c)\\sqrt{bc}} \\sqrt{(b+c-a)\\ell}.\n$$\nParallel arguments yield\n$$\n\\ell_B = \\frac{2T}{(c+a)\\sqrt{ca}} \\sqrt{(c+a-b)\\ell}\n$$\n\nand\n$$\n\\ell_C = \\frac{2T}{(a+b)\\sqrt{ab}} \\sqrt{(a+b-c)\\ell}.\n$$\nHence, using Heron's formula, we obtain\n$$\n\\begin{aligned}\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)} \\\\\n&= \\frac{\\ell^3 (a+b-c)^2 (b+c-a)^2 (c+a-b)^2}{8abc(a+b)(b+c)(c+a)}.\n\\end{aligned}\n$$\nSince $a + b \\ge 2\\sqrt{ab}$, $b + c \\ge 2\\sqrt{bc}$, $c + a \\ge 2\\sqrt{ca}$ by the AM-GM inequality, we deduce from the above that\n$$\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}. \\quad (3)\n$$\nMoreover, since $a$, $b$ and $c$ are side lengths of a triangle, we have\n$$\n\\begin{aligned}\n0 < (a+b-c)(c+a-b) &= a^2 - (b-c)^2 \\le a^2, \\\\\n0 < (a+b-c)(b+c-a) &= b^2 - (a-c)^2 \\le b^2, \\\\\n0 < (b+c-a)(c+a-b) &= c^2 - (a-b)^2 \\le c^2,\n\\end{aligned}\n$$\nand thus\n$$\n0 < (a-b+c)^2(b-a+c)^2(c-a+b)^2 \\le a^2b^2c^2. \\quad (4)\n$$\nFrom inequalities (3) and (4) follows the required inequality. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 42405,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{N} \\to \\mathbb{N}$ be a function satisfying\n$$\nk f(n) \\le f(kn) \\le k f(n) + k - 1\n$$\nfor every $k, n \\in \\mathbb{N}$, where $\\mathbb{N}$ is the set of all positive integers.\n(1) Show that, for every $a, b \\in \\mathbb{N}$,\n$$\nf(a) + f(b) \\le f(a + b) \\le f(a) + f(b) + 1.\n$$\n(2) Show that, if $f$ satisfies\n$$\nf(2007n) \\le 2007 f(n) + 2005\n$$\nfor every $n \\in \\mathbb{N}$, then there exists $c \\in \\mathbb{N}$ such that $f(2007c) = 2007 f(c)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) Let $a, b$ be positive integers. Since\n$$\n\\begin{aligned}\n(a+b)(f(a)+f(b)) &= (a+b)f(a) + (a+b)f(b) \\\\\n&\\le f(a(a+b)) + f(b(a+b)) \\\\\n&\\le af(a+b) + a-1 + bf(a+b) + b-1 \\\\\n&< (a+b)(f(a+b)+1),\n\\end{aligned}\n$$\nwe have $f(a) + f(b) < f(a + b) + 1$. Because $f$ is an integer-valued function, $f(a) + f(b) \\le f(a + b)$. Similarly, since\n$$\n\\begin{aligned}\n(a+b)f(a+b) &= af(a+b) + bf(a+b) \\\\\n&\\le f(a(a+b)) + f(b(a+b)) \\\\\n&\\le (a+b)f(a) + (a+b)f(b) + 2(a+b) - 2 \\\\\n&< (a+b)(f(a) + f(b) + 2),\n\\end{aligned}\n$$\nwe have $f(a+b) < f(a)+f(b)+2$ and hence $f(a+b) \\le f(a)+f(b)+1$.\n\n**Alternative 1.** For a real number $x$, let $\\lfloor x \\rfloor$ denote the largest integer less than or equal to $x$. By the given condition,\n$$\n\\frac{f(n)}{n} \\le \\frac{f(kn)}{kn} \\le \\frac{f(n)}{n} + \\frac{k-1}{kn} < \\frac{f(n)}{n} + 1.\n$$\nTherefore, for any positive integers $k$ and $n$, we obtain\n$$\nf(n) = \\left\\lfloor \\frac{f(kn)}{kn} \\right\\rfloor\n$$\nUsing the above formula, we can prove the desired inequalities as follows:\n$$\n\\begin{aligned}\nf(a+b) &= \\left\\lfloor \\frac{f(ab(a+b))}{ab} \\right\\rfloor = \\left\\lfloor \\frac{f(ab(a+b))}{b(a+b)} + \\frac{f(ab(a+b))}{a(a+b)} \\right\\rfloor \\\\\n&\\ge \\left\\lfloor \\frac{f(ab(a+b))}{b(a+b)} \\right\\rfloor + \\left\\lfloor \\frac{f(ab(a+b))}{a(a+b)} \\right\\rfloor = f(a) + f(b);\n\\end{aligned}\n$$\n$$\n\\begin{aligned}\nf(a+b) &= \\left\\lfloor \\frac{f(ab(a+b))}{ab} \\right\\rfloor = \\left\\lfloor \\frac{f(ab(a+b))}{b(a+b)} + \\frac{f(ab(a+b))}{a(a+b)} \\right\\rfloor \\\\\n&\\le \\left\\lfloor \\frac{f(ab(a+b))}{b(a+b)} \\right\\rfloor + \\left\\lfloor \\frac{f(ab(a+b))}{a(a+b)} \\right\\rfloor + 1 = f(a) + f(b) + 1.\n\\end{aligned}\n$$\n\n**Alternative 2.** By the given condition, we have\n$$\nk f(n) \\le f(kn) = f(nk) \\le n f(k) + n - 1.\n$$\nBy substituting $(k, n)$ with $(a+b, a)$ and $(a+b, b)$, we have\n$$\n(a+b)f(a) \\le af(a+b) + a - 1 \\quad \\text{and} \\quad (a+b)f(b) \\le bf(a+b) + b - 1,\n$$\nrespectively. From the above inequalities, we have\n$$\n(a+b)(f(a)+f(b)) \\le (a+b)f(a+b) + (a+b) - 2 \\\\ < (a+b)(f(a+b)+1),\n$$\nand hence $f(a) + f(b) \\le f(a+b)$.\nSimilarly, by substituting $(k, n)$ with $(a, a+b)$ and $(b, a+b)$, we have\n$$\naf(a+b) \\le (a+b)f(a) + a+b-1\n$$\nand\n$$\nbf(a+b) \\le (a+b)f(b) + a+b-1,\n$$\nrespectively. From the above inequalities, we have\n$$\n(a+b)(f(a+b)) \\le (a+b)(f(a)+f(b)) + 2(a+b) - 2 \\\\ < (a+b)(f(a)+f(b)+2)\n$$\nand hence $f(a+b) \\le f(a) + f(b) + 1$.\n\n(2) Consider the following set\n$$\nS = \\{f(2007n) - 2007f(n) \\mid n = 1, \\dots, 2007\\}.\n$$\nBecause $2007f(n) \\le f(2007n) \\le 2007f(n) + 2005$, the set $S$ has at most 2006 elements. By the pigeonhole principle, there exist distinct positive integers $k, m \\le 2007$ such that\n$$\nf(2007k) - 2007f(k) = f(2007m) - 2007f(m),\n$$\nhence\n$$\nf(2007k) - f(2007m) = 2007(f(k) - f(m)).\n$$\nWithout loss of generality, we may assume $k > m$. Let $k = m + c$. From the above equation, we have\n$$\nf(2007(m+c)) - f(2007m) = 2007(f(m+c) - f(m)). \\quad \\dots\\dots (*)\n$$\nNote that, by the given condition in (1),\n$$\n\\begin{cases} f(2007c) \\le f(2007(m+c)) - f(2007m) \\le f(2007c) + 1; \\\\ f(c) \\le f(m+c) - f(m) \\le f(c) + 1. \\end{cases}\n$$\n\n\\text{Case 1} \\left\\{ \\begin{array}{l} f(2007(m+c)) - f(2007m) = f(2007c); \\\\ f(m+c) - f(m) = f(c) \\end{array} \\right.\n$$\nBy (*), we have $f(2007c) = 2007f(c)$ as desired.\n$$\n\\text{Case 2} \\left\\{ \\begin{array}{l} f(2007(m+c)) - f(2007m) = f(2007c); \\\\ f(m+c) - f(m) = f(c) + 1 \\end{array} \\right.\n$$\nBy (*), we have $f(2007c) = 2007f(c) + 2007$. This contradicts the\ncondition $f(2007n) \\le 2007f(n) + 2005$.\n$$\n\\text{Case 3} \\left\\{ \\begin{array}{l} f(2007(m+c)) - f(2007m) = f(2007c) + 1; \\\\ f(m+c) - f(m) = f(c) \\end{array} \\right.\n$$\nBy (*), we have $f(2007c)+1 = 2007f(c)$. This contradicts the condition\nthat $kf(n) \\le f(kn)$.\n$$\n\\text{Case 4} \\left\\{ \\begin{array}{l} f(2007(m+c)) - f(2007m) = f(2007c) + 1; \\\\ f(m+c) - f(m) = f(c) + 1 \\end{array} \\right.\n$$\nBy (*), we have $f(2007c) = 2007f(c) + 2006$. This contradicts the\ncondition $f(2007n) \\le 2007f(n) + 2005$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42406,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a cyclic hexagon $ABCDEF$, let $BD$ and $CF$ meet at $G$, $AC$ and $BE$ meet at $H$, $AD$ and $CE$ meet at $I$. Suppose that $BD$ is perpendicular to $CF$ and $AI = CI$. Show that $CH = AH + DE$ if and only if $GH \\cdot BD = BC \\cdot DE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $AI = CI$, $\\angle ACI = \\angle CAI$, and so $\\angle ACE = \\angle CAD$. Since $ACDE$ is cyclic, $\\angle CAD = \\angle CED$. Hence $\\angle ACE = \\angle CED$, and so $AC$ and $DE$ are parallel to each other.\n\nNow we will show that if $CH = AH + DE$, then $GH \\cdot BD = BC \\cdot DE$. Let $A'$ be the point on $AC$ with $CA' = DE$. Since $AC$ is parallel to $DE$, $A'CDE$ is a parallelogram. Thus $\\angle AA'E = \\angle ACD$. It follows that $\\angle AA'E = \\angle A'AE$, as $\\angle ACD = \\angle CAE$. Further, the fact that $AH = CH - DE = CH - CA' = HA'$ guarantees that $H$ is the midpoint of the base $AA'$ of an isosceles triangle $EAA'$. Therefore $EH$ is orthogonal to $AA'$. It follows that $\\angle BHC = 90^\\circ = \\angle BGC$, and so the points $B$, $C$, $G$, $H$ are cyclic. Let $K$ be the intersection point of $BG$ and $CH$. Since $\\triangle BKC \\sim \\triangle HKG$,\n$$\n\\frac{BC}{GH} = \\frac{BK}{HK} \\quad (1)\n$$\nSince $DE$ is parallel to $KH$, $\\triangle BDE \\sim \\triangle BKH$. Thus\n$$\n\\frac{BK}{HK} = \\frac{BD}{DE} \\quad (2)\n$$\nCombining (1) and (2), we obtain $\\frac{BC}{GH} = \\frac{BD}{DE}$, and hence $GH \\cdot BD = BC \\cdot DE$. This proves the sufficiency.\n\nConversely, assume $GH \\cdot BD = BC \\cdot DE$. We will show that $CH = AH + DE$. Since $AC$ is parallel to $DE$, $\\triangle BKH \\sim \\triangle BDE$, which implies that\n$$\n\\frac{BK}{KH} = \\frac{BD}{DE} \\quad (3)\n$$\nSince $GH \\cdot BD = BC \\cdot DE$,\n$$\n\\frac{BD}{DE} = \\frac{BC}{GH} \\quad (4)\n$$\nCombining (3) and (4), we obtain $\\frac{BK}{KH} = \\frac{BC}{GH}$.\nLet $G'$ be the intersection point of $KD$ and the circumcircle of the triangle $BHC$. Then $\\triangle BCK \\sim \\triangle HG'K$, and so\n$$\n\\frac{BK}{KH} = \\frac{BC}{G'H}\n$$\nBut, since $\\angle BKC > 90^\\circ$, there is only one point $X$ satisfying $\\frac{BK}{KH} = \\frac{BC}{XH}$. Thus $G' = G$, which implies that $B$, $C$, $G$, $H$ are cyclic. So, $\\angle BHC = \\angle BGC = 90^\\circ$. Since $AC$ is parallel to $DE$, $\\angle DEH = 90^\\circ$. Let $D'$ be the foot of the altitude from $D$ to the line $AC$. Since $H$ is the foot of the altitude from $E$ to the line $AC$, we have\n$$\nCH = CD' + D'H = CD' + DE = AH + DE.\n$$\nThis proves the necessity. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42407,
    "subject": "Mathematics (Multi-modal)",
    "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": "Exactly those integer-coefficient polynomials whose even-degree part is either zero (so the polynomial has only odd-degree terms) or has at least two nonzero even-degree terms; equivalently, exclude polynomials whose even part consists of a single even-degree monomial.",
    "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 odd degree monomials of $f(x)$, $h(x)$ is a polynomial consisting of even degree monomials of $f(x)$ (including the constant term), and $f(x) = g(x) + h(x)$.\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 that $f(x) = h(x)$. Suppose that $f(x) = c x^{2n}$ for a 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 $\\gcd(a, b) = 1$, $\\gcd(a+b, b^{2n}) = 1$. Therefore $a+b \\mid 2c$, which implies that there exist only finitely many such positive integer pairs $(a, b)$.\nNow assume that $f(x) = a_{2n} x^{2n} + t(x)$, 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 $a$ 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 $a+b \\mid f(a)+f(b)$.\nTherefore $f(x)$ satisfies the above property if and only if $f(x)$ contains no even degree monomials (that is, it consists of odd degree monomials only) or contains at least two even degree monomials. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42408,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}$ satisfying\n(i) $f(1) = 2008$,\n(ii) $|f(x)| \\le x^2 + 1004^2$, and\n$$\n(iii) \\quad f\\left(x + y + \\frac{1}{x} + \\frac{1}{y}\\right) = f\\left(x + \\frac{1}{y}\\right) + f\\left(y + \\frac{1}{x}\\right)\n$$\nfor all $x, y > 0$, where $\\mathbb{R}$ is the set of all real numbers and $\\mathbb{R}^+$ is the set of all positive real numbers.",
    "options": [],
    "answer": "f(x) = 2008x",
    "solution": "Let $u := x + \\frac{1}{y}$, $v := y + \\frac{1}{x}$ for $x > 0, y > 0$. Then we have\n$$\nf(u+v) = f(u) + f(v) \\quad (1)\n$$\nHere $uv = xy + 2 + \\frac{1}{xy} \\ge 4$. First we show that for any $u, v > 0$ with $uv \\ge 4$ can be represented as\n$$\nu = x + \\frac{1}{y}, v = y + \\frac{1}{x} \\quad (2)\n$$\nfor some $x > 0, y > 0$. From (2), we deduce\n$$\nu = x + \\frac{1}{y} \\implies x = u - \\frac{1}{y} = \\frac{uy - 1}{y} \\implies v = y + \\frac{1}{x} = \\frac{uy^2}{uy - 1} \\implies uy^2 - uvy + v = 0.\n$$\n\nThe last quadratic equation has a real solution $y > 0$ because $D = (uv)^2 - 4uv \\ge 0$ and $u, v > 0$. (Observe that there's no negative real solution for the equation.) Similarly, there exists real $x > 0$, which satisfies (2) together with the above $y$. Therefore, (1) holds for all $u > 0, v > 0$ $uv \\ge 4$.\nWe verify that (1) is valid for all $u, v > 0$. Indeed, for given $u, v > 0$, we may choose a real number $w > 0$ such that\n$$\n(u + v)w \\ge 4, \\quad u(v + w) \\ge 4, \\quad vw \\ge 4.\n$$\nThen from\n$$\n\\begin{aligned}\nf(u + v + w) &= f(u + v) + f(w) \\\\\n&= f(u) + f(v + w) \\\\\n&= f(u) + f(v) + f(w).\n\\end{aligned}\n$$\nwe may conclude that $f(u + v) = f(u) + f(v)$ for all $u, v \\ge 0$.\nNow we put $h(x) = f(x) - f(1)x = f(x) - 2008x$ and show that $h(x)$ is a zero function. It is clear that $h$ satisfies (1) and\n$$\nh(x + 1) = h(x) + h(1) = h(x) \\quad (3)\n$$\nfor all $x \\in \\mathbb{R}^+$. It follows from the condition $|f(x)| \\le x^2 + 1004^2$ that\n$$\n-(x + 1004)^2 \\le h(x) \\le (x - 1004)^2,\n$$\nwhich implies that $h(x)$ is bounded on $(0,1]$ and hence $h(x)$ is bounded on $\\mathbb{R}^+$ according to (3). Since $h(nx) = nh(x)$ for all positive integer $n$, we must have $h \\equiv 0$. Therefore we have $f(x) = 2008x$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42409,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $m \\ge 2$, define\n$$\nA_m := \\{ m+1, 3m+2, 5m+3, 7m+4, 9m+5, \\dots \\}.\n$$\n(1) Prove that, for any given $m \\ge 2$, there exists a positive integer $a$, $1 \\le a < m$, such that either $2^a \\in A_m$ or $2^a + 1 \\in A_m$.\n\n(2) Assume that, for some $m \\ge 2$, there exist positive integers $a$ and $b$ for which $2^a \\in A_m$ and $2^b + 1 \\in A_m$. Let $a_0$ and $b_0$ be the smallest such $a$ and $b$, respectively, and find a relation between $a_0$ and $b_0$.",
    "options": [],
    "answer": "b0 = (a0 - 1)/2",
    "solution": "(1) An arbitrary element of $A_m$ can be written in the form $m + 1 + k(2m + 1)$, where $k = 0, 1, 2, \\dots$. Putting $\\epsilon = 0$ or $1$,\n$$\nm+1+k(2m+1) = 2^a + \\epsilon \\iff 2^a \\equiv m+1-\\epsilon \\pmod{2m+1}\n$$\n$$\n\\iff 2^{a+1} \\equiv 1-2\\epsilon \\pmod{2m+1}\n$$\n$$\n\\iff 2^{a+1} \\equiv \\begin{cases} 1 & (\\text{mod } 2m+1) \\\\ -1 & (\\text{mod } 2m+1) \\end{cases} \\quad \\text{if } \\epsilon=0, \\text{ if } \\epsilon=1.\n$$\nLet $r$ be the smallest positive integer $t$ satisfying $2^t \\equiv 1 \\pmod{2m+1}$, that is, $r = \\text{ord}_{2m+1}(2)$. Observe that $r > 2$ because $2m+1 \\ge 5$,\n\ni) $r \\le m$:\nSince $1 < r - 1 < m$, we may take $a := r - 1$. Then $1 < a < m$ and\n$$\n2^{a+1} \\equiv 1 \\pmod{2m+1} \\iff \\begin{array}{l} (\\epsilon=0) \\\\ 2^a \\in A_m. \\end{array}\n$$\n\nii) $r > m$:\nSince $2m+1$ is odd, $\\phi(2m+1)$ is even. Furthermore,\n$$\nr | \\phi(2m+1), \\phi(2m+1) \\le 2m \\implies r = \\phi(2m+1).\n$$\nThat is, $2$ is a primitive root (mod $2m + 1$). Thus, $2^{r/2} \\equiv -1 \\pmod{2m+1}$. This is because\n$$\n\\begin{align*}\n\\exists s (0 < s < r), 2^s \\equiv -1 \\pmod{2m+1} &\\implies 2^{2s} \\equiv 1 \\pmod{2m+1} \\\\\n&\\implies r \\mid 2s, 0 < 2s < 2r \\\\\n&\\implies 2s = r.\n\\end{align*}\n$$\nTake $a := (r/2) - 1$. Then, from $4 \\le r = \\phi(2m+1) \\le 2m$, we obtain $1 \\le a < m$ and\n$$\n2^{a+1} \\equiv -1 \\pmod{2m+1} \\quad \\stackrel{(\\epsilon=1)}{\\implies} \\quad 2^a + 1 \\in A_m.\n$$\n\n(2) For a given $m$, we observed that $2^{r-1} \\in A_m$ in (1) above, where $r = \\text{ord}_{2m+1}(2)$. Furthermore, it is clear that the smallest $a$, for which $2^a \\in A_m$, holds, is $a_0 = r - 1$.\n\ni) $r$ is even and $2^{r/2} \\equiv -1 \\pmod{2m+1}$:\nIn this case, $2^{(r/2)-1} + 1 \\in A_m$ holds, and the smallest $b$ for which $2^b + 1 \\in A_m$ holds is $b_0 = (r/2) - 1$.\n\nii) $r$ is even and $2^{r/2} \\not\\equiv -1 \\pmod{2m+1}$:\nLet's assume that there exists an $s$ with $0 < s < r$ satisfying $2^s \\equiv -1 \\pmod{2m+1}$. Then from $2^{2s} \\equiv 1 \\pmod{2m+1}$, we get $r \\mid 2s$. Since $2s \\not\\equiv r \\pmod{2r}$, $2s \\ge 3r$, that is, $s > r$, which is a contradiction. Therefore, in this case, $A_m$ contains no element of the form $2^b + 1$.\n\niii) $r$ is odd:\nLet's assume that there exists an $s$ with $0 < s < r$ satisfying $2^s \\equiv -1 \\pmod{2m+1}$. Then from $2^{2s} \\equiv 1 \\pmod{2m+1}$, we get $r \\mid 2s$ and hence $r \\mid s$ because $r$ is odd. But this is absurd since $0 < s < r$. Therefore, in this case again, $A_m$ contains no element of the form $2^b + 1$.\n\nCombining (i~iii), we may conclude that\n$$\nb_0 = \\frac{a_0 + 1}{2} - 1 = \\frac{a_0 - 1}{2}. \\quad \\square\n$$\n\nLet $x, y$ be the smallest positive integers satisfying\n$$\n2^x \\equiv -1 \\pmod{2m+1}, \\quad 2^y \\equiv 1 \\pmod{2m+1}.\n$$\nClearly, $0 < x < y \\le 2x$. Assume that $y = xq + r$ ($0 \\le r < x$). Since\n$$\n1 \\equiv 2^y \\equiv (2^x)^{q} 2^r \\equiv (-1)^{q} 2^r \\pmod{2m+1}.\n$$\n$q$ should be even. Hence $q = 2$ and $r = 0$. This completes the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42410,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let a quadrilateral $ABCD$ be inscribed in a circle $O$ with $\\angle B$ and $\\angle C$ obtuse. Let $E$ be the point of intersection of the lines $AB$ and $CD$. Let $P$ and $R$ be the feet of the perpendicular lines from $E$ to the lines $BC$ and $AD$, respectively. Let $Q$ be the point of intersection of the lines $EP$ and $AD$, and $S$ be the point of intersection of the lines $ER$ and $BC$. Let $K$ be the midpoint of the line segment $QS$. Prove that the three points $E, K, O$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O_1$, $O_2$ be the circumcenters of the triangles $AED$ and $EBC$, respectively.\n\n**Lemma** Let $H$ be the foot of the altitude of the triangle $ABC$. The line symmetric to the line $AH$ with respect to the bisector of the angle $A$ passes the circumcenter $O$ of the triangle $ABC$.\n\n**Proof** Since the triangle $BOA$ is isosceles and $\\angle BOA = 2\\angle C$, we have\n$$\n\\angle BAO = \\frac{180^\\circ - 2\\angle C}{2} = 90^\\circ - \\angle C.\n$$\nAnd since $\\angle CAH = 90^\\circ - \\angle C$, we have $\\angle CAH = \\angle BAO$ and thus the lines $AH$ and $AO$ are symmetric with respect to the angle bisector of the angle $A$. $\\diamond$\n\nWe have $\\angle BEP = 90^\\circ - \\angle PBE = 90^\\circ - \\angle EDA = \\angle DER$. So by the above lemma $O_1$ lies on the line $EQ$ and similarly $O_2$ lies on the line $ES$.\n\nSince $\\triangle EAD \\sim \\triangle ECB$ we have\n$$\nEO_1 : EQ = EO_2 : ES.\n$$\nSo we have $\\triangle EO_1 O_2 \\sim \\triangle EQS$, and thus $O_1 O_2 \\parallel QS$. Therefore the three points $E$, $K$, $M$ lie on the same line, where $M$ is the midpoint of $O_1 O_2$.\n\nNow we are going to show that the three points $E$, $M$, $O$ lie on the same line. Since $O_1$ is the circumcenter of the $\\triangle AED$ and $O$ is the center of the circun circle of the quadrilateral $ABCD$, the two points $O$, $O_1$ lie on the perpendicular bisector of $AD$. So $OO_1 \\perp AD$ and $EO_2 \\perp AD$ and thus we have $O_1O_2 \\parallel EO_2$. Similarly, since $O_2$ is the circumcenter of the $\\triangle EBC$ and $O$ is the center of circun circle of the quadrilateral $ABCD$, the two points $O$, $O_2$ lie on the perpendicular bisector of $BC$. So $EO_1 \\perp BC$ and $EO_1 \\perp BC$ and thus we have $EO_1 \\parallel OO_2$. So the quadrilateral $OO_1EO_2$ is a parallelogram and the line $OE$ passes the midpoint $M$ of $O_1O_2$, and thus three points $E$, $M$, $O$ lie on the same line. Therefore three points $E$, $K$, $O$ lie on the same line. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42411,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is an $n \\times n$ chess board. Each of the $n^2$ small boxes can present a number from $0$ to $k$, for some positive integer $k$. In each row and column, there is a button and if we push the button in a row (or column), the number in each of the $n$ small boxes contained in that row (or column, respectively) increases by $1$ and the number $k$ is changed to $0$. We call this a \"process\". At first, the number $0$ was presented in every $n^2$ small box. But there were some processes and the numbers were changed. Show that every number in the $n^2$ small boxes can be changed to $0$ by taking at most $kn$ processes.",
    "options": [],
    "answer": "kn",
    "solution": "Let $a_{ij}$ be the number in the box at the intersection of the $i$-th row and $j$-th column in the present phase. 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 pushed, and $d_t$ the number of times the button in the $t$-th column is pushed. Then one may easily show that $a_{st} = c_s + d_t \\pmod{k+1}$.\n\nHence, if $a_{1i} = a_{1j}$, then $a_{wi} = a + wj$ for every $w = 1, 2, \\dots, n$. The same holds for any two rows. Let $a_{11} = \\alpha$ for convenience. For any positive integers $i, j$ ($0 \\leq i, j \\leq k$), let $a_i$ be the number of $i$'s in the first row and $b_j$ be the number of $j$'s in the first column.\n\nNow for each $w$ with $0 \\leq w \\leq k$, consider the smallest number of processes pushing suitable buttons in the columns so that every number in the first row is changed to $w$. Note that the total number of processes to do this is\n$$\n\\sum_{i+j \\equiv w \\pmod{k+1}} j a_i = w a_0 + (w-1)a_1 + \\dots + a_{w-1} + k a_{w+1} + \\dots + (w+1)a_k.\n$$\nSince $\\alpha$ is changed to $w$, the number of times the button in the first column is pushed is $w - \\alpha$ (mod $k + 1$). Therefore the number $i$ in the first row is changed to $w - \\alpha + i$ (mod $k + 1$) during the processes. Now we push the buttons in the columns so that every number in the chess board is changed to $0$. The total number of times the buttons in the rows or columns are pushed during the processes is\n$$\nS_w = \\sum_{i+j \\equiv \\alpha-w \\pmod{k+1}} j b_i + \\sum_{i+j \\equiv w \\pmod{k+1}} j a_i.\n$$\nFor each $w = 0, 1, \\dots, k$, these are possible methods of changing all numbers in the boxes to $0$. Summing up all $S_w$'s for every $w$, we have\n$$\n\\sum_{w=0}^{k} S_w = \\left\\{ \\begin{array}{l} \\sum_{w=0}^{k} \\left( \\sum_{i+j \\equiv \\alpha-w \\pmod{k+1}} j b_i + \\sum_{i+j \\equiv w \\pmod{k+1}} j a_i \\right) \\\\ \\\\ \\approx \\sum_{w=0}^{k} \\left( \\sum_{j \\equiv \\alpha-i-w \\pmod{k+1}} j \\right) b_i + \\sum_{w=0}^{k} \\left( \\sum_{j \\equiv w-i \\pmod{k+1}} j \\right) a_i \\\\ \\\\ \\approx \\frac{k(k+1)}{2} \\left( \\sum_{i=0}^{k} a_i + \\sum_{i=0}^{k} b_i \\right) = k(k+1)n. \\end{array} \\right.\n$$\nTherefore at least one $S_w$ is less than or equal to $kn$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42412,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be the sides of a triangle, and we set\n$$\nA = \\frac{a^2 + bc}{b+c} + \\frac{b^2 + ca}{c+a} + \\frac{c^2 + ab}{a+b}\n$$\n$$\nB = \\frac{1}{\\sqrt{(a+b-c)(b+c-a)}} + \\frac{1}{\\sqrt{(b+c-a)(c+a-b)}} + \\frac{1}{\\sqrt{(c+a-b)(a+b-c)}}\n$$\nProve that $AB \\ge 9$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Clearly\n$$\nB \\ge \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{a}\n$$\nand\n$$\nA - (a + b + c) = \\frac{a^4 + b^4 + c^4 - a^2b^2 - b^2c^2 - c^2a^2}{(a+b)(b+c)(c+a)} \\ge 0.\n$$\nTherefore we have\n$$\nAB \\ge \\left( \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{a} \\right) (b+c+a) \\ge 9.\n$$\nby Cauchy-Schwarz inequality. This completes the proof. □",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42413,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a table, there are $2009$ coins each with one side white and the other side black. At the beginning all coins are aligned in a row with all their white sides up except one with its black side up. In each step we choose a coin with black side up and flip the two adjacent coins. If we choose the outermost coin, then we flip just one adjacent coin. Find all positions of the black coin at the beginning from which we can reach the state where all coins have black sides up.",
    "options": [],
    "answer": "1005",
    "solution": "For each coin in the table we give a positive integer one by one from the leftmost one. We show that if the black coin is in the $s = 1005$th place, then we can make each coin's black side up after some steps. If we choose $s, s-1, s+1$ one by one, then all five coins having numbers from $s-2$ to $s+2$ become black side up.\n\nAssume that all coins with numbers\n$$\ns-k, s-(k-1), \\dots, s-1, s, s+1, \\dots, s+(k-1), s+k\n$$\nbecome black side up. If $k=2\\ell$ then we proceed the steps one by one with black coins having the following numbers\n$$\ns-2\\ell, s+2\\ell, s-2(\\ell-1), s+2(\\ell-1), s-2, s+2, s.\n$$\nand if $k=2\\ell-1$ then\n$$\ns-(2\\ell-1), s+(2\\ell-1), s-(2\\ell-3), s+(2\\ell-3), s-1, s+1.\n$$\nso that we can make all coins with numbers from $s-k-1$ to $s+k+1$ become black side up.\n\nNow assume that all coins whose black sides are up have numbers $T_1, T_2, \\dots, T_k$ in some state. We define\n$$\nM = \\sum_{i=1}^{k} (-1)^{i+1} T_i = T_1 - T_2 + T_3 - \\dots + (-1)^{k-1} T_k.\n$$\nAssume that we choose the coin having a number $T_j$ to proceed the step. First consider the case when $T_j \\neq 1, 2009$. If $T_{j-1} \\neq T_j - 1$ and $T_{j+1} \\neq T_j + 1$ then the value $M$ after the procedure becomes\n$$\n\\sum_{i=1}^{j-1} (-1)^{i+1} T_i + (-1)^{j+1} (T_j - 1) + (-1)^{j+2} T_j + (-1)^{j+3} (T_j + 1) + \\sum_{i=j+1}^{k} (-1)^{i+1} T_i.\n$$\nwhich equals to $M$. For the remaining one may easily check that $M$ is an invariant in each step. If $j=1$ then $M$ is changed by $-M$ whether or not $T_2 = 2$. Finally if $j=2009$, then one may easily check that $M$ is changed by $M + (-1)^{k+1}2010$. Therefore the remainder $r$ or $2010-r$ of $M$ divided by $2010$ is invariant under the procedure.\n\nIf every coin becomes black side up then $M = 1-2+...+2007-2008+2009 = 1005$.\nTherefore only when the coin whose black side is up is in the $1005$th place, we can make every coin's black side up. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42414,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle with $\\angle B < \\angle C$. Let $O$ be the circle tangent to the line $AC$ at the point $C$ and passing through the point $B$. The lines $AB$ and $CO$ meet the circle $O$ at the points $D (\\neq B)$ and $P (\\neq C)$, respectively. Let $E$ be the intersection of $AC$ and the line passing through $P$ and parallel to $AO$. The line $EB$ and the circle $O$ meet at the point $L (\\neq B)$. Let $F$ be the intersection of the bisector of the line segment $BD$ and the line $AC$, and $K$ be the intersection of $LF$ and $CD$. Show that $EK$ and $CL$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K'$ be the intersection of the line $CD$ and the line passing through $E$ and parallel to the line $CL$. Since $\\angle EBD = \\angle DCL = \\angle EK'D$, we have that the four points $B$, $D$, $E$ and $K'$ are on one circle, say $O_1$. Since $AE^2 = AC^2 = AD \\cdot AB$, we have that the line $AE$ is tangent to the circle $O_1$ at $E$.\n\nNow consider the homothety $H$ which sends $O_1$ to $O$. The center of the homothety $H$ is the point $F$ and $H$ maps points $E$ and $O_1$ to $C$ and $O$, respectively. And also $H$ maps the point $K'$ to the point $L$. So three points $F$, $K'$ and $L$ are collinear and thus we have $K' = K$. This completes the proof. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42415,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players play a game on a $(m+1)$ by $m$ grid, which has $m+1$ horizontal lines, $m$ vertical lines, and hence $m(m+1)$ intersection points. One pebble is put on an intersection point. Two players alternatively move the pebble to an adjacent point along an edge, but they are not allowed to use edges used previously by either player. A player loses if he cannot move the pebble any more. Prove that there exists a winning strategy for the first player if the pebble was initially put on a point in the bottom horizontal line.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\{(x, y) : x = 0, 1, \\dots, m, y = 1, 2, \\dots, m\\}$ be the set of all points on the grid. Let $(a, 0)$ be the initial position of the pebble. Let\n$$\nP = (a, 0), \\quad Q = (0, a), \\quad R = (m+1-a, m+1) \\text{ and } S = (m+1, m+1-a).\n$$\n\nLet $X$ be the set of all points of the grid on the lines $PQ$, $QR$, $RS$, $SP$ or the interior of the quadrangle formed by those four lines. The first player initially moves the pebble to $(a, 1)$ and removes the straightline segment joining $(a, 0)$ and $(a, 1)$. We color the point $(x, y)$ red if $x + y - a$ is even, and blue if $x + y - a$ is odd. Then the first player will be given a choice only if the pebble is on the red point, and the second player will be given a choice only if the pebble is on the blue point. Moreover, all blue points in $X$ are not adjacent to any point out of $X$. Furthermore, every red point in $X$ has even number of edges to a blue point. Therefore, whenever the first player is given a choice of moving the pebble, there are odd number of unused edges inside $X$ available for him for moving the pebble to a point in $X$. Since the game must end and the first player can not lose, the first player can always win. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 42416,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $m$, $n$ satisfying the equation $3^m - 7^n = 2$.",
    "options": [],
    "answer": "(m, n) = (2, 1)",
    "solution": "Since $3^m - 2 \\equiv 0 \\pmod{7}$, one may easily show that $m \\equiv 2 \\pmod{6}$. If $m = 2$ then $n = 1$, which is a solution of the equation.\n\nAssume that $m = 2s \\ge 4$. Note that $2 + 7^n$ is divisible by $27$ in this case and\n$$\n7^9 \\equiv 1 \\pmod{27}, \\quad 7^1 \\equiv -2 \\pmod{27}.\n$$\nHence we have $n \\equiv 4 \\pmod{9}$. If we let $n = 9t + 4$, then\n$$\n2 + 7^{9t+4} \\equiv 2 + 7^4 \\equiv 35 \\pmod{37}.\n$$\nFor any positive integer $u$ less than $10$,\n$$\n9^n \\equiv 9, 7, 26, 12, 34, 10, 16, 33, 1 \\pmod{37}.\n$$\nTherefore there does not exist a solution for any $m \\ge 4$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 42417,
    "subject": "Mathematics (Multi-modal)",
    "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 and the inscribed circle. Prove the inequality\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": "Detailed solution",
    "solution": "Put $BC = a$, $CA = b$, $AB = c$, $QR = p$, $RP = q$, $PQ = r$ and let $AP = x$, $BQ = y$, $CR = z$. For short,\nSince $x + y = c$, $y + z = a$, $z + x = b$, we have\n$$\nx = s - a, \\quad y = s - b, \\quad z = s - c \\quad \\left( s = \\frac{a+b+c}{2} \\right).\n$$\nThe Cosine Law applied to the triangles $ABC$ and $ARQ$ thus yields\n$$\na^2 = b^2 + c^2 - 2bc \\cos A = (b - c)^2 + 2bc(1 - \\cos A)\n$$\nand\n$$\np^2 = 2x^2(1 - \\cos A) = 2(s - a)^2(1 - \\cos A).\n$$\nNow, canceling out $1 - \\cos A$ from the above two identities, we may represent $p^2$ in terms of $a, b, c$ as\n$$\n\\begin{aligned}\np^2 &= (s-a)^2 \\cdot \\frac{a^2 - (b-c)^2}{bc} \\\\\n&= \\frac{4(s-a)(s-b)(s-c)}{abc} \\cdot a(s-a).\n\\end{aligned} \\quad (1)\n$$\nMeanwhile, note that\n$$\n4(s - a)(s - b) = (b - c - a)(a - b - c) = c^2 - (b - a)^2 \\leq c^2\n$$\nand similarly that\n$$\n4(s - b)(s - c) \\leq a^2 \\quad \\text{and} \\quad 4(s - c)(s - a) \\leq b^2.\n$$\nThis yields $8(s - a)(s - b)(s - c) \\leq abc$, which in turn, together with (1), yields $p^2 \\leq \\frac{a(s-a)}{2}$, or equivalently,\n$$\n\\left(\\frac{a}{p}\\right)^3 \\geq 2\\sqrt{2}\\left(\\frac{a}{s-a}\\right)^{3/2}.\n$$\nThe parallel arguments yield\n$$\n\\left(\\frac{b}{q}\\right)^3 \\geq 2\\sqrt{2}\\left(\\frac{b}{s-b}\\right)^{3/2} \\quad \\text{and} \\quad \\left(\\frac{c}{r}\\right)^3 \\geq 2\\sqrt{2}\\left(\\frac{c}{s-c}\\right)^{3/2}.\n$$\nThus, denoting by $M$ the left hand side of the given inequality, we obtain\n$$\n\\begin{aligned}\nM &\\geq 2\\sqrt{2} \\left\\{ \\left( \\frac{a}{s-a} \\right)^{3/2} + \\left( \\frac{b}{s-b} \\right)^{3/2} + \\left( \\frac{c}{s-c} \\right)^{3/2} \\right\\} \\\\\n&\\geq \\frac{2\\sqrt{2}}{\\sqrt{3}} \\left( \\frac{a}{s-a} + \\frac{b}{s-b} + \\frac{c}{s-c} \\right)^{3/2}\n\\end{aligned} \\quad (2)\n$$\nwhere the second inequality holds by Jensen's inequality. Moreover, since\n$$\na \\geq b \\geq c \\iff \\frac{1}{s-a} \\geq \\frac{1}{s-b} \\geq \\frac{1}{s-c}.\n$$\nwe have by the Chevyshev inequality and the AM-GM inequality\n$$\n\\begin{aligned}\n\\frac{a}{s-a} + \\frac{b}{s-b} + \\frac{c}{s-c} &\\geq \\frac{1}{3}(a+b+c) \\left( \\frac{1}{s-a} + \\frac{1}{s-b} + \\frac{1}{s-c} \\right) \\\\\n&\\geq \\frac{a+b+c}{\\{(s-a)(s-b)(s-c)\\}^{1/3}} \\\\\n&= \\frac{(a+b+c)s^{1/3}}{\\{s(s-a)(s-b)(s-c)\\}^{1/3}} \\\\\n&= \\frac{1}{2^{1/3}} \\left( \\frac{L^2}{T} \\right)^{2/3}\n\\end{aligned} \\quad (3)\n$$\nNow, we conclude from (2) and (3)\n$$\nM \\geq \\frac{2\\sqrt{2}}{\\sqrt{3}} \\cdot \\frac{1}{\\sqrt{2}} \\frac{L^2}{T} = \\frac{2}{\\sqrt{3}} \\cdot \\frac{L^2}{T}.\n$$\nas required.\n\n□",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42418,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$, $O$, $H$ be the incenter, circumcenter, orthocenter of an acute triangle $ABC$, respectively, and the inscribed circle of $\\triangle ABC$ touch side $BC$ at $D$. Suppose that $\\angle B < \\angle C$, and $AO$ is parallel to $HD$. Show that the four points $E$, $F$, $I$, $O$ lie on a common circle, where $E$ is the intersecting point of lines $OD$ and $AH$, and $F$ is the midpoint of line segment $CI$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $E' (\\neq A)$ be the intersecting point of line $AH$ and the circumscribed circle of $\\triangle ABC$, and $D'$ be the intersecting point of line $OE'$ and side $BC$. We have\n$$\n\\angle OEA' = \\angle D'E'A = \\angle D'HE'\n$$\nsince points $H$ and $E'$ locate symmetrically about side $BC$. And it also holds that\n$$\n\\angle OEA' = \\angle OAE'\n$$\nsince line segments $AO$ and $E'O$ are radii of the circumscribed circle of $\\triangle ABC$.\nHence we obtain that $\\angle OAE' = \\angle D'HE'$, and thus $AO//HD'$. By condition $AO//HD$ in the problem we have that $D = D'$ and $E = E'$, since both points $D$ and $D'$ lie on line segment $BC$.\n\nLet $P (\\neq D)$ be the intersecting point of line $ID$ and the inscribed circle of $\\triangle ABC$, and $Q$ be the intersecting point of line $AP$ and side $BC$. Then by homothety, the inscribed circle touches side $BC$ at $Q$ because that point $A$ is the center of similarity for the inscribed circle and the inscribed circle of $\\triangle ABC$. Thus the midpoint $M$ of line segment $BC$ also becomes the midpoint of line segment $QD$. Now let $P'$ be the intersecting point of lines $OQ$ and $PD$. Since $QM = MD$ and $OM//P'D$, we have\n$$\n2OM = P'D.\n$$\nUsing properties of the Euler line, we have\n$$\n2OM = AH.\n$$\nHence we have\n$$\nP'D = AH.\n$$\nSince $P'D//AH$, rectangle $AP'DH$ is a parallelogram, and thus $AP'//HD$. Therefore points $O$ and $Q$ lie on the line $AP'$. That is, $P = P'$. Hence $2OM = P'D = PD = 2ID$ and $OM = ID$. Therefore, we have\n$$\nOI//BC.\n$$\nBy the relationship between the circumcenter and the orthocenter of a triangle we have that $\\angle BAO = \\angle CAE$ and $\\angle CAE = 90^\\circ - \\angle C$. Thus\n$$\n\\angle OAH = \\angle A - (\\angle CAE + \\angle BAO) = \\angle A - 2(90^\\circ - \\angle C).\n$$\nBy properties of an inscribed angle in the circumscribed circle of $\\triangle ABC$, it holds that $\\angle AEC = \\angle B$. And $\\angle DEA = \\angle OAH$ because $\\overline{AO} = \\overline{OE}$. Thus\n$$\n\\angle DEC = \\angle DEA + \\angle AEC = \\angle A - 2(90^\\circ - \\angle C) + \\angle B = \\angle C.\n$$\nOn the other hand, the point $F$ is the center of the circle that passes through three points $I, D, C$, which means that $FD = FC$. Hence we have\n$$\n\\angle DFC = 180^\\circ - \\angle C.\n$$\nTherefore four points $E, C, F, D$ lie on a circle, and thus $\\angle EDC = \\angle EFC$.\nFrom the fact $OI//BC$ we have that $\\angle EDC = \\angle EOI$, i.e., $\\angle EFC = \\angle EOI$. Hence we finally conclude that four points $E, F, I, O$ lie on a circle. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42419,
    "subject": "Mathematics (Multi-modal)",
    "question": "When there is a link from a webpage $i$ to a webpage $j$, you can move from the webpage $i$ to the webpage $j$ directly. For $n \\ge 2$, there are $n$ many webpages numbered from $1$ to $n$, and for all $i \\in \\{1, 2, \\dots, n-1\\}$, there is a link from the webpage $i$ to the webpage $(i+1)$.\nNow you are allowed to add new links to the webpages so that new links can connect two webpages from $i$ to $j$ when $i < j$. Prove that there exist a way to add at most $3(n-1) \\log_2(\\log_2 n)$ many such new links so that for any two integers $1 \\le i < j \\le n$, you can move from the webpage $i$ to the webpage $j$ using at most 3 links.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $T(n)$ be smallest number of new links so that the property holds. By using mathematical induction we will show that for all $n \\ge 2$,\n$$\nT(n) \\le 3(n-1) \\log_2 \\log_2 n.\n$$\n\nIt is trivial that $T(2) = 0 \\le 3(2-1)\\log_2\\log_2 2$, and $T(3) = 0 \\le 3(3-1)\\log_2\\log_2 3$.\nLet $n \\ge 4$. When $2 \\le m < n$, suppose that\n$$\nT(m) \\le 3(m - 1) \\log_2 \\log_2 m. \\qquad (1)\n$$\nNow we will show that (1) is true when $m = n$. Let $k \\ge 2$ be an integer so that $k^2 \\le n < (k+1)^2$. Since $T$ is an increasing function, $T(n) \\le T((k+1)^2 - 1)$.\nImagine a partition of the webpages $\\{1, 2, \\dots, (k+1)^2 - 1\\}$ into $(k+1)$ many groups of $(k+1)$ many webpages. Only the last group has $k$ many webpages. Precisely, $\\{1, 2, 3, \\dots, k+1\\}$ is the first group, $\\{k+2, k+3, \\dots, 2(k+1)\\}$ is the second group, and $\\{k(k+1)+1, k(k+1)+2, \\dots, (k+1)^2 - 1\\}$ is the last group.\nNow, for each $r = 0, 1, 2, \\dots, k-1$, for all $1 \\le i \\le k-1$, add a new link from the webpage $r(k+1)+i$ to the webpage $r(k+1)+(k+1)$. The total number of new links of this type is $(k-1)k$. Similarly for each $r = 1, 2, \\dots, k$, for all $2 \\le i \\le k$, add a new link from the webpage $r(k+1)$ to the webpage $r(k+1)+i$. The total number of new links of this type is also $(k-1)k$. Also add $\\binom{k}{2}$ many new links between all two webpages whose numbers are multiples of $(k+1)$.\nUsing these new links, any two webpages that belong to different groups are connected by at most 3 links, by taking webpages whose numbers are multiples of $(k+1)$ as webpages in the middle of the link path. From the definition of $T(k)$, if we add $T(k)$ many links to the $k$ many webpages in each group, the required property holds. Hence,\n$$\nT(n) \\le T(k^2 + 2k) \\le 2k(k-1) + \\frac{k(k-1)}{2} + (k+1)T(k).\n$$\nSince\n$$\n3(k^2 - 1) \\log_2 \\log_2 k^2 \\le 3(n - 1) \\log_2 \\log_2 n.\n$$\nto finish the induction argument, it is enough to show that\n$$\n2k(k-1) + \\frac{k(k-1)}{2} + (k+1)T(k) \\le 3(k^2 - 1) \\log_2 \\log_2 k^2. \\qquad (2)\n$$\nBy applying the induction hypothesis on $T(k)$ to (2), it is enough to show that\n$$\n2k(k-1) + \\frac{k(k-1)}{2} + (k+1)3(k-1)\\log_2\\log_2 k \\le 3(k^2-1)\\log_2\\log_2 k^2. \\quad (3)\n$$\n(3) is equivalent to the following, which is true for all $k \\ge 2$.\n$$\n\\begin{align*} & \\iff 2k^2 - 2k + \\frac{k^2 - k}{2} \\le 3(k^2 - 1) \\\\ & \\iff \\frac{k^2}{2} + \\frac{5}{2}k - 3 \\ge 0. \\end{align*}\n$$\nHence (1) holds for $m = n$, which proves the mathematical induction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42420,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a trapezoid $ABCD$ with $AB // CD$, suppose that $A$, $B$, $C$, $D$ lie in the clockwise direction. Let $\\Gamma_1$ be the circle centered at $A$, and passing through $B$. Let $\\Gamma_2$ be the circle centered at $C$, and passing through $D$. Let $P$ be the intersection (distinct from $B$, $D$) of the line $BD$ and the circle $\\Gamma_1$. Let $\\Gamma$ be the circle with the diameter $PD$. Let $X$ be the intersection (distinct from $P$) of $\\Gamma$ and $\\Gamma_1$. Let $Y$ be the intersection (distinct from $D$) of $\\Gamma$ and $\\Gamma_2$. Let $Q$ be the intersection of $\\Gamma_2$ and the circumcircle of $XBY$. Show that $B$, $D$, $Q$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It suffices to show that $X$, $B$, $Y$, $Q$ are cyclic under the assumption that $Q$ is the intersection of the line $BD$ and $\\Gamma_2$. There are six cases depending on the ordering of $P$, $D$, $B$, $Q$. Although we should consider all those six cases, we here give the proof only for the case where the order is $P$, $D$, $B$, $Q$.\n\nLet $\\angle ABP = \\alpha$, $\\angle DPX = \\beta$. Since $AB // CD$, we have $\\angle ABP = \\angle BDC = \\alpha$. For the circle $\\Gamma_1$, the circumference angle for $PB$ equals $\\frac{\\pi}{2} - \\alpha$; hence we have $(\\frac{\\pi}{2} - \\alpha) + \\angle PXB = \\pi$. That is, $\\angle PXB = \\frac{\\pi}{2} + \\alpha$.\n\nOn the other hand, we have\n$$\n\\angle DYQ = \\frac{1}{2} \\angle DCQ = \\frac{\\pi}{2} - \\alpha.\n$$\nFor the circle $\\Gamma$, $\\angle DPX = \\angle DYX = \\beta$. Now it suffices to show that $\\angle DBX = \\angle XYQ$.\n\nFor the triangle $DBX$, we have\n$$\n\\angle DBX = \\pi - \\beta - (\\frac{\\pi}{2} + \\alpha) = \\frac{\\pi}{2} - \\alpha - \\beta.\n$$\nMeanwhile, since $\\angle DYQ = \\frac{\\pi}{2} - \\alpha$ and $\\angle DYX = \\beta$, we have\n$$\n\\angle XYQ = \\angle DYQ - \\angle DYX = \\frac{\\pi}{2} - \\alpha - \\beta. \\quad \\square",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42421,
    "subject": "Mathematics (Multi-modal)",
    "question": "Around a table $2n$ people are seated and $m$ cookies are distributed to them. These people can pass cookies under the following rules:\n\n- One can only pass cookies to his/her neighbors.\n- One can pass a cookie only if he or she eats one cookie.\n\nLet $A$ be one of these people. Find the minimum $m$ such that no matter how $m$ cookies were distributed, there is a strategy to pass cookies so that $A$ has at least one cookie.",
    "options": [],
    "answer": "2^n",
    "solution": "We will prove that the minimum number $m$ of cookies is $2^n$. Let us write $A_{-n+1}, \\dots, A_0(= A), A_1, A_2, \\dots, A_{n-1}, A_n$ to denote the $2n$ people in the counterclockwise order.\n\nFirst let us show that if $m < 2^n$, then there is a way to distribute $m$ cookies so that $A$ cannot get a cookie. Let $a_i$ be the amount of cookies given to $A_i$. Let $N = \\sum_{i=-n+1}^{n} a_i 2^{|i|}$. Note that the value of $N$ is non-increasing when one passes a cookie to the neighbors. So let us initially give all $m$ cookies to $A_n$. Then $N = m2^{-n} < 1$. This means that even if we allow the people to pass cookies under the given rules, we still have $N < 1$ and therefore it is impossible for $A$ to have a cookie.\n\nNow let us show that if $m = 2^n$, then there is a strategy to make $A$ have a cookie. Let $a_i$ be the amount of cookies given to $A_i$. By symmetry, we may assume that $\\sum_{i=0}^{n-1} a_i \\ge \\sum_{i=0}^{n-1} a_{-i}$.\n\nWe ask $A_n$ to pass $\\lfloor a_n/2 \\rfloor$ cookies to $A_{n-1}$ by eating $\\lfloor a_n/2 \\rfloor$ cookies by himself. This allows us to assume that $A_0, A_1, A_2, \\dots, A_{n-1}$ have at least $m/2 = 2^{n-1}$ cookies. (If $a_n$ is even, then this is O.K. because they will have at least $(m-a_n)/2 + a_n/2$ cookies. If $a_n$ was odd, then they will have at least $(m-a_n)/2 + (a_n-1)/2 = m/2 - 1/2$ cookies. Since $m$ is even, we can still conclude that $A_0, A_1, A_2, \\dots, A_{n-1}$ have at least $m/2$ cookies.)\n\nNow we claim that for $k = 1, 2, \\dots, n-1$, if $A_0, A_1, A_2, \\dots, A_k$ have at least $2^k$ cookies, then by passing cookies of $A_k$ to $A_{k-1}$, we may assume that $A_0, A_1, A_2, \\dots, A_{k-1}$ have at least $2^{k-1}$ cookies. We may assume that $a_k > 2^{k-1}$. If $a_k = 2^k$, then we simply let $A_k$ pass $2^{k-1}$ cookies to $A_{k-1}$ by eating $2^{k-1}$ cookies. So we may assume $a_k < 2^{k-1} + 2^{k-1}$. Therefore there exists an integer $i \\in \\{1, 2, 3, \\dots, k-1\\}$ such that\n$$\n2^{k-1} + 2^{k-2} + \\dots + 2^{k-i} < a_k \\le 2^{k-1} + 2^{k-2} + \\dots + 2^{k-i-1}.\n$$\n\nNote that\n$$\n\\sum_{i=0}^{k-1} a_i \\ge 2^k - a_k \\ge 2^k - (2^{k-1} - 2^{k-2} + \\dots + 2^{k-i-1}).\n$$\n\nNow if $A_k$ passes his cookies to $A_{k-1}$ as many as he can, then he can pass at least $2^{k-2} + 2^{k-3} + \\dots + 2^{k-i-1}$ cookies.\n\nThen $A_0, A_1, \\dots, A_{k-1}$ have at least $\\sum_{i=0}^{k-1} a_i + (2^{k-2} + \\dots + 2^{k-i-1}) \\ge 2^{k-1}$ cookies. This proves the claim.\n\nBy the above claim we can inductively make $A_0$ have at least $2^0 = 1$ cookies. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42422,
    "subject": "Mathematics (Multi-modal)",
    "question": "A prime $p$ is called *nice prime* if there exists sequences of positive integers $(n_1, n_2, \\dots, n_k)$ satisfying the following conditions for infinitely many positive integers $k$, but there does not exist such for $k = 1$.\n1. For $i = 1, 2, \\dots, k$, $n_i \\ge \\frac{i+1}{2}$.\n2. For $i = 1, 2, \\dots, k$, $p^{n_i} - 1$ is a multiple of $n_{i+1}$ and $\\frac{p^{n_i} - 1}{n_{i+1}}$, $n_{i+1}$ are prime to each other. Set $n_{k+1} = n_1$.\nShow that $2$ is not a nice prime, but any odd primes are nice primes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us denote by $a|b$ when a positive integer $a$ divides $b$. We write $p^c \\nmid n$ when a positive integer $n$ is a multiple of $p^c$, but not of $p^{c+1}$ for a prime $p$.\n\n**Lemma 1.** For a given prime $q$ and a positive integer $a$ such that $(a, q) = 1$, let $s$ be the smallest positive integer $x$ satisfying $q|a^s - 1$. Any positive integer $y$ such that $q|a^y - 1$ must be a multiple of $s$.\n*Proof.* Put $y = qs + r_0$, where $q, r_0$ are integers and $0 \\le r_0 < s$. Noting $q|a^y - 1 = (a^s - 1)(a^{y-s} + a^{y-2s} + \\dots + a^{r_0}) + a^{r_0} - 1$ leads us to $q|a^{r_0} - 1$. By the minimality of $s$, we get $r_0 = 0$. $\\square$\n\n1. The case of $p = 2$\nIt is immediate to note that all $n_i$s are odd and greater than or equal to $3$. Let $q (\\ge 3)$ be the smallest prime dividing one of $n_1, n_2, \\dots, n_k$. Without loss of generality, we assume $q|n_2$. Let $s \\ge 2$ be the smallest positive integer $x$ satisfying $q|2^r - 1$. Then by Fermat's little theorem and Lemma 1, we get $s|q - 1$. Thus $s < q$. We realize that $s$ has a smaller prime factor than $q$, so it contradicts $q|2^{n_1} - 1$, which implies $s|n_1$. We conclude that there does not exist $(n_1, n_2, \\dots, n_k)$ fulfilling conditions, hence $2$ is not a nice prime.\n\n2. The case of prime $p \\ge 3$\n(1) In the case $k = 1$: we prove that there does not exist a positive integer $n$ such that $n|p^n - 1$ and $\\left(\\frac{p^n-1}{n}, n\\right) = 1$.\nFirst, under the condition $n|p^n - 1$, we show the smallest prime factor $q_1$ of $n$ must divide $p-1$. Let $n = q_1^{e_1} \\cdots q_r^{e_r}$, $q_1 < \\cdots < q_r$. $n|p^n - 1$ implies $q_1|p^n - 1$. Let $s$ be the smallest positive integer $x$ such that $q_1|p^x - 1$. Then it is obvious that $s \\le q_1 - 1$ and $s|n$. If $s > 1$, $s$ is a product of primes smaller than $q_1$, it violates that $n$ cannot have smaller prime factors than $q_1$. We obtain $s = 1$ so $q_1|p - 1$.\nNow let $n = q_1^{e_1} n_0$. In the expression $p^n - 1 = p^{q_1^{e_1} n_0} - 1 = (p^{n_0} - 1) \\prod_{j=0}^{e_1-1} \\frac{p^{q_1^{j} n_0} - 1}{p^{q_1^{j-1} n_0} - 1}$. We easily see the number of prime $q_1$ in $p^n - 1$ is at least $e_1 + 1$. So we get $q_1|\\left(\\frac{p^n-1}{n}, n\\right)$ reducing to contradiction. In sum, for any given prime $p$, there does not exist a positive integer $n$ satisfying all conditions when $k = 1$.\n\n(2) In the case $k \\ge 2$: we are always able to find sequences of positive integers $(n_1, n_2, \\dots, n_k)$ for infinitely many $k$. This proves that any odd primes are nice. There are various ways of taking the sequences and the proposed way below is one of them.\n\n**Lemma 2.** For a positive integer $c$ and a prime $q$, $q^c|p-1$ implies $q^{c+d}|p^{q^d} - 1$. Moreover,\n- When $q$ is odd or $(q = 2, c \\ge 2)$: the smallest positive integer $x$ such that $q^{c+d}|p^x - 1$ is $q^d$ and $q^{c+d}||p^{q^d} - 1$ holds.\n- When $(q = 2, c = 1)$: Let $c'$ be the smallest positive integer such that $2^{c'}||p^2 - 1$. Then the smallest positive integer $x$ satisfying $2^{c'-d}||p^x - 1$ is $2^{d+1}$ and $2^{c'+d}||p^{2d+1}-1$ holds.\n\nProof. Let us show only the first case. When $d=0$, it is true by definition. Let us assume the statement is true when $d=d_0$ and try to show when $d=d_0+1$. We need to find the smallest positive integer $x$ such that $q^{c+d_0+1}||p^x-1$. Since $q^{c+d_0}||p^x-1$ also holds, by the assumption, $x$ is a multiple of $q^{d_0}$ and we set $x=q^{d_0}u$. From the equation $p^x-1 = p^{q^{d_0}u}-1 = (p^{q^{d_0}}-1)(p^{(u-1)q^{d_0}}+p^{(u-2)q^{d_0}}+\\cdots+1)$, $p^{q^{d_0}}-1$ is already a multiple of $q^{c+d_0}$, so it is enough to have $p^{(u-1)q^{d_0}}+p^{(u-2)q^{d_0}}+\\cdots+1$ a multiple of $q$. $p^{(u-1)q^{d_0}}+p^{(u-2)q^{d_0}}+\\cdots+1\\equiv 1+\\cdots+1\\equiv u\\pmod q$ implies $q\\nmid u$ and conversely, we easily check $p^x-1$ is really a multiple of $q^{c+d_0+1}$ when $q$ divides $u$. Thus $x=q^{d_0}u$ is a multiple of $q^{d_0+1}$.\nOn the other hand, $\\frac{p^{q^{d_0}}-1}{p^{q^{d_0}-1}} = p^{(q-1)q^{d_0}} + p^{(q-2)q^{d_0}} + \\cdots + 1$ is only a multiple of $q$ not of $q^2$, whose proof is presented below. In conclusion, we obtain $q^{c+d}||p^{q^d}-1$ by induction.\nThe reason for $q||p^{(q-1)q^{d_0}} + p^{(q-2)q^{d_0}} + \\cdots + 1$: Let $v$ be a positive integer such that $p^{q^{d_0}} = 1+qv$. Then, $p^{(q-1)q^{d_0}} + p^{(q-2)q^{d_0}} + \\cdots + 1 = (1+qv)^{q-1} + (1+qv)^{q-2} + \\cdots + (1+qv) + 1 \\equiv 1 + (q-1)qv + 1 + (q-2)qv + \\cdots + (1+qv) + 1 = q + \\frac{(q-1)q^2}{2}v \\equiv q \\pmod{q^2}$.\nIn the case $(q=2, c=1)$, we are not able to guarantee the term $\\frac{(q-1)q^2}{2}v$ in the almost last equation is a multiple of $q^2$ so we separate the case. $\\square$\n\nNow we define $S = \\{q|q$ a prime dividing $p-1\\}$ and let $q \\in S$. Take $n_1 = p-1$ and let $c_{q,1}$ be an integer such that $q^{c_{q,1}}||p-1$ for $q \\in S$. Similarly, we define $c_{q,2}$ as an integer satisfying $q^{c_{q,2}}||p^{n_1}-1 = p^{n_1-1}-1$. Inductively for $q \\in S, i \\ge 1$, let $c_{q,i+1}$ be an integer such that $q^{c_{q,i+1}}||p^{n_i}-1$ and $n_{i+1} := \\prod_{q \\in S} q^{c_{q,i+1}}$.\nWe define $c_{q,i}$ and $n_i$ in this way accordingly as $i$ gets increased one by one, then we notice that $c_{q,i}$ also gets large as $i$ does so. Fix $q_0 \\in S$ and put $c := c_{q_0,1}$. (When $q_0 = 2, c_{q_0,1} = 1$, let $c = c' - 1$, where $c'$ is in Lemma 2.(ii).)\nSince there exists an integer $i$ such that $q_0^{c_{q_0,i}} > p-1$, we call such $i$ $i_0$. Moreover, there exists a positive integer $k_0$ such that $\\frac{p^{k_0}-1}{(p-1)q_0^{k_0}} \\ge \\frac{p+1}{2}$, which is justified by that the left hand side is a function divergent when $k_0$ goes to infinity. Now let $k$ be any integer greater than $1 + \\max\\{i_0, k_0, 3\\}$. Except the case $q_0 = 2, c_{q_0,1} = 1$, we can, in fact, define $k$ to be any integer greater than $\\max\\{i_0, k_0, 3\\}$. We apply the above definition of $c_{q,i+1}$ and $n_{i+1}$ till $i < k-2$ and let $n_{k-1} := q_0^{c_{q_0,k-1}}$. It is easy to observe $n_{k-1} \\ge p-1 \\ge \\frac{p-1}{2}$.\nFrom our choices of sequences $\\{n_i\\}$, it is clear to verify the following for $1 \\le i < k-1$\n$$\n-1, n_1 = p-1 \\ge \\frac{p+1}{2}, n_{i+1} = p-1 \\ge \\frac{p+1}{2}\n$$\n$$\n-2, n_{i+1} | p^{n_i} - 1, \\left( \\frac{p^{n_i-1}}{n_{i-1}}, n_{i+1} \\right) = 1\n$$\nThe first one comes from $c_{q,i+1} > c_{q,i} \\ge 1$. From Lemma 2, we even know $c_{q,i+1} = c_{q,i} + c_{q,i}$ holds when $q$ is odd or $c_{q,i} \\ge 2$.\nFinally, let $n_k := \\frac{p^{n_k} - 1}{q_0^{c_{q_0,k-1}} \\prod_{q \\in S} q^{c_{q,q}}} = \\frac{p^{n_k} - 1}{q_0^{c_{q_0,k-1}}(p-1)}$, then $(n_k, q_0) = 1$ follows from Lemma 2, hence it satisfies the second condition $(n_k, \\frac{p^{n_k} - 1}{n_k}) = 1$ for $n_k$.\nWe also obtain the first condition $n_k \\ge \\frac{p+1}{2}$ from $\\frac{p^{n_k} - 1}{(p-1)q_0^{c_{q_0,k-1}}} \\ge \\frac{p+1}{2}$. Noting $(p-1, \\frac{p^{n_k}-1}{p-1}) = (p-1, n_k)$ for this $n_k$ and $(q, n_k) = 1$ for $q \\in S$, we establish $(p-1, n_k) = 1$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42423,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are no positive integers $x, y, z$ such that\n$$\nx^2 y^4 - x^4 y^2 + 4x^2 y^2 z^2 + x^2 z^4 - y^2 z^4 = 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove this statement by contradiction. Assume that there are positive integers $x, y, z$ satisfying the equation\n$$\nx^2 y^4 - x^4 y^2 + 4x^2 y^2 z^2 + x^2 z^4 - y^2 z^4 = 0. \\quad (1)\n$$\nIt is easy to check that $x \\neq y$. If there are solutions of the equation (1), we have a solution of the equation (1) such that $\\gcd(x, y) = (x, y) = 1$. Now let $x, y, z$ be a positive integer solution of the equation (1) such that $(x, y) = 1$.\nBy factoring the equation (1), we get\n$$\n\\begin{aligned}\n0 &= x^2 y^4 - x^4 y^2 + 4x^2 y^2 z^2 + x^2 z^4 - y^2 z^4 \\\\\n&= x^2(y^4 + 2y^2 z^2 + z^4) - y^2(x^4 - 2x^2 z^2 + z^4) \\\\\n&= x^2(y^2 + z^2)^2 - y^2(x^2 - z^2)^2 \\\\\n&= (x(y^2 + z^2) + y(x^2 - z^2))(x(y^2 + z^2) - y(x^2 - z^2)) \\\\\n&= ((x - y)z^2 + x y(x + y))((x + y)z^2 + x y(y - x)).\n\\end{aligned}\n$$\nThen\n$$\n(y - x)z^2 = x y(x + y) \\quad \\text{or} \\quad (x + y)z^2 = x y(x - y).\n$$\nBy multiplying both sides of the first equation by $y - x$ we have\n$$\n(y - x)^2 z^2 = x y(y^2 - x^2). \\quad (2)\n$$\nBy multiplying both sides of the second equation by $x + y$ we have\n$$\n(x + y)^2 z^2 = x y(x^2 - y^2) \\quad (3)\n$$\nSince we can solve the equation (3) by the same way in which we solve the equation (2), we are going to solve the equation (2).\nBecause $(x, y) = 1$, $(x, y^2 - x^2) = (y, y^2 - x^2) = 1$. Now the left hand side of the equation (2) is a perfect square of an integer. We know that $x, y$, and $y^2 - x^2$ are perfect squares. So there are positive integers $a, b, c$ such that $x = a^2, y = b^2, y^2 - x^2 = c^2$. Therefore we have the equation\n$$\nb^4 - a^4 = c^2.\n$$\nBut it is well-known that there are no positive integers $a, b, c$ satisfying the above equation. It can be proved by infinite descent. Therefore, we prove the statement by contradiction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42424,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a point on the side $BC$ of an acute triangle $ABC$. Let $H$ be the orthocenter of $\\triangle ABC$ and $D$ be the foot of the perpendicular from $H$ to $AP$. Let $\\Gamma_1, \\Gamma_2$ be the circumcircles of $\\triangle ABD, \\triangle ACD$, respectively. Let $l$ be the line parallel to $BC$ and passing through $D$. $l$ intersects $\\Gamma_1$ and $\\Gamma_2$ at $X$ and $Y$, different from $D$, respectively. $l$ intersects $AB$ and $AC$ at $E$ and $F$, respectively. Two lines $XB$ and $YC$ meet at $Z$. Show that $BP = CP$ if and only if $ZE = ZF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Observation 1.** $A$, $B$, $C$, $Z$ are cyclic.\n$$\n\\angle BAD = \\angle BXD = \\angle ZBC \\text{ and } \\angle DAC = \\angle DYC = \\angle BCZ.\n$$\nHence $\\angle BAC = \\angle BAD + \\angle BCZ = 180^\\circ - \\angle BZC$, that is, $\\angle BAC + \\angle BZC = 180^\\circ$.\n\n**Observation 2.** Both $A$, $E$, $Z$, $Y$ and $A$, $X$, $Z$, $F$ are cyclic.\nBy Observation 1, we have\n$$\n\\angle AEY = \\angle ABC = \\angle AZC = \\angle AZY.\n$$\nSimilarly, $\\angle AFX = \\angle AZX$.\n\n**Lemma 1.** $ZE = ZF$ if and only if $\\frac{BD}{CD} = \\frac{AB}{AC}$.\n\n*Proof.* We can easily see that $\\triangle AEZ$ is similar to $\\triangle ADC$, and symmetrically, $\\triangle AFZ$ is similar to $\\triangle ADB$. Then we have\n$$\nZE = \\frac{AE \\cdot CD}{AD} \\quad \\text{and} \\quad ZF = \\frac{AF \\cdot BD}{AD}.\n$$\nTherefore, we have $ZE = ZF \\Leftrightarrow \\frac{BD}{CD} = \\frac{AE}{AF} = \\frac{AB}{AC}$.\n\nNow let $Q$, $R$, $S$ be the feet of perpendiculars from $A$, $B$, $C$ to $BC$, $CA$, $AB$, respectively. Then $A$, $S$, $D$, $H$, $F$ are cyclic. Since $AD \\cdot AP = AH \\cdot AQ = AS \\cdot AB$, $S$, $D$, $P$, $B$ are cyclic.\nSimilarly, $R$, $D$, $P$, $C$ are cyclic.\nSince $AS : AP = AD : AB$, $\\triangle ASP$ is similar to $\\triangle ADB$.\nNow we have\n$$\n(1) \\qquad PS = BD \\frac{AP}{AB}, \\quad RP = CD \\frac{AP}{AC}.\n$$\n\n**Lemma 2.** Let $P$ be a point on $BC$. Then\n$$\nBP = PC \\Leftrightarrow PR = PS.\n$$\n*Proof.* Suppose $P$ is a midpoint of $BC$. Then $P$, $Q$, $R$, $S$ are on the nine-point circle, so we have\n$$\n\\angle PSR = \\angle RQC = \\angle A = \\angle SQB = \\angle PRS.\n$$\nThat is, $PR = PS$. Hence $P$ is the intersection of the perpendicular bisector of $SR$ and $BC$, so it is unique. Thus $PR = PS$ implies that $P$ is a midpoint of $BC$. $\\square$\n\nNow we are ready to prove that $BP = PC \\Leftrightarrow ZE = ZF$.\n\nSuppose $BP = PC$, then by Lemma 2, $PR = PS$. Then by (1), we have $\\frac{BD}{CD} = \\frac{AB}{AC}$. Then by Lemma 1, $ZE = ZF$.\n\nConversely, assume that $ZE = ZF$, then by Lemma 1, $\\frac{BD}{CD} = \\frac{AB}{AC}$. Then by (1), we have $PS = PR$. By Lemma 2, $BP = PC$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42425,
    "subject": "Mathematics (Multi-modal)",
    "question": "There were $n$ boys $a_1, a_2, \\dots, a_n$ and $n$ girls $b_1, b_2, \\dots, b_n$ in a party. No boys shook hand with each other and no girls shook hand with each other and for each $i \\in \\{1, 2, \\dots, n\\}$, $a_i$ did not shake his hand with $b_i$. We want to divide all $2n$ people into groups satisfying the following two conditions.\n1. In each group, the number of boys is equal to the number of girls.\n2. In each group, no two people shook hands with each other.\nLet $m$ be the number of pairs $(a_i, b_j)$ where $a_i$ shook his hand with $b_j$. Prove that it is possible to make the number of groups less than or equal to $2$ or $\\frac{2m}{n} + 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $G$ be a graph on $2n$ vertices representing $n$ boys and $n$ girls such that two vertices are adjacent if they shook hands with each other.\n\n(1) First let us show that it is true when $m < n$. Let $C_1, C_2, \\dots, C_k$ be components of $G$. Since $G$ has $2n$ vertices and less than $n$ edges, the number of components should be larger than $n$. Then $\\sum_{i=1}^t |C_i|$ for $1 \\le t \\le n+1$ are distinct positive integers less than or equal to $2n$. By the pigeonhole principle, there exist $a < b$ such that $\\sum_{i=1}^a |C_i| \\equiv \\sum_{i=1}^b |C_i| \\pmod{n}$. Therefore $|C_{a+1}| + |C_{a+2}| + \\dots + |C_b| = n$. Let $A = C_{a+1} \\cup C_{a+2} \\cup \\dots \\cup C_b$. Since $|A| = n$, the number of boys in $A$ is equal to the number of girls not in $A$. Thus we can make one group consisting of boys in $A$ and girls not in $A$ and another group consisting of boys not in $A$ and girls in $A$. Then we obtain 2 groups satisfying the conditions.\n\n(2) Now let us assume that $m \\ge n$. Because $a_i$ is not adjacent to $b_i$ for all $i$, there is at least one method to partition people into groups satisfying the conditions. Let $s$ be the minimum number of groups while satisfying the conditions. Let $V_1, V_2, \\dots, V_s$ be the groups partitioning all $2n$ people.\n\nFor all $1 \\le i < j < k \\le s$, if we can merge 3 groups $V_i, V_j, V_k$ into two groups, then this contradicts to the assumption that $s$ is minimum. Thus, by (1), the number of edges among $V_i, V_j$, and $V_k$ is at least $\\frac{|V_i \\cup V_j| + |V_k|}{2}$. Let $S$ be the number of edges among $V_i, V_j$, and $V_k$ for all triples $i < j < k$. Then\n$$\nS \\ge \\frac{1}{2} \\binom{s-1}{2} \\sum_{i=1}^{s} |V_i| = \\frac{1}{2} \\binom{s-1}{2} (2n).\n$$\n\nOn the other hand, each edge is counted $s-2$ times in $S$. By the double counting, $S = (s-2)m$. We deduce that $m \\ge \\frac{s-1}{2}n$. We conclude that $s \\le \\frac{2m}{n} + 1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 42426,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the maximum value of\n$$\n\\frac{1}{a^2 - 4a + 9} + \\frac{1}{b^2 - 4b + 9} + \\frac{1}{c^2 - 4c + 9}\n$$\nwhere $a$, $b$, $c$ are non-negative real numbers satisfying $a + b + c = 1$.",
    "options": [],
    "answer": "7/18",
    "solution": "Note that for $0 \\le x \\le 1$ the inequality\n$$\n\\frac{1}{x^2 - 4x + 9} \\le \\frac{x + 2}{18}\n$$\nholds, where the equality holds if and only if $x = 0$ or $x = 1$. Hence,\n$$\n\\frac{1}{a^2 - 4a + 9} + \\frac{1}{b^2 - 4b + 9} + \\frac{1}{c^2 - 4c + 9} \\le \\frac{1}{18}(a + b + c + 6) = \\frac{7}{18}\n$$\nSince the equality holds for $a = 0$, $b = 0$, $c = 1$, the maximum value we look for is\n$$\n\\frac{7}{18}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42427,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AC < AB < BC$. Choose a point $D$ on the side $AB$. Suppose the circumcircle of $ABC$ meets the internal angle bisector of $\\angle A$ and $CD$ at $E$ ($\\neq A$) and $F$ ($\\neq C$), respectively. Let $K$ be the intersection point of $BC$ and $DE$. Prove that $CK = AC$ if and only if $DK \\cdot EF = AC \\cdot DF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $T$ be the intersection point of $ED$ and the circumcircle of $ABC$ other than $E$. Since $\\angle CTD = \\angle EFD$ and $\\angle CDT = \\angle EDF$, $DCT$ and $DEF$ are similar triangles. Hence\n$$\n\\frac{EF}{DF} = \\frac{TC}{DT}\n$$\nNote that $AE$ is the perpendicular bisector of the line segment $CD$, as $AD = AC$ and $\\angle DAE = \\angle CAE$. It follows that $EB = EC = ED$, which means that $E$ is the circumcenter of the triangle $BCD$. Thus $\\angle BEF = \\angle DEF$. This implies that $CD$ is the angle bisector of $\\angle TCK$. In fact, since $\\angle TCB = \\angle TEB$, $\\angle TCB = 2\\angle BEF$. Then $\\angle TCB = 2\\angle BCD$ comes from $\\angle BEF = \\angle BCD$. In a triangle $CKT$, $CD$ being the angle bisector of $\\angle TCK$ gives rise to the following:\n$$\n\\frac{TC}{DT} = \\frac{KC}{DK}\n$$\n\nFrom this along with the result obtained above, we obtain $\\frac{EF}{DF} = \\frac{KC}{DK}$. Thus,\n$$\nDK \\cdot EF = KC \\cdot DF.\n$$\nNow since $DK \\cdot EF = KC \\cdot DF$,\n$$\nDK \\cdot EF = AC \\cdot DF \\iff AC \\cdot DF = KC \\cdot DF.\n$$\nBut, since $AB > AC$, $D \\neq B$, and so $DF \\neq 0$. Therefore\n$$\nDK \\cdot EF = AC \\cdot DF \\iff CK = AC.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42428,
    "subject": "Mathematics (Multi-modal)",
    "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.) What is the number of shelves if we assign integers to maximize the number of shelves?",
    "options": [],
    "answer": "floor(((n+1)(m+1))/2) - 1",
    "solution": "The answer is $\\left\\lfloor \\dfrac{(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\nClaim 1: If two shelves $R_1$ and $R_2$ do not intersect, then $\\tilde{R}_1 \\cap \\tilde{R}_2 = \\emptyset$.\n\nProof of Claim 1: Let $h_1, h_2$ be the height 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\nClaim 2: If $R_1$ and $R_2$ are distinct maximal shelves smaller than the whole chessboard, then $R_1 \\cap R_2 = \\emptyset$.\n\nProof of Claim 2: Let $h_1, h_2$ be the height 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 to the assumption that $R_2$ is maximal. So Claim 2 is proved.\n\nClaim 3: The number of shelves in the $n \\times m$ chessboard is less than or equal to $\\dfrac{(n+1)(m+1)}{2} - 1$.\n\nProof 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\\sum_{i=1}^{k} \\left( \\left\\lfloor \\dfrac{|\\tilde{R}_i|}{2} \\right\\rfloor - 1 \\right) \\le \\dfrac{(n+1)(m+1)}{2} - k - 1.\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\\max \\left( \\dfrac{m(n+1)}{2} \\cdot \\dfrac{(m+1)n}{2} \\right) - 1 + 1 \\le \\dfrac{(n+1)(m-1)}{2} - 1.\n$$\nIt is easy to see this because $n \\ge 1$ is equivalent to the inequality that $\\dfrac{m+n-1}{2} \\le \\dfrac{(n+1)(m+1)}{2} - 1$. This proves Claim 3.\n\nNow it remains to show that there is an assignment of integers so that the number of shelves is exactly $\\left\\lceil \\dfrac{(n+1)(m+1)}{2} \\right\\rceil - 1$. We proceed by induction on $n+m$. It is trivial if $\\max(n,m) \\le 2$.\n\nNow by symmetry let us assume that $m > 2$. Let us write 1 in each unit square on the second row. For the first row, we write integers larger than 1 obtained by the induction hypothesis on a $1 \\times n$ chessboard. For the remaining rows, we write integers larger than 1 obtained by the induction hypothesis on a $(m-2) \\times n$ chessboard.\n\nNow the number of shelves in this assignment is\n$$\n1 + \\left\\lfloor \\dfrac{2(n+1)}{2} - 1 \\right\\rfloor + \\left\\lceil \\dfrac{(m-1)(n+1)}{2} - 1 \\right\\rceil\n$$\nby the induction hypothesis. This completes the proof. $\\Box$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 42429,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, and $z$ be positive real numbers. Prove that\n$$\n\\frac{2x^2 + xy}{(y + \\sqrt{zx} + z)^2} + \\frac{2y^2 + yz}{(z + \\sqrt{xy} + x)^2} + \\frac{2z^2 + zx}{(x + \\sqrt{yz} + y)^2} \\geq 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "### First Solution\nBy the Cauchy-Schwarz Inequality we have\n$$\n(y + z + z)(y + z + x) \\geq (y + z + \\sqrt{zx})^2 = (y + \\sqrt{zx} + z)^2.\n$$\nTherefore,\n$$\n\\frac{2x^2 + xy}{(y + \\sqrt{zx} + z)^2} \\geq \\frac{2x^2 + xy}{(x + y + z)(y + 2z)} = \\frac{2x}{y + 2z} - \\frac{x}{x + y + z}.\n$$\nApply triples $(y, z, x)$ and $(z, x, y)$ to the above inequality instead of the triple $(x, y, z)$ respectfully, we have\n$$\n\\frac{2y^2 + yz}{(z + \\sqrt{xy} + x)^2} \\geq \\frac{2y}{z + 2x} - \\frac{y}{x + y + z},\n$$\nand\n$$\n\\frac{2z^2 + zx}{(x + \\sqrt{yz} + y)^2} \\geq \\frac{2z}{x + 2y} - \\frac{z}{x + y + z}.\n$$\nAdding the above three inequalities together yields\n$$\n\\frac{2x^2 + xy}{(y + \\sqrt{zx} + z)^2} + \\frac{2y^2 + yz}{(z + \\sqrt{xy} + x)^2} + \\frac{2z^2 + zx}{(x + \\sqrt{yz} + y)^2} \\geq \\frac{2x}{y+2z} + \\frac{2y}{z+2x} + \\frac{2z}{x+2y} - 1.\n$$\nBy the Cauchy-Schwarz Inequality,\n$$\n\\left( \\frac{2x}{y+2z} + \\frac{2y}{z+2x} + \\frac{2z}{x+2y} \\right) (x(y+2z) + y(z+2x) + z(x+2y)) \\geq 2(x+y+z)^2\n$$\nholds, and combining with the inequality\n$$\n(x + y + z)^2 \\geq 3(xy + yz + zx),\n$$\nwe can easily prove that\n$$\n\\left( \\frac{2x}{y+2z} + \\frac{2y}{z+2x} + \\frac{2z}{x+2y} \\right) \\geq 2\n$$\nholds, with an equality if and only if $x = y = z$.\nHence,\n$$\n\\frac{2x^2 + xy}{(y + \\sqrt{zx} + z)^2} + \\frac{2y^2 + yz}{(z + \\sqrt{xy} + x)^2} + \\frac{2z^2 + zx}{(x + \\sqrt{yz} + y)^2} \\geq \\frac{2x}{y+2z} + \\frac{2y}{z+2x} + \\frac{2z}{x+2y} - 1 \\geq 1\n$$\nas desired, and the equality holds if and only if $x = y = z$. $\\square$\n\n\n### Second Solution\nBy the Cauchy-Schwarz Inequality we have\n$$\n(yx + x^2 + x^2) \\left( \\frac{y}{x} + \\frac{z}{x} + \\frac{z^2}{x^2} \\right) \\geq (y + \\sqrt{zx} + z)^2.\n$$\nTherefore, we get\n$$\n\\frac{2x^2 + xy}{(y + \\sqrt{zx} + z)^2} \\geq \\frac{x^2}{xy + xz + z^2},\n$$\nwith an equality if and only if $x = z$.\nApply triples $(y, z, x)$ and $(z, x, y)$ to the above inequality instead of the triple $(x, y, z)$ respectfully, we have\n$$\n\\frac{2y^2 + yz}{(z + \\sqrt{xy} + x)^2} \\geq \\frac{y^2}{yz + yx + x^2}\n$$\nand\n$$\n\\frac{2z^2 + zx}{(x + \\sqrt{yz} + y)^2} \\geq \\frac{z^2}{zx + zy + y^2}.\n$$\n(The equalities hold if and only if $y = x$ and $z = y$, respectfully.)\nHence it is enough to show that\n$$\n\\frac{x^2}{xy + xz + z^2} + \\frac{y^2}{yz + yx + x^2} + \\frac{z^2}{zx + zy + y^2} \\geq 1.\n$$\nLet\n$$\nM = \\frac{x^2}{xy + xz + z^2} + \\frac{y^2}{yz + yx + x^2} + \\frac{z^2}{zx + zy + y^2}.\n$$\nAgain by the Cauchy-Schwarz Inequality we have\n$$\nM((xy + xz + z^2) + (yz + yx + x^2) + (zx + zy + y^2)) \\geq (x + y + z)^2.\n$$\nSince\n$$\n(xy + xz + z^2) + (yz + yx + x^2) + (zx + zy + y^2) = (x + y + z)^2,\n$$\nit is immediate that\n$$\nM = \\frac{x^2}{xy + xz + z^2} + \\frac{y^2}{yz + yx + x^2} + \\frac{z^2}{zx + zy + y^2} \\geq 1,\n$$\nand the equality holds if and only if $x = y = z$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42430,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1, A_2, \\dots, A_n$ be sets. For a subset $X$ of $\\{1, 2, \\dots, n\\}$, let\n$$\nN(X) = \\{i \\in \\{1, 2, \\dots, n\\} - X : A_i \\cap A_j \\neq \\emptyset \\text{ for all } j \\in X\\}.\n$$\nProve that for every integer $3 \\le m \\le n - 2$, there exists a subset $X$ of $\\{1, 2, \\dots, n\\}$ such that $|X| = m$ and $|N(X)| \\neq 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $G$ be a graph on vertices $v_1, v_2, \\dots, v_n$ such that two vertices $v_i$ and $v_j$ are adjacent if and only if $A_i \\cap A_j \\neq \\emptyset$ and $i \\neq j$. For a set $X$ of vertices of $G$, let $N(X)$ be the set of all vertices adjacent to every vertex in $X$.\nSuppose, on the contrary, that $n \\ge m + 2$, $m \\ge 3$ and $|N(X)| = 1$ for all subsets $X \\subseteq \\{v_1, v_2, \\dots, v_n\\}$ with $|X| = m$.\n\n**Solution 1**\nA set of vertices is called a *clique* if every pair of vertices are adjacent in $G$. Let $T$ be a maximum clique of $G$. It is trivial that $|T| < m + 2$.\n\n(1) We claim that $|T| = m + 1$. If not, let $T'$ be a set of $m$ vertices such that $T \\subseteq T'$. Then $N(T') \\neq \\emptyset$ and therefore there is a vertex $v$ adjacent to all vertices in $T$. Then $T \\cup \\{v\\}$ is a clique, contradictory to the assumption that $T$ is a maximum clique.\n\n(2) No vertex $v \\notin T$ has more than 1 neighbor in $T$. Otherwise, if $x, y \\in T$ are adjacent to $v$, then $x, y \\in N((T - \\{x, y\\}) \\cup \\{v\\})$.\n\n(3) Let $v \\notin T$ and let $w_1, w_2 \\in T$ be two vertices non-adjacent to $v$. Since $N((T - \\{w_1, w_2\\}) \\cup \\{v\\})$ can not contain $w_1$ or $w_2$, it must contain $y \\notin T$. Then $y$ is adjacent to at least $m - 1$ vertices of $T$, contradictory to (2) because $m \\ge 3$. $\\square$\n\n\n**Solution 2**\nLet $d_i$ be the degree of the vertex $v_i$. By considering the number of pairs $(X, y)$ of a set $X$ of $m$ vertices and a vertex $y$ adjacent to all vertices of $X$, we deduce\n$$\n\\sum_{i=1}^{n} \\binom{d_i}{m} = \\binom{n}{m}.\n$$\n\n(1) If $d_i \\le m$ for all $i$, then $\\sum_{i=1}^n \\binom{d_i}{m} \\le n$ but $n < \\binom{n}{m}$ because $3 \\le m \\le n - 2$. Therefore there exists $i$ such that $d_i \\ge m + 1$. We may assume $d_1 \\ge m + 1$.\n\n(2) Let $S$ be the set of all neighbors of $v_1$. Then $|S| = d_1 \\ge m + 1$. For a subset $X$ of $S$ with $|X| = m - 1$, since $|N(X \\cup \\{v_1\\})| = 1$, there exists a unique vertex $a \\in S - X$ such that $a$ is adjacent to all vertices of $X$.\n\nIf there are distinct subsets $X_1, X_2$ of $S$ each having $m-1$ vertices such that the corresponding vertices $a_1, a_2 \\in S$ are identical, then $a_1(= a_2)$ is adjacent to all vertices in $X_1 \\cup X_2$ and so a subset of $X_1 \\cup X_2$ with $m$ vertices will have at least 2 common neighbors, contradictory to the assumption.\n\nTherefore no two vertices $a \\in S$ corresponding to some subset $X$ of $S$ with $m-1$ vertices can be identical and so $\\binom{d_1}{m-1} \\le d_1$. But this is a contradiction because $2 \\le m-1 \\le d_1 - 2$. $\\square$\n\n\n**Solution 3**\nWe use (1) of Proof 2 first to show that some vertex has degree at least $m + 1$.\nWe proceed by induction on $m$.\nIf $m=3$, then we use proof 1 to finish the proof.\nIf $m > 3$, then let $S$ be the neighbors of $v_1$. For every subset $X$ of $S$ with $m-1$ vertices, we have $|N(X \\cup \\{v_1\\})| = 1$ and therefore the subgraph $G[S]$ induced on $S$ has the property that every set of $m-1$ vertices has exactly one common neighbor. Since $|S| \\ge m + 1$, such a graph can not exist by the induction hypothesis. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42431,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Prove that there are infinitely many triples of integers $(x, y, z)$ each of which satisfies $nx^2 + y^3 = z^4$, $(x, y) = (y, z) = (z, x) = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\ny^3 = z^4 - nx^2 = (z^2 - \\sqrt{nx})(z^2 + \\sqrt{nx}). \\quad (1)\n$$\nFor $y = s^2 - nt^2$ with $(n, t) = 1$, we have\n$$\n\\begin{aligned}\ny^3 &= (s^3 - 3\\sqrt{n}s^2t + 3nst^2 - n\\sqrt{nt^3})(s^3 + 3\\sqrt{n}s^2t + 3nst^2 + n\\sqrt{nt^3}) \\\\\n&= (z^2 - \\sqrt{nx})(z^2 + \\sqrt{nx}),\n\\end{aligned}\n$$\nso any $(x, z)$ satisfying\n$$\nz^2 = s^3 + 3nst^2, \\quad (2)\n$$\n$$\nx = 3s^2t + nt^3 \\quad (3)\n$$\nsatisfies the equation (1). One can always find $x$ satisfying the equation (3), so let us show that there are infinitely many $z$ satisfying equation (2). Let us divide the problem into two cases:\n\n1. If $3n$ is not a square number, put $s = 1$. Then equation (2) becomes a Pell's equation\n$$\nz^2 - 3nt^2 = 1. \\quad (4)\n$$\nWe know that there are infinitely many solutions of Pell's equation, and if $(z_1, t_1)$ is the pair of smallest positive integers which is a solution of the equation, each $(k$-th) positive integral solution $(z_k, t_k)$ of the equation satisfies\n$$\nz_k + \\sqrt{3nt_k} = (z_1 + \\sqrt{3nt_1})^k. \\quad (5)\n$$\nNow let us prove that there are infinitely many $(z_k, t_k)$ such that\n$$\nx_k = 3t_k + nt_k^3, \\quad y_k = 1 - nt_k^2, \\quad z_k = (1 + 3nt_k^2)\n$$\nare relatively prime to each other.\n$$\n(x_k, z_k) \\quad |(x_k, z_k^2) = (t_k(3 + nt_k^2), 1 + 3nt_k^2) \\\\ = (3 + nt_k^2, 1 + 3nt_k^2) (\\because (t_k, 1 + 3nt_k^2) = 1) = (3 + nt_k^2, -8) = 1 \\text{ or } 2,\n$$\n$$\n(y_k, z_k) \\quad |(y_k, z_k^2) = (1 - nt_k^2, 1 + 3nt_k^2) = (1 - nt_k^2, 4) = 1 \\text{ or } 2,\n$$\n$$\n(x_k, y_k) = (3t_k + nt_k^3, 1 - nt_k^2) = (4t_k, 1 - nt_k^2) = (4, 1 - nt_k^2) = 1 \\text{ or } 2 \\\\\n(\\because (t_k, 1 - nt_k^2) = 1).\n$$\nTherefore, if $n$ is even or $t_k$ is even, we have $(x_k, y_k) = (y_k, z_k) = (z_k, x_k) = 1$. In the equation (5), one can easily check that if $k$ is even, $t_k$ is even. Therefore, there are infinitely many $(z_k, t_k)$ which gives us the solution of $nx^2 + y^3 = z^4$ and $(x, y) = (y, z) = (z, x) = 1$.\n\n2. If $3n = m^2$ for some positive integer $m$, put $s = u^2, z = uv$. Then the equation (2) becomes\n$$\nu^4 = v^2 - 3nt^2 = (v - mt)(v + mt). \\qquad (6)\n$$\nOne can easily check that for each positive integer $k$,\n$$\n\\left( u_k = 2mk + 1, t_k = \\frac{u_k^4 - 1}{m}, v_k = mt_k + 1 \\right)\n$$\nis a triple of integers satisfying the equation (6). Note that $u_k$ is odd, $t_k$ is even, and\n$$\n(u_k, nt_k) \\quad |(u_k, 3nt_k) = (u_k, m u_k^4 - m) = (u_k, m) = 1,\n$$\n$$\n\\therefore (u_k, nt_k) = 1,\n$$\n$$\n(t_k, v_k) = (t_k, mt_k + 1) = 1.\n$$\nNow the triple\n$$\n(x_k = 3u_k^4 t_k + nt_k^3, y_k = u_k^4 - nt_k^2, z_k = u_k v_k = u_k(mt_k + 1))\n$$\nsatisfies the equation (1). We have to show $(x_k, y_k) = (y_k, z_k) = (z_k, x_k) = 1$.\n$$\n(x_k, y_k) = (3u_k^4 t_k + nt_k^3, u_k^4 - nt_k^2) = (4u_k^4 t_k, u_k^4 - nt_k^2) = (4, u_k^4 - nt_k^2) = 1 \\\\\n(\\therefore (u_k^4, u_k^4 - nt_k^2) = (t_k, u_k^4 - nt_k^2) = 1, u_k \\text{ is odd}, t_k \\text{ is even})\n$$\n$$\n\\begin{aligned}\n(y_k, z_k) &= (u_k^4 - nt_k^2, u_k v_k) = (u_k^4 - nt_k^2, v_k) | (u_k^4 - nt_k^2, v_k^2 = u_k^4 + 3nt_k^2) \\\\\n&= (u_k^4 - nt_k^2, 4nt_k^2) = (u_k^4 - nt_k^2, 4) = 1, \\\\\n(\\therefore (nt_k^2, u_k^4 - nt_k^2) = 1, u_k \\text{ is odd}, t_k \\text{ is even})\n\\end{aligned}\n$$\n$$\n\\begin{aligned}\n(z_k, x_k) &= (3u_k^4 t_k + nt_k^3, u_k v_k) = (3u_k^4 + nt_k^2, u_k v_k) \\\\\n&= (3u_k^4 + nt_k^2, v_k) | (3u_k^4 + nt_k^2, v_k^2 = u_k^4 + 3nt_k^2) \\\\\n&= (-8nt_k^2, u_k^4 + 3nt_k^2) = (8, u_k^4 + 3nt_k^2) = 1 \\\\\n(\\therefore (nt_k^2, u_k^4 + 3nt_k^2) = 1, u_k \\text{ is odd}, t_k \\text{ is even}).\n\\end{aligned}\n$$\nSo we have $(x_k, y_k) = (y_k, z_k) = (z_k, x_k) = 1$. Therefore we can find infinitely many triples of integers $(x_k, y_k, z_k)$ with $(x_k, y_k) = (y_k, z_k) = (z_k, x_k) = 1$ which satisfies the equation (1).\n\nFrom above, in any cases, we can find infinitely many integral solutions of the equation $nx^2 + y^3 = z^4$, $(x, y) = (y, z) = (z, x) = 1$. This completes the proof. □",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42432,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the set of positive integers which is not divisible by any prime number greater than $3$. For arbitrary chosen subsets $A_1, A_2, A_3, \\dots$ of $M$, prove that there exist two distinct positive integers $i$ and $j$ such that:\nFor each $x$ in $A_i$, $A_j$ has a divisor of $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Regard $A_n$ as a set of lattice points by\n$$\n2^x 3^y \\mapsto (x, y) \\in \\mathbb{N}_0^2,\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, \\dots \\subset \\mathbb{N}_0^2$, there exist $i, j \\in \\mathbb{N}$ satisfying $\\forall x \\in A_i, \\exists y \\in A_j$ s.t. $y \\le x$.\n\nIn this case we will say that the collection $A_1, A_2, \\dots$ is \"nice\".\n\nA reduced set $A$ is a subset of $\\mathbb{N}_0^2$ satisfying $\\notin x, y \\in A$ s.t. $x < y$. Since a reduced set may have at most one point in $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, \\dots$, 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, \\dots$, if there exists $N \\in \\mathbb{N}$ such that $|A_i| \\le N$ for all $i$, then the collection is nice.\n\n*Proof.* (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}\\}$ be the collection of $A_i$'s containing $p$. From the induction hypothesis, $\\{B_i - \\{p\\} \\mid i \\in \\mathbb{N}\\}$ is nice, which induces easily that $\\{B_i \\mid i \\in \\mathbb{N}\\}$ is nice. Thus the collection $A_1, A_2, \\dots$ is also nice. $\\square$\n\nLet us go back to the original problem. If $A_1, A_2, \\dots$ satisfies the condition in Lemma 1 then it is done. Otherwise $\\exists p$ s.t. $p = (a, b)$ is contained in infinitely many $A_i$'s, then each cardinality of $A_i \\ni p$ is bounded by $|A_i| \\le a + b + 1$ (recall that $A_i$'s are reduced sets). Thus from the corollary 2, the collection of $A_i$'s containing $p$ is nice, and so does the original collection $A_1, A_2, \\dots$, which completes the proof.\n\nNow we only 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 this finitely many $A_i$'s.\n\n1. Suppose $V_h$ is finite for all $h \\in \\mathbb{N}_0$. Among the points in $\\bigcup A_i$ take $(a_0, b_0)$, where $b_0$ is the smallest $y$ coordinate of points in $\\bigcup A_i$ and $a_0$ is the smallest $x$ coordinate of points with $y$ coordinate are all $b_0$. By relabeling, set $(a_0, b_0) \\in A_1$. From the collection $A_1, A_2, \\dots$, remove all the $A_i$'s (except $A_1$) which contribute to $V_0, V_1, \\dots, V_{a_0-1}$. Note that only the finite number of $A_i$'s are removed. Thus the collection of remaining elements (including $A_1$) is nice (by taking $(a_0, b_0) \\in A_1$ as $x \\in A_i$).\n\n2. Suppose there exists $h \\in \\mathbb{N}_0$ such that $V_h$ is infinite. Among these $h$, let $t$ be the smallest element. Since each $A_i$ is finite, infinitely many $A_i$'s contribute to $V_t$. Set $\\{C_i \\mid i \\in \\mathbb{N}\\}$ be the infinitely many collection for $V_t$. Define $H_k := \\{(x, k) \\mid (x, k) \\in \\bigcup C_i\\}$. If all the $H_k$ are finite, then $\\{C_i \\mid i \\in \\mathbb{N}\\}$ is nice. So assume there exists $k \\in \\mathbb{N}_0$ such that $H_k$ is infinite. Among these $k$, let $s$ be the smallest element. Since each $C_i$ is finite, infinitely many $C_i$'s contribute to $H_s$. Set $\\{D_i \\mid i \\in \\mathbb{N}\\}$ be the infinitely many collection for $H_s$. Note that $D_i \\cap V_t \\neq \\emptyset$, $D_i \\cap H_s \\neq \\emptyset$. Thus if $(t, d) \\in D_1$ and $(c, s) \\in D_1$ then the number of set $D_i$'s which contribute to $[t, c) \\times [s, d)$ is finite. So removing these $D_i$'s (except $D_1$) from the collection $\\{D_i \\mid i \\in \\mathbb{N}\\}$ yields an infinite subcollection $\\{D_1\\} \\cup \\{E_i \\mid i \\in \\mathbb{N}\\}$ — of $A_1, A_2, \\dots$. By taking $(t, d) \\in D_1$ or $(c, s) \\in D_1$ as $x \\in A_i$, we can prove that $\\{D_1\\} \\cup \\{E_i \\mid i \\in \\mathbb{N}\\}$ is nice.\n\n$\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42433,
    "subject": "Mathematics (Multi-modal)",
    "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 passing $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": "Detailed solution",
    "solution": "First, we will show that the line $PJ$ passes the midpoint of a side $AB$.\nLet $M$ be the intersection of two lines $PJ$ and $AB$, and $S$ be the intersection of two lines $BP$ and $AI$. By using Menelaus' theorem to the triangle $ABS$ and the line $PJ$, we have\n$$\n\\frac{AM}{BM} \\cdot \\frac{BP}{PS} \\cdot \\frac{SJ}{JA} = 1. \\qquad (1)\n$$\nAlso, by the angle bisector theorem, we get\n$$\n\\frac{SJ}{JA} = \\frac{BS}{AB}. \\qquad (2)\n$$\nIn other hand, since $\\angle JBD = \\frac{\\angle C}{2}$ and $\\angle DEI = \\angle DCI = \\frac{\\angle C}{2}$, four points $J, B, E, D$ are cyclic, therefore $\\angle BEJ = \\angle BDJ = \\frac{\\angle B}{2}$. From the fact $AB \\parallel PE$, we have\n$$\n\\angle BEP = \\angle BEJ + \\angle PEA = \\angle BEJ + \\angle EAB = 90^\\circ - \\frac{\\angle C}{2}.\n$$\nBy the fact $AB \\parallel PE$ again, we know $\\angle EPB = \\angle PBA = \\angle C$, hence we get\n$$\n\\angle PBE = 180^\\circ - \\angle C - \\left(90^\\circ - \\frac{\\angle C}{2}\\right) = 90^\\circ - \\frac{\\angle C}{2},\n$$\nso we obtain $\\angle PBE = 90^\\circ - \\frac{\\angle C}{2} = \\angle PEB$ and therefore\n$$\nPB = PE. \\qquad (3)\n$$\n\nThe fact $AB \\parallel PE$ also implies\n$$\n\\frac{BS}{AB} = \\frac{PS}{PE}. \\qquad (4)\n$$\nFrom (1), (2), (3), (4), we get $AM = BM$, and $M$ is the midpoint of the line segment $AB$, as desired.\nNow let $N$ be the midpoint of the side $AC$. Then since $ABC$ is similar to $ADB$, we have $\\angle AJM = \\angle AIN$, and by the midpoint theorem we get $\\angle AIN = \\angle AA'Q$. We already know that the points $M, J, P, Q$ are collinear, so $\\angle QJA' = \\angle AJM$ holds, and we get $QJ = QA'$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42434,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to [0, \\infty)$ such that, for all real $a, b, c, d$ that satisfy $ab + bc + cd = 0$, the following equality holds:\n$$\nf(a - b) + f(c - d) = f(a) + f(b + c) + f(d)\n$$\n(Here $\\mathbb{R}$ is the set of real numbers and $[0, \\infty)$ is the set of nonnegative real numbers.)",
    "options": [],
    "answer": "f(x) = λ x^2 for λ ≥ 0",
    "solution": "We prove the following\n**Lemma** For all real $p, q, r$ that satisfy $p^2 + q^2 = r^2$, the following equality holds.\n$$\nf(p) + f(q) = f(r)\n$$\n**Proof** Put $a = \\frac{p-q+r}{2}$, $b = \\frac{p-q-r}{2}$, $c = \\frac{p+q+r}{2}$. Then we have $ab+bc+cd = \\frac{1}{2}(p^2+q^2-r^2)$. So if $p^2+q^2 = r^2$ then $ab + bc + cd = 0$ and we have\n$$\nf(r) + f\\left(\\frac{p-q-r}{2}\\right) = f\\left(\\frac{p-q-r}{2}\\right) + f(p) + f(q)\n$$\nThus we have $f(p) + f(q) = f(r)$ and the proof is complete.\n\nFor $(p, q, r) = (0, 0, 0)$, we have, by the Lemma, $f(0) = 0$, and for $(p, q, r) = (p, 0, -p)$ we have $f(-p) = f(p)$ and thus $f$ is an even function.\n\nNow for any $t \\ge 0$ define $g : [0, \\infty) \\to [0, \\infty)$ by $g(t) := f(\\sqrt{t})$. Then we have $g(a+b) = g(a) + g(b)$. And for $a \\ge b \\ge 0$ we have $g(a) = g(a-b) + g(b) \\ge g(b)$. So\n\ng is monotone increasing. So $g(x) = g(1) \\cdot x$ ($x \\ge 0$). Thus $f(x) = f(1) \\cdot x^2$ ($x \\ge 0$) and $f$ is an even function. So $f(x) = f(1) \\cdot x^2$ for any real $x$. And $f(1) \\ge 0$ since $f(x) \\ge 0$. Thus the functions satisfying given equality are of the form $f(x) = \\lambda x^2$, ($\\lambda \\ge 0$) and it is easy to check that those functions satisfy the given equality. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42435,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n \\ge 2$, define the set $T$ by\n$$\nT = \\{ (i, j) : 1 \\le i < j \\le n \\text{ and } i \\ne j \\}.\n$$\nFor nonnegative real numbers $x_1, x_2, \\dots, x_n$ satisfying $x_1 + x_2 + \\dots + x_n = 1$, find the maximum (as a function of $n$) of\n$$\n\\sum_{(i,j) \\in T} x_i x_j.\n$$",
    "options": [],
    "answer": "floor(log_2 n) / (2 (floor(log_2 n) + 1))",
    "solution": "Let $M(n)$ be the maximum of $\\sum_{(i,j) \\in T} x_i x_j$. We will show that\n$$\nM(n) = \\frac{\\lfloor \\log_2 n \\rfloor}{2(\\lfloor \\log_2 n \\rfloor + 1)}.\n$$\nFor $k = \\lfloor \\log_2 n \\rfloor$, set $x_{20} = x_{21} = \\dots = x_{2k} = \\frac{1}{k+1}$, and $x_i = 0$ otherwise. Then we have $\\sum_{(i,j) \\in T} x_i x_j = \\binom{k+1}{2} \\frac{1}{(k+1)^2} = \\frac{k}{2(k+1)}$, thus\n$$\nM(n) \\ge \\frac{\\lfloor \\log_2 n \\rfloor}{2(\\lfloor \\log_2 n \\rfloor + 1)}.\n$$\nConsider an element $(x_1, x_2, \\dots, x_n)$ which yields $M(n)$ and the number of $i$'s such that $x_i = 0$ is maximal. In this case, if $x_a, x_b \\ne 0$ and $a < b$, then we have $(a, b) \\in T$. Suppose $(a, b) \\notin T$, then by setting $x'_a = x_a - \\epsilon$, $x'_b = x_b + \\epsilon$, and $x'_i = x_i$ ($i \\ne a, b$), the equation $x'_1 + x'_2 + \\dots + x'_n = 1$ still holds and the value $\\sum_{(i,j) \\in T} x'_i x'_j$ becomes a linear function of $\\epsilon$. Thus for $\\epsilon = x'_a$ or $\\epsilon = -x'_b$, $\\sum_{(i,j) \\in T} x'_i x'_j \\ge \\sum_{(i,j) \\in T} x_i x_j$, which contradicts the maximality of the number of $i$'s satisfying $x_i = 0$.\nTherefore if $C := \\{i : x_i > 0\\}$, then $i, j \\in C$ and $i < j$ implies $(i, j) \\in T$. Thus if $C = \\{i_1, i_2, \\dots, i_k\\}$ and $i_1 < i_2 < \\dots < i_k$, then due to $i_j \\ge 2i_{j-1}$ we have\n\n$n > i_t \\ge 2^{t-1}i_1 \\ge 2^{t-1}$. This means that $t-1 \\le \\log_2 n$, or equivalently, $|C| \\le k+1$. Applying the Cauchy-Schwarz inequality yields\n$$\n\\sum_{(i,j) \\in T} x_i x_j = \\frac{1}{2} \\left( \\left( \\sum_{i \\in C} x_i \\right)^2 - \\sum_{i \\in C} x_i^2 \\right) \\le \\frac{1}{2} \\left( 1 - \\frac{1}{|C|} \\left( \\sum_{i \\in C} x_i \\right)^2 \\right) \\le \\frac{1}{2} \\left( 1 - \\frac{1}{k+1} \\right).\n$$\nIn particular we have\n$$\nM(n) \\le \\frac{\\lfloor \\log_2 n \\rfloor}{2(\\lfloor \\log_2 n \\rfloor + 1)},\n$$\nwhich completes the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42436,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a triangle $ABC$ let $B_1$, $C_1$ be the excenters corresponding to $B$, $C$, respectively. The line $B_1C_1$ intersects the circumcircle of $ABC$ at $D$ ($\\neq A$). Let $E$ be the intersection of the perpendiculars from $B_1$ to $CA$ and from $C_1$ to $AB$. Let $\\omega$ be the circumcircle of the triangle $ADE$. The tangent to $\\omega$ at $D$ intersects the line $AE$ at $F$. The perpendicular from $D$ to $AE$ intersects $AE$ at $G$ and this line intersects $\\omega$ at $H$ ($\\neq D$). The circumcircle of the triangle $HGF$ intersects $\\omega$ at $I$ ($\\neq H$). Let $J$ be the perpendicular foot at $AH$ from $D$. Show that $AI$ passes through the midpoint of $DJ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first show that $\\angle ADE = 90^\\circ$. $\\angle EB_1C_1 = \\frac{1}{2}\\angle A = \\angle EC_1B_1$, hence we have $EB_1 = EC_1$. Similarly, we have $EB_1 = EA_1$, that is, $E$ is the circumcenter of $\\triangle A_1B_1C_1$. On the other hand, $A$, $B$, $C$ are perpendicular feet of $\\triangle A_1B_1C_1$ because, for example, $C_1$ lies on the bisector of $\\angle ACB$ and $\\angle B_1CC_1 = \\angle B_1CA + \\angle ACC_1 = (\\frac{\\pi}{2} - \\frac{1}{2}\\angle C) + (\\frac{1}{2}\\angle C) = \\frac{\\pi}{2}$. Hence the circumcircle of $\\triangle ABC$ is the nine-point circle of $\\triangle A_1B_1C_1$, which means that $D$ is the midpoint of $B_1C_1$. Therefore, $\\angle ADE = 90^\\circ$.\n\nBecause $AE$ is a diameter of $\\omega$ and $H$ is symmetric to $D$ with respect to $AE$, $HF$ is also tangent to $\\omega$. Let $O_1$ be the center of $\\omega$. Let $O_2$ be the circumcenter of $\\triangle HGF$, then $O_2$ is the midpoint of $HF$. Since\n$$\n\\angle HO_1F = \\angle HO_1D = \\angle HAD = \\angle JAD,\n$$\ntwo triangles $\\triangle HO_1F$ and $\\triangle JAD$ are similar. For these triangles we have\n$$\n\\angle HO_1O_2 = \\frac{1}{2}\\angle HO_1I = \\angle HAI = \\angle JAK,\n$$\nwhere $K = DJ \\cap AI$. Since $HO_2 = O_2F$ in $\\triangle HO_1F$, we have $JK = KD$ in $\\triangle JAD$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42437,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ be relatively prime positive integers and let $a_n$ and $b_n$ be integer sequences satisfying $(a + b\\sqrt{2})^{2n} = a_n + b_n\\sqrt{2}$. Find all primes $p$ such that there is a positive integer $n$ less than or equal to $p$ satisfying $b_n \\equiv 0 \\pmod{p}$.",
    "options": [],
    "answer": "All primes p with p = 2 or p ∤ (a^2 - 2b^2).",
    "solution": "Let $p$ be a prime. First suppose that $p$ is an odd prime dividing $a^2 - 2b^2$. Since $a$ and $b$ are relatively prime, $b_1 = 2ab$ is not divisible by $p$. Suppose that there is a positive integer $n$ such that $b_n$ is divisible by $p$. Let $r$ be the smallest positive integer such that $b_r$ is divisible by $p$. Note that $(a - b\\sqrt{2})^{2n} = a_n - b_n\\sqrt{2}$ because both $a_n$ and $b_n$ are integers, and\n$$\n\\begin{cases} a_n = (a^2 + 2b^2)a_{n-1} + 4ab b_{n-1} \\\\ b_n = 2ab a_{n-1} + (a^2 + 2b^2)b_{n-1}. \\end{cases}\n$$\nThen we have\n$$\n\\begin{aligned}\n0 \\equiv b_r &= 2ab((a^2 + 2b^2)a_{r-2} + 4ab b_{r-2} + (a^2 + 2b^2)b_{r-1}) \\\\\n&= 2(a^2 + 2b^2)b_{r-1} - (a^2 - 2b^2)^2 b_{r-2} \\equiv 2(a^2 + 2b^2)b_{r-1} \\pmod{p},\n\\end{aligned}\n$$\nwhich is a contradiction to the assumption. Therefore $b_n$ is not divisible by $p$ for any positive integer $n$.\n\nNow let $p$ be a prime not dividing $a^2 - 2b^2$. Since $b_1 = 2ab$, we may assume that $p$ is odd and $ab$ is not divisible by $p$. Note that\n$$\nb_n = \\frac{(a + b\\sqrt{2})^{2n} - (a - b\\sqrt{2})^{2n}}{\\sqrt{2}} = \\sum_{k=0,\\ k \\equiv 1 \\pmod{2}}^{2n} 2 \\binom{2n}{k} a^{2n-k} b^k 2^{\\frac{k-1}{2}}\n$$\nFirst assume that there is an integer $m$ such that $m^2 \\equiv 2 \\pmod{p}$. Then we have\n$$\n\\begin{aligned}\nb_n &= \\sum_{k=0,\\ k \\equiv 1 \\pmod{2}}^{2n} 2 \\binom{2n}{k} a^{2n-k} b^k 2^{\\frac{k-1}{2}} \\\\\n&= \\sum_{k=0,\\ k \\equiv 1 \\pmod{2}}^{2n} m^2 \\binom{2n}{k} a^{2n-k} b^k m^{k-1} \\\\\n&= \\frac{(a + bm)^{2n} - (a - bm)^{2n}}{m} \\pmod{p}.\n\\end{aligned}\n$$\nSince $(a + bm)(a - bm) = a^2 - m^2 b^2 \\equiv a^2 - 2b^2 \\not\\equiv 0 \\pmod{p}$, $b_{\\frac{p-1}{2}} \\equiv 0 \\pmod{p}$ by Fermat's little theorem.\n\nNow suppose that $x^2 \\equiv 2 \\pmod{p}$ does not have any integer solution, that is, $2$ is a quadratic non-residue modulo $p$. Clearly,\n$$\n\\binom{p+1}{k} \\equiv 0 \\pmod{p} \\text{ for any } 2 \\leq k \\leq p-1.\n$$\nTherefore by Euler's criterion, we have\n$$\n\\begin{align*}\nb_{\\frac{p+1}{2}} &= \\sum_{k=0,\\ k \\equiv 1 \\pmod{2}}^{p+1} 2 \\binom{p+1}{k} a^{p+1-k} b^k 2^{\\frac{k-1}{2}} \\\\\n&= 2(p+1)a^p b + (p+1)ab^p 2^{\\frac{p+1}{2}} \\\\\n&= 2ab(1 + 2^{\\frac{p-1}{2}}) \\equiv 0 \\pmod{p}.\n\\end{align*}\n$$\nTherefore such a prime is exactly $2$ or relatively prime to $a^2 - 2b^2$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42438,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an one-to-one correspondence $f : \\{1, 2, \\dots, n\\} \\to \\{1, 2, \\dots, n\\}$ for some positive integer $n$, define four sets $A$, $B$, $C$, $D$ as follows:\n$$\nA = \\{i \\mid i > f(i)\\}\n$$\n$$\nB = \\{(i, j) \\mid i < j \\le f(j) < f(i) \\text{ or } f(j) < f(i) < i < j\\}\n$$\n$$\nC = \\{(i, j) \\mid i < j \\le f(i) < f(j) \\text{ or } f(i) < f(j) < i < j\\}\n$$\n$$\nD = \\{(i, j) \\mid i < j \\text{ and } f(i) > f(j)\\}\n$$\nProve that $|A| + 2|B| + |C| = |D|$, where $|X|$ is the number of all elements in $X$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use an induction by the number $|D|$: Firstly, if $|D| = 0$, $f$ is identity and $A = B = C = \\emptyset$. So it holds for $|D| = 0$.\n\nNext, assume that it holds that for all bijections with $|D| < k$ for some positive integer $k$. Given $f$ with $|D_f| = k > 0$, there exists $i$ such that $f(i) > f(i+1)$. Define the function $g$ by\n$$\ng(j) = \\begin{cases} f(i+1), & \\text{if } j = i, \\\\ f(i), & \\text{if } j = i+1, \\\\ f(j), & \\text{otherwise.} \\end{cases}\n$$\nWe are able to define four set $(A_g, B_g, C_g, D_g)$ by the function $g$. Since $|D_g| = |D| - 1 = k - 1 < k$, by the induction hypothesis, we have\n$$\n|A_g| + 2|B_g| + |C_g| = |D_g|. \\tag{5}\n$$\nNow, we divide five cases:\n\ni. If $f(i+1) > i$ and $f(i) > i+1$, then we have\n$$\n(|A_f|, |B_f|, |C_f|, |D_f|) = (|A_g|, |B_g| + 1, |C_g| - 1, |D_g| + 1).\n$$\nii. If $f(i+1) = i$ and $f(i) > i$, then the property\n$$\n|\\{j \\mid j < i \\text{ and } i + 1 < f(j)\\}| = |\\{j \\mid f(j) < i \\text{ and } i + 1 < j\\}|\n$$\nyields\n$$\n(|A_f|, |B_f|, |C_f|, |D_f|) = (|A_g| + 1, |B_g|, |C_g|, |D_g| + 1).\n$$\niii. If $f(i+1) < i$ and $f(i) > i$, then we have\n$$\n(|A_f|, |B_f|, |C_f|, |D_f|) = (|A_g|, |B_g|, |C_g| + 1, |D_g| + 1)\n$$\nin the both of all cases $f^{-1}(i) < i$ and $f^{-1}(i) > i$.\niv. If $f(i+1) < i$ and $f(i) = i$, then the property\n$$\n|\\{j \\mid j < i \\text{ and } i + 1 < f(j)\\}| = |\\{j \\mid f(j) < i \\text{ and } i + 1 < j\\}| + 1\n$$\nyields\n$$\n(|A_f|, |B_f|, |C_f|, |D_f|) = (|A_g| - 1, |B_g| + 1, |C_g|, |D_g| + 1).\n$$\nv. If $f(i+1) < i$ and $f(i) < i$, then we have\n$$\n(|A_f|, |B_f|, |C_f|, |D_f|) = (|A_g|, |B_g| + 1, |C_g| - 1, |D_g| + 1).\n$$\nDue to (5), the results above all cases complete the proof. □",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42439,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ be positive real numbers satisfying $x + y + z = 1$. Prove that\n$$\n\\frac{(1 + xy + yz + zx)(1 + 3x^3 + 3y^3 + 3z^3)}{9(x + y)(y + z)(z + x)} \\ge \\left( \\frac{x\\sqrt{1+x}}{\\sqrt[4]{3+9x^2}} + \\frac{y\\sqrt{1+y}}{\\sqrt[4]{3+9y^2}} + \\frac{z\\sqrt{1+z}}{\\sqrt[4]{3+9z^2}} \\right)^2\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By using $x + y + z = 1$, we have\n$$\n1+xy+yz+zx = (x+y+z)^2+xy+yz+zx = (x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y).\n$$\nWith this equation and Cauchy-Schwarz inequality, we can deduce that\n$$\n\\begin{aligned}\n(LHS) &= \\frac{1}{9} \\left( \\frac{1}{1-x} + \\frac{1}{1-y} + \\frac{1}{1-z} \\right) \\left( (x+3x^3) + (y+3y^3) + (z+3z^3) \\right) \\\\\n&\\ge \\left( \\sqrt{\\frac{3x^3+x}{9(1-x)}} + \\sqrt{\\frac{3y^3+y}{9(1-y)}} + \\sqrt{\\frac{3z^3+z}{9(1-z)}} \\right)^2\n\\end{aligned}\n$$\nTherefore, it is enough to show that for any real number $s \\in (0, 1)$, the inequality\n$$\n\\frac{3s^3 + s}{9(1-s)} \\ge \\left( \\frac{s\\sqrt{1+s}}{\\sqrt[4]{3+9s^2}} \\right)^2\n$$\nholds. If we expand the above inequality, then it is easy to check that this is equivalent to $3(9s^2 - 1)^2 \\ge 0$, thus solving the problem. From the last inequality, we can check that the equality holds when $x = y = z = \\frac{1}{3}$.\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42440,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $ABC$ with $AC = BC > AB$, let $E$ and $F$ be the midpoints of $AC$ and $AB$, respectively. The perpendicular bisector $\\ell$ of $AC$ meets $AB$ at $K$ and the line parallel to $KC$ and passing through the point $B$ intersects $AC$ at a point $L$. For a point $P$ on the line segment $BF$, let $H$ be the orthocenter of the triangle $ACP$. The line segments $BH$ and $CP$ meet at a point $J$ and the lines $FJ$ and $\\ell$ meet at a point $M$. Let $W$ be the intersection point of $FL$ and $\\ell$. Show that $AW = BW$ if and only if the points $B, E, F, M$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $W'$ be the circumcenter of the triangle $ACP$, and let $L'$ be the intersection point of the line $FW'$ and the side $AC$. Note that the points $A, F, C$ and the foot $T$ of the perpendicular from $A$ to $PC$ are on the circle with a diameter $AC$, and thus $\\angle PAT - \\angle PCF$. Since $AC = BC$ and $F$ is the midpoint of $AB$, $\\angle HAF = \\angle HBF$, and so $\\angle PBJ = \\angle HCJ$, which along with $\\angle PJB = \\angle HJC$ induce that the triangles $PBJ$ and $HCJ$ are similar to each other. Therefore, for the feet of the perpendiculars from $J$ to $FP$ and $FC$, $\\frac{XJ}{JY} = \\frac{BP}{CH}$. Let $S$ be the foot of the perpendicular from $W'$ to $AP$. Then $HC = 2W'S$ and\n$$\nBP = BF - PF = AF - (PF - FS) = AF - (AS - FS) = 2FS.\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'$, Thereby $\\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 a diameter $ML'$.\n\nSuppose $AW = PW$, in which case $W$ is the circumcenter of $\\triangle APC$, $AW = PW = CW$. From the fact we proved above, $E$ and $F$ lie on the circle with a 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$, and 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 can 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 a diameter $ML'$, $\\angle MBL' = 90^\\circ$. It can be easily seen that $\\angle MBK = \\angle MEF = \\angle BAO$, and so $BM$ is parallel to $AO$. It follows that $AO$ and $BL'$ are orthogonal to each other, thereby $BL'$ and $KC$ being parallel to each other. Therefore, $BL' = BL$, which implies $L' = L$ and $W' = W$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42441,
    "subject": "Mathematics (Multi-modal)",
    "question": "$n$ students with different names are forming a circle and standing toward the center of the circle. Initially $n$ name tags of the students are distributed to them at random. The students repeat the following process:\n[process] Every student having one's own name tag gets out of the circle, and the remaining students pass their name tags to the right.\nFind the number of initial distributions of name tags such that some students still remain after 4 processes.",
    "options": [],
    "answer": "n! - 4^n + n 3^n - C(n,2) 2^n + C(n,3)",
    "solution": "Let $A$ be a finite set of positive integers and $\\pi$ be a permutation (=bijection) on $A$. A pair $(x, \\pi(x)) \\in A^2$ is called a \"up\" if $x < \\pi(x)$ holds, and $up(\\pi)$ is defined by the number of up's in $\\pi$.\n\nGiven a permutation $\\pi$ on $A = \\{x_1 < x_2 < \\dots < x_m\\}$, define a permutation $\\pi'$ by\n$$\n\\pi' = x_2x_3\\cdots x_{1} \\circ \\pi,\n$$\nwhere $\\circ$ means the functional composition. Then we can easily deduce the following lemma.\n\n**Lemma.** Let $\\pi$ be a permutation on $A = \\{x_1 < x_2 < \\dots < x_m\\}$ without fixed points. Then we have\n$$\nup(\\pi') = up(\\pi) - 1.\n$$\n\nLet $S_0 = \\{s_1, s_2, \\dots, s_n\\}$ be the set of $n$ students and $T_0 = \\{t_1, t_2, \\dots, t_n\\}$ be the set of $n$ name tags. A distribution of name tags can be interpreted as a bijection $\\tau_0: T_0 \\to S_0$. By simplifying $\\tau_0(t_i) = s_j$ to $\\tau_0(i) = j$, we can regard $\\tau$ as a permutation on $n$. Let $F_1$ be the set of fixed points of $\\tau_0$. Define $S_1 = S_0 - \\{s_i \\mid i \\in F_1\\}$ and $T_1 = T_0 - \\{t_i \\mid i \\in F_1\\}$. Then $\\tau_1: T_1 \\to S_1$ represents the state after the exit of each student having one's own name tag. It is obvious that $up(\\tau_0) = up(\\tau_1)$. Also we can easily deduce the following:\n\nGiven a distribution $\\pi$, passing their name tags to the right derives a distribution $\\pi'$.\nTherefore $\\tau'_1$ is the distribution after one process. By the lemma we have\n$$\nup(\\tau'_1) = up(\\tau_1) - 1 = up(\\tau_0) - 1.\n$$\n\nThus, with a single process, the number of up's is decreased by 1. Thus if an initial distribution $\\tau_0$ has at most 3 up's then every student has one's own name tag during 3 processes, i.e., every student gets out of the circle until 4 processes. Therefore the answer of this problem is\n$$\nn! - |\\{\\tau_0 \\in \\text{Sym}_n \\mid \\text{up}(\\tau_0) \\le 3\\}|,\n$$\nwhere $\\text{Sym}_n$ is the set of permutations on $\\{1, 2, \\dots, n\\}$.\n\nLet $B(n, k)$ be the set of $\\{\\pi \\in \\text{Sym}_n \\mid \\text{up}(\\pi) \\le k\\}$ and $b(n, k) := |B(n, k)|$. By considering the position of $n$ in $\\pi \\in B(n, k)$, we can easily deduce the following recurrence relation:\n$$\nb(n, k) = (k + 1)b(n - 1, k) + (n - k - 1)b(n - 1, k - 1), \\quad n \\ge 2\n$$\nwith $b(n, 0) = 1$ and $b(n, k) = 0$ for $k \\ge n$. For $k = 1$ we have\n$$\nb(n, 1) = 2b(n - 1, 1) + (n - 2). \\quad b(1, 1) = 1.\n$$\nThe recurrence relation yields $b(n, 1) = 2^n - n$. For $k = 2$ we have\n$$\nb(n, 2) = 3b(n - 1, 2) + (n - 3)(2^{n-1} - n + 1), \\quad b(1, 2) = 1.\n$$\nThe recurrence relation yields $b(n, 2) = 3^n - n2^n + \\binom{n}{2}$. Finally, for $k = 3$, we have\n$$\nb(n, 3) = 4b(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$$\nThe recurrence relation yields $b(n, 3) = 4^n - n3^n + \\binom{n}{2}2^n - \\binom{n}{3}$. Therefore the answer of this problem is\n$$\nn! - b(n, 3) = n! - 4^n + n3^n - \\binom{n}{2}2^n + \\binom{n}{3}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42442,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles triangle with $AC = BC$. Let $D$ be a point on a line $BA$ such that $A$ lies between $B$ and $D$. Let $O_1$ be the circumcircle of triangle $DAC$. $O_1$ meets $BC$ at point $E$. Let $F$ be the point on the line $BC$ such that $FD$ is tangent to circle $O_1$, and let $O_2$ be the circumcircle of triangle $DBF$. Two circles $O_1, O_2$ meet at point $G$ ($G \\neq D$). Let $O$ be the circumcenter of triangle $BEG$. Prove that the line $FG$ is tangent to circle $O$ if and only if $DG$ is perpendicular to $FO$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first show that both $DB$ and $DE$ are tangent to circle $O$. Since $DFBG$ is concyclic, we have $\\angle FDG = \\angle GBE$. Since $FD$ is tangent to $O_1$, we have $\\angle DEG = \\angle FDG$. Hence $\\angle GBE = \\angle DEG$, which means that $DE$ is tangent to $O$. On the other hand, since $ACED$ is concyclic, $\\angle BAC = \\angle CED$. Since $ABC$ is an isosceles triangle, $\\angle BAC = \\angle ABC = \\angle DBE$. Thus we have $\\angle DBE = \\angle DEB$, that is, the triangle $DBE$ is an isosceles triangle with $DB = DE$, which means that $DB$ is also tangent to $O$.\n\nSince $DB$ and $DE$ are tangent to $O$, the line $FE$ is the polar of the pole $D$ with respect to $O$. By La Hire theorem, $D$ lies on the polar of $F$. Hence $FG$ is tangent to circle $O$ if and only if $DG$ is the polar of $F$, which is equivalent to the fact that $DG$ is perpendicular to $FO$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42443,
    "subject": "Mathematics (Multi-modal)",
    "question": "A continent has a finite number of castles and each castle belongs to exactly one of the two countries $A$ and $B$. Each castle has one general. We say two castles are *neighboring* if there is a path between those two castles. We also say that a castle $P$ and a set of castles $Q$ is *neighboring* if $P$ is neighboring some castle in $Q$. Prove that the following two conditions are equivalent.\n1. If some of the generals from the country $B$ move to some neighboring castles and attack, generals from the country $A$ can move to some neighboring castles so that at each castle from the country $A$, the number of generals from $A$ is greater than or equal to that of generals from $B$.\n2. If $X$ is a set of castles from the country $A$, the number of generals from $A$ who are in $X$ or in castles neighboring $X$ is greater than or equal to the number of generals from $B$ who are in castles neighboring $X$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The implication $(1) \\Rightarrow (2)$ is obvious. We prove the converse. Let $G = (V, E)$ be a graph where $V$ is the set of castles and each edge corresponds to a neighboring relation. We set $A = \\{v_1, v_2, \\dots, v_n\\}$. Now let $P_i \\subseteq V - A$ be the set of castles whose generals decide to attack castles in $A$. We let $Q_i \\subseteq A$ be the set of castles which are either $v_i$ itself or neighboring $v_i$.\nLet us consider the list of sets $Q_1$ ($|P_1|$ times), $Q_2$ ($|P_2|$ times), ..., $Q_i$ ($|P_i|$ times), ..., $Q_n$ ($|P_n|$ times). If we choose a collection $Q_{i_1}, Q_{i_2}, \\dots, Q_{i_k}$ in this list, we have\n$$\n|P_{i_1}| + |P_{i_2}| + \\dots + |P_{i_k}| \\le |Q_{i_1} \\cup \\dots \\cup Q_{i_k}| \n$$\nby (2). Therefore, Hall's Marriage Theorem implies that for each $i = 1, 2, \\dots, n$, we can choose mutually disjoint $D_1, D_2, \\dots, D_n$ such that $|D_i| = |P_i|$ and $D_i \\subseteq Q_i$.\n$\\Box$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 42444,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for all $x, y \\in \\mathbb{R}$\n$$\nf(x^{2015} + f(y)^{2015}) = f(x)^{2015} + y^{2015}.\n$$",
    "options": [],
    "answer": "f(x) = x and f(x) = -x",
    "solution": "Put $x = 0$ in the given equation\n$$\nf(x^{2015} + f(y)^{2015}) = f(x)^{2015} + y^{2015}. \\qquad (1)\n$$\nThen, we have\n$$\nf(f(y))^{2015} = f(0)^{2015} + y^{2015}. \\qquad (2)\n$$\nIt implies that $f$ is a bijective function. By putting $f(x)$ to $x$ in (1), we have\n$$\nf(f(x))^{2015} + f(y)^{2015} = f(f(x))^{2015} + y^{2015} \\qquad (3)\n$$\nand by putting $f(y)$ to $y$ in (1),\n$$\nf(x^{2015} + f(f(y))^{2015}) = f(x)^{2015} + f(y)^{2015}. \\qquad (4)\n$$\nIn (3), we swipe $x$ and $y$ first and then take $f$ on both sides, then we have\n$$\nf(f(f(x))^{2015} + f(y)^{2015})) = f(x^{2015} + f(f(y))^{2015}) = f(x)^{2015} + f(y)^{2015}.\n$$\nSince $f$ is bijective, we have\n$$\nf(f(x)) = x. \\tag{5}\n$$\nBy (2), (3), and (5), we have\n$$\n(f(f(x))^{2015} + f(y)^{2015}) = x^{2015} + y^{2015} = f(f(x)^{2015}) + f(f(y)^{2015}) - 2f(0)^{2015}\n$$\nand, since $f$ is bijective, $f(x + y) = f(x) + f(y) - 2f(0)^{2015}$. Here, by putting $x = y = 0$, we have $f(0) = 2f(0)^{2015}$. Hence $f(0)$ is $0$ or $\\pm(\\frac{1}{2})^{\\frac{1}{2014}}$. By putting $f(y)$ to $y$ in (2), we get $f(y^{2015}) = f(0)^{2015} + f(y)^{2015}$.\nConsider an equation $x^{2015} - x + f(0)^{2015}$. It must have three different real roots $f(-1)$, $f(0)$ and $f(1)$. However, if $f(0) = \\pm(\\frac{1}{2})^{\\frac{1}{2014}}$, this equation does not have three different real roots. Therefore, $f(0) = 0$ and $f(1)$ is either $1$ or $-1$.\nNow we have $f(x + y) = f(x) + f(y)$ and $f(y^{2015}) = f(y)^{2015}$, and $f(1)$ is either $1$ or $-1$. We will prove that only $f(x) = x$ and $f(x) = -x$ satisfy these three conditions. If $f(1) = -1$, then we may consider $g(x) = -f(x)$. It allows us to consider only the case where $f(1) = 1$. By the general Cauchy method, we have $f(x) = x$ for all $x \\in \\mathbb{Q}$.\nNow we need to extend this function to real numbers. To do that, we will prove that $f(x^2) = f(x)^2$.\nLet $\\binom{2015}{k}f(x^k) - f(x)^k = x_{2015-k}$ for $k = 0, 1, 2, \\dots, 2015$. Then, for any integer $m$, by calculating $f((x+m)^{2015}) = (f(x)+f(m))^{2015}$, we have\n$$\n\\sum_{i=0}^{2015} m^i x_i = 0.\n$$\nNow we think of it as a system of infinite number of equations.\n**Claim.** The system has a unique solution $(x_0, x_1, \\dots, x_{2015}) = (0, 0, \\dots, 0)$.\n*Proof.* Suppose there is another solution $(y_0, y_1, \\dots, y_{2015}) = (0, 0, \\dots, 0)$. Let $s$ be the maximum index with $y_s \\neq 0$. By multiplying $\\frac{1}{y_s}$, we may assume that $y_s = 1$. Set $m$ as a number bigger than $|y_0| + |y_1| + \\dots + |y_{2015}|$. Then,\n$$\n0 = \\sum_{i=0}^{2015} m^i y_i = \\sum_{i=0}^{s} m^i y_i > m^s y_s - \\sum_{i=0}^{s-1} m^i |y_i| \\geq m^s - m^{s-1} \\sum |y_i| > 0\n$$\nwhich yields a contradiction. Therefore it has a unique solution $(0, 0, \\dots, 0)$. $\\square$\n\nThe above claim says that $f(x^k) = f(x)^k$ for $k = 0, 1, 2, \\dots, 2015$. Especially, $f(x^2) = f(x)^2 \\ge 0$. Now, for every $x > y$, we have $f(x) - f(y) = f(x - y) = f(\\sqrt{x - y}^2) \\ge 0$, so $f$ is increasing. Therefore, $f(x) = x$, and the candidates of the original problem are $f(x) = x$ and $f(x) = -x$. Indeed, one can check that they are real solutions of the problem by substituting.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42445,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incenter of a triangle $ABC$ whose incircle is tangent to the sides $BC$, $AC$, $AB$ at $D$, $E$, $F$, respectively. Suppose the circumcircle of the triangle $ABC$ intersects the line $EF$ at $P$ and $Q$. If $O_1$ and $O_2$ are the circumcenters of the triangles $IAB$ and $IAC$, respectively, show that the circumcenter of the triangle $DPQ$ lies on the line $O_1O_2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ be the midpoint of the side $BC$. First, we want to show that $M$ is on the circumcircle of the triangle $DPQ$. For the intersection point $X$ of $PQ$ and $BC$, Menelaus theorem guarantees\n$$\n\\frac{BX}{XC} = \\frac{CE}{EA} = \\frac{AF}{FB} = 1.\n$$\nSince $AF = AE$, $BD = BF$, $CD = CE$ and $BX \\cdot CD = XC \\cdot BD$, from which we get $2XB \\cdot XC = (XB + XC) \\cdot XD$ by replacing $BD$ and $CD$ with $XD - XB$ and $XC - XD$, respectively. It follows that $XB \\cdot XC = XM \\cdot XD$, as $\\frac{XB+XC}{2} = XM$. Hence the point $M$ lies on the circumcircle of the triangle $DPQ$.\n\nLet $O'$ be the midpoint of the line segment $O_1O_2$. It suffices to show that $O'$ is the circumcenter of the triangle $DPQ$. first, we will consider the case where $AB \\neq AC$. In this case, we will prove it by showing that $O'$ is on both the perpendicular bisector of $PQ$ and the perpendicular bisector of $DM$. In fact, since $AI$ is perpendicular to both $PQ$ and $O_1O_2$, $PQ$ and $O_1O_2$ are parallel to each other. It follows that $O'$ is on the perpendicular bisector of $PQ$. Let $I_A, I_B$ and $I_C$ be the centers of excircles each of which is tangent to $BC$, $AC$ and $AB$, respectively. It is easy to see that the circumcircle of the triangle $ABC$ is the same as the nine point circle of the triangle $I_A I_B I_C$. Thus, if we let $O_3$ be the circumcenter of the triangle $IBC$ and let $S$ be the intersection point of $OO_3$ and $I_B I_C$, the point $S$ is the midpoint of the line segment $I_B I_C$. Note that the dilatation with center $I$ and ratio 2 sends the triangle $O_1O_2O_3$ to the triangle $I_C I_B I_A$, which implies that $O'$ is the midpoint of the line segment $IS$. Therefore we can conclude that $O'$ is on the perpendicular bisector of $DM$, as both $ID$ and $SM$ are perpendicular to $BC$.\n\nNext, we will consider the case where $AB = AC$. Note that $O'$ is the midpoint of $AI$ in this case. Since $\\angle AEI = 90^\\circ = \\angle AFI$, the points $A, F, I, E$ are on the circle with the center $O'$. Let $R$ be the radius of this circle. Then $O'D^2 - R^2 = DI \\cdot DA = DA^2 - AI \\cdot DA$ holds. The points $B, D, I, F$ being concyclic, we can see $AI \\cdot DA = AF \\cdot AB$. Since $DA^2 = AB^2 - BD^2$,\n$$\nO'D^2 - R^2 = (AB^2 - BD^2) - AF \\cdot AB = AB^2 - BF^2 - AF \\cdot AB = BF \\cdot AF.\n$$\nSimilarly, we can get $O'P^2 - R^2 = PE \\cdot PF$. But $BF \\cdot AF = QF \\cdot PF = PE \\cdot PF$, so\n$$\nO'D^2 - R^2 = O'P^2 - R^2.\n$$\nIt easily follows that $O'D = O'P$. In a similar manner, we can show $O'D = O'Q$. Since $O'P = O'D = O'Q$, $O'$ is the circumcenter of the triangle $DPQ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42446,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $H$ be the orthocenter of an acute triangle $ABC$, and a circle $\\omega$ that passes through all of the points $H$, $A$, $B$ meet the line segment $BC$ at the point $D (\\neq B)$. Let $P$ be the intersecting points of the line $DH$ and the line segment $AC$, and $Q$ be the circumcenter of the triangle $ADP$. Show that the center of the circle $\\omega$ lies on the circumscribed circle of the triangle $BDQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $R$ be the center of the circle $\\omega$, and $E$ be the intersecting points of $BH$ and $AC$. Then we have\n$$\n\\begin{aligned}\n\\angle RBD &= 90^\\circ - \\frac{1}{2}\\angle DRB && \\text{since } RB = RD \\\\\n&= 90^\\circ - \\angle DHB \\\\\n&= 90^\\circ - \\angle PHE \\\\\n&= \\angle EPH \\\\\n&= 180^\\circ - \\angle APD\n\\end{aligned}\n\\qquad (8)\n$$\nOn the other hand, if we let $M$ be the midpoint of $AD$ then $M$, $Q$, and $R$ are collinear, and it holds that\n$$\n\\begin{aligned}\n\\angle MQD &= \\frac{1}{2}\\angle AQD \\\\\n&= \\frac{1}{2}(360^\\circ - 2\\angle APD) && \\text{since } A \\text{ is a circumcenter} \\\\\n&= 180^\\circ - \\angle APD \\\\\n&= \\angle RBD && \\text{by (8)}\n\\end{aligned}\n$$\nThus $R$ lies on the circumscribed circle of $\\triangle BDQ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42447,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a given positive integer $k$, define two sequences $\\{a_n\\}$ and $\\{b_n\\}$ as follows:\n$$\n\\begin{aligned}\na_1 &= k, & a_2 &= k, & a_{n+2} &= a_n a_{n+1} \\quad (n \\ge 1) \\\\\nb_1 &= 1, & b_2 &= k, & b_{n+2} &= \\frac{b_{n+1}^3 + 1}{b_n} \\quad (n \\ge 1)\n\\end{aligned}\n$$\nFor any positive integer $n$, show that $a_{2n}b_{n+3}$ is an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Lemma 2.** For an odd prime $p$ such that $p|k$, $(c_m, p) = 1$ ($-1 \\le m \\le n+1$).\n\n*Proof.* Suppose there is an odd prime $p$ such that $p|(c_m, c_{m+1})$. Then Lemma 1 implies $p|k$, which contradicts Lemma 2. $\\square$\n\n**Lemma 4.** For an odd prime $p$ such that $p^m||c_n$ with $m \\ge 1$, $p^m|(c_{n+1}^3 + k^{3f_{2n+2}})$.\n\n*Proof.* From Lemma 3, $p \\nmid c_{n-1}$. From this computation\n$$\n\\begin{align*}\nc_{n+2} &= \\frac{c_{n+1}^3 + k^{3f_{2n+2}}}{c_n} = \\frac{c_{n-1}^3 c_{n+1}^3 + c_{n-1}^3 k^{3f_{2n+2}}}{c_{n-1}^3 c_n} = \\frac{(c_n^3 + k^{3f_n})^3 + c_{n-1}^3 k^{3f_{2n+2}}}{c_{n-1}^3 c_n} \\\\\n&= \\frac{c_n^9 + 3c_n^6 k^{3f_{2n}} + 3c_n^3 k^{6f_{2n}} + k^{9f_{2n}} + c_{n-1}^3 k^{3f_{2n+2}}}{c_{n-1}^3 c_n} \\\\\n&= \\frac{c_n^9 + 3c_n^6 k^{3f_{2n}} + 3c_n^3 k^{6f_{2n}} + k^{3f_{2n+2}}(c_{n-1}^3 + k^{3f_{2n-2}})}{c_{n-1}^3 c_n} \\tag{9} \\\\\n&= \\frac{c_n^9 + 3c_n^6 k^{3f_{2n}} + 3c_n^3 k^{6f_{2n}} + k^{3f_{2n+2}} c_{n-2} c_n}{c_{n-1}^3 c_n} \\\\\n&= \\frac{1}{c_{n-1}^3} (c_n^8 + 3c_n^5 k^{3f_{2n}} + 3c_n^2 k^{6f_{2n}} + k^{3f_{2n+2}} c_{n-2})\n\\end{align*}\n$$\nthe lemma follows. $\\square$\n\nFor a positive integer $n$, if $2^\\alpha||n$ with a nonnegative integer $\\alpha$, we define $v(n) = \\alpha$. For a positive rational number $q = \\frac{m}{n}$ with integers $m$ and $n$, define $v(q) = v(m) - v(n)$.\n\n**Lemma 5.** If $k \\equiv 0 \\pmod 4$, $v(c_n) = f_{2n}$ for $n \\ge 1$.\n\n*Proof.* We get $v(c_1) = 1 = f_2$ and $v(c_2) = v(c_1^3 + k^3) - v(b_3) = 3v(c_1) = 3 = f_4$. Suppose $v(c_n) = f_{2n}$ and $v(c_{n+1}) = f_{2n+2}$. Since $v(c_{n+1}^3) = 3f_{2n+2} < 3f_{2n+2}v(k) = v(k^{3f_{2n+2}})$, Lemma 1 implies $v(c_{n+2}) = v(c_{n+1}^3 + k^{3f_{2n+2}}) - v(c_n) = 3f_{2n+2} - f_{2n} = f_{2n+4}$. $\\square$\n\n**Lemma 6.** Suppose $k \\equiv 2 \\pmod 4$ and If $n \\equiv 2 \\pmod 3$, then $v(c_n) > f_{2n}$. If $n \\not\\equiv 2 \\pmod 3$, then $v(c_n) = f_{2n}$.\n\n*Proof.* Since $c_{-1} = 1$, $v(c_{-1}) = 0 > f_{-2} = -1$. Since $c_0 = k^3 + 1$, $v(c_0) = 0 = f_0$. From $c_1 = (k^3 + 1)^3 + 1 \\equiv 2 \\pmod 8$, we have $v(c_1) = 1 = f_2$.\nSince $v(c_1) = v(k) = 1$ and $v(b_3) = 0$, $c_2 = \\frac{c_1^3 + k^3}{b_3}$ implies $v(c_2) > 3 = f_4$.\n\nSuppose $v(c_{3m}) = f_{6m}$ and $v(c_{3m+1}) = f_{6m+2}$. Since $v(c_{3m+1}^3) = 3f_{6m+2}$ and $v(k^{3f_{6m+2}}) = 3f_{6m+2}$, we get $v(c_{3m+1}^3 + k^{3f_{6m+2}}) > 3f_{6m+2}$. Thus $v(c_{3m+2}) = v(c_{3m+1}^3 + k^{3f_{6m+2}}) - v(c_{3m}) > 3f_{6m+2} - f_{6m} = f_{6m+4}$.\n\nSuppose $v(c_{3m+1}) = f_{6m+2}$ and $v(c_{3m+2}) > f_{6m+4}$. Since $v(c_{3m+2}^3) > 3f_{6m+4}$ and $v(k^{3f_{6m+4}}) = 3f_{6m+4}$, we have $v(c_{3m+2}^3 + k^{3f_{6m+4}}) = 3f_{6m+4}$. Therefore, $v(c_{3m+3}) = v(c_{3m+2}^3 + k^{3f_{6m+4}}) - v(c_{3m+1}) = 3f_{6m+4} - f_{6m+2} = f_{6m+6}$.\n\nSuppose $v(c_{3m-3}) = f_{6m-6}$, $v(c_{3m-2}) = f_{6m-4}$, $v(c_{3m-1}) \\ge f_{6m-2} + 1$ and $v(c_{3m}) = f_{6m}$. From the equation (9), we have\n$$\n\\begin{align*} \nv(c_{3m+1}) &= v(c_{3m}^3 + k^{3f_{6m}}) - v(c_{3m-1}) \\\\ \n&= v(c_{3m-1}^8 + 3c_{3m-1}^5 k^{3f_{6m-2}} + 3c_{3m-1}^2 k^{6f_{6m-2}} + k^{3f_{6m}} c_{3m-3}) - 3v(c_{3m-2}) \\\\ \n&= v(k^{3f_{6m}} c_{3m-3}) - 3v(c_{3m-1}) = 3f_{6m} + f_{6m-6} - 3f_{6m-4} = f_{6m+2}. \n\\end{align*}\n$$\nThus we prove the lemma. $\\square$\n\nFrom Lemma 5 and Lemma 6, we prove that if $2^m \\mid c_n$, then $2^m \\mid (c_{n+1}^3 + k^{3f_{2n+2}})$. With Lemma 4, it follows that $c_n \\mid (c_{n+1}^3 + k^{3f_{2n+2}})$ and $c_{n+2}$ is a positive integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42448,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ and $k$ be integers satisfying $n \\ge 2$ and $k \\ge \\frac{5}{2}n - 1$. Prove that in every choice of $k$ distinct points among all integer points $(x, y)$ with $1 \\le x, y \\le n$, there exists a circle going through at least four distinct chosen points.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_i$ be the number of chosen points on the line $y = i$. If $a_i \\ge 2$ and $x_1 < x_2 < \\dots < x_{a_i}$ are the x-coordinates of the chosen points on the line $y = i$, then because\n$$\nx_1 + x_2 < x_1 + x_3 < x_2 + x_3 < x_2 + x_4 < x_3 + x_4 < \\dots < x_{a_1-1} + x_{a_1}\n$$\nwe deduce that the number of distinct sums of two x-coordinates among chosen points on $y = i$ is at least $2a_i - 3$. This also holds trivially if $a_i < 2$.\nThere are $2n - 3$ distinct values that can be obtained as a sum of two integers between 1 and $n$. Since\n$$\n\\sum_{i=1}^{n} (2a_i - 3) = 2k - 3n \\ge 5n - 2 - 3n = 2n - 2 > 2n - 3,\n$$\nby the pigeonhole principle, there exist four distinct chosen points $(x_1, y_1)$, $(x'_1, y'_1)$, $(x_2, y_2)$ such that $x_1 + x'_1 = x_2 + x'_2$ and $y_1 \\ne y_2$. There exists a circle going through these four points.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42449,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there are no rational numbers $x$ and $y$ such that\n$$\nx - \\frac{1}{x} + y - \\frac{1}{y} = 4.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that there are rational numbers $x$ and $y$ such that $x - \\frac{1}{x} + y - \\frac{1}{y} = 4$.\nSince $\\left(-\\frac{1}{x}, y\\right)$, $\\left(x, -\\frac{1}{y}\\right)$ and $\\left(-\\frac{1}{x}, -\\frac{1}{y}\\right)$ are solutions of the equation, we can assume that $x > 0$ and $y > 0$.\n\nLetting $u = xy$, we get $x + y = \\frac{4xy}{xy - 1} = \\frac{4u}{u - 1}$. Note that $u > 0$ and $u \\ne 1$.\nNow the quadratic equation\n$$\nT^2 - \\frac{4u}{u-1}T + u = 0\n$$\nhas rational solutions $x$ and $y$. So the discriminant $\\left(\\frac{4u}{u-1}\\right)^2 - 4u$ is a square of a rational number. Then\n$$\n4u^2 - u(u-1)^2 = u(6u - u^2 - 1)\n$$\nis a square of a rational number.\nLet $u = \\frac{q}{p}$ with distinct positive integers $p$ and $q$ such that $(p, q) = 1$. From\n$$\nu(6u - u^2 - 1) = \\frac{q(6pq - p^2 - q^2)}{p^3} = \\frac{1}{p^4} \\cdot pq(6pq - p^2 - q^2),$$\nwe know $pq(6pq - p^2 - q^2)$ is a square of an integer. Because $p, q$ and $6pq - p^2 - q^2$ are pairwise relatively prime, there are positive integers $s, t$ and $w$ such that\n$$\np = s^2, \\quad q = t^2, \\quad \\text{and} \\quad w^2 = 6pq - p^2 - q^2.\n$$\nThus $w^2 = 6s^2t^2 - s^4 - t^4 = (2st)^2 - (s^2 - t^2)^2$. Note that $(s, t) = 1$ and $s \\neq t$.\nWithout loss of generality assume $s > t > 0$. Assume that $(s, t, w)$ is the positive integer solution of $w^2 = 6s^2t^2 - s^4 - t^4 = (2st)^2 - (s^2 - t^2)^2$ which makes $s+t$ minimum.\nBecause $2st$ is even in the Pythagorean triple $\\{w, s^2 - t^2, 2st\\}$, we know $s^2 - t^2$ is even and $s$ and $t$ are odd. So $\\left(\\frac{s^2 - t^2}{2}, st\\right) = 1$. Then\n$$\n\\left(\\frac{w}{2}\\right)^2 + \\left(\\frac{s^2 - t^2}{2}\\right)^2 = (st)^2\n$$\ngives a primitive Pythagorean triple $\\left(\\frac{w}{2}, \\frac{s^2 - t^2}{2}, st\\right)$. So there are positive integers $m$ and $n$ such that $(m, n) = 1$, $s^2 - t^2 = 4mn$ and $st = m^2 + n^2$. We can assume that $m$ is even and $n$ is odd.\nFrom the equality $s^2 - t^2 = (s+t)(s-t) = 4mn$, there are positive integers $A, B, C, D$ which are pairwise relatively prime satisfying\n$$\ns + t = 2AB, \\quad s - t = 2CD, \\quad m = AC, \\quad n = BD.\n$$\nBy plugging these to $st = m^2 + n^2$, we have $2A^2B^2 = (A^2 + D^2)(B^2 + C^2)$. Then we know $C$ is even from the assumption that $m$ is even. Note that $A, B, D$ are odd.\nNow from the equality $2A^2B^2 = (A^2 + D^2)(B^2 + C^2)$, we get $A^2 = B^2 + C^2$, $2B^2 = A^2 + D^2$. Then from $A^2 = B^2 + C^2$, we get $A = a^2 + b^2$, $B = a^2 - b^2$, $C = 2ab$ with\n\npositive integers $a$ and $b$ such that $(a,b) = 1$ and $a > b > 0$. Then $2B^2 = A^2 + D^2$\nbecomes $a^4 + b^4 - 6a^2b^2 = D^2$. Now $(2D)^2 = 6(a+b)^2(a-b)^2 - (a+b)^4 - (a-b)^4$.\nWe have a contradiction from the minimality of $s+t$ since $s+t = 2AB = 2B(a^2 + b^2) > 2a = (a+b) + (a-b) > 0$.\nTherefore, there are no rational numbers $x$ and $y$ such that $x - \\frac{1}{x} + y - \\frac{1}{y} = 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42450,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the maximum value of $(x^2 - yz)(y^2 - zx)(z^2 - xy)$ given that $x, y, z$ are real numbers satisfying $x^2 + y^2 + z^2 = 1$.",
    "options": [],
    "answer": "1/8",
    "solution": "Let $f(x, y, z) = (x^2 - yz)(y^2 - zx)(z^2 - xy)$. Considering $f(x, 0, z) = -x^3z^3$, it is easily seen that maximum of $f$ is positive. Since $f$ is symmetric in $x, y, z$ and $f(x, y, z) = f(-x, -y, -z)$, we may assume that $x \\ge y \\ge z$, $x + y + z \\ge 0$ and consequently $x^2 - yz > 0$. If $f$ takes its maximum at $(x, y, z)$, then\n$$\nf(x, y, z) - f(x, -y, -z) = -2x(x^2 - yz)(y^3 + z^3) \\ge 0.\n$$\nThus we have $y^3 + z^3 \\le 0$. If $z = 0$, then $y$ is also $0$ and thus $f(x, y, z) = 0$. Therefore we must have $z < 0$ and thus $y^2 - zx > 0$. Since $f(x, y, z)$ is positive $z^2 - xy$ is also positive. Now applying the inequality of arithmetic and geometric means, we have\n$$\nf(x, y, z)^{\\frac{1}{3}} \\le \\frac{(x^2 - yz) + (y^2 - zx) + (z^2 - xy)}{3} = \\frac{\\frac{3}{2} - \\frac{1}{2}(x + y + z)^2}{3} \\le \\frac{1}{2}\n$$\nConsequently, $f$ takes its maximum $\\frac{1}{3}$, in the case where $x + y + z = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42451,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of a triangle $ABC$ is tangent to $BC$, $AC$, $AB$ at the points $D$, $E$, $F$, respectively. Suppose the line $EF$ intersects the lines $BI$, $CI$, $BC$, $DI$ at the points $K$, $L$, $M$, $Q$, respectively, where the incenter of $\\triangle ABC$ is $I$. If the line passing through both the midpoint of $CL$ and $M$ intersects $CK$ at a point $P$, show that\n$$\nPQ = \\frac{AB \\cdot KQ}{BI}.\n$$\n\nThe incircle of a triangle *ABC* is tangent to *BC*, *AC*, *AB* at the points *D*, *E*, *F*, respectively. Suppose the line *EF* intersects the lines *BI*, *CI*, *BC*, *DI* at the points *K*, *L*, *M*, *Q*, respectively, where the incenter of *ΔABC* is *I*. If the line passing through both the midpoint of *CL* and *M* intersects *CK* at a point *P*, show that\n$$\nPQ = \\frac{AB \\cdot KQ}{BI}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $BD$ and $BF$ are tangent lines to the incircle of $\\triangle ABC$, $BD = BF$. But the line $BI$ bisects $\\angle DBF$, so $DF$ and $BI$ are perpendicular to each other. Similarly, $DE$ and $CI$ are perpendicular to each other. It follows that $\\angle BKD = \\angle BKF = 90^\\circ - \\angle DFK$ and $\\angle CED + \\angle ECI = 90^\\circ$. Since $AC$ is tangent to\n\nthe excircle of $\\triangle DEF$, $\\angle DFK = \\angle CED$, and so $\\angle BKD = 90^\\circ - \\angle CED$. Thus $\\angle BKD = \\angle ECI = \\angle DCI$, which means that the point $K$ lies on the excircle of the quadrilateral $CEID$. Therefore $\\angle BKC = \\angle IEC = 90^\\circ$. In a similar way, it can be proved that $\\angle BLC = 90^\\circ$.\nApplying Menelaus theorem to $\\triangle DKL$ with respect to the line $MP$ leads to the equality $\\frac{KP \\cdot CJ}{JL \\cdot MK} = 1$, from which we get $\\frac{KP}{PC} = \\frac{MK}{LM}$, as $CJ = JL$. Since $DI$ and $DM$ bisect the internal angle and external angle of $\\triangle DKL$ at $D$, respectively, $\\frac{KQ}{QL} = \\frac{KD}{DL} = \\frac{KM}{ML}$ holds. Combining this with $\\frac{KP}{PC} = \\frac{MK}{LM}$, we get $\\frac{KP}{PC} = \\frac{KQ}{QL}$. It follows that $PQ$ and $CL$ are parallel to each other.\nLet $A'$ be the intersection point of $BL$ and $CK$. Note that $I$ is the orthocenter of the $\\triangle A'BC$. It can easily be seen that $\\angle BA'D = \\frac{1}{2}\\angle BCA$ and $\\angle CA'D = \\frac{1}{2}\\angle ABC$, and so $\\angle BA'C = \\frac{1}{2}(\\angle ABC + \\angle ACB)$. It follows that $\\angle KPQ = \\angle A'CL = 90^\\circ - \\angle BA'C = \\frac{1}{2}\\angle BAC = \\angle IAB$. Since the points $A'$, $L$, $C$, $D$ are concyclic, $\\angle A'KL = \\angle A'BC$, from which we obtain $\\angle PKQ = 90^\\circ + \\frac{1}{2}\\angle ACB = \\angle AIB$. Since $\\angle KPQ = \\angle IAB$ and $\\angle PKQ = \\angle AIB$, $\\triangle KPQ$ and $\\triangle IAB$ are similar to each other, and so\n$$\n\\frac{PQ}{QK} = \\frac{AB}{BI'}\n$$\nfrom which the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42452,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $U$ be a set of $m$ triangles. Prove that there exists a subset $W$ of $U$ satisfying the following conditions.\n\n(i) The number of triangles in $W$ is at least $0.45m^{4/5}$.\n\n(ii) There exist no 6 distinct points $A, B, C, D, E$, and $F$ such that $W$ contains 6 triangles $ABC, BCD, CDE, DEF, EFA$, and $FAB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $U'$ be a subset of $U$ by choosing each triangle with the probability $p$ independently at random. Then the expected number of triangles in $U'$ is $mp$.\n\nA sequence of 6 distinct points $(x_1, x_2, \\dots, x_6)$ is called a *bad configuration* if all 6 triangles $x_1x_2x_3, x_2x_3x_4, \\dots, x_4x_5x_6, x_5x_6x_1, x_6x_1x_2$ belong to $U$.\n\nThe number of bad configurations in $U$ is at most $m(m-1)(3!)^2 \\le 36m^2$, because it is less than or equal to the number of selecting two triangles, selecting $x_1, x_2, x_3$ from the first triangle, and selecting $x_4, x_5, x_6$ from the second triangle.\n\nIf $(x_1, x_2, \\dots, x_6)$ is a bad configuration, then $(x_i, x_{i+1}, \\dots, x_{i+6})$ and $(x_i, x_{i-1}, \\dots, x_{i-6})$ are bad configurations and so $6 \\cdot 2 = 12$ bad configurations form a bunch. Thus the number of bunches of bad configurations in $U$ is at most $36m^2/12 = 3m^2$.\n\nThe probability that a fixed bad configuration of $U$ is contained in $U'$ is $p^6$ and therefore the expected number of bunches of bad configurations in $U'$ is at most $3m^2p^6$.\n\nThus, the expected number of triangles in $U'$ minus the number of bunches of bad configurations in $U'$ is at least $mp - 3m^2p^6$.\n\nTake $p = c m^{-1/5}$. Then\n$$\nmp - 3m^2p^6 \\geq c m^{6/5} - 3c^6 m^2 m^{-6/5} = (c - 3c^6)m^{4/5}.\n$$\nBy taking $c = 1/2$, we deduce $mp - 3m^2p^6 \\geq (\\frac{1}{2} - \\frac{3}{64})m^{4/5} \\geq 0.45m^{4/5}$.\n\nTherefore there exists a subset $U'$ of $U$ such that the number of triangles in $U'$ minus the number of bunches of bad configurations in $U'$ is at least $0.45m^{4/5}$. Let $W$ be a subset of $U'$ by discarding one triangle in each bunch of bad configurations. Then $W$ contains at least $0.45m^{4/5}$ triangles and no bad configurations, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42453,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with $AB < AC$. Points $D, E$ lie on the interiors of $AB, AC$, respectively. Let $P$ be a point satisfying $PB = PD, PC = PE$. Let $X$ be a point on the interior of arc $AC$ of the circumcircle of $ABC$ which does not include the point $B$. The line $XA$ meets again the circumcircle of $ADE$ at $Y$. Show that $PX = PY$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $Z$ be the intersection of the circumcircles of $ABC$ and $ADE$. Then we have $\\angle ZDA = \\angle ZEA = \\angle ZYA$ and $\\angle ZBA = \\angle ZCA = \\angle ZXA$. Hence the triangles $ZBD$, $ZCE$ and $ZXY$ are all similar.\n\nNow let $L, M, N$ be the midpoints of $BD$, $CE$, $XY$, respectively. Then by the similarity of the triangles $ZBD$, $ZCE$, $ZXY$ we have $\\angle ZLA = \\angle ZMA = \\angle ZNA$. Hence $Z, L, M, N, A$ are concyclic. Since $\\angle ALP = \\angle AMP = 90^\\circ$, $P$ lies on this circle. Hence $\\angle ANP = 90^\\circ$. As $N$ is the midpoint of $XY$, we have $PX = PY$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42454,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{R}^+$ be the set of positive real numbers. Let $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$ be a function satisfying the following.\nFor each positive real number $x$, there exists $y \\in \\mathbb{R}$ such that\n$$\n(x + f(y))(y + f(x)) \\leq 4,\n$$\nand the number of such $y$'s is finite.\nProve that $f(x) > f(y)$ for every pair of positive real numbers $x$ and $y$ with $x < y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For each positive real number $x$, let $A_x$ be the set of positive real numbers $y$ satisfying $(x + f(y))(y + f(x)) \\le 4$. The following are easy consequences by the definition.\n(1) If $x \\in A_y$, then $y \\in A_x$.\n(2) For $x < y$, if $f(x) \\le f(y)$ then $A_y \\subseteq A_x$.\nWe prove the following lemma.\n**Lemma.** For $x \\in \\mathbb{R}^+$, there are only finitely many positive real numbers $y (< x)$ such that $f(y) < f(x)$.\n*Proof.* Since $A_x$ is not empty, there exists $z \\in A_x$. For $y (< x)$ with $f(y) < f(x)$, we have $A_x \\subseteq A_y$ by (2). So $z$ also belongs to $A_y$, and $y \\in A_z$ by (1). Therefore, if there are infinitely many $y (< x)$ such that $f(y) < f(x)$, $A_z$ becomes an infinite set, which is a contradiction. $\\square$\nAssume that there exist $a, b \\in \\mathbb{R}^+$ such that $a < b$ and $f(a) \\le f(b)$. Let $S = \\{x \\mid a < x < b, f(x) > f(a)\\}$. By the lemma, there are only finitely many $x$ with $a < x < b$ such that $f(x) < f(b)$. So $S$ is an infinite set. For any $s \\in S$, $A_s \\subseteq A_a$ by (2). Since $A_s$ is not empty, there exists $t \\in A_s \\subseteq A_a$. So, $s \\in A_t \\subseteq \\bigcup_{z \\in A_a} A_z$ by (1). That is, $S \\subseteq \\bigcup_{z \\in A_a} A_z$ which yields a contradiction because $S$ is an infinite set but\n$$\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": 42455,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. There are $n$ boxes $B_1, B_2, \\dots, B_n$ each of which contains some balls. One can perform the following moves.\nChoose positive integers $i$ and $j$ with $1 \\le i \\le j \\le n$, and add exactly one ball to each of the boxes $B_i, B_{i+1}, \\dots, B_j$.\nFor positive integers $x_1, x_2, \\dots, x_n$, let $f(x_1, x_2, \\dots, x_n)$ be the minimum number of moves required to make the number of balls in each of boxes divisible by $3$, starting from $x_i$ balls in $B_i$ for each $i = 1, 2, \\dots, n$. Find the maximum value of $f(x_1, x_2, \\dots, x_n)$.\n(If $3|x_i$ for $i = 1, 2, \\dots, n$, then $f(x_1, x_2, \\dots, x_n) = 0$)",
    "options": [],
    "answer": "ceil((2n+2)/3)",
    "solution": "The answer is $\\left\\lceil \\frac{2n+2}{3} \\right\\rceil$.\n\nFor $i = 0, 1, \\dots, n$, let $y_i = x_i - x_{i+1}$ where $x_0 = x_{n+1} = 0$. Note that $y_0 + y_1 + \\dots + y_n = 0$. The move in the problem is equivalent to the following.\nChoose non-negative integers $i$ and $j$ with $0 \\le i < j \\le n$, and replace $y_i$ and $y_j$ by $y_i - 1$ and $y_j + 1$, respectively.\nWe call this move the $(i, j)$-move. Our goal is to make each $y_i$ a multiple of $3$. Now we prove that for every sequence of integers $y_0, y_1, \\dots, y_n$ with $y_0 + y_1 + \\dots + y_n = 0$, we can make each of $y_i$'s a multiple of $3$ by performing at most $\\left\\lceil \\frac{2n+2}{3} \\right\\rceil$ moves.\n\nWe use induction on $n$.\nOne can easily check that it is possible when $n = 1, 2$.\nSuppose $n > 2$. If $3|y_i$ for some $i$, then we can ignore $y_i$, and so we need at most $\\left\\lceil \\frac{2n}{3} \\right\\rceil$ moves to make all $y_i$'s multiples of $3$. Hence, we may assume that $y_i \\ne 0$ for each $i$. We consider the following four cases.\n\n**Case 1.** There exist $0 \\le i < j \\le n$ such that $y_i \\equiv 1 \\pmod{3}$ and $y_j \\equiv 2 \\pmod{3}$.\nThen, we perform the $(i, j)$-move then $y_i$ and $y_j$ become multiples of $3$. So, in this case, we need at most $1 + \\left\\lceil \\frac{2n-2}{3} \\right\\rceil \\le \\left\\lceil \\frac{2n+2}{3} \\right\\rceil$ moves.\n\n**Case 2.** There exist $0 \\le i < j < k \\le n$ such that $y_i \\equiv y_j \\equiv y_k \\equiv 1 \\pmod{3}$.\nThen, we perform the $(i, k)$-move and the $(j, k)$-move. Then, $y_i, y_j$ and $y_k$ become multiples of $3$. So, we need at most $2 + \\left\\lceil \\frac{2n-4}{3} \\right\\rceil = \\left\\lceil \\frac{2n+2}{3} \\right\\rceil$ moves.\n\n**Case 3.** There exist $0 \\le i < j < k \\le n$ such that $y_i \\equiv y_j \\equiv y_k \\equiv 2 \\pmod{3}$.\nThen, we perform the $(i, j)$-move and the $(i, k)$-move. Then, $y_i, y_j$ and $y_k$ become multiples of $3$. So, we need at most $2 + \\left\\lceil \\frac{2n-4}{3} \\right\\rceil = \\left\\lceil \\frac{2n+2}{3} \\right\\rceil$ moves.\n\n**Case 4.** Neither case 1, case 2 nor case 3 occurs.\nSince $n \\ge 3$, the only possible case is that $n = 3$ and $(y_0, y_1, y_2, y_3) \\equiv (2, 2, 1, 1) \\pmod{3}$. In this case, we perform the (0, 1)-move, the (0, 3)-move and the (2, 3)-move. Then, each $y_i$ becomes a multiple of $3$.\n\nTherefore, by the induction, we can make each $y_i$ a multiple of $3$ by performing at most $\\left\\lceil \\frac{2n+2}{3} \\right\\rceil$ moves.\n\nFinally, for the following cases, we need at least $\\left\\lceil \\frac{2n+2}{3} \\right\\rceil$ moves, which implies that the answer is $\\left\\lceil \\frac{2n+2}{3} \\right\\rceil$.\n\n$$\n(y_0, y_1, \\dots, y_n) = \\begin{cases} (2, 2, \\dots, 2, 2, 2), & \\text{if } n \\equiv 2 \\pmod{3} \\\\ (2, 2, \\dots, 2, 2, 1), (2, 2, \\dots, 2, 1, 2) & \\text{if } n \\equiv 1 \\pmod{3} \\\\ (2, 2, \\dots, 2, 1, 1), & \\text{if } n \\equiv 0 \\pmod{3} \\end{cases}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42456,
    "subject": "Mathematics (Multi-modal)",
    "question": "For an integer $n \\ge 3$ and real numbers $a_1, \\dots, a_n$ and $b_1, \\dots, b_n$, show the following inequality.\n$$\n\\sum_{i=1}^{n} a_i (b_i - b_{i+3}) \\le \\frac{3n}{8} \\sum_{i=1}^{n} \\left[ (a_i - a_{i+1})^2 + (b_i - b_{i+1})^2 \\right] \\\\\n(a_{n+1} = a_1 \\text{ and } b_{n+1} = b_1 \\text{ for } i = 1, 2, 3)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It suffices to prove the following.\n$$\n\\sum_{i=1}^{n} a_i(b_i - b_{i+1}) \\le \\frac{n}{8} \\sum_{i=1}^{n} \\left[ (a_i - a_{i+1})^2 + (b_i - b_{i+1})^2 \\right]\n$$\nBy replacing $b_i$ by $b_{i+j}$ in the above equation and adding up for $j = 0, 1, 2$, we can obtain our desired result. Let\n$$\n\\mathcal{R} = \\frac{1}{2} \\sum_{i=1}^{n} \\left[ (a_i - a_{i+1})^2 + (b_i - b_{i+1})^2 \\right]\n$$\nand consider\n$$\nS_j = \\sum_{i=1}^{n} a_{i+j}(b_i - b_{i+1})\n$$\nwhere $j$ is integer. (We consider any indices as modulo $n$, so that $a_{i+nk} = a_i$ holds for every integers $i, k$.) We can observe\n$$\n\\begin{aligned}\n|S_j - S_{j+1}| &= \\left| \\sum_{i=1}^{n} (a_{i+j} - a_{i+j+1})(b_i - b_{i+1}) \\right| \\\\\n&\\le \\frac{1}{2} \\sum_{i=1}^{n} \\left( (a_{i+j} - a_{i+j+1})^2 + (b_i - b_{i+1})^2 \\right) \\\\\n&= \\frac{1}{2} \\left( \\sum_{i=1}^{n} (a_{i+j} - a_{i+j+1})^2 + \\sum_{i=1}^{n} (b_i - b_{i+1})^2 \\right) \\\\\n&= \\mathcal{R}\n\\end{aligned}\n$$\nand obtain the following as its result:\n$$\n|S_0 - S_j| \\le |j| \\mathcal{R}.\n$$\nMeanwhile we have\n$$\n\\sum_{j=0}^{n-1} S_j = \\sum_{i=1}^{n} \\left( \\sum_{j=0}^{n-1} a_{i+j} \\right) (b_i - b_{i+1}) = \\left( \\sum_{j=0}^{n-1} a_j \\right) \\left( \\sum_{i=1}^{n} (b_i - b_{i+1}) \\right) = 0\n$$\nso for any $-n < k < n$ we have\n$$\nnS_0 = \\sum_{j=-k+1}^{n-k} (S_0 - S_j) \\le \\sum_{j=-k+1}^{n-k} |S_0 - S_j| \\le \\mathcal{R} \\sum_{j=-k+1}^{n-k} |j|.\n$$\nIf $n$ is even then we let $k = n/2$ to obtain\n$$\nnS_0 \\le \\mathcal{R} \\sum_{j=-n/2+1}^{n/2} |j| = \\frac{n^2}{4} \\mathcal{R}\n$$\nand if $n$ is odd then we let $k = (n + 1)/2$ to obtain\n$$\nnS_0 \\le \\mathcal{R} \\sum_{j=-(n-1)/2}^{(n-1)/2} |j| = \\frac{n^2-1}{4} \\mathcal{R} < \\frac{n^2}{4} \\mathcal{R}.\n$$\nIn any cases, we have\n$$\nS_0 \\le \\frac{n}{4} \\mathcal{R} = \\frac{n}{8} \\sum_{i=1}^{n} \\left[ (a_i - a_{i+1})^2 + (b_i - b_{i+1})^2 \\right]\n$$\nthus proving our inequality. $\\square$\nWe will show the following inequality as in the Solution 1.\n$$\n\\sum_{i=1}^{n} a_i (b_i - b_{i+1}) \\le \\frac{n}{8} \\sum_{i=1}^{n} \\left[ (a_i - a_{i+1})^2 + (b_i - b_{i+1})^2 \\right]\n$$\nLet $\\bar{a}$ be the average of all $a_i$. Then above is equivalent to:\n$$\n\\sum_{i=1}^{n} (a_i - \\bar{a})(b_i - b_{i+1}) \\le \\frac{n}{8} \\sum_{i=1}^{n} \\left[ (a_i - a_{i+1})^2 + (b_i - b_{i+1})^2 \\right].\n$$\nBy noting the following (follows from AM-GM)\n$$\n\\sum_{i=1}^{n} (a_i - \\bar{a})(b_i - b_{i+1}) \\leq \\frac{2}{n} \\sum_{i=1}^{n} (a_i - \\bar{a})^2 + \\frac{n}{8} \\sum_{i=1}^{n} (b_i - b_{i+1})^2\n$$\nit suffices to prove\n$$\n\\sum_{i=1}^{n} (a_i - \\bar{a})^2 \\leq \\frac{n^2}{16} \\sum_{i=1}^{n} (a_i - a_{i+1})^2.\n$$\nWe let\n$$\nM = \\max a_i - \\min a_i\n$$\nand we will obtain bounds for both sides using $M$.\n**Lemma 4.** We have\n$$\n\\sum_{i=1}^{n} (a_i - \\bar{a})^2 \\leq \\frac{n}{4} M^2.\n$$\n*Proof.* Both sides of the equation are invariant under adding same constant to all $a_i$, so it suffices to show when $(\\max a_i, \\min a_i) = (M/2, -M/2)$. In that case, we can show:\n$$\n\\sum_{i=1}^{n} (a_i - \\bar{a})^2 = \\sum_{i=1}^{n} a_i^2 - n\\bar{a}^2 \\leq \\sum_{i=1}^{n} a_i^2 \\leq \\sum_{i=1}^{n} (M/2)^2 = \\frac{n}{4} M^2.\n$$\n**Lemma 5.** We have\n$$\n\\sum_{i=1}^{n} (a_i - a_{i+1})^2 \\geq \\frac{4}{n} M^2.\n$$\n*Proof.* Let $a_i, a_j$ be the maximum and minimum among $a_1, \\dots, a_n$ respectively, and assume $i < j$ without loss of generality. Using the Cauchy-Schwarz inequality we have\n$$\nM^2 = \\left( \\sum_{l=i}^{j-1} (a_l - a_{l+1}) \\right)^2 \\leq (j-i) \\sum_{l=i}^{j-1} (a_l - a_{l+1})^2\n$$\nand similarly\n$$\nM^2 = \\left( \\sum_{l=j}^{n+i-1} (a_l - a_{l+1}) \\right)^2 \\leq (n+i-j) \\sum_{l=j}^{n+i-1} (a_l - a_{l+1})^2.\n$$\nThus we have (the last part uses AM-HM)\n$$\n\\begin{aligned}\n\\sum_{l=1}^{n} (a_l - a_{l+1})^2 &= \\sum_{l=i}^{j-1} (a_l - a_{l+1})^2 + \\sum_{l=j}^{n+i-1} (a_l - a_{l+1})^2 \\\\\n&\\le \\frac{M^2}{j-i} + \\frac{M^2}{n+i-j} \\\\\n&\\le \\frac{4M^2}{n}.\n\\end{aligned}\n$$\nCombining those two lemmas yield the desired result of\n$$\n\\sum_{i=1}^{n} (a_i - \\bar{a})^2 \\le \\frac{nM^2}{4} \\le \\frac{n^2}{16} \\sum_{i=1}^{n} (a_i - a_{i+1})^2.\n$$\nWe note that if $n = 3$ then the left hand side (of our original inequality) becomes zero so our problem holds obviously. In this solution, we will prove the following inequality of the Solution 1 for $n \\ge 4$.\n$$\n\\sum_{i=1}^{n} a_i (b_i - b_{i+1}) \\le \\frac{n}{8} \\sum_{i=1}^{n} [(a_i - a_{i+1})^2 + (b_i - b_{i+1})^2]\n$$\nWe will consider sum of the following two inequalities.\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} (a_i - a_{i+1})(b_i - b_{i+1}) &\\le \\frac{1}{2} \\sum_{i=1}^{n} [(a_i - a_{i+1})^2 + (b_i - b_{i+1})^2] \\\\\n\\sum_{i=1}^{n} (a_i + a_{i+1})(b_i - b_{i+1}) &\\le \\frac{\\cot(\\pi/n)}{2} \\sum_{i=1}^{n} [(a_i - a_{i+1})^2 + (b_i - b_{i+1})^2]\n\\end{aligned}\n$$\nThe first one follows easily from AM-GM. For the second one, we consider a $n$-gon whose vertices have coordinates $(a_i, b_i)$. Then we can interpret its the left hand and righthand sides as two times its (signed) area and the sum of squares of its sides respectively. By considering the isoperimetric inequality for $n$-gon and Cauchy-Schwarz inequality, one can show their ratio is maximized for regular $n$-gon, so it suffices to check equality holds for regular $n$-gon case.\nSumming those two gives\n$$\n\\sum_{i=1}^{n} 2a_i(b_i - b_{i+1}) \\le \\left(\\frac{1}{2} + \\frac{\\cot(\\pi/n)}{2}\\right) \\sum_{i=1}^{n} \\left[(a_i - a_{i+1})^2 + (b_i - b_{i+1})^2\\right]\n$$\nso it suffices to show\n$$\n\\frac{1}{2} + \\frac{\\cot(\\pi/n)}{2} \\le \\frac{n}{4}\n$$\nfor $n \\ge 4$. When $n \\ge 6$, we use $\\cot(x) = (\\tan(x))^{-1} < 1/x$ and $\\pi > 3$ to show\n$$\n\\frac{1}{2} + \\frac{n}{2\\pi} < \\frac{1}{2} + \\frac{n}{6} \\le \\frac{n}{4}.\n$$\nFor $n = 4$ and $n = 5$, we can prove it by explicitly calculating $\\cot(\\pi/n)$. ($\\cot(\\pi/4) = 1$, $\\cot(\\pi/5) = \\sqrt{1 + \\frac{2}{\\sqrt{5}}}$)\n\n*Remark.* The 'optimal constant' for this inequality can be given as\n$$\nC_{op} = \\frac{(1 + 2 \\cos(2\\pi/n))}{4 \\sin(\\pi/n)}\n$$\ninstead of $3n/8$. Consider a vector space $V = \\{(x_1, \\dots, x_n) : \\sum x_i = 0\\}$ and an operator $T$ on $V$ defined as $T((x_i)) = (x_{i+1})$. Then our inequality can be expressed as follows. (The absolute value denotes the ordinary Euclidean length induced from $V \\le \\mathbb{R}^n$)\n$$\n\\langle a, (1 - T^3)b \\rangle \\le C (|(1 - T)a|^2 + |(1 - T)b|^2)\n$$\nThe operator $T$ on $V$ is orthogonal, and it can be diagonalized by complex orthogonal basis $v_k = (\\zeta_n^{kj})_{j=1, \\dots, n}$ ($1 \\le k < n$) as $Tv_k = \\zeta_n^k v_k$ ($\\zeta_n = \\exp(2\\pi i/n)$). Thus $1-T$ is invertible, and we can express the above inequality as follows. ($u = (1-T)a, v = (1-T)b$)\n$$\n\\langle (1 - T)^{-1}u, (1 + T + T^2)v \\rangle = \\langle u, (1 - T)^{-1}(1 + T + T^2)v \\rangle \\le C (|u|^2 + |v|^2)\n$$\nOne can see that $C$ can be given as the operator norm of $S = (1-T)^{-1}(1+T+T^2)$. As $S$ is normal operator, its operator norm is given as maximum of absolute value of its eigenvalues $|(1-\\zeta_n^k)^{-1}(1+\\zeta_n^k+\\zeta_n^{2k})|$. One can observe that this obtains maximum $C_{op}$ when $k=1$ or $k=n-1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 42457,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be pairwise coprime positive odd numbers. For positive integers $n$, we define\n$$\nf(n) = \\lfloor \\frac{n}{a} \\rfloor + \\lfloor \\frac{n}{b} \\rfloor + \\lfloor \\frac{n}{c} \\rfloor + \\lfloor \\frac{n}{d} \\rfloor.\n$$\nProve that\n$$\n\\sum_{n=1}^{abcd} (-1)^{f(n)} = 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $a$, $b$, $c$, $d$ are all odd, we have\n$$\n\\begin{aligned}\nf(n) &\\equiv 4n - \\left(a\\lfloor\\frac{n}{a}\\rfloor + b\\lfloor\\frac{n}{b}\\rfloor + c\\lfloor\\frac{n}{c}\\rfloor + d\\lfloor\\frac{n}{d}\\rfloor\\right) \\\\\n&\\equiv \\left(n - a\\lfloor\\frac{n}{a}\\rfloor\\right) + \\left(n - b\\lfloor\\frac{n}{b}\\rfloor\\right) + \\left(n - c\\lfloor\\frac{n}{c}\\rfloor\\right) + \\left(n - d\\lfloor\\frac{n}{d}\\rfloor\\right) \\pmod{2}.\n\\end{aligned}\n$$\nWe can observe that $n - a\\lfloor\\frac{n}{a}\\rfloor$ is the remainder of $n$ divided by $a$—denote it as $(n \\mod a)$. Meanwhile, the Chinese remainder theorem states that the map\n$$\n\\Phi : \\{1, \\dots, abcd\\} \\to \\{0, \\dots, a-1\\} \\times \\{0, \\dots, b-1\\} \\times \\{0, \\dots, c-1\\} \\times \\{0, \\dots, d-1\\}, \\\\\n\\Phi(n) = (n \\mod a, n \\mod b, n \\mod c, n \\mod d)\n$$\nis a one-to-one correspondence, so we can replace our sum in $n$ as a sum in $(x, y, z, w) = \\Phi(n)$ when $(x, y, z, w)$ ranges over all elements of $\\{0, \\dots, a-1\\} \\times \\{0, \\dots, b-1\\} \\times \\{0, \\dots, c-1\\} \\times \\{0, \\dots, d-1\\}$. If $\\Phi(n) = (x, y, z, w)$ then $f(n) \\equiv x + y + z + w \\pmod{2}$, so it follows that\n$$\n\\begin{aligned}\n\\sum_{n=1}^{abcd} (-1)^{f(n)} &= \\sum_{x=0}^{a-1} \\sum_{y=0}^{b-1} \\sum_{z=0}^{c-1} \\sum_{w=0}^{d-1} (-1)^{x+y+z+w} \\\\\n&= \\left(\\sum_{x=0}^{a-1} (-1)^x\\right) \\left(\\sum_{y=0}^{b-1} (-1)^y\\right) \\left(\\sum_{z=0}^{c-1} (-1)^z\\right) \\left(\\sum_{w=0}^{d-1} (-1)^w\\right) \\\\\n&= 1 \\cdot 1 \\cdot 1 \\cdot 1 = 1.\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42458,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n (\\ge 2)$ be a positive integer. Alice distributes $2n$ candies into $4n$ boxes $B_1, B_2, \\dots, B_{4n}$. After checking the number of candies Alice puts in each box, Bob chooses $2n$ boxes $B_{k_1}, B_{k_2}, \\dots, B_{k_{2n}}$ out of the $4n$ boxes satisfying the following, and then takes all candies in the chosen boxes.\n$$\nk_i - k_{i-1} \\in \\{1, 3\\} \\text{ for each } i = 1, 2, \\dots, 2n, \\text{ and } k_{2n} = 4n.\\ (k_0 = 0)\n$$\nAlice gets all candies in the $2n$ boxes Bob did not choose. If Alice and Bob use their best strategies to take as many candies as possible, how many candies can Alice take?",
    "options": [],
    "answer": "n",
    "solution": "The answer is $n$.\nIf Alice puts one candy in each of boxes $B_1, B_2, \\dots, B_{4n-1}$, then Bob can choose at most $n$ out of the $2n$ boxes, so Alice can get exactly $n$ candies.\n\nNow we prove that Bob can take at least $n$ candies. Let $b_i$ be the number of candies Alice puts in $B_i$ for $i = 1, 2, \\dots, 4n$.\n\nSuppose there exists $m = 1, 2, \\dots, n$ such that $b_{4m-2} < 2$. We consider the following two sequences\n$$\np_i = \\begin{cases}\n1 & i=1 \\\\\n1 & i=2, 4, \\dots, 2m-2 \\\\\n3 & i=3, 5, \\dots, 2m-1 \\\\\n3 & i=2m, 2m+2, \\dots, 2n \\\\\n1 & i=2m+1, 2m+3, \\dots, 2n-1\n\\end{cases}, \\quad\nq_i = \\begin{cases}\n3 & i=1, 3, \\dots, 2m-1 \\\\\n1 & i=2, 4, \\dots, 2m-2 \\\\\n3 & i=2m, 2m+2, \\dots, 2n-2 \\\\\n1 & i=2m+1, 2m+3, \\dots, 2n-1 \\\\\n1 & i=2n\n\\end{cases}\n$$\nand let $P_i = p_1 + p_2 + \\dots + p_i$ and $Q_i = q_1 + q_2 + \\dots + q_i$ for $i = 1, 2, \\dots, 2n$. Then, $\\{P_i\\}_{i=1,2,\\dots,2n}$ and $\\{Q_i\\}_{i=1,2,\\dots,2n}$ satisfy the condition in the problem, and furthermore $\\{P_1, P_2, \\dots, P_{2n}\\} \\cup \\{Q_1, Q_2, \\dots, Q_{2n}\\} = \\{1, 2, \\dots, 4n\\} - \\{4m-2\\}$. So,\n$$\n\\sum_{i=1}^{2n} (b_{P_i} + b_{Q_i}) \\ge 2n - b_{4m-2} \\ge 2n - 1\n$$\nwhich implies that either $\\sum_{i=1}^{2n} b_{P_i}$ or $\\sum_{i=1}^{2n} b_{Q_i}$ is at least $n$, so Bob can take at least $n$.\n\nNow we assume that $b_{4m-2} = 2$ for every $m = 1, 2, \\dots, n$. Then, Bob chooses $B_{4k-3}, B_{4k-2}$ for $k = 1, 2, \\dots, n-1$ and $B_{4n-3}, B_{4n}$ where Bob takes $2n - 2 (\\ge n)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42459,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive real number $p (\\le 1)$ such that the inequality\n$$\n\\sum_{i=1}^{2024} x_i (y_{2025-i} - y_{2024-i}) \\ge 1 - p\n$$\nholds for all real numbers $0 \\le x_1 \\le x_2 \\le \\dots \\le x_{2024} \\le 1$ and $0 = y_0 \\le y_1 \\le y_2 \\le \\dots \\le y_{2024} \\le 1$ satisfying\n$$\n\\sum_{i=1}^{2024} x_i = \\sum_{i=1}^{2024} y_i = 2024p.\n$$",
    "options": [],
    "answer": "(sqrt(5) - 1)/2",
    "solution": "If $x_1 = x_2 = \\dots = x_{2024} = y_1 = y_2 = \\dots = y_{2024} = p$, then $p^2 \\ge 1-p$, that is, $p \\ge \\frac{1+\\sqrt{5}}{2}$. So, $p$ must be at least $\\frac{1+\\sqrt{5}}{2}$. We prove that the condition holds for $p = \\frac{1+\\sqrt{5}}{2}$, which implies that the answer is $\\frac{1+\\sqrt{5}}{2}$.\nLet $p = \\frac{1+\\sqrt{5}}{2}$, and let\n$$\nS = \\{(x_1, \\dots, x_{2024}) \\mid 0 \\le x_1 \\le \\dots \\le x_{2024} \\le 1, x_1 + x_2 + \\dots + x_{2024} = 2024p\\}.\n$$\nFor $X = (x_1, x_2, \\dots, x_{2024})$, $Y = (y_1, y_2, \\dots, y_{2024}) \\in S$, we regard $x_0 = y_0 = 0$ for convenience, and let\n$$\nf(X, Y) = \\sum_{i=1}^{2024} x_i (y_{2025-i} - y_{2024-i}) \\\\ = x_1(y_{2024} - y_{2023}) + x_2(y_{2023} - y_{2022}) + \\dots + x_{2023}(y_2 - y_1) + x_{2024}y_1.\n$$\nSince $f(X, Y) = y_1(x_{2024} - x_{2023}) + y_2(x_{2023} - x_{2022}) + \\dots + y_{2023}(x_2 - x_1) + y_{2024}x_1$, $f(X, Y)$ is symmetric with regard to $X$ and $Y$. If $x_1 = x_2 = \\dots = x_{k_1} = a_1$, $x_{k_1+1} = \\dots = x_{k_1+k_2} = a_2$, \\dots, $x_{k_1+k_2+\\dots+k_{m-1}+1} = \\dots = x_n = a_m$ for $0 \\le a_1 < a_2 < \\dots < a_m \\le 1$, we denote $X$ by $(a_1^{k_1}, a_2^{k_2}, \\dots, a_m^{k_m})$ where $k_m = 2024 - k_1 - k_2 - \\dots - k_{m-1}$, and let $l(X) = m$. And in this case, we have\n$$\nf(X, Y) = a_1(y_{2024} - y_{2024-k_1}) + a_2(y_{2024-k_1} - y_{2024-k_1-k_2}) + \\dots + a_m y_{k_m} \\\\ = a_1 Y_1 + a_2 Y_2 + \\dots + a_m Y_m\n$$\nwhere $Y_t = y_{2024-k_1-k_2-\\dots-k_{t-1}} - y_{2024-k_1-k_2-\\dots-k_t} = y_{k_t+k_{t+1}+\\dots+k_m} - y_{k_t+1+\\dots+k_m}$ for $t = 1, 2, \\dots, m$.\nNow we consider $X, Y \\in S$ such that $f(X, Y)$ is minimized, and subject to that $l(X) + l(Y)$ is minimum. Let $X = (a_1^{k_1}, a_2^{k_2}, \\dots, a_m^{k_m})$.\n**Lemma.** There is no $1 \\le i < m$ such that $0 < a_i, a_{i+1} < 1$.\n\n*Proof.* Suppose $0 < a_i < a_{i+1} < 1$ for some $i$.\n**Case 1.** $\\frac{Y_i}{k_i} \\le \\frac{Y_{i+1}}{k_{i+1}}$.\nLet $a = \\frac{k_i a_i + k_{i+1} a_{i+1}}{k_i + k_{i+1}}$. Then, $a_i < a < a_{i+1}$, and for\n$$\nX' = (a_1^{k_1}, \\dots, a_{i-1}^{k_{i-1}}, a_{i-1}^{k_{i-1}+k_{i+1}}, a_{i+2}^{k_{i+2}}, \\dots, a_m^{k_m})\n$$\nwe have\n$$\nf(X, Y) - f(X', Y) \\ge 0\n$$\nand $l(X') = m - 1 < l(X)$, which yields a contradiction to the fact that $l(X) + l(Y)$ is minimum.\n**Case 2.** $\\frac{Y_i}{k_i} > \\frac{Y_{i+1}}{k_{i+1}}$.\nChoose a positive real number $e$ such that $e < a_i - a_{i-1}$ (if $i = 1$, let $a_0 = 0$) and $a_{i+1} + \\frac{k_i e}{k_{i+1}} < a_{i+2}$ (if $i = m - 1$, let $a_{m+1} = 1$). Let $a = a_i - e$ and $b = a_{i+1} + \\frac{k_i e}{k_{i+1}}$, and consider $X' = (a_1^{k_1}, a_2^{k_2}, \\dots, a_{i-1}^{k_{i-1}}, a^{k_i}, b^{k_{i+1}}, a_{i+2}^{k_{i+2}}, \\dots, a_m^{k_m})$. Since $k_i a + k_{i+1} b = k_i a_i + k_{i+1} a_{i+1}$, $X' \\in S$. However, $f(X, Y) - f(X, Y') > 0$, yielding a contradiction. Therefore there is no $1 \\le i < m$ such that $0 < a_i, a_{i+1} < 1$. $\\square$\nThe above lemma also holds for $Y$. Thus, there exist non-negative integers $k_1, k_2, k_3, s_1, s_2, s_3$ and real numbers $0 < a, b < 1$ such that\n$$\n\\bullet \\quad k_1 + k_2 + k_3 = s_1 + s_2 + s_3 = 2024,\n$$\n$$\n\\bullet \\quad x_1 = x_2 = \\cdots = x_{k_1} = 0, x_{k_1+1} = \\cdots = x_{k_1+k_2} = a, x_{k_1+k_2+1} = \\cdots = x_{2024} = 1\n$$\n$$\n\\bullet \\quad y_1 = y_2 = \\cdots = y_{s_1} = 0, y_{s_1+1} = \\cdots = y_{s_1+s_2} = a, y_{s_1+s_2+1} = \\cdots = y_{2024} = 1\n$$\n$$\n\\bullet \\quad k_2 a + k_3 = s_2 b + s_3 = 2024p.\n$$\nWe note that $k_1, k_2, k_3, s_1, s_2, s_3$ could be zero, and we have\n$$\nf(X, Y) = a y_{k_2+k_3} + (1-a) y_{k_3} = b x_{s_2+s_3} + (1-b) x_{s_3}.\n$$\nIf $k_2 = 0$, then $x_1 + x_2 + \\cdots + x_{2024} = k_3 \\ne 2024p$ (since $k_3$ is an integer), so $k_2 > 0$. Thus, $k_3 < k_2 a + k_3 = 2024p$ and $k_1 = 2024 - (k_2 + k_3) < 2024 - (k_2 a + k_3) = 2024(1-p)$. That is,\n$$\nk_1 < 2024(1-p), \\quad k_2 > 0, \\quad k_3 < 2024p,\n$$\nand similarly,\n$$\ns_1 < 2024(1-p), \\quad s_2 > 0, \\quad s_3 < 2024p.\n$$\nSince $y_1 = y_2 = \\dots = y_{s_1} = 0$, $y_{s_1+1} = \\dots = y_{s_1+s_2} = a$, and $y_{s_1+s_2+1} = \\dots = y_{2024} = 1$, in order for $f(X, Y) = a y_{k_2+k_3} + (1-a) y_{k_3}$ to be minimized, the following must hold.\n(i) $y_{k_2+k_3+1} = \\dots = y_{2024} = 1$\n(ii) $y_{k_3+1} = \\dots = y_{k_2+k_3}$\n(iii) $y_1 = \\dots = y_{k_3}$\nSimilarly, the following must hold.\n(i') $x_{s_2+s_3+1} = \\dots = x_{2024} = 1$\n(ii') $x_{s_3+1} = \\dots = x_{s_2+s_3}$\n(iii') $x_1 = \\dots = x_{s_3}$\nBy (i) and (i'), $k_1 \\le s_3$ and $s_1 \\le k_3$.\nIf $k_1 < s_3$, then $y_{k_3+1} = 1$ by (i) and (ii), so $k_1 + k_2 \\le s_3 < 2024p$. Then by (iii), $k_3 = s_2$, $y_{k_3} = b$ and $s_1 = 0$, $k_1 + k_2 = s_3$. So,\n$$\n\\begin{aligned}\na &= \\frac{2024p - k_3}{k_2} = \\frac{2024p - k_3}{2024 - k_1 - k_3} \\ge \\frac{2024p - k_3}{2024 - k_3} = 1 - \\frac{2024(1 - p)}{2024 - k_3} \\\\\nb &= \\frac{2024p - s_3}{s_2} = \\frac{2024p - k_1 - k_2}{k_3} = \\frac{k_3 - 2024(1 - p)}{k_3} = 1 - \\frac{2024(1 - p)}{k_3} \\end{aligned}\n$$\nSince $2024p > k_3 = 2024 - (k_1 + k_2) = 2024 - s_3 > 2024(1 - p)$, we have\n$$\n\\begin{aligned}\nf(X, Y) &= a y_{k_2+k_3} + (1-a) y_{k_3} = a + (1-a)b = 1 - (1-a)(1-b) \\\\\n&= 1 - \\frac{2024(1-p)}{2024-k_3} \\cdot \\frac{2024(1-p)}{k_3} \\ge 1-p. \\end{aligned}\n$$\nSo, we may assume that $k_1 = s_3$ and by the symmetry, we may further assume that $s_1 = k_3$. Then, $k_2 = s_2$, and\n(i) $y_{k_2+k_3+1} = \\dots = y_{2024} = 1$\n(ii) $y_{k_3+1} = \\dots = y_{k_2+k_3} = b$\n(iii) $y_1 = \\dots = y_{k_3} = 0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 42460,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcircle of a triangle $ABC$. Choose a point $D (\\neq A)$ on the extension of segment $BA$ towards $A$. Points $E$ and $F$ are two distinct points on circle $O$ such that lines $DE$ and $DF$ are both tangent to circle $O$. Segment $EF$ intersects side $CA$ at point $T (\\neq C)$. Choose a point $P (\\neq B, C)$ on the arc $BC$ of circle $O$ that doesn't contain point $A$. Let the intersection point of line $DP$ and circle $O$ be $Q (\\neq P)$. Lines $BQ$ and $DT$ intersect at a point $X$ (other than $Q$). Let $Y (\\neq P)$ be the intersection point of line $PT$ and circle $O$. Prove that the points $C, X, \\text{and } Y$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $Z$ be the intersection point of line $CD$ and $O$. For cyclic quadrilateral $BAZC$, $D$ is the intersection point of two sides $BA$ and $CZ$, and $EF$ is the polar line of $D$ with respect to the circle $O$. Therefore, the intersection point of $AC$ and $BZ$ should lie on $EF$, which means the points $B, T, Z$ are collinear.\n\nNow use Pascal's theorem on hexagon $CYPQBZ$, and we can conclude that three points $X' = CY \\cap QB$, $T = YP \\cap BZ$, and $D = PQ \\cap ZC$ are collinear, which implies that $X'$ lies on line $DT$. Therefore, $X'$ is same with $X$, and which implies that $C, Y, X$ are collinear, as desired. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42461,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n (\\ge 4)$ be a positive integer, and $a_1, a_2, \\dots, a_n$ be distinct positive integers less than or equal to $n$. Determine the maximum value of\n$$\n\\sum_{i=1}^{n} |a_i - a_{i+1} + a_{i+2} - a_{i+3}|.\n$$\n($a_{n+i} = a_i$ for $i = 1, 2, 3$)",
    "options": [],
    "answer": "n^2 if n is even; n^2 - 5 if n is odd",
    "solution": "The answer is $n^2$ if $n$ is even, and $n^2 - 5$ if $n$ is odd.\nIt is easy when $n$ is even. Let $k = \\frac{n+1}{2}$. Then,\n$$\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$$\nand the equality holds when $a_{2i-1} = i$ and $a_{2i} = \\frac{n}{2} + 1$ for $i = 1, 2, \\dots, \\frac{n}{2}$. So, the maximum of $\\sum_{i=1}^{n} |a_i - a_{i+1} + a_{i+2} - a_{i+3}|$ is $n^2$.\n\nLet $n$ be odd. Without loss of generality, let $a_1 - a_2 > 0$. If $a_i - a_{i+1} > 0$ for all $i = 1, 2, \\dots, n$, then\n$$\n0 = (a_1 - a_2) + (a_2 - a_3) + \\dots + (a_n - a_1) > 0,\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|a_i - a_{i+1} + a_{i+2} - a_{i+3}| = \\big| |a_i - a_{i+1}| - |a_{i+2} - a_{i+3}| \\big| \\le |a_i - a_{i+1}| + |a_{i+2} - a_{i+3}| - 2.\n$$\nSo $\\sum_{i=1}^{n} |a_i - a_{i+1} + a_{i+2} - a_{i+3}|$ is at most\n$$\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$$\nLet $k = \\frac{n+1}{2}$. Then,\n$$\n2 \\sum_{i=1}^{n} |a_i - a_{i+1}| - 4 \\le 2 \\sum_{i=1}^{n} (|k - a_i| + |k - a_{i+1}|) - 4 = 4 \\sum_{i=1}^{n} |k - a_i| - 4 = n^2 - 5.\n$$\nTherefore,\n$$\n\\sum_{i=1}^{n} |a_i - a_{i+1} + a_{i+2} - a_{i+3}| \\le n^2 - 5\n$$\nand the equality is attained when $a_n = \\frac{n+1}{2}$ and $a_{2i} = i$, $a_{2i-1} = \\frac{n+1}{2} + i$ for $i = 1, 2, \\dots, \\frac{n-1}{2}$. Thus the maximum of $\\sum_{i=1}^{n} |a_i - a_{i+1} + a_{i+2} - a_{i+3}|$ is $n^2 - 5$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42462,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there exists a positive integer $K$ satisfying the following: for any prime $p > K$, the number of integers $1 \\le a \\le p$ such that $a^{p-1} - 1$ is divisible by $p^2$ is less than or equal to $\\frac{p}{2^{2024}}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Define\n$$\nS = \\{1 \\le a \\le p^2 : p^2|a^{p-1} - 1\\}, \\quad S_p = S \\cap \\{1, 2, \\dots, p\\}.\n$$\nWe observe that $S$ is closed in multiplication mod $p^2$: if $a, b \\in S$ and $ab \\equiv c \\pmod{p^2}$ then $c \\in S$.\n\n**Lemma 1.** We have $|S| \\le p-1$.\n\n*Proof.* (Note: actually we have $|S| = p-1$ and there are several ways to show this - applying the Hensel's lemma, or utilizing the existence of primitive root mod $p^2$ etc.. We give a self-contained proof.) If $a, b \\in S$ and $a \\equiv b \\pmod{p}$, then we have\n$$\np^2|a^{p-1} - b^{p-1} = (a-b)(a^{p-2} + a^{p-3}b + \\dots + b^{p-2}).\n$$\nBut as\n$$\na^{p-2} + a^{p-3}b + \\dots + b^{p-2} \\equiv (p-1)a^{p-2} \\pmod{p}\n$$\nis not divisible by $p$, this implies $p^2|a-b$, so $a=b$. This shows that elements of $S$ are all distinct mod $p$, and obviously $S$ does not contain multiples of $p$, so $|S| \\le p-1$. $\\square$\n\nWe start from\n$$\nS_p \\cdot S_p = \\{ab : a, b \\in S_p\\} \\subset S.\n$$\nIf we consider the map $S_p \\times S_p \\to S_p \\cdot S_p$ given as $(a, b) \\mapsto ab$, preimage of an element $m \\in S_p \\cdot S_p$ is of size at most $\\tau(m)$ (number of divisors of $m$), as $a$ should satisfy $a|m$ and $b$ is then uniquely determined. Thus it follows that\n$$\nM|S_p \\cdot S_p| \\ge |S_p \\times S_p| = |S_p|^2, \\quad M = \\max_{1 \\le m \\le p^2} \\tau(m)\n$$\nand we have\n$$\n\\frac{1}{M}|S_p|^2 \\le |S_p \\cdot S_p| \\le |S| < p \\Rightarrow |S_p| < \\sqrt{pM}.\n$$\n\nMeanwhile, we bound $M$ using the following lemma.\n\n**Lemma 2.** For any $\\epsilon > 0$, there exists a constant $C_\\epsilon > 0$ such that\n$$\n\\tau(n) < C_\\epsilon n^\\epsilon\n$$\nfor any positive integer $n$.\n\n*Proof.* Write $n = \\prod p^{e_p}$ be the prime factorization of $n$. Then we have\n$$\n\\frac{\\tau(n)}{n^\\epsilon} = \\prod_p \\frac{e_p + 1}{(p^\\epsilon)^{e_p}}\n$$\nThe sequence $(\\frac{e+1}{k^\\epsilon})_e$ increases if and only if $\\frac{e+2}{k} > e + 1$, or equivalently $e + 1 < \\frac{1}{k-1}$. Thus if $k \\ge 2$ then this sequence always decreases, so its maximum is obtained at $e = 0$, which is 1. If $1 < k < 2$ then this sequence increases if $e < \\frac{2-k}{1-k}$ and decreases if $e > \\frac{2-k}{1-k}$, so it obtains maximum at $e = \\left[\\frac{2-k}{1-k}\\right]$. Denote this maximum as $F(k)$, then we have\n$$\n\\frac{\\tau(n)}{n^\\epsilon} \\le \\prod_{p \\le 2^{1/\\epsilon}} \\frac{e_p + 1}{(p^\\epsilon)^{e_p}} = \\prod_{p \\le 2^{1/\\epsilon}} F(p^\\epsilon)\n$$\nso we can set $C_\\epsilon > 0$ as any constant larger than that.\n\nUsing this lemma gives $M < C_{1/3}p^{2/3}$, which shows\n$$\n|S_p| < \\sqrt{pM} < \\sqrt{p \\cdot C_{1/3}p^{2/3}} = \\sqrt{C_{1/3}p^{5/6}}\n$$\nNow pick $K$ such that $p > K$ implies $\\sqrt{C_{1/3}p^{5/6}} < p/2^{2024}$, thus $|S_p| < p/2^{2024}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42463,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA and $B$ play the following game with $N$ counters. $A$ divides the counters into $2$ piles, each with at least $2$ counters. Then $B$ divides each pile into $2$ piles, each with at least one counter. $B$ then takes $2$ piles according to a rule which both of them know, and $A$ takes the remaining $2$ piles. Both $A$ and $B$ make their choices in order to end up with as many counters as possible. There are $3$ possibilities for the rule:\n\n$R1$ $B$ takes the biggest heap (or one of them if there is more than one) and the smallest heap (or one of them if there is more than one).\n\n$R2$ $B$ takes the two middling heaps (the two heaps that $A$ would take under $R1$).\n\n$R3$ $B$ has the choice of taking either the biggest and smallest, or the two middling heaps. For each rule, how many counters will $A$ get if both players play optimally?",
    "options": [],
    "answer": "[N/2], [(N+1)/2], [N/2]",
    "solution": "Solution:\nAnswers: $[N/2]$, $[(N+1)/2]$, $[N/2]$.\n\nSuppose $A$ leaves piles $n$, $m$ with $n \\leq m$.\n\nUnder $R1$, $B$ can certainly secure $m$ by dividing the larger pile into $1$ and $m-1$. He cannot do better, because if $b$ is the biggest of the $4$ piles, then the smallest is at most $m-b$. Hence $A$'s best strategy is to leave $[N/2]$, $[(N+1)/2]$.\n\nUnder $R2$, if $A$ leaves $a=2$, $b=N-2$, then $B$ cannot do better than $[N/2]$, because if he divides the larger pile into $a$, $b$ with $a \\leq b$, then he takes $a+1$. $A$ cannot do better, because if he leaves $a$, $b$ with $3 \\leq a \\leq b$, then $B$ can divide to leave $1$, $a-1$, $[b/2]$, $[(b+1)/2]$. Now if $a-1 \\geq [(b+1)/2]$, then $B$ takes $b \\geq [(N+1)/2]$. If $a-1 < [(b+1)/2]$, then $B$ takes $a-1 + [b/2]$. But $a-1 \\geq 2$ and $[b/2] \\geq [(b+1)/2] - 1$, so $a-1 + [b/2] \\geq 1 + [(b+1)/2]$, or $B$ takes at least as many as $A$, so $B$ takes at least $[(N+1)/2]$.\n\nUnder $R3$, $A$'s best strategy is to divide into $[N/2]$, $[(N+1)/2]$. We have already shown that $B$ can secure $[(N+1)/2]$ and no more by following $R1$. He cannot do better under $R2$, for if he divides so that the biggest pile comes from $[N/2]$, then the smallest does too and so he gets $[(N+1)/2]$. If he divides so that the biggest and smallest piles come from $[(N+1)/2]$, then he gets only $[N/2]$. But one of these must apply, because if he divided so that the smaller from $[N/2]$ was smaller than the smaller from $[(N+1)/2]$, and the bigger from $[N/2]$ was smaller than the bigger from $[(N+1)/2]$, then $[N/2]$ would be at least $2$ less than $[(N+1)/2]$ (which it is not).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42464,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven three arbitrary infinite sequences of natural numbers, prove that we can find unequal natural numbers $m$, $n$ such that for each sequence the $m$th member is not less than the $n$th member.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nGiven any infinite sequence of natural numbers, we can find a non-decreasing subsequence (proof below).\n\nSo suppose the three sequences are $a$, $b$, and $c$. Take a non-decreasing subsequence of $a_i$. Suppose it is $a_{i_1}$, $a_{i_2}$, $a_{i_3}$, ... . Now consider the infinite sequence $b_{i_1}$, $b_{i_2}$, ... . It must have a non-decreasing subsequence. Suppose it is $b_{j_1}$, $b_{j_2}$, ... . Now consider the infinite sequence $c_{j_1}$, $c_{j_2}$, ... . It must have a non-decreasing subsequence $c_{k_1}$, $c_{k_2}$, ... . Each of the three subsequences $a_{k_1}$, $a_{k_2}$, ... , $b_{k_1}$, $b_{k_2}$, ... , $c_{k_1}$, $c_{k_2}$, ... is non-decreasing. So we may take, for example, $m = k_2$ and $n = k_1$.\n\n[Proof that any infinite sequence of natural numbers has a non-decreasing subsequence: if the original sequence is unbounded, then we can take a strictly increasing subsequence. If not, then since there are only finitely many possible numbers not exceeding the bound, at least one of them must occur infinitely often.]",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42465,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n120 unit squares are arbitrarily arranged in a $20 \\times 25$ rectangle (both position and orientation is arbitrary). Prove that it is always possible to place a circle of unit diameter inside the rectangle without intersecting any of the squares.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIf a circle with unit diameter intersects a unit square, then its center must lie inside an area $3 + \\pi /4$, namely an oval centered on the square and comprising: the original square, area $1$; four $1 \\times 1/2$ rectangles on the sides, total area $2$; and four quarter circles at the corners, total area $\\pi /4$. So if it does not intersect any of the $120$ unit squares, then it must avoid ovals with a total area of $120 \\times (3 + \\pi /4) = 454.2$. Of course, for many arrangements of the squares, these ovals might overlap substantially, but the worst case would be no overlap.\n\nThe circle is also required to lie inside the rectangle, so its center must lie outside a strip $1/2$ wide around the edge, and hence inside an inner $19 \\times 24$ rectangle, area $456$. The total area of ovals is less, so they cannot cover it completely and it must be possible to place a circle as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42466,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven a rectangle $ABCD$ with $AC$ length $e$ and four circles centers $A$, $B$, $C$, $D$ and radii $a$, $b$, $c$, $d$ respectively, satisfying $a + c = b + d < e$. Prove you can inscribe a circle inside the quadrilateral whose sides are the two outer common tangents to the circles center $A$ and $C$, and the two outer common tangents to the circles center $B$ and $D$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $O$ be the center of the rectangle. Let $r = (a + c)/2 = (b + d)/2$. The required circle has center $O$, radius $r$. Let an outer common tangent touch the circle center $A$ at $W$, and the circle center $C$ at $X$. Let $P$ be the midpoint of $WX$, then $OP$ is parallel to $AW$ and $CX$ and has length $r$, hence the circle center $O$ touches $AW$ at $P$. Similarly for the other common tangents.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42467,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that any 39 successive natural numbers include at least one whose digit sum is divisible by 11.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $n$ be the smallest number in the sequence and $m$ the smallest with last digit $0$. $m$ and $m+10$ have different digit sums unless (possibly) the penultimate digit of $m$ is $9$, but in that case $m+10$ and $m+20$ have different digit sums. So two of $m$, $m+10$, $m+20$ are sure to have different digit sums. Hence at least one has a digit sum not congruent to $1 \\pmod{11}$. Adding the appropriate final digit gives a number whose digit sum is divisible by $11$. This number lies in the range $m$ to $m+29$ and $m \\leq n+9$. Hence the result. $n=999981$ shows it is best possible.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42468,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. Given a quadruple $(a, b, c, d)$ of positive reals, transform to the new quadruple $(ab, bc, cd, da)$. Repeat arbitrarily many times. Prove that you can never return to the original quadruple unless $a = b = c = d = 1$.\n\nb. Given $n$ a power of $2$, and an $n$-tuple $(a_1, a_2, \\ldots, a_n)$ transform to a new $n$-tuple $(a_1 a_2, a_2 a_3, \\ldots, a_{n-1} a_n, a_n a_1)$. If all the members of the original $n$-tuple are $1$ or $-1$, prove that with sufficiently many repetitions you obtain all $1$s.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nLet $Q_0$ be the original quadruple $(a, b, c, d)$ and $Q_n$ the quadruple after $n$ transformations. If $abcd > 1$, then the products form a strictly increasing sequence, so return is impossible. Similarly if $abcd < 1$. So we must have $abcd = 1$. Let the largest of the four values of a quadruple $Q$ be $M(Q)$. If a member of $Q_1$ is not $1$, then $M(Q_1) > 1$. $Q_3$ consists of the elements of $Q_1$ squared and permuted, so $M(Q_3) = M(Q_1)^2$. Hence the sequence $M(Q_1)$, $M(Q_3)$, $M(Q_5)$, ... increases without limit. This means no return is possible, because a return would lead to the values cycling.\n\nb.\nAfter $r < n$ transformations, the first number of the $n$-tuple is the product $a_1^{(r\\choose 0)} a_2^{(r\\choose 1)} \\ldots a_{r+1}^{(r\\choose r)}$, where $(r\\choose i)$ denotes the binomial coefficient. [This is an easy induction.] Hence after $n = 2^k$ transformations it is $a_1^2$ times the product $a_2^{(n\\choose 1)} \\ldots a_n^{(n\\choose 1)}$. So it is sufficient to prove that $(n\\choose i)$ is even for $n$ a power of $2$ and $0 < i < n$. But observe that $(n\\choose i) = (n-1\\choose i) \\cdot n/(n-i)$ and $n$ is divisible by a higher power of $2$ than $n-i$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42469,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven an $m \\times n$ array of real numbers. You may change the sign of all numbers in a row or of all numbers in a column. Prove that by repeated changes you can obtain an array with all row and column sums non-negative.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe array has $mn$ entries. Call an array that can be obtained by repeated changes a reachable array. A reachable array differs from the original only in that some or all of the signs of its $mn$ entries may be different. There are at most $2^{mn}$ different reachable arrays. For each reachable array calculate the sum of all its entries. Take the reachable array with the largest such sum. It must have non-negative row and column sums, because if any such sum was negative, changing the sign of that row or column would give another reachable array with strictly greater total sum.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42470,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven $n > 1$ points, some pairs joined by an edge (an edge never joins a point to itself). Given any two distinct points you can reach one from the other in just one way by moving along edges. Prove that there are $n - 1$ edges.",
    "options": [],
    "answer": "n - 1 edges",
    "solution": "Solution:\nEvery point must have at least one edge. We show that there is a point with just one edge. Suppose the contrary, that every point has at least two edges. We now construct a path in which the same edge or point never appears twice. Starting from any point $b$, move along an edge to $c$. $c$ is not already on the path, because otherwise the edge would join $b$ to itself. Now suppose we have reached a point $x$ not previously on the path. $x$ has at least two edges, so it must have another one besides the one we used to reach it. Suppose this joins $x$ to $y$. If $y$ is already on the path, then we have two distinct ways of moving along edges from $x$ to $y$: directly, or by backtracking along the path from $x$ to $y$. But this is impossible, so $y$ is not already on the path and we may extend the path to it. But this procedure allows us to construct a path containing more than the $n$ distinct points available. Contradiction.\n\nThe result is now easy. Induction on $n$. Take a point with just one edge. Remove it and the edge. Then the remaining $n - 1$ points satisfy the premise and hence have just $n - 2$ edges.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 42471,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven any natural numbers $m$, $n$ and $k$. Prove that we can always find relatively prime natural numbers $r$ and $s$ such that $rm + sn$ is a multiple of $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $d = (m, n)$, the greatest common divisor of $m$ and $n$. Let $r = n/d$, $s = nhk - m/d$, where $h$ is any integer sufficiently large to ensure that $s > 0$. Now $rm + sn = mn/d + nnhk - mn/d = nnhk$, which is a multiple of $k$. If $e$ divides $r$, then it also divides $rdhk = nhk$. So if $e$ divides $r$ and $s$, then it also divides $s - nhk = -m/d$. But $n/d$ and $m/d$ are relatively prime, so $e$ must be $1$. Hence $r$ and $s$ are relatively prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42472,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABCD$ is any convex quadrilateral. Construct a new quadrilateral as follows. Take $A'$ so that $A$ is the midpoint of $DA'$; similarly, $B'$ so that $B$ is the midpoint of $AB'$; $C'$ so that $C$ is the midpoint of $BC'$; and $D'$ so that $D$ is the midpoint of $CD'$. Show that the area of $A'B'C'D'$ is five times the area of $ABCD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCompare the triangles $A'B'A$ and $ADB$. The base of $A'B'A$ can be taken as $A'A$, which is the same length as $AD$. The height of $A'B'A$ is $AB'$ times $\\sin B'A'A'$, which is twice $AB$ times $\\sin BAD$. So $\\text{area } A'B'A = 2 \\times \\text{area } ADB$.\n\nSimilarly, $\\text{area } B'C'B = 2 \\times \\text{area } BAC$, $\\text{area } C'D'C = 2 \\times \\text{area } CBD$, and $\\text{area } D'A'D = 2 \\times \\text{area } DCA$.\n\nSo adding, the $\\text{area } A'B'A + \\text{area } C'D'C = 2 \\times \\text{area } ABCD$, and $\\text{area } B'C'B + \\text{area } D'A'D = 2 \\times \\text{area } ABCD$.\n\nBut $ABCD = A'B'A + B'C'B + C'D'C + D'A'D + ABCD$. Hence result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42473,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$AB = BC$ and $M$ is the midpoint of $AC$. $H$ is chosen on $BC$ so that $MH$ is perpendicular to $BC$. $P$ is the midpoint of $MH$. Prove that $AH$ is perpendicular to $BP$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTake $X$ on $AH$ so that $BX$ is perpendicular to $AH$. Extend to meet $HM$ at $P'$. Let $N$ be the midpoint of $AB$. $A$, $B$, $M$ and $X$ are on the circle center $N$ radius $NA$ (because angles $AMB$ and $AXB$ are $90^{\\circ}$). Also $MN$ is parallel to $BC$ (because $AMN$, $ACB$ are similar), so $NM$ is perpendicular to $MH$, in other words $HM$ is a tangent to the circle. Hence $P'M = P'X$. $P'B$. Triangles $P'XH$ and $P'HB$ are similar (angles at $P'$ same and both have a right angle), so $P'H / P'X = P'B / P'H$, so $P'H \\cdot P'H = P'X \\cdot P'B$. Hence $P'H = P'M$ and $P'$ coincides with $P$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42474,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe triangle $ABC$ satisfies $0 \\leq AB \\leq 1 \\leq BC \\leq 2 \\leq CA \\leq 3$. What is the maximum area it can have?",
    "options": [],
    "answer": "1",
    "solution": "Solution:\nIf we ignore the restrictions of $CA$, then the maximum area is $1$, achieved when $AB$ is perpendicular to $BC$. But in this case $CA$ satisfies the restrictions.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42475,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven unequal integers $x$, $y$, $z$ prove that $\\left(x - y\\right)^{5} + \\left(y - z\\right)^{5} + \\left(z - x\\right)^{5}$ is divisible by $5(x - y)(y - z)(z - x)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nPut $x - y = r$, $y - z = s$. Then $z - x = - (r + s)$, and\n$$\n\\left(x - y\\right)^{5} + \\left(y - z\\right)^{5} + \\left(z - x\\right)^{5} = r^{5} + s^{5} - (r + s)^{5}\n$$\nExpand $(r + s)^5$:\n$$\n(r + s)^5 = r^5 + 5r^4s + 10r^3s^2 + 10r^2s^3 + 5rs^4 + s^5\n$$\nSo,\n$$\nr^5 + s^5 - (r + s)^5 = r^5 + s^5 - [r^5 + 5r^4s + 10r^3s^2 + 10r^2s^3 + 5rs^4 + s^5]\n$$\n$$\n= -5r^4s - 10r^3s^2 - 10r^2s^3 - 5rs^4\n$$\n$$\n= -5rs(r^3 + 2r^2s + 2rs^2 + s^3)\n$$\n$$\n= -5rs(r^2 + rs + s^2)(r + s)\n$$\nTherefore,\n$$\n\\left(x - y\\right)^{5} + \\left(y - z\\right)^{5} + \\left(z - x\\right)^{5} = -5rs(r + s)(r^2 + rs + s^2)\n$$\nRecall $r = x - y$, $s = y - z$, $r + s = x - y + y - z = x - z = -(z - x)$.\n\nThus, $rs(r + s) = (x - y)(y - z)(z - x)$ (up to sign), and the expression is divisible by $5(x - y)(y - z)(z - x)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42476,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven $a_0$, $a_1$, ..., $a_n$, satisfying $a_0 = a_n = 0$, and $a_{k - 1} - 2a_k + a_{k + 1} \\geq 0$ for $k = 1, 2, ..., n-1$. Prove that all the numbers are negative or zero.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe essential point is that if we plot the values $a_r$ against $r$, then the curve formed by joining the points is cup shaped. Its two endpoints are on the axis, so the other points cannot be above it. There are many ways of turning this insight into a formal proof. Barry Paul's was neater than mine: $a_{r + 1} - a_r \\geq a_r - a_{r - 1}$. Hence (easy induction) if $a_s - a_{s - 1} > 0$, then $a_n > a_s$. Take $a_s$ to be the first positive, then certainly $a_s > a_{s - 1}$, so $a_n > 0$. Contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42477,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a fixed circle $C$ and a line $L$ through the center $O$ of $C$. Take a variable point $P$ on $L$ and let $K$ be the circle center $P$ through $O$. Let $T$ be the point where a common tangent to $C$ and $K$ meets $K$. What is the locus of $T$?",
    "options": [],
    "answer": "The pair of fixed tangents to the circle that are perpendicular to the given line.",
    "solution": "Solution:\n\nLet the common tangent meet $C$ at $S$. Let $X$ be the intersection of $C$ and $OP$ lying between $O$ and $P$. $PT = PO$, hence $\\angle POT = \\angle PTO$, so $\\angle OPT = 180^{\\circ} - 2\\angle POT$. But $PT$ and $OS$ are parallel, because both are perpendicular to the common tangent. Hence $\\angle POS = 2\\angle POT$, so $\\angle SOT = \\angle XOT$. Hence $TX$ is tangent to $C$, in other words $T$ lies on the (fixed) tangent to $C$ at $X$. Conversely, it is easy to see that any such point can be obtained (just take $P$ such that $PO = PT$). Thus the required locus is the pair of tangents to $C$ which are perpendicular to $L$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42478,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven integers $a_0$, $a_1$, ..., $a_{100}$, satisfying $a_1 > a_0$, $a_1 > 0$, and $a_{r + 2} = 3 a_{r + 1} - 2 a_r$ for $r = 0$, $1$, ..., $98$. Prove $a_{100} > 2^{99}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAn easy induction gives $a_r = (2^{r} - 1)a_1 - (2^{r} - 2)a_0$ for $r = 2$, $3$, ..., $100$. Hence, in particular, $a_{100} = (2^{100} - 2)(a_1 - a_0) + a_1$. But $a_1$ and $(a_1 - a_0)$ are both at least $1$. Hence result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42479,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that there are no integers $a$, $b$, $c$, $d$ such that the polynomial $ax^3 + bx^2 + cx + d$ equals $1$ at $x = 19$ and $2$ at $x = 62$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIf there were such values, then subtract the equation with $x = 19$ from the equation with $x = 62$ to get:\n$$\na(62^3 - 19^3) + b(62^2 - 19^2) + c(62 - 19) = 1.\n$$\nBut the left hand side is divisible by $62 - 19 = 43$, contradiction.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 42480,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven an $n \\times n$ array of numbers. $n$ is odd and each number in the array is $1$ or $-1$. Prove that the number of rows and columns containing an odd number of $-1$s cannot total $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIf we change a $-1$ to $1$, we affect the total number of rows and columns (containing an odd number of $-1$s) by $0$, $2$ or $-2$. After changing all the $-1$s we have total of $0$. Hence the starting total must be even. So it cannot be $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42481,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven the lengths $AB$ and $BC$ and the fact that the medians to those two sides are perpendicular, construct the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $M$ be the midpoint of $AB$ and $X$ the midpoint of $MB$. Construct the circle center $B$, radius $BC/2$ and the circle diameter $AX$. If they do not intersect (so $BC < AB/2$ or $BC > AB$) then the construction is not possible. If they intersect at $N$, then take $C$ so that $N$ is the midpoint of $BC$. Let $CM$ meet $AN$ at $O$. Then $AO/AN = AM/AX = 2/3$, so the triangles $AOM$ and $ANX$ are similar. Hence $\\angle AOM = \\angle ANX = 90^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42482,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven four positive real numbers $a$, $b$, $c$, $d$ such that $abcd = 1$, prove that\n$$\na^2 + b^2 + c^2 + d^2 + ab + ac + ad + bc + bd + cd \\geq 10.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nApplying the arithmetic/geometric mean result to the 10 numbers gives the result immediately.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42483,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven a fixed regular pentagon $ABCDE$ with side $1$. Let $M$ be an arbitrary point inside or on it. Let the distance from $M$ to the closest vertex be $\\mathbf{r}_1$, to the next closest be $\\mathbf{r}_2$ and so on, so that the distances from $M$ to the five vertices satisfy $\\mathbf{r}_1 \\leq \\mathbf{r}_2 \\leq \\mathbf{r}_3 \\leq \\mathbf{r}_4 \\leq \\mathbf{r}_5$. Find\n\na) the locus of $M$ which gives $\\mathbf{r}_3$ the minimum possible value,\n\nb) the locus of $M$ which gives $\\mathbf{r}_3$ the maximum possible value.",
    "options": [],
    "answer": "a) The five midpoints of the diagonals. b) The five midpoints of the sides.",
    "solution": "Solution:\nLet $X$ be the midpoint of $AB$ and $O$ the center of $ABCDE$. Suppose $M$ lies inside $AXO$. Then $ME = \\mathbf{r}_3$. So we maximise $\\mathbf{r}_3$ by taking $M$ at $X$, with distance $1.5590$, and we minimise $\\mathbf{r}_3$ by taking $M$ as the intersection of $AO$ and $EB$ with distance $0.8090$. $AXO$ is one of $10$ congruent areas, so the required loci are\n\na) the $5$ midpoints of the diagonals,\n\nb) the $5$ midpoints of the sides.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 42484,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven a number with 1998 digits which is divisible by 9. Let $x$ be the sum of its digits, let $y$ be the sum of the digits of $x$, and $z$ the sum of the digits of $y$. Find $z$.",
    "options": [],
    "answer": "9",
    "solution": "Solution:\n$x \\leq 9 \\times 1998 = 17982$. Hence $y \\leq$ the greater of $1 + 7 + 9 + 9 + 9 = 35$ and $9 + 9 + 9 + 9 = 36$. But $9$ divides the original number and hence also $x$, $y$ and $z$. Hence $z = 9$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42485,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven 5 circles. Every 4 have a common point. Prove that there is a point common to all 5.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet the circles be $a$, $b$, $c$, $d$, $e$. Let $A$ be a point common to $b$, $c$, $d$, $e$, let $B$ be a point common to $a$, $c$, $d$, $e$ and so on. If any two of $A$, $B$, $C$, $D$, $E$ coincide then the coincident point is on all 5 circles. Suppose they are all distinct. Then $A$, $B$, $C$ are on $d$ and $e$. Hence $d$ and $e$ coincide (3 points determine a circle). Hence $D$ is on all 5 circles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42486,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn infinite arithmetic progression contains a square. Prove it contains infinitely many squares.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet the square be $a^{2}$ and the difference $d$, so that all numbers of the form $a^{2} + nd$ belong to the arithmetic progression (for $n$ a natural number). Take $n$ to be $2a + dr^{2}$, then $a^{2} + nd = (a + dr)^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42487,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCan we label each vertex of a 45-gon with one of the digits $0, 1, \\ldots, 9$ so that for each pair of distinct digits $i, j$ one of the 45 sides has vertices labeled $i, j$?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n$10 \\times 5 > 45$, so some digit $i_0$ must appear less than $5$ times. But each occurrence can give at most $2$ edges $i_0, j$, so there are at most $8$ edges $i_0, j$, which is one too few.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 42488,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real $p$, $q$, $a$, $b$ such that we have $$(2x - 1)^{20} - (a x + b)^{20} = (x^{2} + p x + q)^{10}$$ for all $x$.",
    "options": [],
    "answer": "a = ±(2^{20} − 1)^{1/20}, b = −a/2, p = −1, q = 1/4",
    "solution": "Solution:\nComparing coefficients of $x^{20}$, we must have $a = (2^{20} - 1)^{1 / 20}$ (note that we allow either the positive or the negative root).\n\nSet $x = 1 / 2$. Then we must have $(a x + b)^{20} = 0 = (x^{2} + p x + q)^{10}$, and hence $a x + b = 0$ and $x^{2} + p x + q = 0$. So $b = - a / 2$, and $1 / 4 + p / 2 + q = 0$.\n\nSet $x = 0$. Then we get $q^{10} = 1 - b^{20} = 1 / 2^{20}$, so $q = 1 / 4$ or $- 1 / 4$, and $p = - 1$ or $0$ respectively. Comparing the coefficients of $x^{19}$, we must have $p = - 1$ and $q = 1 / 4$. So, if there is a solution, then it must be: $a = (2^{20} - 1)^{1 / 20}$, $b = - a / 2$, $p = - 1$, $q = 1 / 4$. This is indeed a solution because with these values, the $lhs = 2^{20}(x - 1 / 2)^{20} - (x - 1 / 2)^{20} a^{20} = (x - 1 / 2)^{20} = (x^{2} - x + 1 / 4)^{10} = rhs$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42489,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWe place labeled points on a circle as follows. At step 1, take two points at opposite ends of a diameter and label them both $1$. At step $n > 1$, place a point at the midpoint of each arc created at step $n - 1$ and label it with the sum of the labels at the two adjacent points. What is the total sum of the labels after step $n$?\n\nFor example, after step 4 we have: $1$, $4$, $3$, $5$, $2$, $5$, $3$, $4$, $1$, $4$, $3$, $5$, $2$, $5$, $4$, $3$, $4$.",
    "options": [],
    "answer": "2 * 3^(n - 1)",
    "solution": "Solution:\n\nAnswer: $2 \\cdot 3^{n - 1}$.\n\nTrue for $n = 1$. The new points added at step $n + 1$ have twice the sum of the points after step $n$, because each old point contributes to two new points. Hence the total after step $n + 1$ is three times the total after step $n$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42490,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven an isosceles triangle, find the locus of the point $P$ inside the triangle such that the distance from $P$ to the base equals the geometric mean of the distances to the sides.",
    "options": [],
    "answer": "The arc BC of the circle through B and C that is tangent to AB and AC.",
    "solution": "Solution:\nLet the triangle be $ABC$, with $AB = AC$. Take the circle through $B$ and $C$ which has $AB$ and $AC$ as tangents. The required locus is the arc $BC$.\n\nSuppose $P$ lies on the arc. Let the perpendiculars from $P$ meet $BC$ in $L$, $AB$ in $N$ and $AC$ in $M$. Join $PB$ and $PC$. The triangles $PNB$ and $PLC$ are similar ($\\angle PNB$ and $\\angle PLC$ are both $90^\\circ$, and $\\angle NBP = \\angle LCP$ because $NB$ is tangent to the circle). Hence $PN / PL = PB / PC$. Similarly, triangles $PMC$ and $PLB$ are similar and hence $PM / PL = PC / PB$. Multiplying gives the required result $PL^2 = PM \\cdot PN$.\n\nIf $P$ is inside the circle and not on it, take $P'$ as the intersection of the line $AP$ and the arc. We have $PL < PL'$, but $PM > PM'$ and $PN > PN'$, hence $PL^2 < PM \\cdot PN$. Similarly, if $P$ is outside the circle and not on it, then $PL^2 > PM \\cdot PN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42491,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n8 players compete in a tournament. Everyone plays everyone else just once. The winner of a game gets 1, the loser 0, or each gets $\\frac{1}{2}$ if the game is drawn. The final result is that everyone gets a different score and the player placing second gets the same as the total of the four bottom players. What was the result of the game between the player placing third and the player placing seventh?",
    "options": [],
    "answer": "The third player won against the seventh player.",
    "solution": "Solution:\n\nThe bottom 4 played 6 games amongst themselves, so their scores must total at least 6. Hence the number 2 player scored at least 6. The maximum score possible is 7, so if the number 2 player scored more than 6, then he must have scored $6 \\frac{1}{2}$ and the top player 7. But then the top player must have won all his games, and hence the number 2 player lost at least one game and could not have scored $6 \\frac{1}{2}$. Hence the number 2 player scored exactly 6, and the bottom 4 players lost all their games with the top 4 players. In particular, the number 3 player won against the number 7 player.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42492,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. The two diagonals of a quadrilateral each divide it into two parts of equal area. Prove it is a parallelogram.\n\nb. The three main diagonals of a hexagon each divide it into two parts of equal area. Prove they have a common point. [If $ABCDEF$ is a hexagon, then the main diagonals are $AD$, $BE$ and $CF$.]",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na. Let the quadrilateral be $ABCD$ and let the diagonals $AC$, $BD$ meet at $E$. Then area $ABC = AC \\cdot EB \\cdot \\sin CEB / 2$, and area $ADC = AC \\cdot ED \\cdot \\sin CEB / 2$, so $E$ is the midpoint of $BD$. Similarly, it is the midpoint of $AC$. Hence the triangles $AEB$ and $CED$ are congruent, so $\\angle CDE = \\angle ABE$, and hence $AB$ is parallel to $CD$. Similarly, $AD$ is parallel to $BC$.\n\nb. Let the hexagon be $ABCDEF$. Let $BE$, $CF$ meet at $J$, let $AD$, $CF$ meet at $K$, and let $AD$, $BE$ meet at $L$. Let $AK = a$, $BJ = b$, $CJ = c$, $DL = d$, $EL = e$, $FK = f$. Also let $KL = x$, $JL = y$ and $JK = z$. Consider the pair of diagonals $AD$, $BE$. They divide the hexagon into 4 parts: the triangles $ALB$ and $DLE$, and the quadrilaterals $AFEL$ and $BCDL$. Since area $ALB$ + area $AFEL$ = area $DLE$ + area $BCDL$, and area $ALB$ + area $BCDL$ = area $DLE$ + area $AFEL$, the two triangles must have the same area (add the two inequalities). But area $ALB = 1/2 \\cdot AL \\cdot BL \\cdot \\sin ALB$, and area $DLE = 1/2 \\cdot DL \\cdot EL \\cdot \\sin DLE = 1/2 \\cdot DL \\cdot EL \\cdot \\sin ALB$, so $AL \\cdot BL = DL \\cdot EL$ or $de = (a+x)(b+y)$. Similarly, considering the other two pairs of diagonals, we get $bc = (e+y)(f+z)$ and $af = (c+z)(d+x)$. Multiplying the three inequalities gives: $abcdef = (a+f)(b+y)(c+z)(d+x)(e+y)(f+z)$. But $x$, $y$, $z$ are non-negative, so they must be zero and hence the three diagonals pass through a common point.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42493,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe natural numbers $m$ and $n$ are relatively prime. Prove that the greatest common divisor of $m + n$ and $m^2 + n^2$ is either $1$ or $2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIf $d$ divides $m + n$ and $m^2 + n^2$, then it also divides $(m + n)^2 - (m^2 + n^2) = 2mn$ and hence also $2m(m + n) - 2mn = 2m^2$ and $2n(m + n) - 2mn = 2n^2$. But $m$ and $n$ are relatively prime, so $m^2$ and $n^2$ are also. Hence $d$ must divide $2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 42494,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven a circle $c$ and two fixed points $A$, $B$ on it. $M$ is another point on $c$, and $K$ is the midpoint of $BM$. $P$ is the foot of the perpendicular from $K$ to $AM$.\n\na. Prove that $KP$ passes through a fixed point (as $M$ varies).\n\nb. Find the locus of $P$.",
    "options": [],
    "answer": "a) KP always passes through X, where Y is the point on the circle such that angle ABY is a right angle and X is the midpoint of BY. b) The locus of P is the circle with diameter AX.",
    "solution": "Solution:\n\na.\nTake $Y$ on the circle so that $\\angle ABY = 90^\\circ$. Then $AY$ is a diameter and so $\\angle AMY = 90^\\circ$. Take $X$ as the midpoint of $BY$. Then triangles $BXK$ and $BYM$ are similar, so $XK$ is parallel to $YM$. Hence $XK$ is perpendicular to $AM$, and so $P$ is the intersection of $XK$ and $AM$. In other words, $KP$ always passes through $X$.\n\nb.\n$P$ must lie on the circle diameter $AX$, and indeed all such points can be obtained (given a point $P$ on the circle, take $M$ as the intersection of $AP$ and the original circle). So the locus of $P$ is the circle diameter $AX$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42495,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the smallest value $x$ such that, given any point inside an equilateral triangle of side $1$, we can always choose two points on the sides of the triangle, collinear with the given point and a distance $x$ apart.",
    "options": [],
    "answer": "2/3",
    "solution": "Solution:\nAnswer: $2/3$.\n\nLet $O$ be the center of $ABC$. Let $AO$ meet $BC$ at $D$, let $BO$ meet $CA$ at $E$, and let $CO$ meet $AB$ at $F$. Given any point $X$ inside $ABC$, it lies in one of the quadrilaterals $AEOF$, $CDOE$, $BFOD$. Without loss of generality, it lies in $AEOF$. Take the line through $X$ parallel to $BC$. It meets $AB$ in $P$ and $AC$ in $Q$. Then $PQ$ is shorter than the parallel line $MON$ with $M$ on $AB$ and $N$ on $AC$, which has length $2/3$.\n\nIf we twist the segment $PXQ$ so that it continues to pass through $X$, and $P$ remains on $AB$ and $Q$ on $AC$, then its length will change continuously. Eventually, one end will reach a vertex, whilst the other will be on the opposite side and hence the length of the segment will be at least that of an altitude, which is greater than $2/3$. So at some intermediate position its length will be $2/3$.\n\nTo show that no value smaller than $2/3$ is possible, it is sufficient to show that any segment $POQ$ with $P$ and $Q$ on the sides of the triangle has length at least $2/3$. Take $P$ on $MB$ and $Q$ on $AN$ with $P$, $O$, $Q$ collinear. Then $PQ \\cos POM = MN - QN \\cos \\pi/3 + PM \\cos \\pi/3$. But $PM > QN$ (using the sine rule, $PM = OM \\sin POM/\\sin OPM$ and $QN = ON \\sin QON/\\sin OQN$, but $OM = ON$, $\\angle POM = \\angle QON$, and $\\angle OQN = \\angle OPM + \\pi/3 > \\angle OPM$), and hence $PQ > MN \\sec POM > MN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42496,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. A $6 \\times 6$ board is tiled with $2 \\times 1$ dominos. Prove that we can always divide the board into two rectangles each of which is tiled separately (with no domino crossing the dividing line).\n\nb. Is this true for an $8 \\times 8$ board?",
    "options": [],
    "answer": "a: yes; b: no",
    "solution": "Solution:\n\na.\nWe say a domino bridges two columns if half the domino is in each column. We show that for $0 < n < 6$ the number of dominoes bridging columns $n$ and $n+1$ must be at least $2$ and even.\n\nConsider first $n = 1$. There cannot be $3$ dominos entirely in column $1$, or it would be separately tiled. So there must be at least one domino bridging columns $1$ and $2$. The number must be even, because it must equal the number of squares in column $1$ (even) less twice the number of dominoes (entirely) in column $1$.\n\nNow suppose it is true for $n < 5$ and consider column $n + 1$. There must be at least one domino bridging columns $n + 1$ and $n + 2$, or columns $1$ thru $n + 1$ would be separately tiled. The number must be even, because it must equal the number of squares in column $n + 1$ (even) less the number bridging $n$ and $n + 1$ (even) less twice the number entirely in the column.\n\nSo in total there are at least $5 \\times 2 = 10$ dominos bridging columns. By the same argument there are at least another $10$ bridging rows, but there are only $18$ dominoes in total.\n\nb.\nNo. For example:\n1 2 3 3 1 1 2 2 1 2 1 2 2 3 3 1 3 3 1 3 1 2 4 1 1 2 2 3 1 2 4 3 1 3 3 2 2 1 2 3 3 2 1 1 4 1 2 1 3 2 3 2 4 3 3 1 1 1 3 2 1 1 2 2",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42497,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven a set of $n$ different positive reals $\\{a_{1}, a_{2}, \\ldots , a_{n}\\}$. Take all possible non-empty subsets and form their sums. Prove we get at least $\\frac{n(n + 1)}{2}$ different sums.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nAssume $a_{1} < a_{2} < \\ldots < a_{n}$. We have the following collection of increasing sums:\n\n$a_{1} < a_{2} < \\ldots < a_{n}$  ($n$ sums)\n\n$a_{1} + a_{n} < a_{2} + a_{n} < \\ldots < a_{n - 1} + a_{n}$  ($n-1$ sums)\n\n$a_{1} + a_{n - 1} + a_{n} < a_{2} + a_{n - 1} + a_{n} < \\ldots < a_{n - 2} + a_{n - 1} + a_{n}$  ($n-2$ sums)\n\n$\\vdots$\n\n$a_{1} + a_{2} + \\ldots + a_{n}$  (1 sum)\n\nA total of $1 + 2 + \\ldots + n = \\frac{n(n + 1)}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 42498,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven a triangle $ABC$. Let the line through $C$ parallel to the angle bisector of $B$ meet the angle bisector of $A$ at $D$, and let the line through $C$ parallel to the angle bisector of $A$ meet the angle bisector of $B$ at $E$. Prove that if $DE$ is parallel to $AB$, then $CA = CB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe idea is to find an expression for the perpendicular distance $h$ from $D$ to $AB$. Let $\\gamma = \\angle ACB$, $\\alpha = \\frac{1}{2} \\angle CAB$, and $\\beta = \\frac{1}{2} \\angle ABC$. We have $h = AP \\sin \\alpha$.\n\nUsing the sine rule on $APC$, we have $AP = AC \\frac{\\sin (\\gamma + \\beta)}{\\sin (\\alpha + \\beta)}$, so $h = AC \\frac{\\sin \\alpha \\sin (\\gamma + \\beta)}{\\sin (\\alpha + \\beta)}$. Similarly, the perpendicular distance $k$ from $E$ to $AB$ is $BC \\frac{\\sin \\beta \\sin (\\gamma + \\alpha)}{\\sin (\\alpha + \\beta)}$.\n\nWe also have that $\\frac{AC}{BC} = \\frac{\\sin 2\\beta}{\\sin 2\\alpha}$, and hence $\\frac{h}{k} = \\frac{\\sin 2\\beta \\sin \\alpha \\sin (\\gamma + \\beta)}{\\sin 2\\alpha \\sin \\beta \\sin (\\gamma + \\alpha)}$. Using the fact that $\\sin (\\gamma + \\beta) = \\sin (2\\alpha + \\beta)$, and the expression for $\\sin 2\\theta$, we get $\\frac{h}{k} = \\frac{\\sin (2\\alpha + 2\\beta) + \\sin 2\\alpha}{\\sin (2\\alpha + 2\\beta) + \\sin 2\\beta}$ and hence $h = k$ iff the triangle is isosceles.\n\nFor some reason the geometric solution took me longer to find. Let $ED$ meet $BC$ at $X$. Then $XCD$ and $XBE$ are isosceles, so $BC = BX + XC = DX + XE = DE$. Similarly, $AC = DE$. Hence $AC = BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42499,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the triangle $ABC$, the length of the altitude from $A$ is not less than $BC$, and the length of the altitude from $B$ is not less than $AC$. Find the angles.",
    "options": [],
    "answer": "90°, 45°, 45°",
    "solution": "Solution:\n\nLet $k$ be twice the area of the triangle. Then $k \\geq BC^2$, $k \\geq AC^2$ and $k \\leq AC \\cdot BC$, with equality in the last case only if $AC$ is perpendicular to $BC$. Hence $AC$ and $BC$ have equal lengths and are perpendicular. So the angles are $90^\\circ$, $45^\\circ$, $45^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 42500,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA circle center $O$ is inscribed in $ABCD$ (touching every side). Prove that $\\angle AOB + \\angle COD$ equals $180$ degrees.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $AB$ touch the circle at $W$, $BC$ at $X$, $CD$ at $Y$, and $DA$ at $Z$. Then $AO$ bisects angle $ZOW$ and $BO$ bisects angle $XOW$. So $\\angle AOB$ is half angle $ZOX$. Similarly $\\angle COD$ is half angle $XOZ$ and hence $\\angle AOB + \\angle COD$ equals $180$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  }
]